UNIT 1 ELECTRIC CHARGE, FORCE AND FIELD

1.1 Introduction

Objectives

1.2 Properties of Electric Charge

Types of Charges

Unit of Charge

Conservation of Charge

Quantization of Charge

1.3 Coulomb's Law

1.4 Principle of Superposition

1.5 The Electric Field

Calculating the Electric Field

Electric Lines of Force

.- 1:6 Summary

' 1.7 Terminal Questions

1.8 Solutions and Answers

1.1 INTRODUCTION

During hot humid days, you might have observed that rain is often accompanied by lightning and thunder. Do you h o w the cause of the lightning and thunder? Benjamin Franklin was the first to prove through his experir ent that roaring clouds possess electric charge. These charged clouds, when discharged in the atmosphere, give rise to a giant spark. This spark is called lightning. Can you

I imagine that the amount of current during discharge of the cloud is about 20,000 amperes!

When there is lightning due to an electric discharge, 'a great amount of heat is produced. During one millionth of a second, the temperature rises to 15000°C which is about two and a half times the temperature of the sun. The lightning

. flash develops in a zone 20cm wide. Due to excessive heat produced in this zone, the air molecules move very fast and cause sound. This is called thunder. When this sound is reflected by clouds or hills or any other obstacle you hear the roaring of clouds.

This spectacular event of nature - which has drawn the attention of most ancients - is associated with charging and discharging phenomena. The most noticeable thing about electric charges is that the forces between them are , extremely large. This force, known as electrostatic force (or electric force), is responsible for holding electrons to nuclei to form atoms and for holding the groups of atoms together to form molecules, solids and liquids. In this unit, you will learn about the nature of dharges and the electrostatic force between them. Around every charged body, there is a region where the electric force can be detected. This region is called electric field. You wi l l learn to calculate the electric field due to .different charge configurations. In the next unit, a mare easy and elegant method will be used to determine the electric field due to various charge distributions. . .

Objectives

After studying this unit, you should be able to:

@ distinguish between the two types of electric charge,

9 show that the total electric charge in an isolated system is conserved,

@ infer that any electric charge is always an integral multiple of the charge on the electron,

e~ use Coulomb's law to find the electrostatic force between two charges,

o state the principle of superposition of forces and calculate the resultant force due to more than two charges,

s calculate the resultant electric field due to an arbitrary distribution of charges, I

and

s draw the electric lines of force. -

1.2 PROPERTIES OF ELECTRIC CHARGE

The word 'electricity' o r 'electric' is derived from the Greek word 'elecktron' which means amber. In about 600 B.C., the Greek philosopher Thales discowered that when amber (a natural resin) is rubbed with fur, the amber becomes capable of attracting small particles of matter. This discovery did not attract much attention until 1600 A.D. William Gilbert showed that many substances, such as glass, ebonite and.resin, when rubbed with silk, flannel or other suitable materials acquire similar property as the rubbed amber. The substances in such a state are said to be electrified or to have acquired electric charges or are simply referred to as charged bodies. This section will be a quick recapitulation of what you have learnt so far in your school. i 1.2.1 Types of Charges Consider two pith balls suspended by a metal thread from metallic supports at a , I

short distance. The metal wire connects the two supports. A piece of rubber is I taken and it is rubbed with a piece of glass rod so that both of them get charged. I

When the charged rubber is touched to the metal support, the charge distributes I itself over the pith balls. As a result, the two balls move apart as shown in Fig. l. la. Next, when the charged glass rod is touched to the metal support the balls again move apart as in Fig. I . lb. But when the metallic connection between the pith balls is removed as shown in Fig. 1, lc, and if one of the supports is touched. with charged glass rod while the other with charged rubber, then the two balls move towards each other. I

I

I

1

(1) @) (4 ' -

Ell. 1.1: Pith ball demonstration of like charges- repelling and unlike charges attracting. (a) Repulsion produced between two balls when'charged rubber is touched to the metal'support. @) kepulsion produced between two balls when charged glass rod is touched to the metal support.

In (a) and @) a metal wire connects the two supports. (c) Attraction produced between two balls when the metallic connection between the pith balls is

removed and one is touched with charged rubber and the other with charged glass rod.

"d

What do you infer from these observations? In case (a), both the balls acquire the M C w , F-aid

same kind of charge and move apart. In case (b), the two balls again acquire the same kind of charge and move apart.

In case (c), the two balls do not acquire the same kind of charge otherwise the balls would have repelled as happened in the earlier two cases. It means that the charge on the ball suspended from that support which is being touched by glass rod is different from the charge on the ball suspended from the support which is being touched by rubber. In other words, there are only two types of electric charge. Bodies carrying the same kind of charge repel one another, whereas bodies carrying different types of charge attract one another. 'The origin of the two types of charges is explained as follows :

You all know from your previous classes that an atom consists of a positively charged nucleus with negatively charged electrons around it. The nucleus consists of protons and neutrons. The neutron is uncharged (neutral) while the electron and proton have equal but opposite charges (negative and positive respectively). As the prqtons and neutrons are in the nucleus, they are held together very tightly by a nuclear force. This force is so strong that protons are unable to move away from the atomic nucleus, whereas the force holding the electrons to the atomic nucleus is much weaker. So that the electrons are more free to move away from the atom as compared to protons.

When two different materials are brought into contact and rubbed together, the electrons (being more free) get transferred from one material to the other. Since some materials tend to hold their electrons more strongly than others, the directipn of transfer of electrons depends on the materials concerned and is always the same for any two materials. For example, when a plastic ruler is rubbed with a woollen cloth, as shown in Fig. 1.2, electrons flow from wool to plastic. This process will leave an excess of electrons on the plastic, so that it carries a net negative charge, whereas the wool, with a deficit of electrons, carries a positive charge of equal magnitude. In a similar way, amber or ebonite or rubber rod when rubbed with wool or fur acquire negative charge whereas wool or fur becomes positively charged. This method of charging the bodies by means of rubbing them together is called charging by friction (though friction actually has nothing to do with the charging process). Table 1.1 lists some materials in the triboelectric series (tribo means friction), which ranks materials according to their tendency to give up their electrons. Materials towards the top of the list become positively charged when placed in contact and rubbed with those lower on the list:

Table 1.1

The triboelectric series:

rabit fur glass wool cat's fur silk felt cotton wood cork rubber celluloid

SAQ 1

We have two charged bodies X and Y which attract each other. X repels a third charged body 2. Will Z attract or repel Y?

, ,

1.2.2 Unit of Charge In the Systeme Internationale (SI), the unit of charge is Coulomb (abbreviated C) which is defined in terms of ampere. You all must be familiar with the definition of the ampere which is as follows:

a. 1.1: Charging by friction. When a plastic ruler is rubbed with a woollen cloth, electrons flow from the wool to the plastic. This process will leave .n excess of etcctrons on the plastic, so that it carries a net negative charge, whereas the wool, with a deficit of electrons, carries a positive charge of equal magnitude.

~ k t m @ a ~ c s h F m Space " ~ n ' ampere is the current, which when maintained in two straight parallel wires placed one metre apart in vacuum, would produce between these wires B force .equal to 2 x 10-' N per metre of length."

The defdtion of ampere involves force between currents, which we shall discuss later in Block 3.

Using ampere, the unit of charge is defined as:

"A Coulomb is the amount of charge that flows through any cross-section of a wire in one second if there is a steady current of one ampere in the wire." In symbols,

q = It ...( 1.1)

where q is in Coulombs, if is in amperes and t is in seconds. .

The reason for defining Coulomb in terms of ampere is that it is easy to maintain; control and measure a current through a conductor rather than the amount of charge.

1.2.3 Conservation of Charge In the rubbing process of a plastic ruler by woollen cloth as shown in Fig. 1.2, no new charges are created. The algebraic sum of the individual charges, i.e., net charge always remains constant. Let us see how? Before the process of rubbing, both the plastic and wool are neutral (having .no charge). So the net charge is zero. After rubbing, the plastic ruler gets negatively charged and the woollen cloth acquires positive charge of equal magnitude. Now the algebraic sum of the equal and opposite charge on the plastic and wool is zero. So the net charge is again zero.

This shows that electric charge is a conserved quantity. Conservation of charge implies that the total charge in an isolated system never changes. It does not mean that the total amount of positive or negative charge in a system is fiied, rather it

Here isolatecr means that no implies that for every additional positive charge created, there is always an equal mattu is allowed to cross the amount of negative charge created. An example of conservation of' electric charge boundary of the system. is beautifully illustrated in the process known as 'pair production' as shown in Fig.

1.3. Here, a thin-walled box in a vacuum is an isolated system. When this box gets exposed to a gamma ray photon (carrying no charge), we find the existence of an electron and positron (a particle having same mass as electron but having a charge -' . equal and opposite to that of an electron) inside the box. Although, two electrically charged particles have been newly created, but the net change in the total charge in the box is zero. The charge conservation law may be stated as follow: The total electric charge in an isolated system, that is, the algebraic sum of the positive and negative charge present at any time, never changes.

Before After

Fig. 1.j: " Par production" exhibiting cansmation of chitge. In this process, a,garqma ray phiton is. converted into an electron (with a negative charge denoted as -e) and a posit& (a particle with thc same mass as an electron but with charge + e).

SAQ 2

Complete the following equations using the principle of conservation of charge. The notation for writing these equations is : ,xA where X represents the dhemical symbol of an element, Z is the atomic number (number of electrons) afld A is the

10 - mass number (number of protons + number of neutrons).

(i) 9 2 ~ 2 3 8 , 7 ~ h U 4 + *He4 (radioactive decay) Ekxbak Charge. Fum and Fkld

(ii) 20~a44 + IP' - 21S~44 + ?nl (nuclear reaction)

The smdlest charge that is passible to obtain is that of an electron or proton. (Both electron and proton have the same magnitude of charge but electron is '

negatively charged while proton is positively charged). The magnitude of this charge is denoted by 'e'. It was first measured by Milbkan in his famous oildrop experiment by observing the motion of a charged oil drop under the combined influence of gravitational field and externally applied electric field.

4E A

Atomizer I

ltarwgg

Microscope rl al

(c)

Fig. 1.4: (a) The Millikan oil-drop apparatus for measuring the elementary charge e. (b) An oil drop falls at terminal speed v in a field-free region. Its weight is balanced by an upward drag force. (c) An eIcctric field force acts upward on the drop, which now rises with a terminal speed v ' . The drag fom. which always opposes the velocity. now acts downward.

Fig. 1.41 shows the apparatus for the measurement. The oil droplets introduced by the atomizer in the chamber A are either positively or negatively charged. !,a us consider a drop in the chamber C which has got into it through a small Erole in plate PI. In-the absence of the electric field, two forces act on the drop: its weight rng and an upwardly directed viscous force F as shown in Fig. 1.4b. The magnitude of F is proportional to the speed of the falling drop. The drop acquires a constant terminal speed v when the gravitational force gets just balanced by the viscous '?he cxnstence of charged particla force. In the prevnce of electric field, a third force qE acts on the drop. IT q is C J I ~ , quarks, whose electric negative, this force will act upward and the drop will now move upward. The new charazr come in multiples of e/3,

drag force will act downward as it has to point in the direction opposite to that in 'lot alter fa* that charge is quantized- it would which the drop is moving. As shown in Fig. 1.4c, when the upward electric force reduce the size of the,

q& is just balanced by the weight mg and the new drag force F', the drop acquires from to e/3. , a new terminal speed v'. By measuring v and v', the charge q is found. Millikan made observations on a large number of drops and found that charges on different drops were an integral multiple of a number which is 1.6 x 10-"c, i.e., ektronic charge. In fact, a charge smaller than e has not been found (see margin remark on quarks). 1f one determines the amount of charge on any charged body @ke a charged sphere or charged drop) or any charged particle (like positron, a-particle) pr any ion, then its charge is always found to be an integral multiple sf e, i.e., e, /&, 3e; 4e, ..... . No Charge will be fractional multiple of e like 0.7e or 2 .h . 'This is not only true for negative charges but also for positive charges. Mathematically, it can be expressed as:

q = ne . . . (1.23

where n is an positive or negative integer. Thus, the charge exists in discrete packets rather than in continuous amounts, Whenever a physical quantity possesses discrete values instead of continuous values, then that physical quantity is said to be quantized. Hence, charge is 'quantized'.

SAQ 3

A conductor possesses 3 . 2 ~ lo-" Coulomb positive charge. How many electrons does it have in excess or deficit? 11

Eketrostnh in Free Space

1.3 COULOa9S LAW

- 41 r12 q 2

@I Fig. 1.5: Quantities and forces involved in Coulomb's Law.

(a) F,z is the electrostatic force on charge q, due to ql. The separation between the charges is r,,. The unit vector r12 conveys directional information only, allowing the law to be written in vector notation. F2, is the force that charge q, exerts on ql. (b) r,, is a vector originating from the position of q2 and ending at the position of q,. The distance betwecn the positions of these two charges represents its ma~nitude.

The unit vectors along the positive x, y and z - ~ c s are denoted by & 1 and k respectively.

We have already seen in the last section that like charges repel while unlike charges attract one another. The quantitative study of electrostatic force of attraction or repulsion between two point charges at rest was first done by a French physicist Charles Augustin de Coulomb in 1785. He observed that electrostatic force depends on the magnitude of charges. Specifically, it is proportional to their product. The force also depends on the separation between the charges. It is inversely proportional to the square of the separation between them. Here the distance between the charges is large compared to their dimensions so that the charges are treated as point charges.

To express this mathematically, suppose we have two point charges ql and 9, plaFed at a distance r12 as shown in Fig. 1.5. According to Coulomb's law, the force, F12, acting on the charge ql due to the presence of q2 can be written as

where lF121 represents the magnitude of force F12 and r12 d e n o t ~ the distance between the point charges. We should be able to write Eq. (1.3) as vector equation, since it involves force, which is a vector quantity. In order to indicate that the right hand side of the equation is also a vector quantity, we introduce the unit vector i12 (read as r12 cap) which is a vector of unit magnitude and has the direction of the vector joining the position of charge 2 to charge 1.

Mathematically, if r12 denotes the vector originating from position of charge 2 and ending at the position of charge 1, then,

where Jr12( is the magnitude of the vector r12 or the distance between the charges. With this unit vector notation, we rewrite Eq. (1.3) as follows:

The constant of proportionality is normally written as 1/4sco where eo is called the permittivity of free space. The value of 1/4re0 in SI system is 9x 10' N mZ C-'. The constant of proportionality depends upon the system of units. In another commonly used system called CGS, the constant is set equal to unity. However, we shall not employ the CGS system here.

SAQ 4

Write down the equation similar to Eq. (1.4) for the force F2,.

The proportionality constant introduced as 1/4xc0 has important physical significance. If the charges are placed in different medium, it is found that Eq. (1.4) always holds except that the constant of proportionality (permittivity) varies from medium to mkdium. It is found that the maximum electrostatic force between two charges separated by a fixed distance is obtained when two charges are placed in .vacuum and decreases when the charges are placed in any other medium, We can infer that the permittivity of free space is minimum. The ratio ,of permittivities for a medium to that of vacuum is known as di/eleetric constant or specific inductive capacitance. This ratio for air is about 1.005. For electrostatic experiment done in a medium, Coulomb's law may be written as

where E is the permittivity of the medium,

A

Let us now Solve a problem using Coulomb's law. Electric Charge, Fom and Field I Example 1

A charge ql = 5.OpC is placed 30cm to the west of another charge q2 = -12pC. What is the force exerted by the positive charge on the negative charge? klso calculate the force experienced by the positive charge due to negative charge.

'I Solution 9-5.0 PC q2 =-12/.~C

Refer to Fig. 1.6, Coulomb's law gives force on negative charge due to positive F12

charge as follows : i - 30cm --------gJ

. Fig. 1.6: Example 1

where we write i for iZ1 because a unit vector pointing from the positive charge ql towards the negative charge q2 is in the positive x-direction. The minus sign shows that the force is actually in the negative x-direction or towards west, that is, attractive.

The force on positive charge due to the negative charge is:

Here the unit vector i12 becomcs - 1 because a unit vector from negative charge q2 tawards positive charge ql is in the negative'x-direction. The two minus signs multiply to a plus sign showing that force is in positive x-direction or towards east, that is, attractive.

Thus Newton's third law is explicitly satisfied, i.e., the two charges exert equal but opposite forces on each other. r,

SAQ 5

Hydrogen atom consists of an electron and a proton separated by an average distance of 5,3 x 10-"m. Find the electrical force between the electron .and proton

t and compare it with gravitational force acting between them. (Charge of electron 1.6 x 10-l9 C, mass of electron = 9.1 x kg, mass of proton = 1.7 x lomz7

kg, G = 6.7 x lo-" N m2 kg-2, and 1/4re0 = 9 x 10' N m2 C2. )

SAQ 6

Two point charges QI and Q2 are 3m apart and their combined charge is 20pC. If one repels the other with a force of 0.075N, what are the two charges?

1.4 PRINCIPLE OF SUPERPOSITION

In Section 1.3, we had considered the forces (electrostatic) between two point charges. Suppose we have more than two charges, say three charges ql, q2 and q3, placed as shown in Fig. 1.7.

Then how do we calculate the electrostatic force on any charge say ql due to the presence of other two charges. We canlstill calculate the force between different paif of charges by making use of Coulomb's law, The total force~on ql will be the vector sum of forces on ql due to q2 and q3 independently. This is the principal of

Eleclmstatlcs 1. Free Space

Fig. 7.7: Illus:r.~~ing the principle of superposition. Charges q2 and q3 are at a distance of r,, and rls respectivdy fram the charge ql. F12 is the force on g1 due to the charge q2. Because q, and q2 are like chmgcs, the force will be repulsive and it will act away from q2 along the line joining q2 and ql. Si;nil,lrly. the force Fx3 on q1 due to q, will act along the line joining q3 and q, and away from qp Accord'mg to principle of supcrposition, the resultant forcc F1 on q, is the vector sum of the forces F12 aad I,j which may be found by drawing a paraUelogram of for-.

superpositicn. The fact that elec~ric forces add vectorially is known as snprposition principle.

Thus, the total force F1 on charge ql, due to charges q2 and q3 at distances r12 and rl, respectively, will be given by F1 = F12 + F13

Here Fl1 is the force acting on ql due to q2 and F13 is the force acting on ql due to q3. The unit vector i12 and iU have the directions of the lines from q2 to g1 and q3 to ql respectively. To illustrate this principle let us find the solution of thg following example.

In Fi. 1.8, g1 = - l.OpC, q, = 2.OpC and q3 = 4.OpC. Find the electrostatic force on q1 owing to other two charges. Here rn = lOcm and r13 = 20cm. Express your result both in unit vector notation and as a magnitude and direction.

\ -. \ .

= +4.0*

I - P : FI3

I--------2k,)- ----- - a. 1.8: Charges of -I.*C, + 2.0@ and + 4.W are located at the corners of a rightaugle triansk.

Solution

This problem can be solved using the superposition principle. The force on ql due PO the charge q2 is given by:

I 471 92 . Ftz = - - 112 4% 4 2

The unit vector 3 becomes (-j) because it points from q2 to q l in the negative y- direction. Positive sign shows that force F12 is in positive y-direction, that is, attractive. Similarly, the force on ql due to q3 is :

Force FI3 is in positive x-direction, that is, attractive. According to the superposition principle, the force F1 acting on q, is the vector sum of the forces due to q2 and q 3 . That is,

Magnitude of the force F1 is

1.8 and it makes an angle 8 = tan -' - = tan-' 2 = 63.5" with positive x-axis.

0.9

Answer

F1 = (0. 90; + 1.83)~; F, = 2.01N, making an angle of approx. 63.5' 4 t h positive x-axis.

Now you would like to work out an SAQ on superposition principle.

SAQ 7

If the individual forces acting on a given charge due to the presence of five charges are represented by the sides of a closed pentagon, what will be the resultant force on the test charge? (This is a tricky problem involving the principle of superposition; if the actual calculation takes more than 30 seconds, you are doing wrong-)

1.5 THE E.LEC

An electric field is a region in which the electric charges experience an electric force. By knowing the electric field, one can calculate the forces on electric charges and then, via Newton's law, one can know the motion of these charges. Since all matter contains electrically. charged particles, an understanding of electric field will help to know the structure and behaviour of matter. Moreover, one can build devices in which electric fields accelerate charged particles in useful ways. For example, in your television sets, the electric fields in the TV tube accelerate electrons toward the front of the tube where their energy is converted to the light that we see.

In this section, you will learn to calculate electric field due to a single point charge and due to simple charge distribution including continuous charge distribution. In order to picturise the entire electric f ields, the concept of electric lines of force will also be introduced.

~eclmstatirs ID F ~ W Space 1.5.1 Galcullating the Electric Field

We have learnt in Section 1.3 about the electrostatic force acting between two charges, say ql and 92. If one of the charges, say 92, is a unit charge, then the force exerted on this unit charge due to the presence of charge ql is defined as the electric field at the location of unit charge.

In other words, if the force experienced by a test charge q located at a point in the electric field be F then, according to definition, the electric field (also called electric field intensity) E at that point is given by

In fact, to measure electric field in a given region, one has to introduce a test charge and measure the force on it. However, test charge exerts forces on the charges that produce the field, so it may change the configuration of these charges. So, in principle, the test charge should be so small as to have no appreciable effect on the charge configuration that produces the field.

Eq. (1.7) shows that the electric field is measured in Newtons coulomb-' (NC-I).

Since F is a vector quantity, E will also be a vector. If g is positive, the electric field E has the same direction as the force acting on the charge. If q is negative, the direction of E is opposite to that of the force F.

In the case of a point charge, calculation of the electric field is particularly simple. We already know from Coulomb's law that if we place a point charge ql at a distance r from another point charge q, the force on ql will be

with i a unit vector pointing from g towards ..the location of q,. Since the electric field is defined as the force per unit charge, we divide the force in Eq. (1.8) by the charge q, to obtain the field due to q a t the location of q,. That is

This equation gives the field arising due to the charge q at any location v)hich is at a distance of r from q. In Eq. (1.9), the unit vector i points from the charge q (due to which an electric field exist) to the location at which the electric field is being determined.

Now, what would be the electric field due to two or more point charges? Since the electric force obeys the superposition principle, so 'does the electric field (since it is the force per unit charge). Therefore, the field at A given point due to two or more charges is the vector sum of the fields of individual charges. The fields of individual point charges is given by Eq. (1.9). Hence, the electric field E due to n charges may be written as:

where E) are the electric fields due to the point charges'qj that are located at distances rjs from the point where we are evaluating the field.

Example 3

An electric field is set up by two point charges ql and q, such that q, = -92 = 12 x 1 0 - ' ~ and separated by distance of 0. lm as shown in Fig. 1 ;9. Find the electric field at the points marked as A and B.

-e-- 0.04m- -I- -- -0 .05~1 - *-- - - --O.lm -- ---.. ,

Fig. 1.9 : Example 3

i) At A, the electric field El due to ql is

At A, the electric field E2 due to q2 is

Therefore, net electric field FA) at A

ii) At B, the electric field El due to ql is

(Here i+ points diagonally upward to the right)

El is directed in the same direction as i+, as shown in the diagram.

Also the field E2 due to q2 is

(Here i- points diagonally upward to the left)

Eaectrlc Charge, F o m and Field

The minus sign shows that the electric field points diagonally downward to the light.

NOW we:must add the two forces vectorially. If we resolve El and E2 into components along x-axis and y-axis, it is clear from the figure that y-components of vectors $ and & cancel out and those along x-axis, i.e., BO, add. The angle between either vector and the x-direction is 60' because the triangle formed by R, ql and -q2 is an equilateral triangle. The direction of the resultant field is, therefore, along BO and its magnitude is given by

EB = (1-08x lo4 cos 60° +1.08 x lo4 cos 60")

Try to solve the following SAQ so that you can work out the problems on your owfi.

SAQ 8

See Fig, 1.9. Find the electric field at the point C which is at a distance of 0.05m from ql . Till now we have considered electric field due to simple kinds of charge distribution, viz., an isolated point charge and an arrangement of two or more point charges. Now suppose that the charge is continuously distributed over a . region as shown in Fig. 1.10. Such situations occur frequently, for example, when we give some charge to a metallic body, it gets distributed over its surface.

charge distribution

Flg. 1.10: The electric field at piont P is the sum of the vectors dE arising from all the individual charge elements dq in the entire charge distribution. The appropriate distance r and unit vector P both vary from one charge element to another.

To calculate the electric field at point P due to this continuous distribution of charge, we consider the charged region to consist of many small charge elements dq. The charge element is chosen small enough so that every point in the element can be treated as equidistant from P. Now each dq will produce a small electric field dE in accordance with Eq. (1.9) as follows:

Then, in analogy with Eq. (1.10), the vector sum of all the dE'swill give the total electric field E due to the whole charged region. Here, we lpave made an approximation that, by placing together all the charge elements. we get the charged region. However, in order to improve the approximation, we have to make the charge eleinents dq infinitesimally small and for the continuous distribution, the vector sum will become an integral. Thus I

The limits of this integral are so chosen that it includes the entire region over which charge is distributed.

The region over which the charge is continuously distributed may be a line, an area or a volume as shown in Fig. 1.11.

Ekclrlc Chrgc, Force and field

Line clrarge A Surface charge , Volume charge p

(4 @I (4 Fig. 1.1k (a) Line charge @) Surface charge (c) Volume charge

In such distributions, instead of charges, we speak of the density of charges, viz., line charge density denoted by A, surface charge density o and volume charge density p. These quantities describe the amount of charge per unit length, per unit area and per unit volume respectively. They have units ~ r n - ' , and

Consider Fig. 1.1 la. Here the charges are distributed along a line. Consider a small element of length dl. The charge element dq will be equal to A dl, where h is line charge density. Replace dq in Eq. (1.12) by dq = h dl. Thus, the electric field of a line charge is

If you consider Cartestan ca-ordiqate, then in Eq. (1.13) the integration is performed with respect to single variable either x- or y- or z- axis depending on whether the line charge distribution is along x- or y- or z- +xis. Since a surface is defined in two dimensions, the integration in Eq. (1.14) is perfonntd with respect lo any two variables x and y; or y and z; or z and x. Similarly. a volume is defined in three dimensions and hence, in Eq. ( I . IS), the integration is performed with respect to three varieblcs x, y and z.

See Fig. l.llb. Here, the charges are spread over a surface. If dS is a surface dement, then the charge element dq will be udS where a is surface charge density. By replacing dq by dq =ads in Eq. (1.12), the electric field for a surfam charge is:

Similarly, in Fig. 1.1 lc, the charges are distributed over a volume having volume charge density p. If dV be a volume element then dq = p dV. Hence, the electric field due to volume charge is:

The vector integrals in Eq. (1.13), Eq. (1.14) .and Eq. (1.15) are called line integral, surface integral and volume integral respectively. If you have offered the course on

I Mathematical Methods in Physics-I (PHE-M), these terms will not be new for you , and you would be knowing their mrming. However, within this course, we have

defined and explained these terms as and when you encounter them in detail. Surface and volume integral has been explained in Unit 2, whereas tine integral has

1 been explained in Unit 3. Note that the charge densities, viz., A, o or p need not be constant. We can allow these charge densities to be function of position. Thus, in Eq. (1.13) to Eq. (1.15), X, o and p can themselves depend upon r. For example, in the case of a volume charge distribution p ( r l ) over a volume Y' (Fig. 1.12), the expression for electric field at the point P at r is

Fig. 1.12: Electric field at P due to a volume charge density p(r'). The vect~r r varies during integration, but r is fixed.

Electrostalfcs In Free Space

Another common way of representing the vector nature of equations lik,:: Eq. (1.4) is to write it as

where r12 is now a vector with t h e magnitude of the distance r12. The above equation has been obtained by using the relation

The rt2 in the denorninater of Eq. (1.4) has been changed to ri2 to compensate for this. Hence the meaning is identical with that of Eq. (1.4). However we shall use the Unit-vector notation.

Fig. 1.13: A pair of charge elements dq on either side of the origin contributes to a net field dlE in the y-direction.

p(r') (r - r') dV' E(r) = -

Here, the dashed (or primed) coordinates refer to the source point and r is the point where field E is evaluated.

Example 4

An infinitely long uniformly charged rod shown in Fig. 1.13 coincides with the x-axis and carries a line charge density X Cm-'.. What is the electric field at a point P on the y-axis?

Solution

Let the point P be at a distance of y along the perpendicular bisector of the rod. Consider a small length dx of rod containing charge dq located at a distance x to the right of the origin. Then

dq = Xdx

The magnitude of the electric field at P due to this element of chargc dq will be given by:

To determine the net field at P, we would write this dE in terms of its x- and y-components, arm then integrate each component over the entire line. Remember that for each dq to the right of the origin, there is a corresponding dq the same I

distance to the left. So, the x-components of the fields from such a pair cancel, while the y-components are the same. y-component of dE will be

1 dq Y = - - '.' cos 8 = - 4ueo r 2 r r

1 So that dE, = - ~ d 4

2 3 / 2 4mo (x2 SY )

Addition of the two equal y-components then gives the net electric field m,,:

The net electric field due to the whole rod will be: X=a

x = O

Although the line extends from - a to + a, we integrate over only half the line because the expression we are integrating is already the field of a charge pair dq. Substituting the expression for dq (dq' = X dx) and bringing constant out of the integral, we get

Put x = y tan 6' so that dx = y sec29 dB. Since tho rod is very long, as x ranges

9r from 0 to a, B ranges 'frqrn 0 to -. Then we have

. 2

Electric Charge, Fom and Fkld

- X -- X [sin 8]f2 S = -

2 r c g 2xeO J'.

Thus, the electric field due to infinitely long positively charged rod points radially outward'from the rod and its magnitude decreases inversely with distance.

As you have seen, in practice, the determination of electric field due to continuous charge distribution by above method requires more difficult calculations. There is a much easier and more elegant way to determine the electric field for such distribution which will be discussed in the next unit.

Line of Force

1.5.2 Electric Lines of Force In the last subsection, you learnt how to calculate the electric field (both magnitude and direction) at any point due to various charge distributions. In fact, the electric field extends throughout space. There is a useful method for representing the entire electric field visually by means of lines of force (also called electric field line or electric lines of force). This representation, though not good for quantitative purposes, serves a very useful purpose by allowing us to know the general features of the electric field in the entire region at a glance. The line of force is a line drawn in such a way that the tangent to it at any point shows the direction of electric field at that point as shown in Fig. 1.14. These lines are continuous and extends throughout space depicting the electric field.

Let us see how the lines of force provide information regarding the strength of electric field. Fig. 1.15 shows the lines of force due to a positive point charge. For a positive charge, the field at any point is directed away from the charge because a positive test charge would be repelled in that direction. So the lines of force are straight line pointing radially outward from the point charge. The lines' start on the charge and extend outward up to infinity. You would observe that the lines of force spread apart as they extend farther from the charge. Now you know that, according to Coulomb's law, the electric field decreases as you move away from the charge. So in Fig. 1.16, the electric field must be stronger at region A than at B meaning thereby that the electric field is stronger where lines of force are close together and weaker where they are farther apart.

Suppose, you have a charge q and another charge 24 and if you are told to draw the lines of force of these two charges, then can you draw as many lines of force as you want. No, it is not so. To make'the lines of force picture useful, we associate a fixed number of lines of force with a charge of given magnitude. So, if 6 lines of force emanate from a charge q, then a charge 24 will be represented by 12 l ies of force. The above statement is consistent with the Coulomb's law. Because, from Coulomb's law we know that electric, field is proportional to the magnitude of charge (Eocq). Therefore, the number of lines of force originating or ending (in case of negative charge) on charges is proportional to the magnitude of eachcharge. Fig. 1.17 shows the lines of force for two equal unlike charges and two equal like charges. It should be remembered that lines of force are not real and they do not actually exist as threads in space; they are simply a device t@ help our thinking about the field.

Fig. 1.14: An electric line of I force.

Line of force is also defined as a path along which a free, positive, point charge would travel in an :lectric field. Hence a line of force is always provided with an arrowhead indicating the direction of traveI of the positive charge.

.Fig. 1.15: Lines of force due to a positive charge.

(Region A) \ A

+ \ (Region B )

Fig. 1.16: The field is greater at region A than at B because in the region A, the lines of f,.ir:e are close together wherca~ 1 * ,he region B they are fartllrr - r I r.

5. 1.17 : The nature of lines of force. We have adopted the convention of drawing 18 lines per cbarge. (a) two unlike charges.

(b) two like positive charges.

SAQ 9

A charge of t 3pC and -lac are fured at a distance of 2cm from each other. Sketch the lines of force.

I Let us now sum up what we have learnt in this unit.

I 0 Only two types of electric charge mbst and they are arbitrarily cdect posttive and negative. Like charges repel and unlike charges attract each other.

I 0 In S1 system, unit of chaqe is Coulomb (C).

Charge is always conserv@.'.~hat is, the algebraic sum of charges in a closed region never changes.

I 0 Electric charge is quantized, occurring only in discrete amounts.

0 ~ a e form two charges is proportional to the product of their charges and inversely proportional to the square of the distance between them. The force acts along the line joining the two charges.

The value of 1/4rg is 9 x 10% m2 c - ~ , 0 The electric force on a charge due to the presence of two or more charges is

simply the vector sum of the farws caused by the individual charges. This important property of the electric force is known as superposition principle.

The electric field at a point in space is defied as the electric force exerted on a test charge'placcd at that point.

a The electtic field of a point charge q is given by

where i is a unit vector pointing from the point charge q to the location at which the el& fleld is b e b calculated.

a Thc electric field due to a distribution of charges, according to superpadtion principle, is the vector sum of the fields of the individual charges making up the distributibn:

With continuous distributions of charge, the sum becomes an integral over the entire charge distribution as foilows :

As an example, for volume charge distribution p(rt) over a volume V ' :

9 Electric lines of force are visual way of representing an electric field. It is a line in an electric field such that the tangent to it at any point shows the direction of the electric field at that point. The lines are close together where the fieEd is strong. tines of force. always begin or end on electric charges.

1) Two point charges $4 and +e are fuped at a distance of 'a'. A third charge q is placed cm a straight line joining these two charges so that q iS in equilibrium. Find the position of q. Under what circumstances will Phis equilibrium be 'stable' and 'unstable'?

2) ABCD is a square of 0.04 metre side, charges of 16 x f 0-', -I6 x f W9 and 32 x 1W9 Coulomb are placed at the points A, C and D respectively. Find the intensity of the electric field at point B.

3) A small object canying a charge of -5 x 1 O-'C experiences a force of 20x f 0 - ' ~ in the negative x-direction when placed at a certain point in an electric field. (a) What is the electric field at this point? (b) What would be the '

magnitude and direction of the force acting on a proton placed at t h i s point?

4). Two identical balfmns Piled with helium are tied with a load of 0.005 kg and are suspended in air in equilibrium. If each balloon carries a charge of q Coulomb, calculate the value of g. The length of each thread i s f m and the distance between their centres is O.Sm.

(Hint :

1 ~leetroststiar in Free Space As shown in Fig. 1.18, following forces act on the system in equilibrium

1 i) upward thrust, W ii) electrostatic .force, F

I iii) tension T iv) load

I In equilibrium, the moment of forces about the point C will be zero.)

1.8 SOLUTIONS AND ANSWERS

SAQ 1 .

Z will attract Y.

SAQ 2

According to the principle of conservation of charge, the amount of charge present before the radioactive decay (or any nuclear reaction) is equal to the amount of charge present after the decay.

In (i) the charge present before the decay is 92e and the charge present after the decay is 2e. Hence Eq. (i) will be

9 2 ~ 2 3 8 - w ~ h 2 3 d + z ~ e 4

Eq. (ii) is

&a4" + ,PI - 2 1 ~ ~ 4 4 + fll on1 is neutron.

SAQ 3

Since the conductor is positively charged, it is short of electrons.

Let it be short of n electrons, then from the relation q = ne, the number of electrons can be found out. Here q = 3.2 x 10-"C

and e = 1.6 x IO-'~C

The conductor is short of 200 electrons.

SAQ 4

If the force on q2 is to be fouild, it is only necessary to change every subscript' 1 to 2 and every 2 to 1.

1 9291 Force Fzl = - - 4sc0 r$ hi

where i2, is the unit vector from ql to gz.

SAQ 5

According to Coulomb's law, the magnitude of force acting between two charges a t a distance r apart in air is given by

1 Here - = 9x 10' N m2 c-~.

~ ' K Q

q, = q2 = 1.6~ IO-'~C (Since-the magnitude of charges on electron and protog .. are equal)

r = 5.3 x 10-"m

According to Newton's law, the gravitational force between electron and proton is given by

Here rn, = mass of electron = 9.1 x 10"' kg

m2 = mass of proton = 1.7 x kg

r = 5.3 x lo-" m G = 6 . 6 ~ lo-" N m2 kgg-'

You may observe that electrical force is lo3' times stronger t h a ~ gravitational force. Therefore in such problems gravitational forces could be neglected.

SAQ 6

Q1 + Q2 = 20pC or Q2 = (20 - Ql) pC

Since force is repulsive, the two charges are of same type. , According to Coulomb's law

Substituting the value of Q2 we get

QI (20 - QI > = 75 or Ql - 20 Ql + 75 = 0 or Q1 = 5 and 15.

Therefore, the charges are 5 and 15 pC.

1 SAQ 7 The solution to this problem is obtained by using the principle of superposition. This means that to calculate the force on the test charge, you can first calculate the force F1, due to the first charge alone (ignoring all the others); then you calculate the force Fz due to the second charge alone; and so on. Finally, you take the vector sum of all the individual forces,viz. F, + Fk + Fg ....... which gives the resultant force on the test charge. It is given that these individual forces are represented t y the sides of a closed pentagon. You know that if the vectors representing various forces form a closed geometrical figure then the resultant of these vectors is zero. This implies that resultant of these forces is zero. Therefore, the resultant force on the test charge is zero.

SAQ 8 The electric field at C due to ql is

El E . 4.03 X lo4 (-f) NC-' The electric Weld at C due to q2 is

EIectrlc Charge, Fom and Field

Electric forks dominate on a small -scale structure whereas gravitational forces dominate on 11 a largescale structure. I

I I1 I

The electric fields are oppositely directed so the magnitude of the net electric field E, at C is E, = El - E2, where El and E2 are the rniizitudes of the respective electric fields. ... E, = (4.03 - 0.48) x lo4 NC-'

= 3.55 x lo4 NC-I and it is directed towards negative x-direction.

SAQ 9

Fig. 1.19 shows the lines of force due to two charges: + 3pC and -1pC. We have adopted the convention that 12 lines of force emanate from 1pC charge. Then, a total of 36, lines must emaxlate. from the + 3pC charge and 12 of these must terminate on' the -1pC charge. Very near each of these point charges, the lines of force are radially directed, pointing outward from the positive charge and inward to the negative. To complete the pattern, 12 of the lines emanating from the positive charge are connected to the 12 lines of force terminating on the negative +

charge in such a way that no lines intersect. The connecting links are shown as dashed line. 24 lines of forced extend to infinity. Hence at a sufficiently large distance from this charge distributibn,'the field pattern is the same as that due to a net charge of +2pC.

Terminal Questions

Let the charge q be positive, As shown in Fig. 1.20, let us locate it at some point x to the right of +4e, so that x < a. At this point, the charge q is at a distar'ce (a - x) from + e, so that the force on q due to + e is

where (-1) is a unit vector from + e to q and is in the negative x-direction.

Tie charge q nes at a distance x from +4e so that the force on q due to +4e is :

where ! is a unit vector from +4e to q and is in the positive x-direction. For q to be in equilibrium, the two forces must cancel, that is,

we then have

or' the + sign we have

For the - sign we have

Here, 2a/3 is the only possible value of x because the charge g is placed to the right of -1-49 and x< IX. Therefore, for the equilibrium, the charge q is to be placed at a distance of 2a/3 from +4e.

We have assumed that q is positive. If q is slightly displaced (say towards right) from its equilibrium position, then the value of Fqrle will decrease and Fqe will increase, This is because of the fact that the electrostatic force between two charges is inversely proportional to the square of the distance between them. Hence, a net force (IF,, - Fqk) will act on q towards left due to which the charge will again return to its equilibrium position. Hence, it is clear that the equilibrium of q is stable.

If q is negative, then Fq4e and I?,, will be attractive forces and their direction will be as shown in Fig. 1.21.

+ 4e . . -4 +e ---- 4-- -------4

%-

1

The charge q will also be at equilibrium at a distance of 2a/3 from i-4e. If -9 is slightly displaced (say towards right), then Fqk will decrease while F,, will increase. ' Hence, a net force (F,, - Fq& directed towards right will act on -9. Therefore, -q will move more. towards right. Hence, it is clear that equilibrium of -q is unstable.

-

Refer to Fig. 1.22. The electric field at the point B due to the chargat at A, C and D are referred to as EA, Ec, and ED respectively. Therefore, the electric field E at B will be yector sum of EA, Ec and ED.

FJectric Charge, Force and Fidd

Electrostatics in F m Space ' l q Here EA = - -

4m0 r2

EA is directed along AB.

Similarly Ec = ( 9 x lo9 ~m~ c - ~ ) (-16 x C) (0.04m)

= 9 x lo4 NC-'.

Ec is directed along BC.

ED is directed 'along DB.

Now resolving ED along x- and y-axes, we get

component of ED along x-axis = ED cos 45"

component of ED along y-axis = ED sin 45"

Net component to E along x-axis = EA + ED cos 45" and

Net component to E along y-axis = Ec - ED sin 45" . :. Magnitude of resultant electric field E at B is given by

E = [(EA + ED cos 4S0)' + (Ec - ED sin 4 ~ " ) ~ ] "

= [(E; + E; + E; + (EA cos 45" - E~ sin 45' )I " = fl EA (because cos 45' = sin 45" = 1/G)

= m x 10' NC".

If B is the angle which E makes with the horizontal then,

Ec - ED sin 45' . tan tl = EA + ED cos 45'

8 = tan-' (0.171)

Hence, the resultant electric field, E makes an angle of 9" 45 ' with the horizontal.

3) Electric field E is given by

F E = - 9

Here F = -20x fN and q = - 5 x C.

.'. E = -20 x 10-~i N

= 4r NC" -5 x c

he direction of E is in the positive xbdirectibn - opposite to the force on the I

negative charge. Force F acting on any charge q placed in this field is given by the -relation'

L L I

2 8 P'= gE

Here q = charge on a proton = 1.6 x lo'-'' C .

and E = 4 NC-I

The force on proton is in the same direction as the field, i.e., towards positive x-direction.

4) A and B are'two balloons and each is acted upon by an upward thrust W

For equilibrium

2 W = load

or 2W = 0.005 kg

or 2 W = 0.0025 kg = (0.0025 x 9.81) N .

if F be the force of repulsion between the balloons then

As is clear from Fig. 1.18, following forces act on the system in the equilibrium position.

(i) Upthrust W, (ii) Force of repulsion F, (iii) Tension T in thread, and (iv) load.

In equilibrium, the moment of forces about C will be zero. So, taking moment of forces about C, we get

F x C D - W x A D + T x O - loadxO = 0

or F x CD - (0.0025 x 9.81)N i A D = 0

AD or F = 0.0025x9.81 Nx-

CD

q 2 But from Eq. ,(i), we have F = 9 x lo9 ~ r n ~ C-' x - (0.5m)'

0.0025 x 9.81 x0.5 x0.5 x0.25 AD = 0.25m and or q2 =

[(i12 - ( 0 . 2 5 ) ~ ~ ~ 9 x 109 C2 [.: co = ( l 2 -(o.251211/2 IXI I '

UNIT 2 GAUSS'S LAW

Structure

2.1 Introduction Objectives

2.2 What Gauss's Law is All About Counting Lines of Force Electric Flux

2.3 Gauss's Law

2.4 Gauss's Law - Some Applications Spherical Symmetry Line Symmetry Plane Symmetry A Charged Isolated~Conductor

2.5 Differential Form of Gauss's Law

2.6 Summary

2.7 Terminal Questions

2.8 Solutions and Answers

' 2 . INTRODUCTION

Michael Faraday carried out an interesting experiment. A large metallic box was mounted on insulated supports. Faraday went into the box and'got the box charged with a powerful electrostatic generator. He was completely safe. Can you believe? No. But it is true. Let us find out the reason for this strange phenomenon. Well, after going through this unit, you will bk able to find out the reason for it yourself because the explanation of this phenomenon is facilitated by the Gauss's law, which we discuss in this unit.

Gauss's law is a consequence of Coulomb's law, so though it contains no additional information, its mathematical form enables us to solve many problems of elGctric field calculation far more conveniently than through the use c~f Coulomb's law. In the preceding unit, you learnt that electric field at any point is given by t!g for& experienced by a unit positive charge placed at that point. In this unit, we will develop the concept of flux of an electric field and then arrive at the Gauss's law. We will also see how this law allows us to calculate the electric field far more easily than we could using Coulomb's law.

In mechanics, in addition to the conc pt of force, we introduce the concepts of 4 work and energy, Similarly, in electrostatic phenomena, we would discuss notions of work and energy. For this purpose, we need to discuss the concept of elcctric potential which provides a link between the concepts of electric field, work and energy. Therefore, the next unit deals with electric potential.

Objectives After studying this unit, you should be able to:

e appraise yourself that the number of lines of force crossing a closed surface is proportional to the net charge enclosed by that surface,

@ relate the electric flux through any surface to: (i) the field strength, (ii) surface area, and (iii) orientation of surface relative to the field,

e write the relation between-the electric flux and the charge enclosed within the surface,

@ compute the electric flux through any closed surface placed in the electric field,

a use the Gauds law to compute electric fields in case of spherical, linear and planar symmetry, "

@ discover how the Gauss's law enables the calculation of the electric field due to an infinitely long charged wire in a much easier way than the method of the preceding unit,

@ explain that if a conductor carries, a net charge, this charge must be distributed over its surface and there is no electric field inside a closed metallic box, and

@ show that Gauss's law can be written in terms of divergence of an electric field.

2.2 WHAT GAUSS'S LAW IS ALE ABOUT

Gauss's law expresses the relation between an electric charge and the electric field that it sets up. In the preceding unit, you learnt that the electric field of any charge distribution extends throughout space. You also saw how the electric field is visually represented by means of electric lines of force. It might strike you that there exists some relation between the charge and the number of lines of force. Do you remember that in Unit 1, we said that "number of lines of force" is a vague term because we can draw as many lines of force as we want? Here we shall define a quantity called flux of the electric field which is a very precise mathematical notion and can be physically thought of in terms of lines of forces. So, while the concept of lines of force is useful for picturising the electric field, the notion of electric flux is very useful for reformulating the laws of electric field.

2.2.1 Counting Lines of Force Before proceeding, let us do some analytical exercises. Fig. 2.1 shows some charge distributions.

Fig. 2.1: In all cases, the number of lines of force emerging from a closed surface is proportional to the net charge enclosed.

Those surfa& are c l o d for which there can be a clear distinction between points that are inside the surface, on the surface and outside the surface. In other words, a closed surface has an enclosed volume. For. such 8 surface, you can tell i t s inside from its outside; there is no ambiguity.

Ehxtmstrllcs la Free Space For each distribution, a number of closed surfaces are indicated by coloured lines. Count the lines of force and say how many lines of force cross each surface. Adopt the convention that a line of force crossing the surface from inside to outside is counted as positive and the one crossing from outside to inside as negative. Further, we draw by convention 8 lines emerging from charge q, 16 from 2q and so on.

Consider Fig. 2.la. For surfaces 1 and 2, the answer is eight. Surface 3 is a little ambiguous but the number of lines of force crossing this surface is also eight. This is because one line of force (shown bold) which crosses the surface three times actually does so twice while going out and once while going in for a net gain of one crossing. In fact, any closed surface you might draw that encloses the charge q would have eight lines of force crossing it. For surface 4; two lines of force cross it while going in and two while going out, making zero net crossings. So, no lines of force cross surface 4. Although, surface 4 lies in the field of the charge q but q is not inside the surface.

Figure 2.lb is identical except that now the surfaces 1, 2 and 3 enclose the charge 2q and hence sixteen lines of force cross these surfaces. Surface 4, which does not enclose the charge, still has zero net crossings. Figure 2.lc is similar-to Fig. 2.la except for the sign of the charge. So, -8 lines of force cross the surfaces enclosing the charge -4.

In Fig. 2.ld, surfaces 1 and 2 each enclose the charge q and hence eight lines of force cross each of these surfaces. But how many lines of force cross surface 3. Your answer should be sixteen. See what is the total charge enclosed by surface 3. The charge is q + q = 29 which is expected. The surface 4 has zero net lines of force crossings. The answer is obvious because it encloses no charge.

Let us see Fig. 2.le, which shows two charges equal in magnitude but opposite in sign. Eight lines of force cross the surface 1. It is obvious because it encloses charge q. Similarly, surface 2 encloses -q and has - 8 lines of force crossing it. Surface 3 is interesting. It encloses both charges q and - q. The net enclosed charge is zero. Count the lines of force. As many go out as come in, so the total number of Iines of force crossing the surface is zero.

This simple exercise of counting lines of force crossing the surfaces for various distributions tells you a simple statement about the electric field: The number of lines-of force croasing a closed surface is proportional to the net cha'rge enclosed by that surface.

SAQ 1

What would happen if we were to bring an enormous charge Q close to the surface 3 in Fig. 2.le? How many lines of force will cross the surfaces 1, 2 and 31

To describe rigorously this very interesting observation about lines of force crossing a surface, we will develop a 'new concept, that of the electric flux.

2.2.2 Electric Flux

In the last subsection, we considered closed surface. Let us now consider an open surface such as a flat sheet of area S placed in a uniform electric field E which is represented, say, by means of four lines of force as shown in Fig. 2.2a.

NOW if we have another uniform electric field 2E, then such electric field will be represented by eight lines of force according to our earlier convention. Now bring the same flat sheet into this field (shown in Fig. 2.2b) and count the lines of force crossing this area. Answer will be eight. In Pig. 2.2c, the field is the same as in Fig. 2.2b, but the area of the sheet is halved and so the number of lines of force crossing the sheet is halved. If the flat sheet of Fig. 2.2b is turned to position shown in Fig. 2.2d, then it remains no longer perpendicular to the Iines of force and hence the number of lines of force crossing the sheet is reduced.

We see that the number of lines of force crossing any surface depends on three things: Field strength E, surface area S and orientation of surface relative to field.

The surfaces shown in Fig. 2.2 are open surfaces because they do not define an enclosed volume.

Fig. 2.2: (a) The number of lines of force crossing a flat surface depends on, (b) the field strength, (c) the surface area, and (d) the orientation of the suiface relative to the field direction.

To specify the orientation of the surface, we draw a perpendicular to the surface. If 8 is the angle between the electric field and the perpendicular as shown in Fig. 2.3, then the number of lines of force passing through the surface range from maximum to minimum depending on 8. That is,

when 8 = 0" ; lines of force crossing the surface is maximum

when 8 = 90" ; lines of force crossing the surface is zero.

Clearly the number of lines of force crossing a surface is proportional to the projection of the field on to the perpendicular to the surface, i.e., cosine of 8 . [Note cos. 0" = 1 and cos 90' = 01. Putting together the three quantities on which the number of lines of force depends gives

number of lines of force crossing a surface aES cos B.

By including cos 8 in the dot product you can write

number of lines of force crossing a surface a E S . . .(2.1)

where E is the electric field vector and S is a vector whose magnitude is equal to the area of the surface and whose direction is that of the perpendicular to the surface.

The quantity on the left side of Eq. (2.1) is a vague term because we can draw as many lines of force as we like. But the quantity on the right side of the equation

rl . has a definite value and is called electric flux denoted by +. Hence,

+ = E*S ...( 2.2)

Eq, (2.2) shows that flux being the scalar product of two vectors, is itself a scalar quantity. Since E is measured in NC-', SI unit of flux is IVrn2~''.

To find the total flux through any closed surface placed in a non-uniform electric field as shown in Fig. 2.4, we divide the surface into many small patches so that the electric field is nearly uniform over each patch. Then from Eq. (2.2), the flux diP through each patch will be

- C-A-

Fig. 2.3: The numher of lines of force crossing the area is proportional to cos 9, where B is the angle between the field and the perpendicular to the surface.

Dot (scalar) product of two vector quantities A and B is given by A.B = AB cos 0 where 9 is the angle between the vectors A and B.

where E is the electric field at the patch and dS is a vector whose magnitude is the area dS and the direction is that of the outward drawn normal to surface.

-

Fig. 2.4: (a) A closed surface of arbitrary shape immersed in an electric field. Its surface is divided into small patches of area dS. (b) The electric field vectors E and the area vectors a'§ for three representative patches marked x, y and z.

Total flux through the surface will be obtained by adding the fluxes through all the patches. Suppose each patch becomes infinitesimally small and the number of such patches is arbitrarily large, then the sum .becomes a surface integral. Hence, the total flux is given by:

If the surface is closed, one often indicates this in the integral as

The circle on the integral sign indicates that the integration is to be taken over the In order to.simplify the writing entire (closed) surface. Before explaining how the above expression leads to the of integrals, surface integrals are Gauss's law, you would like to know the meaning of the surface integral. written only with one integral sign. It will be understood that A surface Integral of any vector function F. over a surface S means just this: when the integral is taken over Divide S into small Datches (surface elements). Each ~ a t c h is a vector auantitv. -. an area dSp surface jntegra' represented by a vector pjoin;ing towards outkard noimal and having magnitude or double integral is implied.

equal to the patch area. At every patch, take the scalar product of the patch area vector and the vector function F. Sum all these products. The limit of this sum, as the patches shrink, is the surface integral.

Let us see how Eq. (2.4) is used to find out the total flux through any closed surface.

Example 1

Fig. 2.5 shows a closed surface S in the form of a cylinder of radius R im'inersed in q uniform electric field F, the cylinder axis being parallel to the field. What is the flux cP of the electric field through this closed surface?

ds

ds

34 Fig. 2.5: A cylindrical surface, closed by end caps, is immersed in a uniform electric field. The cylinder axis is parallel to the field direction.

Solution

We can write the total electric flux through the surface S as the sum of three terms, an integral over the surfaces: S1, i.e., the left cylinder cap, SZ, i.e., the cylindrical surface, and S j the right cap. Thus, from Eq. (2.4) we have

* = j E . &

= S E . d S + J E . & + {E.&

Sl sa 'S3

For the left cap, angle t9 for all points is 180°, E is constant, and all the vectors a5 are parallel. Thus,

E dS = E (cos 180") dS = -E d~ = - = E R ~ S S Sl S because =R2 is the cap area. Similarly, for the right cap,

J E . ~ s = + TER'

s, the angle t9 for all points being zero there. Finally, for the surface S2,

' j ' E - & = O ,

s2

the angle t9 being 90" for all points on the cylindrical surface. Total flux through the cylindrical surface S becomes

- T E R ~ + o + TER' = 0.

Therefore, the net outward flux of the electric field through this closed surface is zero.

2.3 GAUSS'S LAW

'In the last sectioq, we found two simple results: (i) the number of lines of force crossing through any closed surface is proportional to the net charge enclosed by that surface, and (iiythe concept of flux which' quantifies the physical notion of lines of force crossing a surface. The net result is: The electric flux through any closed surface is proportional to the net charge enclosed by that surface. Mathematically,

* 01 genclosed . . .(2.5) or

* = j' E . a qe"c,owi ...(2 .ti)

To evaluate the proportionality constant in Eq. (2.5) or (2.6), consider a positive point charge q placed in free space and a spherical surface of radius R centered on q (Fig. 2.6). The flux through any surface is given by Eq. (2.4), i.e.,

where B is the angle between the direction of the electric field and the outward drawn normal to the surface. You have seen in Unit 1 that the magnitude of the

. electric field at a distance R due to a point charge q is given by

where eo is permittivity of free space. The field points radially outward so that the electric field is everywhere parallel to the outward drawn normal to the surface.

Pi8. 2.6: The y i c field of a point charge 4 as the same magnitude over a spherical surface centered on the charge and is everywhere perpendicular to the surface.

~ ~ ~ t i o s t a t i c s in ~ r e e Space Then 8 = 0, so Jhat cos 0 = 1. Putting the values of E and cos 8 in Eq. (2.7), the flux through the spherical surface of radius R becomes

sphere

The expression. for the magnitude of electric field has been taken outside the integral sign because it has the same value (or in other words it is constant) everywhere on the spherical surface. The remaining integral is just the sum of the areas of all the infinitesimal elements, dS, on the surface of the sphere - in other words the remaining integral is the surface area of the sphere, i.e., 4x~'. Then the flux becomes

Comparison of Eq. (2.8) and Eq. (2.5) shows thay the proportionality constant is l/eo. The value of eo is 8.85 x lo-'' CZ N-' m- '. Sb Eq. (2.6) becomes

This is known as Gauss's law. It tells us Chat the! electric flux through the sphere is propoftional to the charge and independent of the radius of the surface. In order to prove Eq. (2.9) for any arbitrary closed surface, we will first define what is meant by solid angle.

A solid angle is the space included inside a conical surface, as shown in Fig. 2.7a. Its value is expressed in steradians (abbreviated sr). Its value is obtained by drawing, with arbitrary radius R and centre at the vertex 0 , a spherical surface and applying the relation

where S is the area of the spherical cap intercepted by the solid angle. Since the surface area of a sphere is 47r~', we concludk that the complete solid angle around

, a point' is 4a steradians.

. .

0) (b) (4

Fig. 2.7: Solid angle

When the solid angle is small (Fig. 2.7b), the surface area S becomes dS and is not necessarily a spherical cap, but may be a small plane surface perpendicular to OP so that

In some instances, the surface dS is not perpendicular to OP, but its normal N makes an angle 8 with OP as shown in Fig. 2 .7~. Then it is necessary to project dS on a plane perpendicular to OP, which gives us the area dS' = dS cos 8. Thus

.The definition of solid angle puts no limitation on the shape of the cone. What matters is the area of the projected surface area, not the shape of the projected area.

Now let us consider a charge q inside an arbitrary closed surface S as shown in Fig. 2.8. The electric field E at every point, of the surface is directed radially outward from the charge. Let us consider any sufficiently smalls area dS on the surface for which E can be considered to have same magnitude and direction. If 8 be the angle between E and the outward normal to the surface dS, then according to Eq. (2.3), the flux d@ through the area dS is given by

Gauss's LRW

Here dScos t31r2 is the solid angle 6n subtended by the surface element dS as FIB. 2.8: ~ h c clecrrlc flux viewed from the charge q. Hence, through a closed surface

surrounding a charge is independent of the shape of the surface.

I !

To obtain the total flux through the surface S, integration is done over the entire I I ' 1 ,closed surface as follows: I

~he'total solid angle around any point is 4 ~ . heref fore,

Q . 4~ = -- @ = $ E . & = - 4reo eb

This result is the same as the previous result for a spherical surface concentric with the charge. Thus, the relation expressed by Eq. (2.9) is valid for any closed surface irrespective of the position of the charge within the surface,

SAQ 2 Can we use Gauss's law for the surfice shown in Fig. 2,2? Give reasons.

Nqte that q in Eq. (2.9) is the net charge, taking its algebraic sign into account. If a surface encloses equal and opposite charges, the flux is zero. Charge outside the surface makes no contribution to the value of p. However, E on left side of Eq. (2.9) is the electric field resulting from all charges, both those inside and outside the surface.

SAQ 3

Fig. 2.9 shows three objects each carrying an electric charge and a coin carrying no charge. The cross-sections of two surfaces S1 and S2 are indicated. What is the flux of the electric field through each of these surfaces? Assume q, = +3.1 nC, 9 2 = - 5.9 nC and 9, = - 3.1 nC.

Instead of assuming point charges to be contrained within the closed surface, let us assume that the charge is continuously distributed throughout the volume enclosed by the surface. Let the chargedensity be p. l f dV be a small volume element then the charge contained within this volume elelment will be p dV. Therefore, total charge enclosed within the entire volume is jJjp dV. Then Gauss's law is expressed as:

For simplicity, also written as

Flp. 2.9: Three objects each ' i 1 I f wrying an electric charge; and a , 1,

coin which carries no charge.

Instead of wnting multiple integral signs, for simplicity, integration is represented by onc inteiral sign. When the integral is taken over a volume dV, it will be implled as volume Integral or lrlple Inlegmi.

T&lrostntics In Free Spare

Fig. 2.10: For a'sphericalIy symmetric, charge distribution. electric field vectors at a given radius, all have the same magnitude and point in the radial direction.

/ 4 4 Gaussian

/ , surface I , / r \ / ' I . ,

I I

I d - - - c E \

I I

1 I Flg.'2.11: A spherical ~aussiar! surface centered on a point

where j indicates volume integral through the.region enclosed within the closed surface. 'Eq. (2.10) is an integral form of the statement of Gauss's law. '

Now you would like to know the meaning of volume iategral. Let the region enclosed within the closed surface be V. It is a region in three dimensional space. Let p be a quantity, say, charge which is defined for unit volume. Divide the region V into n elementary volumes dVl , dVB ...., dV,. Then the product of the quantities p and dV l will give the charge enclosed within elementary volume d V , . Sum all these products, i.e.,

n

Then the limit of this sum, as n tends to infinity and the dimensions of each sub- division tend to zero, is called the volume integral of p over the region V. It represents the total charge enclosed within the entire region V and is denoted by SSS P d V .

The solution of Eq. (2.10) is often difficult to perform mathematically, although the physical meaning of the law in this form is more comprehensive. A differential form of Gauss's law is useful for the solutibn of many problems. We shall arrive at this form of Gauss's law a t a later stage in Section 2.5. Now let us see how Gauss's law is useful in determining field distribution about objects having symmetrical geometfl. We shall discuss a few such cases in the next section.

2.4 GAUSS'S LAW - SOME APPLICATIONS

Gauss's law applies to any hypothetical closed surface (called a Gaussian'surface) and enclosing any charge distribution. However, evaluation of surface integral becomes simple only when the charge distribution has sufficient symmetry. In such situation, Gauss's law allows us to calculate the electric field far more easily than we could using Coulomb's law. Since Gauss's law is valid for an arbitrary closed surfgce, one uses this freedom to choose a surface having the same symmetry as that of charge distribution to evaluate the surface integral. We will illustrate the use of Gauss's law for three important symmetries. .

2.4.1 Spherical Symmetry A charge distribution is spherically symmetric if the charge density (that is, the charge per unit volume) at any point depends only on the distance of the point from a central point (also called centre of symmetry) and not on the direction. Fig. 2.10 represents a spherically symmetric distribution of charge such that the charge density is high at the centre and zero beyond r. Spherical symmetry of charge distribution implies that the magnitude of electric field also depends on the distance r from the centre of syrnnietry. In such situation, the only possible direction of the field consistent with the .symmetry is the radial direction-outward for a positive charge (Fig. 2.10) and inward for a negative charge. The examples of spherically symmetric charge distributions are: (i) a point charge, (ii) a uniformly charged sphere, and (iii) a uniformly charged thin spherical shell.

(I) E of a point charge

Fig. 2-11 shows a positive point charge q. Using Gauss's law, let us find out electric field a t a distance of r from the charge q. Draw a cancentric spherical Gaussian surface of radius r. We knolw from symmetry that E points radially outward. If we divide the Gaussian surface into differential areas dS then, both E and dS will be at right angles to the surface, the angle 8 between them being zero. Thus, the quantity E. dS becomes sinlply E dS and Gauss's law (see Eq. 2.9) becomes

Because E has the same magnitude for all points on the Gaussian surface, we can factor it out of the integral leaving

,

However, the integral in Eq. (2.11) is just the area of the spherical surface, i.e., 4xr2, SO that the equation becomes

rt

Gauss's Law

. . or

1 q E = - - 4?r go r 2

which is Coulomb's law in the form in which we have written in Unit 1. This shows that Gauss's law and Coulomb's, law are not two independent physical laws but the same law expressed in different ways.

(ii) E of a spherical charge distribution . ~

I . ! ; . Suppose a total charge Q is spread uniformly throughout a sphere of radius R as shown in Fig. 2.12. Let us find the electric field at some point such as P1 outside ~ i g . 2.12: The electric field of a the distribution and at point P2 inside it. If you use Coulomb's law to find out the spherical charge distribution. The '

field, you have to carry out an integration which would sum the electric field surface S, encloses the entire charge Q; while surface S2 vectors at PI arising from each elementary volume in the charge distribution. Let's encloses only some of the charge.

try a different method using Gauss's law.

(a) Field for.points outside the charge distribution: Let us draw an imaginary Gaussian surface SI of radius rl through the point P I , where we wish to find electric field. Let the magnitude of the field be denoted by El. Because of the spherical symmetry, the electric field is same at all points on this Gaussian surface. Also at any point on the Gaussian surface, the field is radially directed, i.e., perpendicular to the surface so that cos 9 = 1. (Here it is assumed that the sphere *

has net positive charge; if there is net negative charge, the field will point radially inward and cos 8 = -1). Then the flux through this Gaussian sphere SI becomes

.-" . because f dSl.is just the surface area of the sphere S1, i.e., 41.r:.

4 According to Gauss's law, the flux through the sphere S1 is given by -, where q '4

is the net charge endosed by the sphere S1. Equating the flux.in Eq. (2.13) to

- gives €0

Q 4s r f El = - (because charge enclosed within the sphere S1 is Q) €Q

so that

This shows that the beld at all points on surface S1 is the same as if all the ' charges' within the surface S, were concentrated at the centre.

(b) Field for points inside the charge distribution: The field inside the charge distribution depends on how charge is distributed, This is because 'ay Gaussian

' sphere with r < R, such as surface S2 of Fig. 2.12, does not enclose the entire charge Q. The charge enclosed depends on the charge distribution. Suppose, a Gaussian sphere Sz of radius r2 is drawn passing through 'the pojnt Pa. where we

. wish to find the electric* field. Let the field be denoted by E2. Inside the sphere S2 Eq. (2.9) For the flux still holds, but now the oharge enclosed is some fraction of

Eledrostntics in Free Space

Fig. 2.13: Field strength versus radial distance for a uniformty charged sphere of radius R. For r > R, the field has the inverse- square dependence of a point charge field.

4

Flg. 2.14: A charged spherical shell. Any Gaussian spherc inside the shell encloses zero net charge and the field inside is zero.

Mg. 5-15: The net charge enclosed by the sphere is zero, but the field within the sphere is not zero. Here the charge distribution - a dipole - is not spherically symmetric, so that Eq. (2.13) is not a valid expression for the flux.

40

47r Q. The volume of the charged sphere is - R~ and it contains a total charge Q.

3 Since charge is spread uniformly throughout the sphere, the volume charge density p is constant and is given by:

Therefore, the charge enclosed by the sphere S2 will be just the volume of that sphere multiplied by the volume charge density, that is,

From Eq. (2.9). we have

so that I

The electric field inside the charge distribution increases linearly with distance from the centre (E CY r) whereas outside the charge distribution, the electric field falls

1 off as - as clear from Eq. (2.14). Fig. 2.13 shows the combined results for the

r 2

fields both inside and outside the sphere.

(iii), Field of a thin spherical shell

Consider a thin spherical shell of radius R carrying a total charge Q distributed uniformly over its surface as shbwn in Fig. 2.14. Since this distribution is spherically symmetric, we already know that the electric field outside the shell is the point charge field of Eq. 2.14. To find the electric field inside the shell, a Gaussian sphere is drawn inside the shell. The charge enclosed within this Gaussian surface is zero. Equating the flux from Eq. (2.13) to this zero enclosed charge gives: 4% r 2 ~ = 0 .

Since r f 0, the field is zero everywhere inside the shell. Remember that the zero field inside the shell did not follow only from the fact that the charge enclosed within it is zero; but it was possible because of the spherical symmetry which leads to Eq. (2.13). Now consider a spherical surface S ground a dipole as shown in Fig. 2.15. Here the net charge enclosed by the surface is zero but field within the surface is not zero. This is because here the charge distribution- a dipole - is not spherically symmetric, so that Eq. (2.13) cannot be obtained although the flux (i.e., net number of lines of force) crossing the surface is zero.

The above examples have shown that, to calculate the electric field using Gauss's law, the following steps are needed:

1) Study the symmetry to see if you can construct a Gaussian surface on which the field magnitude and its direction relative to the surface are constant. If this is not possible, then Gauss's law, although true, will not provide a simple calculation of the field. Gaussian surface can be of any size or shape as long as it is closed.

2) Evaluate the flux, The choice of the Gaussian surface makes the term Ecos 0 constant so that this term can come outside the flux integral leaving an integral equal to the suiface area.

3) Evaluate the enclosed charge. If the ~auss ian surface is within the charge distribution, then the enclosed charge will not be the same as total charge.

4) Equate the flux to qenClod/@~ and solve E. The direction of E can be determined from the symmetry,

2.4.2 Line Symmetry Gauss's LBW

A charge distribution has the cylindrical symmetry when it is infinitely long and has a charge density that depends only on the perpendicular distance from a line I called symmetry axis (Fig. 2.16). By symmetry, the electric field will point radially outward from the axis and its magnitude will depend only on perpendicular distance from the axis. (Here, we assume positive charge, for negative charge the field points inwards.) Let us find an expression for E at a distance r from the line charge (say a wire).

, . Draw a Gaussian surface which is a circular cylinder of radius r and length i closed r- at each end by plane caps normal to the axis as shown in Fig. 2.16. To calculate T: the flux, note that electric field also has the cylindrical symmetry, which implies its magnitude at a point depends only on the perpendicular distance of the point from the symmetry axis, and its direction has to be radially outwards. (You should try

- out other possibilities for field direction and convince yourself that there is only j; one possibility compatible with the cylindrical symmetry.) The flux through the cylindrical surface is .. - * = !E.& = J E ~ S = E i d s = 2xrlE ... (2.16)

where 2nd is the area of the curved surface. The flux through the end of the cylinder is zero because the field lines are parallel to the plane caps of the Gaussian surface. Mathematically, the vector E and dS are perpendicular, so that cos 9 = 0 in the dot product E. dS. Therefore, the only flux is through the curved Fig. 2,16: A cylinder of length I part of the cylinder given by Eq. (2.16). Gauss's law tells us that the flux is and radius r is the Gaussian proportional to the chargC enclosed within the cylindrical Gaussian surface, i.e., surface. ~t encloses a portion of

an infinitely long line charge. qenclosed 2arlE = ---

€0 - so that E = qenclosed .. . (2.17)

2n ~$1

If the line charge density is A, then the charge enclosed by the Gaussian cylinder of length 1 is XI. Using this expression for qenclOsd in Eq. (2.17) gives:

...( 2.18)

This is the same-result we found in Example 4 of Unit 1 through a tedious Coulomb's law method. Notice how much simpler is the Gauss's law.

In order to find the electric field inside the wire, we consider two cases:

i) Suppose the charge is distributed uniformly within the wire and charge density is p. Let the radius of the wire is R. To find E at an inner point P, a distance

fj$$ .--++

r apart from the axis of the wire, draw a Gaussian cylinder of radius r and -

length I passing through P as shown in Fig. 2.17. As explained earlier, the flux is due to the curved surface only. Hence, from Gauss's law

S Q ' E.dS = E2nri = - Ng. 2.17: Enlarged view of the

Eo wire is shown to calculate the electric field at any point within

The charge q' inside this Gaussian surface = ?rr2pl. it.

ur2pl E 2rrl = -

€0

rP or E = - .. .(2.19) 2 ~ 0

Thus, the electric field at a point inside an infinite uniformly charged wire is radially directed and varies as the distance from its axis.

ii) When the charge is on its surface only, the electric field at any point inside it is zero because the net charge in the Gaussian surface through this point is zero. 41

+ + t

- - t

--.- + t t +

- _ 1 -

. ,

.:; Gaussian 6- I . .\ Furtace .,.I

_ . I a; ::,; ,E

., . .,

Eiectrostatics in Free Spnce . Eqs. (2.18) and (2.19) show that the electric field due to a charged wire or cylinder does not depend upon its radius. Hence, we can say that it is same as though the charge on the wire or cylinder were concentrated in a line along its axis. (Remember it while solving Terminal Question 4.)

2.4.3 Plane Symmetry E

When the charge density depends only on the perpendicular distance from a plane, the charge distribution is said to have plane symmetry. The electric field is everywhere normal to the plane sheet as shown in Fig. 2.18 pointing outward, if positively charged and inward, if negatively charged. To find the electric field at a distance r in front of plane sheet, it is required to construct a Gaussian surface. A convenient Gaussian surface is a closed cylinder of cross-section area S and length 2r. The sides of Gaussian surface are perpendicular to the symmetry plane and the

t

IC ends of the surface are parallel to it. Since no lines of force cross the sides, the flux through the sides is zero.'But the lines of force cross perpendicular to the

rig. 2.18: A charge distribution ends. so that E and the area element vector dS on the ends are oarallel. The cos B with plane symmetry showing the in th, product E, dS is 1 over both ends (-1 i f charge is negativk). Since the flux electric field. The area of the sheet enclosed by the Gaussian

through the sides is zero, the total flux through our Gaussian surface then becomes

surface is the same as the area S of its ends. @ = 1 E d S = 2Es.

both ends

The factor 2 arises because there are two ends. Then Gauss's law gives

If a is the surface charge density, then the charge enclosed is US, then

2.4.4 A Charged Isolated Conductor Gauss's law permits us to prove an important theorem about isolated conductors.;

" If there are any unbalanced, static charges on a conductor, they must reside on the surface of the conductor."

I

Consider a solid metallic conductor such as the one shown in Fig. 2.19a carrying a charge q. The dotted line shows a Gaussian surface that lies just below the actual surface of the conductor.

52; Gaussian surface

Fig. 2.19: An insulated solid metallic conductor carrying a charge q. (a) A Gaussian surface is drawn within the metal, just bdow the actual surface. (b) The conductor has an internal cavity. A Gaussian surface lies within the metal close to the cavity wall. (c) A cylindrical Gaussian surface pierces the surface of 'the conductor. It contains a charge oS.

Before proving the theorem, you should realize that, inside the charged metal object, the electric field is zero everywhere. We can see why this must be true without a formal calculation. Suppose, there were an electric field in the interior of the object. Then the charges inside the object that are free to move (electrons in the case of met& would do so under the influence of the field and internal' currents would be set up. But no currents are observed in a charged conducting object except for a short time after 'the charge is placed on it; since some energy is needed to maintain an electric current, a supply of energy would be needed for currents to continue in such an object. The only conclusion is that the interior'of an isolated conducting object is always free of electric field.

If electric field is zero everywhere inside the conductor, it must be zero for all points on the Gaussian surface because that surface, though close to the surface of the conductor, is definitely inside it. This means that the flux through the Gaussian surface must be zero. Then. according to Gauss's law, the charge inside the Gaussian surface must also be zero. It follows that if a net charge does reside on the body, it can be distributed only over the surface layer of that body.

Since the interior of the solid conductor contains no unbalanced charges, we could scoop out some of the material, leaving a hollow cavity as shown in Fig. 2.19b. Draw a Gaussian surface surrounding the cavity but inside the conducting body as shown in Fig. 2.19b. There can be no flux through this new Gaussian surface because E = 0 inside the conductor. Therefore, from Gauss's law, that surface can enclose no net charge. We conclude that there is no charge on the cavity walls; it remains on the outer surface of the conductor as in Fig. 2.19a.

This result has immense practical implications. It tells us that we can shield an object from the influence of electrostatic fields by simply enclosing it within a rnnductor sheath.

SAQ 4

Go through the introduction of this unit once again and find out the reason of the strange phenomenon mentioned in it.

Now you would like to know the electric field outside the charged conductor. Let us find out in the next paragraph.

Electric field near a charged conductor

Suppose, magnitude of the electric field at the point P, just outside the charged conductor, be E. As shown in Fig. 2.19c, draw a cylindrical Gaussian surface such that the end caps are parallel to the surface, one lying entirely inside the conductor and the other entirely outside but very close to the surface. The qrea of its two end caps is S and the point at which electric fie!d is lo be required is within the cap. The cylindrical walls are perpendicular to the surface of the conductor. The direction of the electric field just outside a charged isolated conductor is perpendicular to the surface. This is because if E had a component parallel to the surface, electro~~s on the surface would be in constant motion. Because they are not so, E must be perpendicular to the surface. The flux through the exterior end cap of the Gaussian surface will be ES. The flux through the interior end cap is zero because E = 0 for all interior points of the conductor. The flux through the cylindrical walls is also zero because the direction of E is parallel to the surface, so ' they cannot pierce it. The charge enclosed by the Gdussian surface is US where u is the surface charge density at the point near which we are to find electric field. Hence, after summing up;the total flux through the entire Gaussian surface is ES and the charge enclosed by that surface is US. Gauss's law then gives

Gauss's Law

Hectrostaiieu in FEW Space At the end of the Section 2.3, we mentioned that Gauss's law can be stated in two forms: integral and differential. In that section, the Gauss's law was written in integral form. Before ending this unit, let us see how differential form is obtained.

2.5 DIFFERENTIAL FO OF GAUSS'S LAW

Consider an infinitesimal element of volume AV with sides Ax, Ay and Az parallel to the axes of x, y and z respectively as shown in Fig. 2.20. Let the electric field be E at the middle of this element and suppose it has components of magnitude Ep. E, and E, along the x, y and z-axes respectively.

Fig, 2.m : An infinitesimal element of volume A V having sides h, Ay and Az parallel to thew, y and z-axes.

Consider the two faces of the volume element each with area LLV Az perpendicular to the y-axis. On the middle of the right-hand face of the volume element which is at a distance Ay/2 away from the middle of the element, the approximate value of electric field is given by

Since the face area is infinitesimally small, hence the above value may be taken as the value all over the face. he second term in the above expression is the space rate of change of Ey as we move in the positive y-direction. For your convenience, we tell you how the above expression for electric field is obtained.

It is often necessary to find out what is the increment (or change) in any quantity when the variable on which it depends is changed by a small amount. This can be found at once by Taylor's theorem: In'differential calculus, you must have read about the Taylor's theorem.

If f(x) is a quantity which depends on the variable x and, if x changes by a small amount dx, then the quantity f (x ) will change to f (x+dx) and, according to Taylor's theorem its value will be given by

If dx is small enough, then second and higher powers of dr can be neglected and we have

Thus the change, df (x), in the quantity f (x) is given by :

Thus, the derivate -!!&!?- tells you how rapidly the function f(x) varies when x dx

changes by a tiny amount dx. In other words, if x changes by an amount dx, then f(x) changes by an amount df (x).

AY df(x) by %. ind & by - . Replacing f (x) by E, ; - dx ay 2

We get the electric field a t the right-hand face as:

Tlre partial differential coefficient of f ( x , y ) with respect to x is the ordinary differential coefficient off (x, y ) when y is regarded as a constant. I t is written as aJ/ax.

Notice that, instead of ordinary differentiation, we have used padhP differenaiatisn, I *. This is because E, is a function of x, y and r; and we have to differentiate ar

E,, with respect to only y while keeping x and z as constant.

It is hoped that you have understood the mathematical portion given above. Let us now move further.

The value of the electric field on the left-hand face of the volume element is i l

The flux through the right-haxid face of the element ip the y-direction is given by I ' l l I I / 1

And the flu through the left-hand face of the element in the y-direction is given by

Hence, the net outward flux through these 'two faces of the element in the y-direction is given by :

Likewise, for the other two faces of the element perpendicular to x-direction, the net outward flux is:

and for the top and bottom faces, the net outward flux is

Therefore, total net outward electric flux -&rough the volume element AY will be the sum of Eqs. (2.221, (2.23) and (2.24), i-e.,

=(%+- aEy + 5) A Y (because A V = br Ay Az) .. .(2.25) a~ az

4f As already stated, a derivative of the type - is s u p p e d t o tell us how fast the dx

quantity f varies if the variable x on which the quantity depends is changed by a small amount. Suppose f is quantity depending on x, y and z, then according to the theory of partial derivatives

d

Elleetrostatlcs In Free Space

I In the definition of v id Eq. (2.27), the unit vectors are written to the left instend of to the right just because one may not confuse

a f with - which would be zero,

ax

since is constant.

This rule tells us how f varies as we go a small distance (dx, dy, dz) away from the point (x, y, z). If dx, d y and dz are the components of a vector dl along x, y and z-axes, then Eq. 2.26 is .reminisceat of a dot product as follows:

= (Vf, (dl)

a a a where V (called 'del') = I' - + - + -

ax au az v is a vector operator which acts on a scalar or vector function via the dot product or cross product. Suppose we want to operate V on a vector function E via the dot product, then from the definition of V, as given in Eq. (2.27), we have

v -E has been given the name divergence. It is also written as div E. The divergence of a vector function (here' it is electric field) is a scalar quantity and it measures how much the vector E spreads out (diverges) from the point in question. If the vector function E has a large (positive) divergence at any point P as shown in Fig. 2.21a, then it is spreading out. (If the arrows pointed in, it would be a large negative divergence.) On the other hand, the electric field vector in Fig. 2.21b has zero divergence at P, which means it is not spreading out at all.

Now using Eq. (2.28), we can write Eq. (2.25) as follows:

Total electric flux through volume element AV = (div E) AV ...( 2.29)

(a) (b) Fig. 2.21 : (a) Positive divergence at P (6) Zero divergence at P.

Thus, according t o Eq. (2.25), the flux through a small volume is given by (div E) AK Using the original definition of the flux given in Eq. (2.4), we may write

If p is the 'volume charge density, then the total charge enclosed within the volume element AV will be given by:

4 = pAV ...( 2.31)

Now, see Section 2.3 and write down Eq. (2.9)

Substitute the value of E . dS and q,,,,,, from Eqs. (2.30) and (2.31) respectively, we get:

Eq. (2.32) is the alternative statement of Gauss's law. This defines the Gauss's Law in differential form. The divergence of the electric field represents the nbt amount of flux coming out of a unit volume element. If theldivergence of the electric field is positive (or negative) at any point, then the electric flux is emanating (or terminating) from (or on) the closed surface enclosing the charges at that point. Also div E is zero in any region in space having no net charge, Thus, the existence of finite positive value of the divergence at a point shows that there must be a positive charge at the concerned point.

Gauss's Theorem

Eq. (2.32) can be integrated throughout any arbitrary volume V enclosed by a surface S. Thus

' Using Eq. (2.10)' we get

Eqs. (2.34) defines the divergence theorem (or Gauss's theorem as distinguished from Guass's law). This theorem is helpful in expressing surface integral of a vector as a volume integral and vice ver@sa. Differential form of Guass's law is important since it is a starting point for more advanced treatments.

Let us now sum up what we have learnt in this unit.

2.6 SUMMARY

e The number of lines of force crossing a closed surface is proportional to the net charge enclosed by that surface.

The concept of electric flux quantifies the notion "number of lines of force crossing a surface." Flux 9 is defined as the surface integral of the electric field E over a surface as follows:

where dS is an infinitesimal vector whose direction at any point is towards outward drawn normal to the surface at that point and its magnitude being the area of the surface.

Gauss's law and Coulomb's law, although expressed in different forms, are the ways of describing the relation between charge and electric field. Remember that Gauss's law can be derived from Coulomb's law, but not vice versa. So, Gauss's law is not a complete restatement of Coulomb's law.

-

Elecirostallcs in Free Space e Gauss's law is

in which q is net charge inside an imaginary closed surface (called a Gaussian surface) and eo is permittivity of free space. Gauss's law expresses an important property of the electric field. Unlike Coulomb's law, Gauss's law is not sufficient to determine electric field in all cases.

e The integral form of Gauss's law is:

where p is the volume charge density and dV an infinitesimal volume element.

e The electric field outside a spherically symmetrical shell.with radius R and total charge q is directed radially and has magnitude

The charge behaves as if it were all concentrated at the centre of the sphere.

e The field inside a uniformly charged spherical shell is exactly zero:

e The electric field due to an infinite line of charge with uniform charge per unit length, A, is in a direction perpendicular to the line of charge and has magnitude

X E = - 2 r 6 r

The electric field due to an infinite sheet of charge is perpendicular to the plane of the sheet and has magnitude f

where u is the surface charge density.

In equilibrium, an excess charge on an insulated conductor is entirely on its outer surface

The electric fiela near the surface of a charged conductor is perpendicular to u

the surface and has magnitude E = -. €0.

P Differential form of Gauss's law is div E = -. €0

Gauss's theorem states that div E dV = E dS. S v S S .' 2.7 TERMINAL QUESTIONS

1) The electric field in a certain space is given by E = 200 i . How much flux passes through an area A if it is a portion of (a) the xy plane, (b) the m plane, (c) the yz plane?

2) A point charge is placed at the centre of a spherical Gaussian surface. Is flux changed: (a) if the surface is replaced by a cube of the same volume, (b) if the sphere is replaced by a cube of one-tenth the volume, (c)'if the charge is moved off-centre in the original sphere but still remaining inside, (d) if the charge is moved just outside the original sphere, (e) if a second charge is placed near and outside the original sphere, and (f) if a second charge is p l d inside the Gaussian surface?

_-.. - .

3) Suppose that a Gaussian surface encloses no net charge. (a) Does Gauss's law require that E is zero for all points on the surface? (b) if the converse of this statement is true, that is, if E equals zero everywhere on the surface, does Gauss's law require that there be no net charge inside?

4) A thin-walled copper pipe 30.0m long and 2cm in diameter carries a net charge q = 5.8pC, distributed uniformly. What is the electridfield 5.0 mm from the pipe axis? 8cm from the axis? Assume in both cases that the point where you are evaluating the field is not too close to the ends of the pipe (E,, = 8.9 x 10- l~ C ~ N - ' m-2).

5) A flat sheet of area 50cm2 carries a uniform surface charge density cr. An ' electron 1.5cm from a point near the centre of the sheet experience a force of 1.8 x 10-12N directed away from the sheet. Find the total charge on the sheet.

6) Is Gauss's law useful in calculating the field due to three equal charges located at the corners of an equilateral triangle? Explain.

2.8 SOLUTIONS AND ANSWERS

SAQ 1

The pattern of the lines of force would certainly change. The charge Q will make no contribution to the total number of lines of force crossing any of the surfaces because Q lies outside all three surfaces that we are considering.

SAQ 2

No. Gauss's law deals only with closed surfaces. The surfaces shown in Fig. 2.2 are open surfaces, because they do not define an enclosed volume.

SAQ 3

For surface S1, the net enclosed charge is qt. The uncharged coin makes no contribution even though the positive and negative charges it contains may be separated by the action of the field in which the coin is immersed. Charges q2 and q3 are outside the surface S,. From Eq. (2.8); we then have

The plus sign indicates that the net charge within the surface is positive ana alw that the net flux through the surface is outward.

For surface S2, the net enclosed charge is q1 + q2 + q3 so that

The minus sign shows that the net charge within the surface is negative and that the net flux through the surface is inward.

SAQ 4

On charging the metal box, the total charge resided on the external surface of the box and hence Faraday was safe:

Terminal Questions

1) See Fig. 2.22

a) * = E. A = 200A (P . k) = 0 (direction of dA is towards 3 I '

Gauss's I;$w

Elerlrostatior lo Free S p a n b) @ = 200A ( i . j ) = O .-.

2) (a) No (b) No (c) No (d) Yes (e) No (0 Yes

. Mux depends only on charge enclosed. It does not depend Flux @ = - €0

4 on position of charge or charges. If q = 0, @ = 0. Remember for charge, external to the surface, as .much flux enters the'surface as that .leaves the surface as shown in Fig. 2.23.

3). a) When the Gaussian surface contains no net charge, Gauss's law becomes , . E . dS = 0 which does not mean E = 0. Here E and dS may be at right

Fig. 2.23: The net number of lines emerging from a volume angles. due to a charge outside the volume is zero. (The number of

b) When E equals zero everywhere on the surface Gauss's law requires that lines entering must just equal the there be no net charge inside. number emerging.) 4) Here, we do not have a truly infinite line. Rut for points close to the pipe and

sufficiently far from the ends, the contribution to the field from distant charges becomes very small, so the field becomes approximately that of an infinite line. A point 5.0mm from the axis lies inside the 2cm diameter pipe. A Gaussian cylinder entirely inside the pipe encloses zero net charge. Therefore, the field is zero everywhere inside the pipe.

For a point outside the pipe, Gaussian cylinder will enclose the entire pipe. Therefore, in Eq. (2.17), length I = 30.0m and the enclosed charge q is 5.8pC. Putting r = 8.0cm, we get

5) The sheet looks effectively infinite for the point which is 1.5cm from it and ,

far from its edges. So the field is given by Eq. (2.20). From the definition of electric field (see Eq. (1.7) of Unit l), the force F on the electron is just -eE. Here e is the electronic charge. Then we can write

(because u is the total charge, q, on the sheet divided'by the sheet area A).

so that

The minus sign shows that the sheet carries a negative charge. This is expected because the electron is repelled by the sheet.

6) Gauss's law is not useful for calculating the field due to three equal charges located at the corners of an equilateral triangle'because it is difficult to find a surface of appropriate symnietry over which the electric field can be taken constant, and thus to evaluate the integral.

UNIT 3 ELECTRIC POTENTIAL

Structure 3.1 Introduction

Objectives 3.2 Mathematical Background

Gradient of a Scalar Line Integral of a Vector

3.3 Work Done in Moving a Charge Line Integral of Electric Field path Independence of Line Integral of Electric Field Consequence of Path Independence

3.4 Electric Potential Potential due to a System of Charges Potential Difference

3.5 Relation Between Electric Field and Electric Potential

3.6 Electric Field and potential of an Electric Dipole and Quadrupole

3.7 Dipole in an Electric Field

3.8 Summary

3.9 Terminal Questions

3.10 Solutions and Answers

Appendix

3.1 INTRODUCTION

In Unit 1 of this Block, you have learnt that the force between any two charges is explained by the Coulomb's Law. The force experienced by a unit positive charge also defines the strength of the electric field E at a point. The computation of E directly or through the use of Gauss's law was the topic of discussion in the first two units of this block.

In most problems in electrostatics, our aim isdo calculate E. Since E is a vector quantity, its computation requires calculation of each-of its component. Many a time, to make this computation easier, we first calculate a scalar quantity known as the electrostatic potential 4, from which E can be calculated by a simple relation. Since 4 is a scalar, its computation in most cases is not so difficult as in the case of electric field. The concept of potential is also important because potential is closely linked to the work done by the charged particles and their energies. The study, of potential, its difference between any two points, and its connection with E are the main topics of discussion of this Unit. However, for developing the concept of 4 and for obtaining its relation with E, we introduce two new mathematical concepts in this Unit. These are: (i) the gradient of a scalar function, and (ii) the line integral of a vector.

'I_

In the next Unit, we will further discuss various related topics like potential due to continuous charge distributions, equipotential surfaces and electrostatic potential energy.

Objectives +

After studying this unit, you should be able to:

0 compute the work done in taking a charge q from one point to another,

0 show that the line integral of the electric field over a closed path is equal to zero,

e compute the electric potential at a point due to a single charge,

g relate the electric potential and electric field, and thereby compute the electric field at a point knowing the electric potential;

@ compute the electric potential at a point due to a dipole and a quadrupole, and

e compute the torque experienced by an electric dipole in a uniform electric field.

3.2 MATHEMATICAL BACKGROUND

The concepts of gradient of a scalar function and of line integral of a vec'or have been used in this Unit. These concepts are useful in describing the physical topics of this Unit. Let us, therefore, understand these concepts. If you have recently mastered the concepts given in the course of Mathematical Methods in Physics-I (PHE-04), you may skip this section.

3.2.1 Gradient of a Scalar You know there are three different products involving vectors: the multipiication of a vector A by a scalar (kA), the scalar or dot product of two vectors (A-B) and the vector or cross product of two vectors (Ax B). Similarly, there are a number of differential operations involving vectors, all with different uses. The simplest of these is the differentiation of a radius vector with respect to a scalar

at such as time denoted as -. The other differential operations connected with

lit

vectors involve the vector differential operator, del denoted as V. This, in expanded form, in Cartesian coordinates, is:

a a a Here -, -, - are partial derivatives. If, say, f(x, y, z) is a function of x, y

ax ay az

af ,-

and z, then - is a partial derivative off with respect to x, keeping both y and z ax

af as constant. Similarly, - is a partial derivative off with respect to y, when both a~

x and z are kept constants.

You may note that V is a vector, but is also an operator which has to operate on something which appears on its right. You may remember that since v is an operator, it cannot appear on the right of something on which it is supposed to operate.

When v operates on a scalar field or function f(xJ y, z), it determines its derivative with respect to the space coordinates x, y and z.

The gradient of a scalar field, say, potential 4 gives a vector field, as follows:

The magnitude of V+ determines the maximum spatial rate of change of 4. The direction of V+ is that for which there is maximum change in 4. It is always perpendicular to surfaces of constant +. To understand the physical meaning of gradient operation, consider a temperature in a room that is being heated' by sun3rom one side. The temperature in the room varies from point to.po-nt and can be written as a function of coordinates (xJy,z).

Let this be denoted by T(x , y, z ) . Now suppose we ask aboubits rate of vaiiation in space. Clearly, in different directions, it varies with different magnitudes. If we move from a point (xo, y& zo) through a distance Ar, the change in temperature AT will be different in different directions. If Ar is small, we can use Taylor's expansion in three variables to calculate the temperature difference. In Unit 2,

,, Section 2.5, Taylor's expansion of a function of a single variable has been mentioned. If the function depends on three variables, as in this case, we can write.

T(xo + AX, yo + AY, ZO + aT aT

= T (xi, YO, 20) + Ax - (XO, Yo, 20) + AY - (xo, YO, zo) 8x0 ~ Y O

aT + &- (xO, yo, zo) + ............ azo

As AX, Ay and & are very small, the higher order terms are neglected in this expansion.

Note that Ax, Ay, Az form the vector components of the displacement vector Ar, so we may write

T(xo + AX, yo + Ay, zo + Az) = T(xo, YO, 20) + Ar ' V T (XO, YO, 20)

Or more compactly

T ( rO + Ar) = T (rO) + Ar - VT(ro) ...( 3.4)

Thus, the knowledge of V T enables us to know temperature variation in an arbitrary direction.

From this equation, we can clearly understand the meaning of the direction of the gradient operation. Suppose we ask, in which direction does the T-field varies most. From Eq. (3.4), it is clear that AT = T(ro + Ar) - T(ro) is maximum when Ar is parallel to V T . Thus, V T has the direction in which the variation of T field is most rapid. Another point about gradient is the following. In the above example, we can draw surfaces in the room on which the temperature is constant. For example, through ro, we can find a surface along which the temperature has a fixed value which is the value at ro. Such surfaces are called, in general, equipotential surfsce (isothermals for temperature field, isobars for pressure field). Now, by definition, the component of V T along the surface in any direction is zero. As AT = Ar, . V T = 0 where r , is along the surface V T must be perpendicular to the equipotential surface.

To make certain that you can really calculate the gradient of a scalar function, we are giving a solved example.

Example 1

Find,grad 4, if 4 = AX^ yz3, where A is a constant,

Solution

a a a grad 4 = f - (AX' yz3) + f (Ax2 yz3) + ); - (AX' yz3)

ax az a

Now, for finding - (xZ yz3) , we assume yz3 as constant, and find the derivative ax

of x2 with respect to x. We follow a similar procedure for the other two terms. This gives:

V 4 = ?.AX yz3 f + AX' z3 j + 3 ~ x 2 yz2 1; Note that V + is a vector.

Electric Potential

Eleetroetstics in Free Space 3.2.2 Line Integral of a Vector

A line integral is an important vector field operation. It means integral along a curve (or line), that is, a single integral as contrasted to an area integral over a surface or a voIurne integral over a volume. The essential point to understand I

*P" about a line integraI is that only one independent variable needs to be varied for moving along a curve because in any dimensions the equation of a curve can be written in terms of a single independent variable.

* -/

Suppose we want to evaluate the line integral of a vector field E from a point A to

A a point B along a curve C in a particular region of space. See Fig. 3.1. Divide the I

I R curve C into a large number of small parts Arl, Arz, ... Ar3, ...., Ar,. These Ar's \ , are aIong tangents to the curve C at various points. Take the scalar product of the \-:-

, vector field E at the jth part and the displacement vector Ar,, of the jth part, and I I then sum over all parts, i.e.?

Fig. 3.1: Line integral of a vector field. The direction of vector Ar, is along the tangent to the point under consideration on the curve C.

E (rl) . Ar, + E(rz) Ar2 + ..... + E(rn) . Ar, n

Since the curve C is continuous, the limit of this expression when n -- oc and IArj[ - 0 is called the line integral of the vector field E along the path C, i.e.,

3 E . dr = Limit E(rj) A q A n- w , Arj-0

C j = l

A line integral is thus an integral over the product of the component of E along the direction of dr (i.e., along the tangent to the curve) and dr. +

The value of line integral depends on the relative orientation of E and dr which, in general, varies in space. If they are parallel at each point of space, then 8 = 0,

and j:E dr is simply equal to lE( Jdr 1 . If, at each point, they are

C C 5," perpendicular t o each other, i.e., 9 = 90°, then E . dr = 0. s: SAQ 1

Let A and B refer to any two points in a vector field E due to a point charge. Show that the line integral of the vector field E from B to A is negative of the' from A.to B, i.e.,

The most common use of a line integral is the calculation of work by a force I?. The workk(W) done by a force F in taking an object from a point C to a point D is commonly written as:

D

2 w = s c F - d r

*. s.i: Path AB for motion of . where dr is a displacement vector, You will learn about it in next section.

Example 2

Calculate the work done by a force F = -(2yf + 6 4 ) N in moving a; object

54 along a straight line from A (0,O) to B(2,l) in the xy plane' as shown in Fig. 3 -2.

Solution Electric Potential

We know that

Work done, W = F. dr i: = j:(2yf + d). ( d . f + d y j )

where dr = du ! + dy in Cartesian coordinates. a s f . i = j . j = I

:. W = {I(ly dr + *y dy) a n d i - j = j . f = O

Since the equation of straight line joining A and B can be written by calculating its slope, we get y = x/2 and

-

~ u b h t u t i n ~ these in the equation of bork done and writing the integral in one variable, i.e., x, we get :

Here, the limits on the values of x are those given by the x coordinates of points A and B, respectively. Solving the integrals, we get: . .

SAQ 2

Calculate the wdrk done by a force F = xy f - y2 j in moving an object from xL

(0, 0) to (2,l) along the parabolic path y = -. 4

3.3 WORK DONE IN MOVING A CHARGE

You have seen in Unit 1 that a single charge (or a combination of charges) produces an electric field in its vicinity. The electric field E at a point is defined as the force experienced by a unit positive charge placed at that point. If, instead of a unit positive charge, we place a charge q' at that point, then force F experienced by the charge q' in the electric field E is given by F = q'E. If we move this charge q' against the force F from a point A to a point B through a small distance dr as shown in Fig. 3.3, we have to do work against this force. This work done may be written .as:

d b = -F . dr = - IF1 ldrl cose ...( 3.8)

and hence

Electrostatics in Free Space Here, F cosB determines the component of F along the displacement vector dr. Notice that in order to move the charge q' against the force F produced by another charge, an external force Fext has to be applied. This F,, is equal in magnitude but opposite in sign to F;i.e., F,, = - F.

Fig. 3.3: Movement of charge in an electric field, produced by Q.

3.3.1 Line Integral of Electric Field Now let us extend the foregoing discussion and move the charge q' from A to B along the path shown in Fig. 3.4. The path AB lies within the region of electric field E. Divide this path into a very large number of small segments each with a length dr. The vector dr represents the direction as well as the length of any segment'of this path. The work done in carrying the charge q' from A to B (see Fig. 3.4) is the negative of the scalar product-ofthe7orce F and the displacement vector dr, i.e., -F dr.

Fig. 3.4: Movement of charge q' from A tcr B within an electric field E.

If we add the work done in all the infinitesimal segments between A and B, we get the total work done in moving sthe charge q' from A to B. In the limit, where the number of these segments approaches infinity, work W done in moving the charge q' from A to B is written as

B B !

W = - [ F . d r = - q ' [E.& ...( 3.10) J J A A

Now, what happens if instead of the charge q' , we move only a unit positive charge between A and B? You can easily see that the work W' done in that case woilld be obtained simply by dividing W by q ' , i.e.,

B

The integral on the right-hand side is called the line integral of the electric field. So the line integral of the electric field along any path is equal in magnitude to the work done in taking a unit positive charge along that path.

Let us now see if the integral given in Eq. (3.11) depends on the path between A and B, or is independent of that. The reason why we are interested in finding it out will be clear to you later in Secs. 3.4 and 3.5.

3.3.2 Path Independence of Line Integral of Electric wield ' Electric Potcntlnl

Let us first consider the field due to a charge q. Let there be two points A and B at distances r~ and r~ from the charge q. Let us try to carry a unit positive charge from A to B along the path shown by the coloured line in Fig. 3.5(a). Eq. (3.11) in that case may be written as:

Fig. 3.5: Work done in carrying a unit positive charge from A 'to B along the path shown as colourcd.

The first integral on the right-hand-side represents the work done .in taking a unit charge from A to A' along the arc of a circle of radius r ~ . The second integral represents the work done in taking the same charge from A' to B along the radius of a bigger circle. The first integral E+ dr is equal to zero as E and dr are perpindicular to each other as shown in Fig. 3.5(b). The second integral

B B .

[ E . dr = idrl as both E and dr are parallel to each other along A'B.

Can you tell why it is so? This is because, in the.case of first integral, cos'8 = 0 and in the second integral cos 8 = 1.

Let us now work out the second integral in more detail. We know from Unit 1 that the electric field E for a charge q at a distance r is given by

1 -- i, where i is a unit vector along the direction of r. Using this, and 4.r €', r2

replacing dr by t dr, (since dr is a vector element along the direction of i), we get:

Therefore, from Eq. (3.12)

...( 3.13)

along the.path as shown between A and B.

Let us now consider another arbitrary path (shown coloured) to carry a unit positive charge. from A to B within the same field as above (See Fig. 3.6.)

mmstoties In Frsc Space

. Fig. 3.6: Work done along the path (shown coloured) between A and B in carrying a unit positive

charge.

We can calculate the work done in the same manner along this path. The unit charge dong the path goes for a while along the arc of a circle, then radially for a while, then again along an arc, again radially, and so on. We may note that every time when the unit charge goes along the arc, no work is done. However, whenever the unit charge goes along the radial path, work is done. So, we may write:

From Eq. (3.13), the work done along each of these radial stritches is: A,

and B

Adding all these together, we get B

Comparing Eqs. (3.13) and (3.15). we note that the work done in carrying a unit positive charge in the field of a point charge q is the same along the two different paths.

From the above representative example, we can conclude that we get the same result for any path between A and B. It implies that the line integral of the electric 5eld is independent of the path between A and 19.

SAQ 3

Calculate the work done in moving a unit positive charge through a distance d in a uniform electric field parallel to the field direction.

3.3.3 Consequence of Path Independence We have seen just now that the work done in moving a unit charge against the electrostatic force is independent of the path taken between the initial and final positions. Using Eqs. (3.11) and (3.15), we can write for the work done in moving a charge q' from A to B in the field of another fixed charge q as

W'= (UB- uA) . . where U = --- qq' (+) 47r €0

U is the electrostatic potential energy. In such a situation, we can express the work done in terms of difference in potential energies of a charge at two points. Under this condition, we store the energy used in changing the position of the charge and this energy is recovered by allowing the charge to return to .its initial position. The potential energy of a charge in an electric field, about which we are talking, is a measure of the energy stored in it by virtue of its position relative to the charges which give rise to the electric field. You will read more about the electrostatic potential energy and the consequence of path'independence in the next unit.

3.4 ELECTRIC POTENTIAL

In the last section, we have seen that the work done in moving a unit positive charge from A to B in an electric field E of a point charge q is independent of the path between A and B, and depends only on the end points A and B. We can, therefore, represent it as a difference between two numbers (or scalars). These

(-&) and --&- (-&) 'respectively. If we numbers, from Eq. (3.15), are - 4r Eo

denote these numbers by cbB and +,., respectively, then their difference +BA is

cbaA=bB-(bA=- E - d r s" ...( 3.17)

A Let us now see what happens if our initial point A is at infinity. Following the discussion prior to Eq. (3.131, we can write

, ' The scalar +B = in Eq. (3.18) is usually referred to as the dectmstatlc 4r €0 rB

Elntroatitles in Free Space potential per unit charge (or simply potential) at a point distant rs from a charge q. You also know that the negative of the line integral of the electric field gives the work done in taking a unit positive charge from one point to another in an electric field. So, the electric potential at any point B at a distance rs from a charge q (see Fig. 3.7) is usually defined as the work done in bringing a unit positive charge from infinity up to that point. Therefore, SI unit for potential is the Joule/Coulomb. This combination occurs so often that a special unit, the volt (abbreviation V), is used to represent it.

Electric potential 4, due to a point charge q at a distance r from it is:

4 4 r = - 4 r €0 r

Ng. 3.R ~otcntial at a point B due to a charge q.

Before moving further, try to solve the following SAQ using the expression given above.

SAQ 4

What is the electric potential at the surface of a gold nucleus? The radius of a gold nucleus is 6.6 x 10-Ism and the atomic number of gold is Z = 79. Assume, the nucleus acts as a point charge, and electronic charge e = 1.6 x 10-I9C.

Have you figured out why we have involved infinity in our definition of electric potential? If you have not, let us tell you. If, for rs we insert oo in Eq. (3.18), we find C # I ~ = 0. This means that we define the potential at any point relative to a point at where the potential due to any charge is equal to zero.

From Eq. (3.18), we note that, for a positive charge, the potential at a point'is positive; while for a negative charge, it is negative. Do you know why? This is because, for a positive charge, the work has to be done in bringing the unit positive charge from infinity against the repulsive force of a positive charge, and hence it is positive. For a negative charge, on the other hand, the work is done by the electric field while bringing the test charge from infinity (this work done is negative).

You can notice from this discussion that, when work is done against the force (in this case electric field), potential (energy) of the system increases. This can be easily understood tiy taking an example in the case of gravitational field. When a body of mass 'm' is raised to a height 'h', without giving any acceleration, against the force of gravity mg acting downwards, then the potential energy of the body increases. Here, work is done against gravity. When work is done by the force of gravity as in the free fall of a body, the potential (energy) decreases. The difference in potential gets converted into kinetic energy of the freely falling object.

3.4.1 Potential due to a System of Charges If, instead of a single charge, we have a system of charges, we have to use the superposition principle. That is, the resultant potential +p at a point P due to a system of charges ql, q2 ,.,., q~ is equal to the .sum of the potentials due to the individual charges a t that point. If rl, r2 ,.., r~ are the distances of the charges 91, 92 ,..., q~ respectively from the point P, the potential at that point is:

q1 + 92 4~ = +...a + 9.N 4r io rl 4+ €0 rz 4.r €o rhr

You should note that here &ch charge is acting as if the other charge is not present. The potential at point P may be written in a summation form as:

...( 3.20)

60

AS a caution, you may, keep in mind that the sum given in Eq. (3.20) is an algebraic sum and not ,a vector sum as the potential at a point is a scalar quantity.

Example 3

The following point charges are placed on the x-axis: 2pC at x = 2&m, -3pC at x = 3&m, -4pC at x = 40cm. Find the potential on the x-axis at x = 0.

Solution

We know that potential is a scalar and using the superposition principal, it is written as:

On substituting the numerical values of qi and r i , ~ e ' ~ e t

If the charge distribution is continuous (like that on a charged sphere) rather than being a collection of various charges, the sum in Eq. (3.20) gets replaced by an

'

integral. In that case, we may write: n

...( 3.21) r

where p is the volume density of charge, dV is the element of volume, and r is a variable giving the distance of each point in the volume element to the point where the potential is being calculated. Eq. (3.21) can, however, be evaluated only when explicit expressions for the charge density and position for the entire charge distribution are available.

3.4.2 Potential Difference The way we have defined the potential at a point allows us to define another very useful quantity called the potential difference (or the difference of potential) between two points.

Let us write down the amount of work done in bringing a unit positive charge from infinity first to point A and then to point B (see Fig. 3.8) within the field, of charge q. We have from Eq: (3.18)

4A = wl = 4 4r €0 r~

and

The difference of these two (i.e., 4B - 4A) is the work dont in M u g a unit chvge from A to B, and it & called the potential dlfferencc between the two points B and A, It is written as:

/

This is so because the work done in carrying the charge in an electric field is independent of path. It is just this path independence that enables us to define the concept of potential. If, instead of the unit positive charge, we trahsport a charge q between A and B, then the work W done is given by:

FIB. 3.8: Work done in taking a unit charge from A to B.

Electrostatics in Free Space

The potential difference is a very important concept in the field of electrostatics and current electricity. Its knowledge helps us in determining the exact value of the current which flows between any two points in an electric circuit, provided the resistance between the two points is known.

SAQ 5 How much work is required to transport an electron from the positive terminal of a 12V battery to its negative terminal?

3.5 RELATION BETWEEN ELECTNC FIELD AND ELECTRIC POTENTIAL

We have seen in Eq. 3.17 that the electric potential q5 is related to the electric field E through a line integral. Thus, by knowing the electric field and evaluating the line integral, the potential at a point can be found out. We are now looking for a reverse relation: where knowing 4, E can be found out. This we do in the discussion which follows:

Wc have seen in Eq. (3.22) that the difference of potential 4 B ~ between two points B and A in a field of charge q is given by:

The right-hand side of this equation is also equal to the negative of the line integral of the electric field E between the same two points (see Eq. 3.15). So, we get:

B r

Now, using Eq. (3.8), we can write the potential difference dq!J between any two points separated by dr as

dtp = - E e dr

or dtp = - E cos 0 ldrl

Because of the presence of cose factor, we find that.'the electric field is a special kind of derivative of the potential. We call it he directional derivative. I In sub-section 3.2.1, you have studied how the variation of temperature in a room in different places in different directions can be written by using the gradient operator. In a similar manner, we can write the difference in electric potential, dtp, between two neighbouring points in an electric field in terms of a gradient operator. This can then be related to the electric field E at a point. Thus

= v+ . dr

Comparing this equation with Eq. (3.25), ,we get

Electric Potential

The components of E aloni x, y, z directions are

a4 E , = - - E Y = - - 84 and E, = - - a4 ax ar az

Thus, the electric field E is the negative of the gradient of the potential 4 at any point. Since by gradient, we mean the slope, the value of the electric field (a vector) is found by evaluating the rate of change of potential (a scalar) along the direction of the field. In the plane polar coordinate system, we use (r, 0) coordinates. This coordinate system will be used in finding the electric field due to a .dipole in the next section. Therefore, the gradient (del) operator in this coordinate system has been derived in Appendix at the end of this unit. It is better that you go through this derivation for a better insight; however, you will not be examined on it.

Example 4 q

The electric potential at a point is given by the relation 4 = Ax + By - Cz where A, B and C are constants. Find the value of electric field at that point.

Solution

Since 4 = Ax + By - Cz

Now apply this method yourself to SAQ 6.

SAQ 6

The potential at any point is given ,by 4 = x(Y - 4x2). Calculate the electric field E at that point.

To further illustrate the relation between the electric field and potential, let us discuss in detail the case of a dipole and quadrupole.

3.6 ELECTRIC FIELD AND POTENTIAL OF AN ELECTRIC DIPOLE AND QUADRUPOLE

A pair of equal and opposite charges, kg, separated by a vector distance a is called a dipole (Fig. 3.9). The vector a, which is also along the axis of the dipole, is drawn from the negative to the positive charge. A molecule consisting of a positive and negative ion is an example of electric dipole in nature. An atom consists of equal amounts of positive and negative charges whose centres coincide; hence, it is neutral for all points outside the atom, In the presence of an external electric field, the centres of positive and negative charges get separated. It then becomes a dipole. The electric field and potential in the vicinity of a dipole forms the first step in understanding the behaviour of dielectrics under the influence of an external electric field. By arranging two oppositely directed dipoles in a line, we get a linear quadrupole. Its field and potential are more complicated as compared to those of a dipole. Let us first study the electric field and potential due to a dipole.

a) Electric field at a point P along the ads of the dipole

Let the distance between the mid-point of the dipole and the point P which is along the axis be equal to r (see Fig. 3.9). We shall evaluate the electric field at P. The electric field at P due to +q is

Electroslntic8 la Free Space 4 E, = - P 4aeo ( r - a121

and that due to -q is

-4 P E- = - 4?rco Ir + a/2)2

The resultant field at P is

Fig. 3.9: Electric dipole AS with centre C and axis a (separation E m 2p for r > > a between positive and negative 47r co r3 charges). The point P is along the axis. where p = qaP

Here, in the denominator, we have neglected a2/4 as compared to rz, since a < < r in actual physical problems. In Eq. (3.30a), qaf has been replaced by p known as the dipole moment. In atomic and molecular dipoles, a = 10-'~rn and r > > a.

(b) Electric field at a point P on the perpendicular bisector of the dipole axis

Let the distance between the centre of the dipole and the point P in this case be equal to r. See Fig. 3.10.

Flg. 3.10 : Electric field due to n dipole at a point P.

The electric field due to + q at P is along PF and of magnitude

E+ =

The electric field'due to 14, at P is along PA and of magnitude

In order to find the resultant electric fidd; draw PD perpendicular to CP. Now, if you resolve these two electric fields along PD and in the direction perpendicular to PD, you will notice that the components E+ sin0 and E- sin 8 cancel each other, whereas the components E+ cos 0 and E- cos 8 add up along PD (shown in Fig. 3.10). As both are equal in magnitude, the resultant field is along PD and has

64 magnitude

a/2 where cos 8 =

( r2 + a2/4) ' I 2

The resultant field at P is a l o n g g p ~ which is antiparallel to p. Thus, the vector field is given by:

B 1 - ------ - for r > > a 4 5 ~ ~ ~ r 3

1

(c) Potential due to'a dipole

The electric field can also be evaluated from the electrostatic potential +. To do this, let us evaluate the potential 4 at P at a .distant r from the mid-point C of the dipole. (See Fig. 3.1 1). The line joining P to C makes an angle 8 with a. The potential at P is evaluated by using Eq. (3.19) for the two charges -q and + q of the dipole.

Ng. 3.11: Calculation of potential and field around dipole. BS, AT are perpendiculars to PC.

The distances of P from -q and +q are A P and BP respectively. a

From the geometry, you will notice that BP = SP = PC - CS = r - cose

.Thus, the potential at P is equal to:

a2 a2 When P is far away, r2 is lafge compared to - cos28 and neglecting - cos28 in the denominator, we can write 4 4

Eleclrostatlcs In Free Space p cos e or'& = ': p P = qaP . f = qacos0

4x EO r I To find the electric field at P we-use the gradient operator in polar coordinates. (See Appendix at the end of this Unit). Thus,

The resultant E is in the direction PR shown in Fig. 3.11 and has magnitude

E makes an angle a! with P given by:

1 . tan a! = - tan0

2 ... (3.34)

For 8 = 0, point P lies along the axis of the dipole; in this case, only the radial component is present and we get the same result as in Eq. (3.30). For 0 = x/2, point P is on the perpendicular bisector of the dipole axis, the radial component is

X + s zero, cos - = 0 and we get same result as in Eq. (3.31) as sin - = 1 and 0 is

2 2

antiparallel to p for this point. From these you can conclude that:

. . i) the dipole potential varies as 1/r2 and the field as 1/$ as compared to a point charge for which the potential varies as l / r and the field as I/$. Thus, the potential and field decreases more rapidly with r for a dipole than for a point charge.

ii) the dipole potential vanishes on points which lie on the perpendicular bisector of the dipole axis; hence, no work is done in moving a test charge along the perpendicular bisector.

Let .us now find the electric potential and electric field due to a quadrupole.

Consider the arrangement of 4 charges as shown in Fig. 3.12.

Fig. 3.12: Potential at P due to a quadrupolc.

It is one form of a quadrupole. The potential at P due to this quadrupole is:

In the denominator, one can neglect d as compared to $ provided r > > a. This gives;

29 a2 4 = ...( 3.35) 4 s E,, r 3

The electric field in this case has only radial component. hence it is given by

This can also be proved by direct computation. In the case of a quadrupole, the potential decreases more rapidly, i.e., as I/r3 and the field also dekreases rapidly as 1/P compared to that of the dipole and a point charge.

3.7 DIPOLE IN AN ELECTRIC FIELD

After having discussed the field and potential due to a dipole, we now turn our attention to the effect of an external electric field on a dipole.

Let us imagine a dipole in a uniform external electric field E. A uniform electric field means that its maknitude and direction are the same everywhere. Let p makes an angle 0 with respect to the field direction as shown in Fig. 3.13.

' ~ 1 ~ . 3.13: Torque experienced by a dipole placed in a uniform electric field E.

Due to the external electric field, the charge + q experiences a force IF = Eq, while the charge -q experiences an equal arid opposite force -Eq. Since the field is uniform, the resultant force on the dipole is zero, i.e.,

As the resultant force is zero, the dipole is not accelerated, that is, there is no effect on its translatory motion. Does it mean that the external electric field has no effect on the dipole? No, it is not so.

The dipole still experiences a turning effect due to the torque which acts on it. This torque is there because the two equal and opposite forces, which cancel each other as free vectors, are acting at different points. They provide a turning effect. From Fig. 3.13, you can see that this torque has a magnitude JFI a sin8 = q IEl a sine. It has a clockwise turning effect; hence, it can be written as p x E, Thus, the torque r , which acts on a dipole, is given by:

The unit of torque is clearly Newton metre (N m). Under the action of the torque, the dipole aligns itself along the field direction with dipole moment vector p parallel to E vector.

In this position, the torque on the dipole is zero. The system being in a stable position, the potential energy of the dipole for this position is minimum. Therefore, in rotating the dipole from this position, the work done by an external agency is stored in the form of potential energy in the dipole.

Let us choose the potential energy of a dipole to be zero in an external electric field when angle 0 = 90". This is an arbitrary choice to make the final result simpler. The potential energy' U for any other orientation 0 is thus:

3 n

where W(8} is the work done in turning the dipole from its reference orientation to angle 8. Evaluating the integral, we get:

Writing this in vector form, the potential energy of a dipole is

Electrostatics in Free Space U(8) = - p . E ...( 3.40) Eq. (3.40) shows that U is minimum (most negative) when the dipole is aligned along the field direction (i.e., 6 = O0), and is maximum (most positive) when it is opposite to the field direction (i.e., 9 = 180").

Let us now sum up what we have learnt in this unit.

3.8 SUMMARY

9 The line integral - E . dr of the electric field E is equal to the work W done 3 A

in taking a unit positive charge from the point A to B, i.e., .=-s"..* A

e The work done in taking a unit positive charge from one point to another in an electric field is independent of the path chosen between the two points.

e The work done in carrying a unit positive charge from infinity to some point against the electric field is known as the potential at that point.

(I The potential 4, at a point at a distance r from a point charge q is given as

@ The difference of potential +BA between two poirits B and A is equal to the work done in taking a unit positive charge from A to B. If a charge q is taker. from A to B, then the work done is:

w = 4R-I = ( 4 ~ - @A)

e The unit of potential difference is volt. The potential difference between points A and B is 1 volt when the work done in carrying unit positive charge between these two points is equal to one Joule.

8 The electric field E at a point is the negative gradient of the potential + at that point:

e The electric field of a dipole at a point along the axis of the dipole is given by:

E = 2p . for r > > o 4a eo. r3

And at a point on the perpendicular bisector of the dipole axis is given by:

E = -* , > > a 4r ro r3 . ., . .

where r is the distance of the polnt from the centre of dipole and p is dipole moment ,vector.

0 The potential at any point P on a line which makes an angle 8 with the axis ot a dipole is .given by:

where i is a unit vector from the centre'of dipole to the point I? where field is to be determined.

I e A dipole experiences a turning effect in a uniform electric field. The torque T

experienced by the dipole is given by:

r = p X E 68

8 ..

The potential energy of the dipole is given by U = - p.E. It is'minimum when p is parallel to E.

INAL QUESTIONS

1) Show that the line integral of the electric field E over a closed path is equal to zero.

2) Show that, in a pair of oppositely charged plane parallel plates, the electric field E is equal to the potential difference between the plates divided by their separation. You may assume that the electric field is canfined to the between the plates as shown in Fig. 3.14.

Fig. 3.14 Flf. 3.15

3) Find the potential at two points A and B at distances of 10cm and 50cm from a charge of 2pC as shown in Fig. 3.15. Also find the work needed to be done in bringing a charge of O.OSpC from B to A.

4) Compute the potential difference between points A and B assuming that a test charge go is moved without acceleration from A to B along the path shown in Fig. 3.16.

3.10 SOLUTIONS AND ANSWERS I

E, I - - . v w . w-

Fh. 3.16 SAQ 1 See Fig. 3.17. The line integral of the electric field E due to a point charge

q from a point A to another point B is given by B B \

I n

E-dr = Limit E (rj) . ~r~ n- m

A 9 - 0 J= l

where A rjJs are along the tangents to the paths from'A to B. The dot product for a typical line segment of E (rj) and Atj at any point P is given by 4

E (rj) . Atj = IE(rj)l (Arjl cose

where 8 is the angle between E ( 9 ) and Ar,. Now, if the line integral from B to A is to be evaluated, then the dot products will be negative of the corresponding dot products of the first line integral. This is because the

f A'

infinitesimal segments will be making angles equal to 180" m inu the corresponding angles of the first line integral dot products as shown in Fig. Flf. 3.17 69

Electroiltotics in Flee Space 3.17 for a particular line segment. Thus A

SAQ 2 The work done by force F is equal to:

(xy dx - y Z dy)

C

where C is the parabolic path y = x 2/4 from (0,O) to (2,l). Substituting for y and dy in terms of x and dx, the integral is given by:

Here the integral is evaluated between the lirhits 0 to 2 for the variable x.

SAQ 3 Let the electric field be E and element of path length be dr. Since both E and ah. are parallel, the angle 8 between the two vectors is zero. Then work done

. SAQ 4 Charge on the nucleus q = Ze = 79 x 1.6 x 1 0 - l ~ ~ and r = 6.6 .x 1 0 ~ ~ ~ r n

\ *

1 q ( 9 x 1 0 ~ ~ m ~ ~ - ~ ) ( 7 9 ~ 1 . 6 ~ 1 0 - ~ ~ ~ ) 4 = - - - - 4r ~,-j r 6.6 x 10-'~m

SAQS In going from the positive terminal of a battery to the negative terminal, the electron (a negatively charged puticle) gcss from r: point at a higher potential to a point at a Iower potential, Since

q = - 1.6 x 10-19c

and

- +B = - 12v

.'. Work done, W = q (h - +B)

= ( ~ 1 . 6 x 10-19c) (-12V) = 1.92 x 10-"5 -

where 9 = x (y2 - 4x2)

and

v

Terlninal Questions

Electric PotenUial

1) Let us consider a closed path starting from and ending at A as shown in Fig. 3.18. Let B be some point on this closed path. If 9A and 4B are potentials at A and B respectively, we can write

- E . d r = 9 ~ - $ ~ s" A

along L '

0 - E - d = 9. - 9, A

along L'

Since B

j E - d r = - / E . d r A B along L' along L'

:.- J ~ . t = + ~ - h

B along L'

Adding Eqs. (3.41) and (3.43), we get

along L alons t '

That is, along a closed path, the line integral of the electric field is equal to zero.

Alternative method: One may also use the concept of path independence

along L along L'

B B

O ' . ~ E . m - S E . & = O

A' A along L along L '

,-

Electroslatln in Free Space

A E .along L along L '

(L + L' implies closed path).

2) Let A and B be two oppositely charged plates separated by a distance d. Let E be the uniform electric field between the two plates. We then have

A where +A and 4B are the potentials at the plates A and B respectively.

In the present case, writing . { dx, and noting that both E and A

XI I' dx are parallel, we can write

That is, the magnitude of electric field between two oppositely charged parallel plates is equal to the difference of potential between them divided by their separation.

3) The potential 4, at a point distant r from a charge q is given by 4

4r = - .. .(3.20) 41r co r

\

- 9 x lo9 Nm2 c - ~ and r = 0. 10m where q = 2pC = 2 x '10-6~, - - and 0.50m . 4x €0

Using Eq. (3.20), we get

Work done, W = q' ($o.lo - Q)o.so), where q' = 0.05 x .'. W = (0.05 x 1 0 - 6 ~ ) (1.8 x lo' - 0;36 x 1 0 ' ) ~

= 7.2 X ~ o - ~ J .

4) We can write

For path C

Therefore,

C A

to B, E and ah. are perpendicular to each other. B B

S E dr = (El dr cos 90" = 0. Thus, - E . dr C C S

This gives

4B - ,A = - [E . dr

Eleclric Potential

For path A to C, the angle between E and dr = 135".

... E dr = E dr cos 135'

E E = - - j d r = - -

fi (AC) = - -

fi - a d a

since AC = d/cos 45" = a d

:.,.-, = - S E . dr

A

= Ed.

You may note that this is also the value obtained via the direct path from A to B.

Polar Coordinates

In addition to the Cartesian coordinates, you will be required to use polar coordinates in two dimensions. The polar coordinates are useful in describing the problems having circulqr symmetry. In Mechanics, you would have studied circular motion. The mathematical treatment of this problem becomes much simpler if one uses polar coordinates instead of Cartesian coordinates.

A point P (x, y) in two dimensions is represented in polar coordinates in terms of r and 8. Here r is the distance of P from the origin 0, and 0 is the angle, the line joining 0 to P makes with the positive x-axis. See Fig. A.1.

Fig. ~ . l : polar coordinates and unit vectors t and / as a sum of two vectors. . OP is known as the position vector of P. Projection of OP on x-axis gives the x coordinate of P and its projection on y-axis gives the y coordinate of the point P. The relations between (x, y) and (r, 8) are

, x' &!,r cod y = r sine . . . (A. 1) 73

Elec(mstnlics b Free Space The vector OP (r) can be written as the sum of two vectors, i.e.,

y' Let us define unit vectors in the direction of increasing r at P as P and in the direction of increasing fl as 8 as shown In Fig. A. 1. The direction of 6 is the same as the tangent to the circle of radius r at P with 0 as the centre. P and 6 are unit vectors; these can be written as the sum of two vectors. Projecting ; in the x direction, one gets cos0 as the x component of ; and in the y direction, sin0 as the

h y component of ;. See Fig. A.1. Thus

; = i COSB + j sine ...( A.3)

Similarly, projectink 6 in thex and y direction, one gets -sin@ as the x component .X and cose as the y component. See Fig. A.1 Thus

Flg. A.1: Unit vectors at two different points. 8 = - l" sin0 -F*J cose . 4A.4)

Note the ; . 8 = 0, P P = 1 = b 8. Refening to Fig. A.2, you will see that the unit vectors ; and 8 at P are not equal to the unit vectors r and 8 at Q. These unit vectors change directions as you go from one point to another. If you differentiate the relations (A.3) and (A.4) of P and b partially with respect to 0, you get:

and

Any vector A in (x, y ) coordinates is written as a sum of vectors in the x, y di i t ions in terms of its components along these directions as follows :

A = ~ A , + ~ A ,

Similarly, if A, and Ae are the components in the (r, 8 ) coordinates. then we can write A as:

A = PA, + $ A , In order to write the del (V) operator in the (r, B) coordinates, we need the infinitesimal changes in length in r and 0 directions. The infinitesimal change in length in P direction is dr and in 6 direction, it is (rdi8). Here de, being only a change in angle, has to be multiplied by r to get the irSnitesima1 change in length in the 6 direction at P (r, 6). See Fig. A.3a.

(4 (6,

a. A& a) l%mcnts of lqths in C and 6 ctWom. b) Displacenmt a¶ as a sum of ,f (0 and d (&I

An infmitesimal displacement dl in (x, y ) coordinates is written as :

Similarly, dl in (r, d l coordinates (see Fig. A.3b) can be written as:

It is now pasibre to obtain the del operator in (r, 6) coordinates. Consider a scalar function +. Suppose ( r ~ , 81) is its value at P (rl 6,) and & (rz, 4) is its value at Q ( r t 82). If JQ = dl is an infitesimal displacement, then

The points (r,, 81) and (rz, 02) being close to each other, we can write dd, as

Writing V4 in (r, 8 ) coordinattesJwe get

where (v4), and ('7410 are r, 8 components,respectiuely.

Hence

a4 a+ Comparing with d4 = - dr + - &, we get at ae

a4 1 a4 (p#), = - and ( V 4 ) @ = - - at- r a8

Hence, the del operator in polar coordinates is given by:

Example

Just to illustrate the convenience in handling a problem using polar coordinates, Iet us consider the motion of a particle In a circIe at constant speed. Let the origin be chosen at the centre of the circle of radius r and the speed of the particle be v. Then we can write

Differentiating first of these equations with respect to time, we get

an4 differentiating the second equation, we get

"These show that velocity Y is at right angles to r at every point and the acceleration a is at right angles to v. This means that acceleration a is either

s parallel to r or antiparallel to r. Differentiating with respect to time r v = 0 equation, we get

As v2 is positive, cod < 0 which means 8 = z as r and a are either parallel or antiparael. From this, it is clear that acce1eration a is opposite to r and its magnitude is

Hence, the 'acceleration is directed towards centre.

-

UNIT' 4 POTENTIAL FOR CONTINUOUS CHARGE DISTRIBUTIONS AND ENERGY

Structure 4.1 Introduction

Objectives

4.2 Potential due to Continuous Charge Distributions

Line Charge Charged Circular Disc

4.3 Equipotential Surfaces

4.4 Electrostatic Potential Energy

4.5 Nature of Electrostatic Force

. 4.6 Method of Images

4.7 Summary

4.8 Terminal Questions

4.9 Solutions and Answers

4.1 INTRODUCTION

In the previous Units, you have calculated the electric field E and potential $I due to discrete charge distributions While calculating 4 from E, you had to evaluate a line integral. On the other hanh: you can also calculate E from r#~ by a simple differentiation. In this Unit, we shall extend these ideas to evaluate + for continuous charge distributions. After computing potential and electric field due to selected continuous charge distributions, you will study equipotential surfaces.

The concept of potential and potential difference have already been introduced in Unit 3. The electrical appliances which we use in our homes work on a potential difference of 220 volts. This concept is also important because physicists do interesting experiments using high voltage sources. If a charged particle is allowed to fall through a potential difference, it accelerates and its kinetic energy increases, Several machines called 'particle accelerators' have been designed to produce high energy charged particles. These high energy charged particles are used in "atom smashing" experiments for studying nuclear structure. You will learn more about this in the nuclear physics course. In this unit, you will also learn the concept of electrostatic energy and the nature of the electrostatic force. These are basic concepts which will help you in understanding not only the nuclear physics course but many other undergraduate level courses.

In the next Block of this course, you will learn about the macroscopic and microscopic properties of the dielectrics in an electric field. There you will come across problems involving potentials and electric fields in dielectrics.

Objectives After- studying this unit you should be able to:

0 obtain expressions for potential due to continuous symmetric charge distributions,

0 sketch the electric field lines knowing the equipotential sbrfaces, 0 calculate the electrostatic potential energy for a given charge distribution, and

76 0 prove that the electrostatic force is conservative. 1

PotcnUnl for Qnllnuous Charge 4.2 POTENTIAL DUE TO CONTINUOUS CHARGE Distributlaas and Energy

DISTRIBUTIONS

In the last unit, you have learnt about electrostatic potential and its relation with the electric field E. Both electrostatic potential and electric field are very important quantities. In electrostatics, most of the time, we are interested in calculating either of these two, because knowing any one of them, the other can be determined easily. In this section, we will discuss the evaluation of potential due to infinite line charge and uniformly charged circular disc.

4.2.1 Line Charge In Unit 2 of this Block, we have calculated the electric field at a point near an infinitely long charged wire (or a line charge). It is given by:

Here, X is the charge per unit length on the wire; r is the perpendicular distance of the point from the wire, €0 is the permittivity of free space, and P is a unit vector along the direction of increasing r (Fig. 4.1).

The question now is: What is the potential due to this wire at any point P? You have seen in Unit 3 of this block that the negative of the line integral of the electric field between infinity and any point gives the value of the potential at that point, i.e.,

We shall evaluate this integral by first taking a finite distance rl instead of infinity and then letting rl go to infinity. Here r1 is the distance of the point Q from the wire (see Fig. 4.1). This integral then gives us the difference in potentials between P and Q, i.k.,

J rl

Inserting the expression for E from Eq. (4.1), we get

Or - Or, = - rl

Since P and dr are in the same direction, we get

Let us'now try to evaluate the potential with respect to infinity by letting rl go to infiaity. We notice from Eq. '(4.3) that tpr, anywhere in the vicinity of the linear charge distribution (r finite), goes to infinity. This is because the assumption of a uniform and finite charge per unit length over an infinitely long line really amounts to an infinite amount of charge. Therefore, the sum of finite contributions from each part of an infinite amount of charge leads to an infinite potential. However, this does not cause any problem because only the difference in potential enters in practical situations. The choice of infinity for zero potential is only for convenience. It is important to note that only potential differences have any real significance. The absolute value of potential does not have any bhysical significance. Notice that Eq. (4.3) gives finite values of potential differehces ful

finite distances of r and r,. -

Fig. 4.1: Potential at a point P due to an infinitely long charged wire.

Electrostatics in Free Space 4.2.2 Charged Circular Disc

I I I the foregoing discussion, we have calculated the potential at a point near an infinitely long charged wire using the value of E and calculating its Line bitegral. The wire is a one-dimensional system. Let us now consider a continuous charge distribution in two-dimensions, namely, the uniformly charged circular disc. Wllat is the potential at any point on the a x i s of such a disc? We will now calculate the potential at any point on the axis of a uniformly charged circular disc by directly using the general formula for potential due to a point charge derived in the last Unit, i.e.,

For computing potential, we shall divide the disc into a large number of concentric circular strips and then add the potential due to each of them.

Consider a uniformly charged thin circular disc of radius 'a' having a surface charge density u (charge per unit area) as shown in Fig. 4.2.

4.2: Potential at pbint P on the axis of a uniformly charged circular disc of radius a.

Let us find the potential at some point P lying on its axis. The point P is at a distance r from the centre 0 of the disc and the line joining P to 0 is perpendicular to the plane of the disc. For calculating the potential, first consider a narrow circular strip of thickness dx at a distance x from its centre, and write the value of the potential at the point P due to this strip.

Area of the circular strip is equ* Let the charge on this strip be dQ, where to the product of the circumference and thickness. dQ = ( ~ T X d ~ ) a ...( 4.5)

Here, in Eq. (11.9, 27rx dx is the area of the strip. Notice, from Fig. 4.3, that all parts of the strip are equidistant from the point P. The charge dQ on this strip can

n be written as a sum of a large number of point charges, 6qi such that dQ =g 6qi,

'

n being very large. i= I

Fig. 4.3: Distances of different parts of strip from point P.

The distance between all the point charges on this strip and the point P is The potential d~$ at P due to the charge dQ (i.e., due to the whole strip) using Eq. (4.4),and the principle of superposition is given as:

n

As pointed out earlier, the total potential a t point P due to the whole disc is obtained by dividing the disc into a very large number of similar but concentric strips and adding their contributions. As this addition contains a large number of terms (infinitesimals), the summation can be replaced by integration. Thus integrating Eq. (4.6) over all the concentric strips, we get the total potential C$ due to all the charges on the disc as follows:

a

In Eq. (4.7), a and r are constants for a given charge density (disc) and point P, respectively. The limits of x are from 0 to 'a', as we go from the centre of the disc where x = 0, to the edge of the disc where x = a. Carrying out the integration, we obtain

Equation (4.8) clearly shows that, for a point at the centre of the disc for which au

r = 0, r$ reduces to C$ = -. 2 t 0

For points far off from the centre for which r > > a, the quantity may .be approximated using Binomial expansion as follows:

Thus, inserting Eq. (4.9) in Eq. (4.81, we get

Multiplying both numerator and denominator by T, we get

Where q = ?ra2 is the total charge on the disc. Compare Eq. (4.10) with Eq. (4.4). You can see that, for points far off on the axis, the disc behaves like a point charge. Further, you may note that the potential 4 at a point is inversely proportional to r, i.e., as r increases, $ decreases.

Do you know why we have considered points only on the axis of symmetry? For points off the axis, the evaluation of definite integr,al in Eq. (4.7) is complicated and beyond the scope of this course.

The two SAQs that follow should convince you that, by using the above ideas, you cah solve any related problems.

h

SAQ 1

From the expression of the potential, i.e., Eq. (4.8), obtained above, calculate the value of the electric field near a charged circular disc, Hence, deduce the electric field for an infinite sheet of charge.

SAQ 2

Derive expressions for potentials at points outside and inside of a uniforrdy

Potential kr Coatinnons Charge Distributions and Energy

Let X2 + r2 = y; on differentiation, we get 2x dx = dy

a

Electrostntfcs la ~ r e e Spree charged spherical shell (hollow) of negligible thickness. See Fig. 4.4.

Fig. 4.4: Spherical shell.

[Hint : Divide the shell into a large number of p&allel rings such as AB shown in Fig. 4.4. The plane of the ring is perpendicular to the line joining the centre C of the shell to the point P where potential is required. The contributions to the potential a t P by these parallel rings can be summed up by integration. Express the contribution of ring AB in terms of one variable 'x' to do the integration.]

We have so far discussed the electric field and potential which give a detaiIed quantitative description of the electrostatic forces. For a qualitative description, the concepts of lines of force and equipotential surfaces are very useful. These give. a geometrical interpretation of the field. In the next sectioh, we shall give a description of the equipotential surfaces.

4.3 EQUIPOTENTIAL SURFACES

The locus of all points having the same potential is defined as equipotential surface. For a point charge far away from all other charges, the potential 4, at a distance r is given as

From this, you will see that if, on a surface, r is constarit, 9, is same everywhere on this surface. Thus the locus of points having the same value of r is a spherical surface (for which r is constant) with the point charge as centre. For a different value of r, we get a different spherical surface. See Fig. 4.5a. Notice that the electric field lines are everywhere perpendicular to the equipotential surface.

For a uniform infinite line charge, the electric potential is the same for points Potentid far ~anci.uoils Charge equidistant from the line of charge. Therefore, equipotentials are cylindrical with mtdbutlolur aud E m the line charge as axis of the cylinder. See Fig. 4.5b. Another example of an equipotential surface is a conducting surface. An ideal conducting surface must be an equipotential surface; if any potential difference exists, then charges move from higher to lower potential points until the potential everywhere becomes equal. You will see later in Unit 6 on "capacitors" that this property helps us to compute the field and potential in the space between the plates of a capacitor easily.

Since the equipotential surfaces are "constant potential" surfaces, the potential difference between a?y two points on them is zero. This implies that the work done in taking a unit charge from any one point to another on' such a surface is also zero.

Using the discussion of Unit 3, you may now say that, since the potential difference between any two points on an equiootential surface is zer.0, the line integral of the electric field between any two points over such a surface is also equal to zero. See Fig. 4.6. Writing this mathematically, it means that

B

A

where +A and +B are potentials at A and B respectively.

This is true only when the electric field E and the small displacement vector bt are perpendicular to each other. Since dr is an infinitesimal displacement on the equipotential surface, E is at all points perpendicular to such a surface (see 4.6).

Fb. 4.6: Direction'of electric field vector E relative to equipotential surfaces. PQRS and P'Q'R'S' are part of quipotential surfaw.

It is for this reason that we have drawn the electric field lines as perpendicular to the equipotential surfaces in Hg. 4.5. For an arbitrary charge distribution, the equipotentials may look like the ones drawn in Fig. 4.7.

Conventionally, the equipotential surfaces are drawn such that there is a constant difference of potential, say A+, between the adjacent surfaces (see Fig, 4.7).

Fb. 4.2 Separation of equipotcntial surfaces for arbitrary distribution of charges. Portions of four a equipotcntial surfaces arc shorn.

These surfaces may or may not be parallel to each other. They are relatively closer .where IEI is large, and are relatively far apart where (El is small. The reason for

this is that the spacing A[ betwekn the equipotential surfaces at any point is given by:

where dl is along the normal to the equipotentials.

You have seen earlier in Unit 3 that the electric field E and the potential + at a point are related through the relation

The negative sign in Eq. (4.13), along with the observation that the electric field E is always perpendicular to the equipotential surfaces, indicates that E is always towards the equipotentials of decreasing $. This is elaborated in Fig. 4.8.

Flg. 4.1: D i d o n of el& fidd E from cquipotential surfaces.

For the equipotentials shown in Fig. 4.8, we find that the resultant electric field vector E is along AB because.in this direction, the decrease in t,b is the fastest as determined by the relative ratios

A# - and - hl Ai*

You may thus remember that the resultant vector E is always along the direction of maximuin (or the steepest) decrease of +. Along AC, the magnitude of electric field is given by IEl cos9, where 0 is the angle between AB and AC.

Swnming up, you may note from the foregoing discussion that a sketch of the equipotential surfaces gives us a visual picture of both the direction and the magnitude of E in a region of space con- a single charge, a group of charges, or a charge distribution of some mcdar fonn (or shape). So far, we

WIpotentials have described the electrostatic field in terms of electric field vector E, potential and equipotential surfaces. We shall now discuss in the next section the energy d a t e d with assembling of charges, both discrete and continuous. ]Before moving to the next d o n , you would like'to do an SAQ.

SAQ 3

a) Suppose you are given a sketch of electric field lines due to a group of chzuges and are asked to draw the equipotential surfaces. List the various points which you will keep in mind while attempting to draw equipotential surfaces.

. Charged metal object b) The equipotentials for a charged solid metal object are shown in Fig. 4.9.

Draw the electric fidd lines.

Work done in assemtiling charges is stored as potential energy of the charges. 82 Suppose, there are two charges ql and q2 which initially very far apart.

Suppose ql is f ied at rl and q2 is brought from infinity to a position r2 (see ~ i g . 4.10).

Mg. 4.10: Assemblage of three charges q,, q, and q, .

You may ask how much work has been done in bringing q2 from infinity to a position r2. From Unit 3, you know that it is equal to the charge q2 multiplied by the potential at r2 due to ql. that is

This, in effect, is equal to the work done in assembling the two point charges ql and q2 at rl and r2 by bringing them close together. The work done is stored in the system and is usually interpreted as the electrostatic potential energy (or simply as potentid energy) of the system of two charges. If, however, these charges are pulled apart such that q2 is taken back to infinity, an equd amount of energy is supplied back by the system.

You may be wondering as to where this potential energy is stored. Is it at the location rl of q,. or at r2 of q2? This energy is neither at rl nor at rz, but in the system as a whole. It is not located at any particular point.

When both the charges are either positive or negative, they repel each other. In that case, work has to be done in assembling the system together. This work m y be recovered by allowing the charges to move apart on account of repulsion. The stored potential energy in that case gets converted into kinetic energy of the charges. When one charge is positive and the other is negative, work has to be done in separating &em from each other. In that case, work done can be recovered by allowing the charges to come close togettier due to attraction.

If the product of the two charges q, and qz is positive, i.e., their polarities are same, the potential energy is positik. A positive potential energy means that work has to be done to assemble the Like charges' together. If the product ql xq2 is negative, i.e., their polarities are opposite, the potential energy is negative. A negative potential energy means that work has to be done to pull the charges away from each other. It is then easy to say that a positive potential energy corresponds to repulsive electric forces while a negative potential energy corresponds to attractive electric forces.

Imagine what happens when one has to assemble a system of many charges instead of just two. To begin with, start with just three charges q,, q2 and g3 which have to be assembled at positions rl, r2 and r3 as shown in Fig. 4.10. The assembling may be done step by step. First bring q, to rl and 92 to r,. For this, work will have to be done as given in Bq. (4.14). Now bring q3 to r3 against the force that 61 and 92 exert on it. The work done for -this stage! is:

dW' = 9 3 Q1 42 + 93. ...(4 -15)

4 ~ . 6 lr3 - r11 4~ Q lr3 - rzl

~0ic.w for a m t i n l ~ o l r s I I ~Ilorn €mi%-

~lcrtmtaties in FRC Space This is because the total force on q3 is equal to the sum of two individual forces. The total work done including the first stage is then W = d W + d W 1

gj qk = C 4 ~ . , r j - r k ,

f o r j = 1 t o 3 a n d k = 1 t o 3 b u t j ~ k all pairs

The last step in Eq. (4.16) has been written with a factor 1/2 before the summation sign to make sure that the contribution from each pair of charges is ipcluded only once. For example, for pair ql and q2 we get contributio when j = 1 and k = 2; and similarly when k = 1 and j = 2. The factor of 1/2 t 2 us reduces this double contribution to a single contribution. Further, you may note that we have written j#k below the second summation sign. This is to avoid the force between a charge with its ownself.

Generalising the above discussion for the assemblage of N point charges q I , 92 ,...., qN at r,, r2 ,..., r ~ , the expression for electrostatic potential energy (P.E.) may be written as:

I N N P.E. = - C C Qj 9k

2 j=l k= l 4 r Q Irj - t k I

I # k

You may note in Eq. (4.17) that for each value of j (as fixed by the first' summation), the summation on k avoids that value of k which is equal to j. This amounts to considering the potential at charge qj by all the other charges except its own. In terms of potentials +j at the position of the charge qj, Eq. (4.17) may be written as

P.E. = - j= 1

This equatioq implies that, for calculating the electrostatic potential energy for a group of point charges, one may consider each charge turn by turn, and the corresponding potential at its position due to all other charges except the one in question. I

Now suppose, we take a simple example of adding point charges on an isolated conductor gradually in .stepehen the work done can be evaluated as follows. Let the charge on the conductor at a given time be q. Then the potential 9 of this

I charge is proportional to g. The work done G W in adding an additional charge 6q on q is then F sw = $89. w e can write 9 as 4 = kq.

I where k is the constant of proportionality. Hence

, c SW = kg 69. i '

i A s we go on adding more and more charges to this conductor, the total work done

1 is stored as potential energy in the charged body. This total work can be found by "

t integration (equivalent to summation). Thus, the potential energy is given as % s follows (if Q is the final charge on the body): I : , 1 : i ' P . . = Saw= ! k q 6 9 = : [ $ ] ;

I e2 Q = k - = -

2 z Q ' 84 where 9, = k Q is the final potential of the charged body.

A

For continuous charge distributions, summation has to be replaced by integration. Po'cnm tor Continmom I

If in an infinitesimal volume dV, we assemble point charge such that the volume Mstrlbatlom a d Ercrty

charge density is p and the potential is 4, then the potential energy may be written as

For a charge distribution,on a surface if a is the charge per unit area on the element of surface area dS, then

For a line charge distribution, if X is the charge per unit length, then the potential energy P.E. is

P.E. = - 2

line

Example 1

Three charges are arranged as shown in Fig. 4.1 1. What is their electrostatic potential energy? Assume q = 1.0 x 10-'~, and d = 0. lm.

Solution: The total potential energy (P.E.) of the system is the algebraic sum of the potential energies of all pair of charges, viz.,

P.E. =

- - - (9.0 x lo9 ~rn'c-') (10) x (1.0 x lom5 C)' 0. lni

= - 90J

SAQ 4 I

After drawing a diagram, estimate the number of terms that will contribute to the electrostatic potential energy for a system of five point charges.

So far, we have discussed the electrostatic potential, equipotential surfaces and electrostatic potential energy. We now discuss the nature of the electrostz!ir f~ rce .

ESectnwbtles in Free S p w

4.5 NATUH9E OF ELECTROSTATIC FORCE

Ng. 4.12: Work done around this loop is zero.

You have seen in Unit 3 that the work W done in moving a charge q from A to B in the region of the electric field E is written as

where F is the electrostatic force on q. You have also seen in the same unit that B

the line integral of the electric field, i.e., E .&, is. independent of the path

A betweell A and B. This implies that the line integral of the electrostatic force, viz.,

B

[ F C, is also independent of the path between A and I?. That is the work done J A

on a charged particle in moving it against the electrostatic force F is independent of tlie path between A and A and depends only on the end points A and B. This also implies tha/t work done in taking a path around a closed loop (Fig. 4.12) is zero. If. this was not so, then one can find a loop, traversing which yields negative work, i.e., energy to us. Thus, one could recover any amount of energy by going around this. That this does not happen is related to conservation of energy. Thus, path independence of work done in an electrostatic field and concept a f potential are essentially related to the fact of energy conservation. Hence, the electrostatic force is conservative just as gravitational force. The conservation of energy holds good for a conservatibe force.

4.6 METHOD OF

The concept of equipotential surface is useful in solving a few problems involving charges and conducting surfaces. This is best illustrated in the method of images. We shall discuss these while evaluating the force on a charge Q placed in front of an infinitely large grounded conducting plate. By grounding, the conducting plate is kept at zero potential.

A charge Q placed at a distance r from an infinitely large grounded conducting plate experiences a force because of the induced charges on the conducting plate. See Fig. 4.13. You may ask a question that why does induced charges appear on the conducting surface facing charge Q? These induced charges on conducting plate are a must for ensuring the absence of electric field inside the conductor. These induced charges are negative if the charge Q is positive.

4=0 To evaluate the force on Q, we must know the distribution of induced charges on

4.13: induced cKarges on a the conducting plate. Once we know the charge density of the induced charges on grounded conducting plate due to the conducting plate, we can use Coulomb's law for finding the force on Q. a point charge Q. However, we can arrive at the solution in a very easy manner. Suppose, we place

an equal and opposite charge -Q on the other side of the conducting plate as shown in Fig. 4.14 at an equal distance r from the place. This charge -Q is like the mirror image of Q produced by regarding the conducing plate is a mirror. Since every point on the conducting plate is equidistant from the two charges, it is an equipotential plane. As far as the field produced on the right side of the conducting plate is concerned, the field produced by this dipole and the earlier field produced by the point charge Q and the induced charges on the plate is identical.

As seen by the point charge Q the induced charges of the metal plate produce

Fig. 4.14: The field of a dipole resulting in an identical field to the right of the conducting plate.

exactly the same field as would a point charge -Q placed at a distance 2r away. from Q. Hence, the force between conducting plate and charge Q is obtained by applying Coulomb's law between the charges Q and -Q. This is given by :

This method of images can be used in a number of similar situation. Instead of using the method of images, if we had computed the charge density on the. conducting plate at all points and then used Coulomb's law for evaluating the force, the problem would have been very tedious. You may wonder why we have grounded the conducting plate in the above problem. By grounding, we ensure that the potential $ of the conducting plate is kept constant and only the induced charges contribute to the force between the plate and the point charge Q.

Let us now sum up what we have learnt in this unit.

o The potential difference between two points at distances r and r, from an infinitely long charge wire is given by

where X is the charge per unit length of the wire.

@ The potential 4 at a point at a distance r on the axis of a charged circular disc of radius 'a' it given by:

4 - " J ~ T - 2 - r ) 2?r €0

where a is the charge per unit area on the disc.

@ Equipotential surfaces are surfaces on which the potential at each point is constant.

o The electric field E is always directed perpendicular to an equipotential surface. It is always along the direction of the fastest decrease of the electric potential 4.

O Equipotential surfaces are close together 'in regions of strong electric field and are relatively far apart in rggions of weak electric field.

o 'The electrostatic potential energy is the energy stored in a system of char~es. It

Eleeirostntics in Free Space is equal-to the amount of work done in assembling the system togethh by bririging the charges from infinity. This eaergy is recoverable in the form of kinetic energy of the charges, if the charges move away from each other on account of repulsion.

a The electrostatic potential energy for a group o.f charges is written a$:

where #J~ is the potential at the position of charge qj due to all the charges except the qj.

B) The electrostatic force is conservative, which is a consequence of the f a a that the work done in taking a charge around a closed path is zero.

INAL QUESTIONS

I ) If electric field E equals zero at a given point, must 4 (potential) equal zero for that point? Give one example to prove your answer.

2) An infinite charged sheet has a surface charge density a of 1.0 x lo-' ~rn-'. How far apart are the equipotential surfaces whose potentials differ by 5.0 volts?

3) Show that the electric potential at a point distant x on the axis of a ring of charge of radius a is

+4 + = - 1 9 4a eo

4) Derive an expression for the work required to put the four charges together as indicated in Fig. 4.15.

I

a 5) Calculate the gain or loss of electrostatic energy when a droplet of radius R

carrying a charge Q splits into two equal sized droplets of charge Q/2 and radius R'i Assume droplets are repelled to a large distance compared to R'

- 9 a +4 because of electrostatic repulsion.

Fig. 4.15 6) There are two charged conducting spheres of radii a and b. Suppose, they are

connected by a conducting wire? Using the result from this arrangement, explain why charge density on sharp and poinJed ends of a conductor is higher than on its flatter portions.

(Hint: Charges redistribute till potentials are same. Sharper ends have smaller radii, while flatter ends have larger radii.)

7) Devise an arrangement of three point charges, separated by finite distances, that has zero potential energy.

8) Derive expressions for potentials at points outside and inside a uniformly charged non-conducting sphere.

(Hint: Divide the sphere into concentric shells and use the results of a spherical shell.)

- -- 4.9 SOLUTIONS AND ANSWERS

Potential for Continuous Charge Distributions and Energy

By symmetry one can see in the case of a infinitely long charged wire that the electric field is

a4 along r. Hence E = - ? - I

ar

eauation has to be used. See ~&end ix given at the end of For an infinite sheet of charge the radius 'a' of the disc goes to infinity. Then the Unit of this block. . electric field is given as

u r E = P - as

4 7 7 7 -- 0 when a -- a 260

(This is the same expression as obtained earlier in Unit 2 for E at a point near an infinitely charged plane sheet.)

SAQ 2 For point P outside the shell

The spherical shell of radius 'a' has a uniform charge density u (charge per unit area). To find the potential at P distant r from the centre C of the shell, we divide the shell into a large number of thin rings such as AB. The axis of this ring is along CP. The contribution to potential at P from this ring of charge is first evaluated. Then the contributions from all other parallel rings are added to evaluate the potential due to the spherical shell. This addition can be done by integration.

The ring AB has a thickness, a dB and radius a sine (which is AD in Fig. 4.4). Thus, the surface area of ring is 2 s (a sin 0) ad0 and the total charge on this ring is

(2a a2 sin 0 dB) u I All the points of this ring are at the same distance x from P. Therefore, the potential dC#J at P due to this ring of charge is

To get the potential 4, we integrate this for the whole of the shell. In order to integrate easily, we rewrite it in terms of one variable. Using the geometry, one can see that in triangle ACP

k d x = 2ar sin fl dB

Thus, substituting for asinddfl, we get

This expression can be easily integrated over the whole of the spherical shell. The lower limit of x is (r - a) for the ring at E and the upper limit of x is r + a for the ring at J. Thus, for including the contributions from all the rings of the shell, we integrate d+ and obtain

Multiplying the numerator and denominator by 4a, we get

u4a a2 Q + = - = - 4a ear 4 s E$

Electrostatics in ~ r n Space where Q is the total charge on the shell. The expression is the same as that for a point charge Q at C.

For point P inside the shell

When P is inside the spherical shell, the limits of x vary from (a+r), i.e., PF. See Fig. 4.16. Then the potential at P is:

(a - r ) , i.e.$

~ l g . 4.16: Potential due to a Thus, potential is constant inside the shell and is equal to its value at the surface. spherical shell at a point P inside the shell. SAQ3 a) i) Equipotentials are always perpendicular in the electric field lines.

ii) Equipotentials never cross each other. iii) Separation between the equipotentials depends on the strength of

the electric field. b) Electric field lines for the charged metal object are shown in Fig. 4.17.

SAQ 4 ?he diagram is shown in Fig. 4.18. Since each pair of charge has a potential energy and there are 10 pairs between 5 point charges, 10 terms would be contributing to the potential energy of 5 charges.

Rale : If there are n charges, the number of terms (pairs) contributing to the n (n - 1)

potential energy is fi. 4.17 2

Terminal Questions

d4 shows that for J E ~ = 0.4 has to be a mnstmt. ~t is not 1) (El = -- &

necessary that 4 be'equal to zero when JEl = 0. Consider, for example, two equal charges separated by a distance 20. At the mid-point between the charges. .

1 9 IE( = 0, but + = - -. 4,.

21rg a

2) The electtic field intensity I El near an infinite charged sheet is gjven by (see Unit 2).

where cr is the surface charge density. Therefore, in the problem under consideration,

- 1 1 , The spacing Al between the equipotential surface is given by 2 !

where A 4 is the potential difference between the adjacent surfaces, With I I A+ = 5.0V

3) A ring is just like a strip which we have cut-out from a charged circular disc in Sub-section 4.4.2. Follow similar steps assuming a small portion of the ring as an element of charge.

4) The work required to assemble four charges together as shown in Fig. 4.15 is equal to the potential energy of the system. It may be obtained by considering the charges in pairs. It is

4 5) Total volume of 2 droplets after splitting = 2 x - ?r RI3

3 4

Volume of original droplet = - r R~ 3

Since volumes have to be equal

Electrostatic energy (E.E.) of original droplet with charge Q

Total electrostatic energy of 2 droplets after splitting

Using Eq. (I), E.E. =

Q2 .'. Loss in electrostatic energy after splitting = - 8 r eo.R

6) When two charged conducting spheres are connected by a wire as shown in Fig. 4.19, the charges redistribute themselves till both spheres are at the same potential, i.e.,

where q, and q2 are charges on spheres of radii a and b mpedvely. This gives

The surface charge densities ul and u2 on these spheres are:

Dividing one by the other we get

a, 9, b2 - = - . - . . .@) a2 92 a2

Potential for ConUnuoos Charge DisMbnUons and Energy

Eleetrostatles in Free Space Combining Eqs. (1) and (Z), we get

low a That is, the surface charge densities are inversely proportional to their radii. high a

For sharp and pointed ends, the radii are small, resulting in high surface charge densities. For flatter ends, the radii are larger. These result in low surface charge densities. See Fig. 4.20.

7) If we devise an arrangement as shown in Fig. 4.21, the potential energy (P.E.)

Fig. 4.20 turns out to be zero

1 (-9) 4 + (-q) (9) + (q) (4) P.E. = - - = 0 4r €0 2a 2a a

A-q 8 ) Potential due to a non-conducting sphere (uniforml y charged)

A Let p be the charge density (charge per unit volume) of the uniformly charged non-conducting sphere:The radius of the sphere is equal to 'a ' . See Fig. 4.22.

B C +q . a +q

d

Fig. 4.21 I

Mg. 4.22: Potential due to a non-conducting sphere. I i I

I

For points outside the sphere 1 For points outside the sphere such as PI (distance r from the centre C), the whole charge behaves like a point charge at the centre. This can easily be deduced from the derivation of the potential due to a spherical shell. We can divide the non- conducting sphere into a large number of thin concentric shells. For each of these shells, the charge can be regarded as concentrated at the centre C for points outside the shells. Thus, the whole charge can be regarded as a point charge at the centre C. Hence, for points outside the sphere, the potential $J is given as

where Q is the total charge and r is the distance from the centre.

i For points inside the sphere

Let the point be at a distance r from the centre C (P2 in the Fig. 4.22). If we i I

I divide the spheie into a large number of concentric shells with centre C as before, 1 1 then for shells with radii s r, the point P2 is outside and for shells which have 1 radii between r and a, the point P2.is inside. For shells with radii less than or t i equal to r potential at P2 is given as:

To evaluate the contribution t o potential by shells for which Pz is inside, consider a shell of radius x and' thickness dx. See Fig. 4.22. For this shell, the total charge is equal to volume times charge density, i.e.,

47r x2 dxp '

This charge contributes a (constant) potential d@ at any inside point .given by

For adding the contributions from all such sRells, we integrate this expression for x varying from r to a. This gives the potential d2 at P2 due to shells for which P2 is inside as

a

r Thus, the potential at P2 due to the whole non-conducting sphere is:

PotenHal for Continuous Charge Distributions and Energy

where Q is the total charge on the sphere.

UNIT 5 MACROSCOPIC PROPERTIES OF DIELECTRICS

Structure

5.1 Introduction

Objectives

5.2 Simple Model of Dielectric Material

5.3 Behaviour of a Dielectric in an Electric Field

5.4 Gauss's Law in a Dielectric Medium

5.5 Displacement Vector

5.6 Boundary Conditions on D and E

5.7 Dielectric Strength and Dielectric Breakdown

5.8 Summary

5.9 Terminal Questions

5.10 Solutions and Answers

SAQ's .

TQ's

5.1 INTRODUCTION

In Unit 4 of Block 1 of the present course, you have learnt the concepts of elecrric field, electrostatic energy and the nature of the electrostatic force. ~ow&er, for reasons of simplicity we confined our considerations of these concepts for charges that are placed in vacuum. For example, Coulomb's law of electrostatic force is the ,

electric field due to a distribution of charges given in Unit 4; refer to the situation in which the surrounding medium is vacuum. Of equal importance is the situation in which the electrical phenomenon occurs in the presence of a material medium. Here we must distinguish between two different situations, as the physics of these sittiations is completely different. The first situation is when the medium consists of -

insulating materials i.e., those materials which do not conduct electricity. The second situation corresponds to the case when the medium consists of conducting materials,

1 i.e. materials like metals which are conductor of electricity. The conducting materials contain electrons which are free to move within the material. These electrons move under the action of an electric field and constitute current. We shall study conducting materials and electric fields in conducting mateFials at a later stage.

I In the present unit, you will study the electric field in the presence of an insulator. In these materials there are practically po free electrons or number of such electrons is so small that the conduction is not possible. In 1837, Faraday experimentally found

I 1

that when an insulating material, also called dielectric (such as mica, glass or polyesain etc.) is introduced between two plates of a capacitor, it is found that the capacitance is increased by a factor which is greater than one. This factor is known

I as dielectric constant (K) of the material. It was also found that tliis capacitance is independent of the shape and size of the material but it varies from material to material. In the case of glass, the value of the'dielectric constant is 6, while for water it is 80. All the electrons in these materials are bound to their respective atoms or molecules.

When a potential difference is applied to the insulators no electric current flows; however, the study of their behaviour in the presence of an electric field gives us very useful information. The choice of a proper dielectric in a capacitor, the understanding of double refraption in quartz or calcite crystals are based on such

. studies. Natural materials, such as wood, cotton, natural rubber, mica are some

Dielectric substances are insulator (or non-conducting) rubaanm~ an they do not dlowr conduction of e l d c i t y lhrough them.

Electrostatics in Medium popular examples of electric insulators. A large number of varities of plastics are also good dielectrics.

In this unit first of all we will study a simple model of dielectric material and deduce a relationship between applied field E qnd the dipole moment p of a mlecule/atom. You will learn about elecmc polarisation in a dielectric material and define polarisation vector P. In Unit 2, you have studied Gauss's law in vacuum. You will now apply it p a dielectric medium. Here we will also introduce you to a new vector known as the electric displacement vector D. After that we will discuss the continuity of D and E at the interface between two dielectrics.

In recent years dielectric materials have become important specially due to their large scale use in electric and electronic devi.ces. There are high demands for the improvement of operating reliability of these devices. Reliability of these devices is measured to a great extent by the quality of electrical insulation. In the last section you will study the dielectric strength and break down in dielectrics.

J

In the next unit you will study about the details of capacitors, specially the capacitance of a capacitor, energy stored in a capacitor, capacitor with a dielectric and different forms of the capacitors etc.

Objectives

After going through this unit, you will be able to

B' explain the behaviour of a dielectric in an electric field,

c deduce Gauss's law for a dielectric medium,

e define dielectric polarisation and classify dielectrics as polar and nonpolar,

o explain Displacement Vector (D) and relate it to the electric field strength (E); . .

e define dielectric constant,

e state and derive the boundary conditions on E and D,

e explain dielectric strength and dieIectric breakdown.

5.2 SIMPLE MODEL OF THE DIELECTRIC MATERIAL

You must be aware that:

o every material is made up of a very large number of atoms/molecules,

e an atom consists of a positively charged nucleus and negatively charged particles, with electrons revolving around it,

a the total positive charge of the nucleus is balanced by the total negative charge of the electrons in the atom, so that the atom, as a whole, is electrically neutral w.r.t. any point present outside the atom, ,4

e a molecule may be constituted by atom of the same kind, or of different kinds.

To understand the poldsatibn we shall consider a crude model of the atom. A simple crude model of an atom is shown in Fig. 5.1.

Plg. 5,l : Model of an Atom.

The nucleus is ar ihe centre and the various electrons revolving around it can be thought of as a spherically symmetric cloud of electrons. For points outside the atom this cloud of electrons can be regarded as concentrated at the centre of the atom as a point charge.

In most of the atoms and molecules the centres of positive and negative charges coincide with each other, whereas, in some molecules the centres of the two charges are located at different points. Such rnolecules are called polar molecules.

Further, we note that in dielectrics, all the elections are firmly bound to their respective atoms and are unable to move about freely. In the absence of an electric field, the charges inside the molecules/atoms occupy their equilibrium positions. The arrangement of the molecules in a dielectric material is shown in Fig. 5.2.

Flg. 5.2: The arrangement of the atoms In a dldectrlc materlal.

The charge cet?tres are shown coincident at, the centre of the sphere. Koxping this picture of,a dielectric in mind we shall proceed to study its behaviour in an'electric field in the next section,

5.3 BEHAVIOUR OF A DIELECTRIC IN AN ELECTRIC FIELD

You have seen in Section 5.2 that in a dielectric material, the centres of positive and negative charges of its atoms are found to coincide at the centre of the sphere. It is shown in Fig. 5,3.

Fig. 53: Atoms In whlch the Fentres of charges are colnddent wlth the centre of the spheres.

,In Unit 1, you have studied that a charge experiences a'force in the presence of an electric field. Therefore when a dielectric material is placed in an electric field, the positive charge of each atom experiences a force along the direction of the field and the negative charge in a direction opposite to it. This results in small displacement of

1 charge centres of the atoms orpolwules, This is also true of molecules whose charge

i centres do not coincide in the absence of an electric field. The separation of the charge centres due to an applied field E is shown in'Fig. 5.4.

Flg. 5.4 : The separation of the chnrge eentrcs duo to an applied fleld E.

Macroscopic ProporUes of Diclectrics

Electric dipole moment per unit volume is known an polarisetion

Electroslatics in Medium This phenomenon is called polarisation. Thus when an electrically neutral molecule is placed in an electric field, it gets polarised, with positive charges moving towards one end and negative charges towards the other. The otherwise neutral atom thus becomes a dipole with a dipole moment, which is proportional to electric field. The dipole and its dipole moment was discussed in Unit 3.

Now we consider another kind of molecule in which. the charge centres do not coincide as shown in Fig. 5.5.

Flg. SJ: A dlelectrlc materlal In which'charge centres do not colndde.

Due to this reason the molecule already possesses a dipole moment. Such materials are called polar materials. For such materials, let the initial orientation of the dipole axis be AOB as shown in Fig, 5.6.

Fig, 5.6: Mdccule Possessing a dlpole moment.

Now an electric field E is applied, This field pulls the charge centres along lines parallel to its direction. Thus the electric field exerts a torque on the dipole causing it to reorient in the direction of the field. Recall our discussion of the torque on a dipole in Unit 3. In the absence of an electric field these polar materials do not have any resultant dipole moment, as the dipoles of the different molecules are oriented in random directions due to thermal agitation. When an electric field is applied, each of these molecules reorients itself in'the*direction of the field, and a net polarisation of the material results. The reorientation or polarisation of the medium is not perfect again due to thermal agitation. Thus polarisation depends both on field (linearly) and temperature.

SAQ 1

What are dielectrics? In what respects do they differ from conductor?

5.3.1. .Nan-poIar and Polar Molecules . +

We have considered two types of molecules. One in which the centre of positive charges coincide with the centre of negative charges. The molecule as a whole has no

These are molecules in which there is resuimt charg?. Molecules of this type are called~on-polar. Examples of Non-polar

elecidoal neutrality and the cams of molecules are &, hydrogen, benzene, carbon, tetrachloride etc. The second type is ~ ~ s i t I v ~ and neealive chareel lie at the one in which the centre of positive charges and the centre of negative charges do - - one and the sane point. not coincide. In this case the molecule possesses a permanent dipole moment. This

type of molecule is called a pol& Molecule. ~xambles of polar-molecules are water, In suchmolecules the charge centres lie at different points and glass, etc. consequently there is an inherent dipole moment associated with the Thus we see that, a Non-polar molecule acquires a Dipole ~ o & e n t dnly in the moIeeuIu. presence of an electric field: wheieas in a Polar Molecule the already existink dipole

moment orients itself in the direction of the external electric field. &en in $lax 8 molecules, there is some induced dipole moment due to additional separation of

charges, however this effect is comparatively much smaller than the ~eorientation effect and is thus" ignored for polar molecules.

5.12 ~olarisstion Vector P

LetFus study the effect of an electric field on a dielectric material by keeping a dielec?ric slab between two parallel plates as shown in Fig. 5.7. The electric field is

Fig. 5.7: Effect of an Electrlc ticlo on n alclecrrlu nlrrwrd'by keeping a dlelcctric slab between two porallcl plates.

set up by connecting the plates to a battery. We limit our discussion to a homogeneous and isotropic dielectric. A homogeneous and isotropic dielectric is one in which the electrical properties are the same at all points in all directions. The applied electric field displaces the charge centres of the constituent molecules of the dielectric. The separation of the charge centres are shown in Fig. 5.7. We find that the negative charges of one molecule faces the positive charges of its neighbour. Thus within the dielectric body, the charges neutralise. However, the charges appearing on the sufface of the dielectric are not neutralised. These charges are known as Polarisation Surface Charges. The entire effect of the polarisation can be accounted for by the charges which appear on the ends of the specimen. The net surface charge, however, is bound and depends on the relative displacement of the charges. It is reasonable to expect that the relative displacement of positive and negative charges is proportional to the average field E inside the specimen.

' From Fig. 5.7, we find that these polarisation charges appear only on those surfaces of the dielectric which are perpendicular to the direction of the field. No surface charges appear on faces parallel to the field. Such a situation occurs only in the special case of a rectangular block of dielectric kept between the plates of a parallel plate condenser. It is shown later in this section that surface density of bound charges depends on the shape of the dielectric material.

The polarisation of the material is quantitati;ely discussed in terms of dipole moment induced by,the electric field. Recall that the moment of a dipole consisting of charges q.and q separated by a displacement d is given by P = -qd. It is known from experiments that the induced dipole moment (p) of the n~olecule increases with the increase in the average field E. We can say that p is proportional to E

or p = aE (5.1)

where a is the constant of proportionality known as Molecul~r/Atomic Polarisability. Let us now define a new vector quantity which we shall represent by P and shall calf it polarisation Of the dielectric or just poldsation. Polarisation P is defined as the ele~tric dipole moment per unit volume of the dielectric, It is important to note that '

the term polarisation is used in a general sense to describe what happens in a dielectric when the dielectric is subjected to an external electric field. It is also used in this specific sense to denote the dipole moment per unit volume.

Let us first consider a special case of n polarised molecules each with a dipole moment p present per unit volume of a dielectric and let all the dipole moments be parallel to each other. Then from the definition of P

P = n p . - From the above definition, units of P are

?heae are the charges that appear on-' the fape of a dielecvic material when it is subjected to an external fleld l h u e charges are fouid on the faces lhrt are pcrpadicular to the direction of thc field.

It is the ratio of the induced dipole mmcnt of the molecule to the applied electric field.

Coulomb m C ~ u l ~ m b = (., m-z Units of P = - - m3 m2 ,

E1:lsctrupta~lcs In Medium In g e n d , P is a point function depending upon the coordinates. In such cases, where the ideal situation mentioned above is not satisfied, we would consider an infiitesimal volume V throughout which d l the p's can be expected to be parallel and write the equation '

" Pi P = Lim - (N is the number of dipoles in volume V) (5.la)

AV+O inl v

Here V is large compared to the molecular volume but small compared to ordinary volumes, Thus, although p is a point function, it is a space average of p. The direction of p will of course, be parallel to the vector sum of the dipole moment of the molecu l~ within V. In such a case where the p's are not parallel, as in a . dielectric hat has polar molecules, Eq. (5.la) still holds as the defining equation for p,

SAQ 2

Show that the dipole moment of a molecule p 'md the dipole moment per unit volume are related by

where n is the number of molecules per unit volume of the dielectric.

To understand the physi~al meaning of P, we consider the special ccas$. of a rectangular , block of a dielectric material of length L and cross-sectional area A. Fig. 5.8 represents such a block.

Fig. $.a: Surface polnrhUon cbuger on a rectangular block of dielectric.

Flg. 5.81: Surfnce pdarlsation charges. AcWol displacement of charge on right la dx cos 0. ..

Let p be the surface density of polarisation charges, viz., the number of charges on a unit area or eharge/unit area on the surface. The total number of polarisation charges appearing on the surface = Aa ,

Induced dipole moment = A a L . . (5-2)

Volume of the slab = AL

By definition dipole moment per unit volume = P

Induced dipole moment = PAL . . . . . (5.3)

Now we can compare the magnitudes of Eqs. (5.2) and (5.3) to obtain the magnitude ,

p of the polarisation vector to be

;'P. = $ (5.4) t

I

Thus, the surface density of charges appearing on the faces perpendicular to the field Macroacopk Properties

is a measure of P, the polarisation vector. Eq. (5.4) is true for a special geometry e l Didcarla

when the dielectric material is a rectangular block. For a block' shown in Fig. 5.8a -the surface on right is not perpendicular to P. The normal unit vector (a) to the surface makes an angle 0 with P. If the charges are displaced by a distance dx the effective displacement is d.x cos0 for the surface on the right. If n is the number of charged particle and q i s 6he charge on each particle, then the surface charge density a is given by

I

where q is the positive charge on each atom/molecule and Pn is the component of P 1 normal to the surface on the right. This also shows why no charges n p p dn the

i surfaces parallel to the applied field (0 = 90') and on the left side of the block the angle between P and n, unit vector normal to the surface is 180" the surface charge density is negative.

For an ideal, homogeneous and isotropic dielectric, the polarisation P is proportional to the average field E, i.e.,

Where x = P / E ~ E and is known as electrical susceptibility. This relation is related to Eq. (5.1), Eq. (5.1) refers to one molecule, whereas Eq. (5.6) refers to the material. F m SAQI p = np using Bq. ( 5 4 , Thus the latter is a macroscopic version of Eq. (5.1). The constant E, is included for ,, 1' . the purpose of simplifying the later relationships. ,

up " "P The relation (5.6) requires that P is linearly related to the average (microscopic) field. This average field would be the external applied field as modified by the polarisation momem ltam in "' surface charges. The susceptibility is a characteristic of lhc material and gives the c*aop*q ,dXCMe *

measure of the ease with which it can be polarised, it is simply related to a for the nonpolar materials.

5.4 GAUSS9 LAW IN A DIELECTRIC.

In Unit 2, you have studied Gauss law in vacuum. Here, we shall modify and generalise it for dielectric material. Consider two metallic plate as shown in Fig. 5.9. Let E, be the electric field between these two plates. Now, we introduce a dielectric material between these two plates. When he dielectric is introduced, there is a reduction in the electric field, which implies a reduction in the charge per unit area, since, no chiyge has leaked off from the plates, such a reduction can be only due to the induced charge appearing on the two surfaces of the dielectric. Due to this reason, the dielectric surface adjacent to the positive plate must have an induced negalive charge, and the surface adjacent to the negative plate must have an induced positive charge of equal magnitude. It is shown in Fig, 5.9. I

Fig. 5.9 : Induced charges od the faces of a dielectric In an external ffeld.

' For the sake of simplicity, you consider the charge on the surface of dielectric material as shown in Fig. 5.9a. Now we apply Gauss' flux theorem to a region which is wholly within the dielectric such as the Gaussian volume at region 1 of Fig. 5.9a.

1 i

Electrostatics in Medium

Fig. 5.9a: Gaussian volumcs at 1 and 2 Inslde a dlelectrfc. Tbe displacement of charges Bt the ~ rmrfaces perpendicular to the applied fleld are shown 1

The net charge inside this volume is zero even though this material is polarised. The positive charges and negative charges are euqal. For this volume the flux of field through rhe surface is zero. We can wiite

E.dS '= 'I E, x P.dS = 0 surface at 1 '1

This shows that "lines" of P are just like lines of E except for a constant (E, ). Instead of this Gaussian volume suppose we take another one at region 2. In this Gaussion volume one su~face is inside the dielectric and the other is outside it, The. curved surface is parallel to the lines of field (E or P). For the surfwe of this Gaussian volume outside the material P is nonexistent. However, lines of P must terminate inside the Gaussian volume. Hence the net flux of P is finite and negative as shown h Fig.5.9a since the component of P normal to the surfack, i,e. PnSand o the surface charge density are equal to each other in magnitude, the surface integrd

- - - qp.

Where % is the charge inside the Gaussian volume. Thus the flux of P is equal to the negatlve of the charge included in the Gaussian volume., Notice the difference in . the flux of P and flux of E. .

Note : See Kip's book for a good account of the generalised Gauss' Law.

Now we can generaiise Gauss' flux theorem. Since the effects of polarised matter can be accounted for by the polarisation surface charges, the electric field in any region can be related to the sum of both free and polarisation charges. Thus in general

1 I E . ds =- (q/ t q,) closed surface eo

where qf represents free charges and q, the polarisation charges.

SAQ 3 I

Two parallel plates of area of a cross section of 100m2 are<given equal and opposite . charge of 1.0 x C.,The space between the plates is filled with a dielectric material, and the electric field wilhin the dielectric is 3.3 x ~ o ~ ~ / m . What is the dielectric constant of the'dielectricand the surface charge density on the plate?

Using Gauss' theorem for vectors this surface integral can be converted into a volume integral. Thus he above equation becomes

1 (V.E) dV =- (p,t pf) dV Eo

(5.10) v ,

- 12

where pf and pp are repectively the free and bound charge densities. As this is true for any volume, the integrands can be equated. Thus

< E V.E = pf + pp (5.1 1)

The flux of p through the closed surface is given by (See equation 5.8)

which can-'be written using Gauss' flux theorem

v. P = pp

E, V. E = p, - V. P

;. co V. E + V. p = pf

V. (cO E + P) V = pf

V. D = p,

where D = e0 E + P

is known as electic displacement vector. '

(Note that 5.12 is already Gauss's Law.)

SAQ 4

Show that Eq. (5.12) reduces to Eq. (5.11) when P = 0.

The dimension of D is the same asthat of P.

The units of D are C . m-2.

From Eqs. (5.12) and (5.10) we observe that the source of D is the free charge density p,, whereas the source of E is the total charge density pf + pp . When we write P = q, Eo (see Eq. 5.5)

We have D = (1 + 2) eO E

Where E, = (1 + X ) is known as' the relative permittivity. Another usual form of elecaic displacement vector D is given by

D = e E ' (5.14)

where e = E,E,

Eq. (5.14) provides the relation between Electric displacement D and electric field E

SAQ 5

Consider two rectangular plates of area of a cross section of 6.45 x m2 each are kept parallel to each other. The sepaiation between, them is 2 x 109m, and a voltage of 10V is applied across these plates. If a material of dielectric constant 6.0 is inlroduced within the region between the two plates, calculate:

1) Capacitance

2) The magnitude of the charge stored on each plate.

Macrd.scopic Properties of Dielertrics

3) The dielectric displacement D

Electraatalks In Medium -r

' 5.5 DISPLACEMENT VECTOR D

In Section 5.4, we introduced a new vector D and called it Disglacenlent Vecto~ (or) Electric Dhglacement. .

It is one of the basic vectors for m We found (see Sec. 5.4) that the electric displacement is defined by D = &o E + P; elocuic lhm dcpen& only On Gauss' law iq dielectric is given by D . dS = q#V. For isolated charge q, kept at magnltuda of free charge md its distribution. the centre of a dielectric sphere of radius r , we find that the Gauss' flux theorem

gives (being a case of spherical summetry)

which gives

:. D n e E we get E = qrj416 e r2 (5.16)

From (5.16) it follows that the force F, between two charges q,, and q2, kept at a distance r in a dielectric medium is given by

and the exptessian for the potential + st a'distance r from q is

When we compare Eq, 5.16 with the corresponding expression for E in free space, 4. 5.17 and 5.18 shows sirnilu expressions for Coulomb force and potentials (see Unit 3). We may find that in d l these expressions, q, has been replaced by e i? a dielectric medium.

SAQ 6

- Two farge metal plates each of 8n-a 1 sq. metre face each other at a distance. (One 1

metre apart they carry equal and opposite charge on their surface.) If the electric intensity betwesn the plates is 50 newton per coulomb, calculate the charge on the plates.

With this background, we may wrongly conclude that D for a dielectric medium is same as E for free space. It is therefore important to clearly distinguish between these two vector quantities: ,

e E is defined as the force acting on unit charge, irrespective of whether a dielecvic medium is present or not. It is to be calculated taking into account the free or external charges as well as the induced charges of the medium. On the other hand P is defined by Eq. (5.10), viz., D = e,, E + P, and it is a vector like electric field, but is determined only by free or external charges. Note from E q s , (5.15) and (5.16) that the value of D does not depend upon the dielectric constant while the value of E as well as the force between the charges involve e,

\

r The quantity D . dS in usuall$ referred to as the electric flux through ihe

elemcgll, of area dS. For this reason D is also known as electric flux density. From g ~ e integral form of Gauss' law in dielectrics, we find that the total flux is q, through an aiea surrounding a charge q, and this flux is unaltered by the presence of a dielectric medium. This is not rrpe in the case of total flux of electric intensity, since

J E . dS = (q/e) 5

* Since P is a vector, we may draw lines of displacement in the same way as we

14 draw the lines of force. The number of lines of displacement passing through I

I !

unit area is proportional to (D). These lines of disp:;;ement hgin and end only on free charges, since the origin of D is the conduction charges/charge density (see Section 5.4).

Again by using Gauss' law it can be shown easily that the lines of displacement are continuous in space containing no free charges. In other words, at the boundary of two dielectrics, if there are no free lines of charges D are continuous, while the lines of E are not continuous because lines of electric force can end on both free and polarisation charges. This behaviour of D and E is dealt with in greater detail in the next section. These rules are contained in two Boundary Cond,itions at the interface between two dielectxic media.

5.6 BOUNDARY CONDITIONS ON D AND E

We wish to determine the relationships that E and D must satisfy at the interface between two dielectrics, Here, we will assume that lhcre are only polarisation charges at the interface i.e., since the dielectrics are ideal they have no free electrons, and thus there is no conduction charge at the interface. Laler, these boundary conditions will be useful for proving laws of reflection and refraction of elec~romagnetic waves. Now we will determine the boundary condition for vector D. . Boundary Condition for D:

We apply the Gauss' law for dielectrics to a small cylinder in the shape of g pill box which intersects fhe boundary between two diclecuic media and whose axis is normal . to the boundary.

Fig. ,5.10 shows the cylinder, let the height of Ihe pill box be very small compared to '

its cross sectianal area. The contribution to D. dS comes from Ihc components of D

normal to the boundary. That is I

Flg. 5.10: Boundary coldition for D bctwcon two diclcclrlc nlcdin.

where D,,, Dn2 are the normal cpmponents of D in media 1 and 2 respectively.

D,, is opposite to the direction of the normal to dS in the medium (el)

D. dS = 0 since there are no frce charges on the boundary surface,

\. "'.Pal = Dn2 (5.20)

. Thus the normal components of electrical displacement vectors arc continuous across the boundary (having no free charges).

Boundary condition for E

We shdl make use of the.conservativc nature of the electric field in this case. To otjlain the boundary condition for P, we calculate the workdone in taking a unit cdarge around a rectangular loop ABCDA, Fig. 5.11 shows such a loop. The sides BC

Macroscopic Proportics. of Dlelwtria

~ o u n & r ~ conditions give the way in which the basic vector8 -hange when rhey are incident on the surfaee of discontinuity in dielectric behaviour.

D.dS = D.n d where n is the unit vector along the outward drawn normal to the area dS. This representa~ion gives the bolrnllary condition as

which gives Eq. (5.20). Otherwise the boundary condi~ions becomes

where 8, and 0, arc? the angles between n and I), and n, nnd D, rcspectivcly.

Electrostatics In Medium and AC of the loop are very small. As the work done in taking a unit charge round a closed path is zero (conservative force)

(5.21) ABCDA

Pig. 5.11 : Boundary condition for E betwwn two dielectric medin.

Let Ell and E, be the tangential components of E in the media 1 and 2 respectively as shown in Fig. 5.11. Then

I E . d l = I ~ t l d l - J ~ 1 2 d l ABCDA AB CD

= Etl 1 - Et2 1 4.

where I = AB = CD.

Using Eq. 5.21 in Eq. 5.22 we get

Eq. 5.23 states that tangential component of electric: field is continuous along the boundary. Note ,that to calculate work done, we need force which is related to the 'electic field.

The boundary condition contained in Eq. (5.23) may be written in the vector form as

where El, E, are the corresponding electric fields and n is the unit vector normal to the boundary.

SAQ7 .

Prove Eq. 5.23a using equation 5.23, Using the vector identity.

f E . dl = ( V x E ) . n dS = -1 V(n >: E ) dS SUlface

Note on Eq. 5.23a

We write Eq. (5.23a) as

E,sin 9, = E,sin 8,

where O1 and 8, are angles between n and E, and n and E, respectively in the media 1 and 2.

This is yet another form of thz boundary condition. We write Eq. 5.23b as

D2 sin 0, = - sin e2 €1 . €2 .

D, sin 0, - €1 - - Dz sin 0, €2

Eq. (5.23~) implies that the tangential component of D is not continuous across the boundary.

SAQ 8

Show that the normal component of E is discontinous across a dielectric boundary.

5.7 DEEECTRIC STRENGTH AND BREAKDOWN

We have seen that under the influence of an external electric field, polarisation results due to displacement of the charge cenues. In our discussion, we have treated the phenomenon as an elastic process. A question that arises in our minds is, "what would happen if the applied field is increased considerably? One thing that is certain is that the charge centres will experience a considerable pulling force. If the pulling force is less than the binding force between the chage centres, the material will retain the dielectric property and on removing the field the charge centres will return to their equilibrium positions. If the pulling force just balances the binding force, the charges will just be able to overcome the strain of the separation and any slight imbalance will loosen the bonds between the electrons and the nucleus, A further increase of the applied field will result in the separation of the charges. Once this happens the electrons will be accelerated. The fast moving electrons will collide with the other atoms and multiply in number. This will result in the flow of conduction current. The minium potential that causes the charge separation'is known as the BREAKDOWN POTENTIAL and the process is known as the DIELECTRIC BREAKDOWN.

Breakdown potential varies from substance to substance. It also depends on the thickness of the dielectric (thickness measured along the direction of the field). The field strength at which khe dielectric is about to break down is known as the Dielectric Strength. It is measured in kilo voltas per metre. Knowledge of the breakdown potential is very important for practical situations, as in the use of ,

capacitors in electrical circuits.

5.8 SUMMARY

0 When an electric field is applied to an insulating material, it gets polarised. This means that a dipole'moment is created in the material. This dipole moment is also exhibited as a surface charge density.

e Electric dipole moment per unit volume is known as polarisation.

e At atomic level polarisation of the medium takes place in two ways, as there are two kinds of molecules polgr and nonpolar. In nonpolar molecules the cehtres of positive and negative charges lie at one point and their inherent dipole moment is zero.

9 In polar molecules the positive and negative charge cenues lie at different points and consequently there is an inherent dipole moment associated with the molecules, though the net charge of the molecule is zero.

For a dielectric medium, it is convenient to introduce another vector related to E and P. This is called the displacement vector D defined in

D = % E + P

0 For the analysis of dielectric behaviour, the relation between the polarisation vector P and the total electric field E is important. For an ideal, homogeneous and isotropic dielectric, the relation is expressed as

When a dielectric is subjeded to a gradually incmeaing c lec~ ic potential a aragc will reach when the electron of the constituent molecule is tom away from the nucleus. Now the dielectric breakadown, viz., loses its dielectric properties, and begins to conduct electricity.

It is the applied poteqtial differcnce per unit thickness of the dielectric when the dielectric just breaksdown.

The constant X , is known as the electric susceptibility of the medium.

Electrostatics in Medium e The constant a, cotresponding to the susceptibility x,, is known as the atomic (or molecular) polarisability when we consider the polarisation of a single atom (or molecule.

e In a polarised piece of a dielectric, volume charge density pp(= -div P) and surface charge density 6, are given by P . n or P, .

e, The presence of dielectric leads to the modification of the Gauss' law. It's , modification is

where q is the total unit free or external charge

or div D = p

where D, depends only on the magnitude of free charge and distribution.

e The general relation between the vectors D, E and P can be used to define the dielectric constant K and permittivity E,, of dielectric medium. Using the permitivity E, the relation between D, P and E can be expressed in the lir~ear form

e . The vectors E and D satisfy certain boundary conditions on the interface between two dielectric media. These conditions are:

i. the tangenrial component of E is the same on each side of the boundary, i.e., E,, = E, and

ii. the ilormal component of D is same on each side of the bouqdary, i.e.,

Dnl = En2

Dielecuic strength is the applied potential difference per unit thickness of the dielectric when Lhe dielectric just breaks down.

5.8 TERMINAL QUESTIQNS

1) Calculate the relative displacement of the nucleus of the molecule, modelled in Fig. 5 1 2 (spherically symmetric molecule) when it is subjected to an external electric field and hence its polarisability.

2) Suppose two metallic conducting plates are kept as shown in Fig. 5.13.

Fig. 5.124

Rg. 5.12: Model of atom.

The area of cross section of each plate is 2.0 m2 and are separated apart, The potential dirference between them in vacuum, vo is 3000 volts, and it decreases to I

1009 volts when a sheet of dielectric 1 crn thick is inserted between the plates. Calculate the followings:

a) The relative permittivity K of the dielectric

b) its permittivity, E,

c) its susce.ptibility x d) the electric intensity between the plates in vacuum (here it is gi"een that

Intensity = Voltage across the plate/Area of Cross section

e) the resultant elektric itensity in the dielectric

f ) the electric itensity set up by the bounded charges

a) Two conducting plates without dielectric

(b) Two conducting plates with dielectirc

Pig. 5.13: Two metnllic conducting plates (n) and (b) with dlelcctric mabrinl.

3) Consider two isotropic dielectric medium I and 2 separated by a charge free

Macroseopk Properties of' Dldmtrles

Fig; 5.w Line of force ncross the houndnry between two dldcctrics.

Now, a electric vector E, goes from medium 1 and entres into the medium 2. If i is the angle of incidence and r is the angle of rellcction, prove that

tan i 'E l = - tan r E2

4) Show that the polarisation Charge dcnsity at ~hc interface bciween two dielectrics is.

"--.- 5.9 SOLUTIONS & ANSWERS

SAQs

SAQ 1) Please see tcxt.

S.4Q 2) The dipole moment per molecule = P, Thc numbcr of rnolcculcs pcr unit volumc = n 19

From vector analysis we have

E . d l = I ( V x E ) . ndS = -I V . (n x 'E )dS surface

For V. (n x E) dS to be zero, the integrand V . (n x E) has to be to zero. I Again, in as much asV.(n x E) represents a space derivative operation we can set (n x E) to be either a constant or zero. If we set n x E = 0 then a trivial result foIIows, So it is better to choose

n x E = a constant

Applying this to Fig. 5.11, we get

which is Eq. (5.23a).

SAQ 8) he integral from of Gauss' law in dielectrics is

total free I D. dS = charge

S U ~ ~ W C enclosed

(Refer to the Pig. 10).

(n . D,,, - n . ?,,)ds = o+fi

where a, is the surface charge density on the interface between the dielectrics and n the unit vector along the outward drawn normal to the surface Dd, are the normal components of displacement vectors in media 2 and 1 respectively.

when a = 0, we get n . D,, = n . D, Now D,,, = el En, and D, = ~2 En2

Thus we f i d that the normal component of E is discountinuous.

Answers

1) Let the applied electric field be E, the relative displacement of the necleus be x, the radius of the electron cbud be R and the charge of the nucleus be q. The electron cloud is equivalent to a uniform sphere of charge with the charge density given by

The density of this charge = - 4 (4 1 3 ) z R 2

the total charge of the electron .cloud = +

we find the field at a distance x from the centre of the sphere using Gauss' law (see Unit 2). This gives

Force on the nucleus, F = qE2, when it is displaced by an amount x. (F is the coulomb restoring force on the necleus). Now

External force on the nucleus = E q

This balances the coulomb restoring force

:. E q = -F

or

and

41c€$' x = E

4

Resulting dipole moment per molecule

The dipole moment is proportional to .

The molecular polarisability (a) = 4 ~ a $ ~

2) a) The dielectric constant K is given by

The bound charges of the dielecuic set up which opposes the electric field so due to the plate charges. The new field E is the resdtant of the two

= 2 x 1 0 5 ~ ~ 1

3) The tangential component of E at the boundary is continuous. 'Thus E, sill i = E2 sin r.

The normal component of D is continuous. Here we will use D = E E and write

E~ B1 cos i = c2E2 cos r

tan i tan r . -=- " E, €2

or tan i El = - tan r Ez

4) The polarisation charges appear on the surfaces of the dielectric, perpendicular to the direction of the electric field. We wfite Eq. (5,4), viz., a, = P in the vector form as

where n is the unit vector normal to the face on which polarisation charges appear and P the Polarisation vector. Let Pl and P, be the polarisation vector in the two media. At the interface between the two dielectrics, the surface density of polarisation charg o, is

From the Boundary condition for D we have

a: P , = E ~ E ~ - D ~ andP2=eoE2-D2

using (iv) in (i) we get

%'= n . ( E ~ Ez - D2) - (q El -Dl)

= e0 (n . E, - n . El) in view of (ii)

6 = E, ( ~ ~ 1 % n , El - n . E,) = E, (el/% - 1) n . El

= e, [(el - %)/%I n . E

Structure

6.1 Introduction

Objectives

6.2 capacitance

6.3 Parallel Plate Capacitor or Condenser

Energy Stored in a Capacitor

6.4 Parallel Plate Capacitor with Dielectrics

Voltage Rating of Capacitor

6.5 Capacitance of Cylindrical Capacitor

6.6 Capacitors in Series and Parallel

Combination of Capncitor in Parallel

Combination of Capacitor in Series

6.7 Energy Stored in Dielectric Medium

6.8 Practical Capacitors

Eixd Capacitor

Ceramic Capacitor

Electmly tic Capacitor

Variable Air CapacitorIGang Capacitor

Guard Ring Capacitor I

6.9 Summary

I Solution & Answers

6.1 INTRODUCTION

I You have studied in your earlier classes thal the potential of a conductor increases as the charge placed on it is increased. Mathematically we write

Where C is the proportionality constant.

We call this constant C as the capacity or capacitance. We also call any device that h'as capacitance as the capacitor (condensoi). You are already familiar with this device.

i I We change the capacitance in our radio-transistor while operating the 'tuning' knob

and get the radio station of our choice. Capacitors are used in many electrical or 1 electronic circuits, they provide coupling between amplifier stages, smoothen the output of power supplies. They are used in motors, fans, in combination with inductances to produce oscillations which when transmitted 'become radio signdWTV signals etc. Besides, these capacitors have a variety of applications in electric power transmission.

In the present unit, we shall learn about capacitance, capacitors of different forms, energy stored in a capacitor, working principle of a capacitor. We have studied the

Electrostatics In Medium macroscopic properties of dielectrics in Unit 5. Here we will study the effect on the capacitance of a capacitor, when .a dielectric is placed between the two plates of a capacitor. Then we will introduce some practical 'capacitors.

In next unit we will study the microscopic properties of the dielectrics.

Objectives:

After going through this unit you will be able to:

e define capacitance of a capacitor,

e describe capacitors of different geometries and obtain mathematical expression for their capacitance,

e able to calculate the energy stored in a capacitor,

describe the effect of introducing a dielectric material in a capacitor,

e obtain expressions for vthe effective capacitance of grouping a number of capacitors in series and in parallel,

0 describe practical capacitors such as a guard condenser and an electrolylic capacilor.

6.2 CAPACITANCE

A capacitor or a condcnser is an eleclronic device for storing electrical energy by allowing chargcs LO accumulale on metal plates. This clectrical energy is recovered whcn ~hcsc charges are allowed to move away from thcsc plates into the circuil of which thc capacitor forms a part. Any (levice which can store 'charges is a capacitor. For example, an insulated conducting spherical shell of radius R can store charges; hence it can be used as a condcnser. ' let us see how it works as a capacitor. If a chsuge Q is placed on it the outer surface of the shell becomes an equipotential surfacc. The potential of the outer surface of the shell (see Unit 4) is given by

with ,infinity as zero potential. Instcad of infinity we can regard the ground (carlh) as zero potential. Then thc capacitance of this shell (w.r.t. ground) is

C = Q / $ = 4 n a R Coulomb

volt

The unit of capacitance C in SI sysLcm is &ad.

farad = Coulomb volt

If R = 100 crn in thc above spherical shcll ils caipacily in farads is

(4nE.O) 100 = 1.1 X 10-lo far:ld.

Thus il is dear from this qlat if a capacitor is to bc rnadc will1 one unit (farad) capacity it has to havc huge dimensions (10l0 m in the above case). Practical form of condensers have small dimensions and smaller units such as picofarad (10-l2 farad) and microfarad fiuad) are more co~nmonly used. The symbolic rcprescntation of a capacitor is o j t o ,

Thc above example of a spherical conductor as a capacitor is given only to illustrate thc concept. Howevcr, thc most commonly bsed practical form of condensers always has a sys1c111 of two ~nelal shects (circular, cylindrical or rectangular) kepi close to aich olhcr wilh an insulator scparaling the two shects. This system has the abilily to have larger capacily wilhout having the corrcspondi~~g kuger dir~icnsions. You will lcnrn lnorc about this in dctail in the next section,

26

6.3 PARALLEL PLATE CAPACITOR OR CONDENSER Capacitor ,

This is the simplest and most commonly used form of a condenser. A parallel plate condenser consists of two rectangular or circular sheets (plates) of a metal arranged parallel to one another separated by a distance d. The value of d is us&lly very small and an insulating material is normally inserted between the two sheets. See Fig. 6.1. A charge Q (positive) placed on the upper plate disuibutes equally on this plate to make it an equipotential surface. The lower plate is shown grounded (earthed. h e symbol used for showing the grounding). The lower plate is therefore at ground potential (zero potential). Because of electrostatic induction an equal amount of negative charge appears on the upper side of the lower plate. This induced negative charge pulls up almost all the positive charge placed on the upper plate to the lower side of the upper plate. Thus the electric field now gets confined to the space between the two plates: the positive charge acting as sources and the negative charge as sink (the lines of force originate on the positive charges and end on negative charges). The induced negative charge is equal to the amount of positive charge because of the zero field requirement inside the material of the conducting sheets. Besides, both the metal sheets are equipotential surfaces. The lines of force field lines are normal to these sheets except at edges. See Fig. 6.1. Since all the field lines originate front the upper plate and end on the lower plate, the value of the electric field, E is uniform in the space between the plates except at the edge. The edge effects are negligible if the area of the plates, A, is large compared to d. Since E is uniform the potential difference between the upper and the lower. plates is ghen by

, - A

4 - d

f

Plg. 6.1: Parallel plate condense. A rind B arc the mctal plates scpnrntcd nt n distance 'd ' .

where @2, $1 refer to the potentials of upper and lower plates respectively. As the lower plate is earthed,

To evaluate E let us use Gauss's theorrn. Suppose we evaluate the electric flux for a closed cylindrical surface EF'GH of base area S with its axis normal to the plate. See Fig. 6.2

I 4 Fig. 6.2: Cussiarl closed cyli~~dcr EFCH. i I

One of the horizontal surraces is inside Lhc meld and thc other in the spabe bctween the plates, the c i i ~ e d filccs are parallet to the field lines. Thcrc is no flux through EF as the field inside the conducting surrace is zero. Similarly, thcre is zero flux through '

EH and FG as the curved surfaces of the Gaussian cylindcr .ate pmllcl LO ~ h c field . lines,

27

Electrostatics in Medium The flux through the surface HG of area S is equal to ES. Since E is dong the normal .to the area, hence, we can apply Gauss' theorem. According to Gauss theorem

where cris the charge per unit area on the condenser plate. The potential @ of the -

upper plate is Ed from Eq (6.1). The total charge Q is a. A.

By keeping a small vaIue for dl the capacity C can be increased. In the above derivation we have taken the medium between the plates to be vacuum. The above arrangement has the advantage of the electric field being unaffected by the presence of other charges or conductors in the neighbourhood of the capacitor. Moreover, if the area A of the plates is much greater than d the correction for the capacitance due to the nonuniform field at the edges is negligible.

SAQ 1

Suppose we have the distance of separation between the plates, what happens to the capacitance?

SAQ 2

Find the charge on a 1000 pF capacitor when charged to a voltage of 24 V. 1

In next subsection you will learn about the energy stored in capacitor.

6.3.1 Energy Stored in a Capacitor

In Unit 3, it was shown that the work done, W in assembling a charge Q by adding Sincs,the potential i s &Tied as the infinitesimal increments of charge is given by work done per unit charge, the work done in moving a small charge 6q W = 1 D Q $ (6.8) against a charge potential 4 will be work done = 46q Where $ is the final potential of the charged body. In the case of a capacitor of

4 capacitance C, this work done in placing a charge Q on the capacitor must also be But + = - c given by similar expression, i.e., The total wok done in charging a capacitor to Q coulombs is given by w = I n Q @

This can be written in terms of the capacitance C = Q/$ as 4c4 q2 Total work done = - = - C 2 C W = 112 C 92 = QZ/2C joules

0

Q2

This work is stored up in the electric field as potential. energy. =2C

SAQ 3

Show that in a parallel plate capacitor of area A and the separation of plates by a distance d in vacuum the energy stored in the (space) volume of the electric field between the plates is given by 1/2 Q $.

'6.4 PARALLEL PLATE CAPACITOR WITH DIELECTRICS ,

When a dielectric slab is inserted between the parallel plates of a condenser the capacity increases. The polarised dielectric slab ABCD (see Fig. 6.3) reduces h e electric field E inside the dielectric by a factor (11~) where E, is the relative permittivity as discussed in the last unit: This can be proved by computing the ,

electric field by using Gauss' law for electric displacement, D inside the dielectric ABCD. Recall the Gaussian cylinder us+ in evaluating E in Section 6.2. The flux of D is now given by (only free charges contribute to the flux)

Fig. 63: Dlelcctricslab bctwccn capacitor plates.

D S = a S (6.11)

as the bound surface charges do not contribute to this flux and

D = a & E (6.12)

for an isotropic uniformly polarised dielectric. Thus the field

The potential difference between the plates is equal to E.d Where d is now the thickness of the slab filling the entire space between the plates. The capacitance now becomes

,The value of the capacitance'c increases by the factor Er which is relative permittivity of the dielectric inaterial.

From Eq. (6.14) we note that the capacitance of a parnllel plate capacitor increases with the increase in surface area (A) of the plates and also with the decrease of the distance separating the plates.

The effect of introducing a dielectric in between the plates increases the capacitance (-: e, > I). Thus inclusion of a dielectric enables the capacitor to hold more charges at a given potential difference between the plates.

We rewrite Eq. (6.14) as

and compare it with Eq. (6.7). We frnd that a dielectric of thickness d has an equivalent free space thickness (dl&). This observation, will be useful la@r when we deal with the capacitor in which the space in bctwwn the plates is only partially filled

. with a dielectric. I ! I 1 SAQ 4 I , i Findathe capacitance of the parallel plate capacitor consisting of two parallel plates of

area 0.04 m2 each and pJaced lk3 apart in free space.

1 A capacitor is shown in Fig. 6.4 in which a dielectric slab of thickness r is inserted between the plates kept apart at a disthce d. We write the capacitance of this capacitor, on the basis of the equivalent free space thickness of the dieleclric. We fuld the free space thickness between the plates = (d-t) where t is the thickness of the dielectric material. This t is equivalent to t/er in free space. The, capacitor of Fig. 6.4

Elcctrostatlcs in ~cdlurn is equivalent to a capacitor with free space between the plates, with the separation of - (d-t + t/er).We wrile the expression for the capacitance as

Flg. 6.5: Shows the equivalent capacitor.

Now we will obtained Eq. (6.16) with another simple methods. the voltage across the capacitor which is shown in Fig. 6.4 is V:When,a dielectric of thick 't' is introduced between the two plates of the capacitor, the distance between the positive plate of capacitor and the upper surface of the dielectric is say dl and from lower surface of dielectric to negative plate of the capacitor is dz. Now assume that the voltage between positive plate and upper surface of the dielectric is V1, the voltage between upper and lower surface of the dielectric is Vz and lower plate of the dielecuic to negative plate of the capacitor is V2. The total voltage V across capacitor is the sum of these three voltages i,e.,

Let E be the field inside the dielectric. Then

From 'he figure

From thc above equation we get

V = (d - t)E + E . t/€r

Using Eq. (6.5). we gct that in Lhis case

d = [(d - t) + C / E ~

From Eq. (6.151,. w e gct

, . We can also find hat the. ratio of the capacitance with dielectric belween the plates to b c capacitance wilh free space belwecn Lhe plates is equal to lhe relative permittivity, viz.,

Capacitance wilh dielectric between the plates e, =

Capacitance with free space between the plates

TabIe 6.1: Relative permittivity (E,) of some common materials

1.0006

Castor Oil 4.7

Mica

Glass

Bakelite

Paper

I Porcelain 5.5 I

SAQ 5:

A dielectric of-relative permittivity 3 is filled in the space between the glntes of a capacitor. Find the factor .by which the capacitance is increased, if the dielectric is only sufficient to fill up 314 of the gap.

6.4.1 Voltage Rating of a Capacitor .K

Capacitors are designed and manufactured to operate at a certain maximum voltage which depends on the distance between the plates of h e capacitor. If the voltage is cxceeded, the elcctrons jump across he space between the plates and this can result in permanent damage to the capacitor. The .maximum safe voltage is called the working voltage. The capacity and the working voltage (WV) is marked on the capacitor in the case of bigger capacitors and indicapd by the colour code (similar to that of resistance) 'in he case of capacitors h'aving low values of h e capacitance.

,, . ~ $ 4 ' ~ - 6.5 CAPACITANCE OF A CYLINDRICAL CAPACITOR

In Section 6.3, wc have calculnlcd thc capacimncc of a parallcl plat$ capacitor. Another important form or capacilor is a cylindrical c~paccitor. his is shown in Fig. 6.6a. A scction of his capacitor is show11 in Fig. 6.6b. I t is made up of lwo hollow coaxial cylindrical conduclors of radii a and b. Thc space betwcen the cylindcrs is iilled wilh a dielcc~ic of rcl:~live pcrmiltivity e,. ~raclicil iorms of such capacitors arc

i) a coaxial cablc, in which lhc inncr conductor is a wirc and:the oulcr conductor is nor~naIIy a lncsh or conducling wire scparatcd iro~n thc in~icr conductor by an insulnlor (usually. plastic)

Fig. 6.6: (a) Cylil,drlcal cnpncitor (h) cross scclloll of the c y l l ~ ~ d r ~ n l e:tprrcltor.

i In Fig. 6.6b direction of the field 1 lines is radial, viz.,.normal to ~ h c I

surface of the cylinder. Small lines in between the two cylinders, show Ihe direction of fixed line.

ii) thc submarine cable, in which b coppczr conduclor is c o v ~ d by polysl.yrcnc (lhk outcr conductor is scri water), Since bolli'll~c inncr pnd oulcr cylindcrs arc conductors, they ruc cquipotcntid surfaccs (scc Unit 4). The ficld is radial , ,(normal to Lhe surl'acc o l lhe cylinder): Bccause of cylindrical spmmctry wc conclude hat lhc, capncilimcc is proportioni~l to lhk Icnglh or (he cylind& (as the

length will increase, the area of the plot will increase). We shall now find the capacitance per unit length of the capacitor.

- --- Ag. 6.7: Gausslan surface ABCD.

Let the charge per unit length placed on the inner cylinder of the capacitor be A . The outer cylinder is grounded. An equal and opposite amount of charge appears on inner side of the outer cylinder. This is because of the zero field in the'conductor. To evaluate the field l'et us consider a coaxial closed cylindrical surface ABCD of unit. length and of radius r. See Fig. 6.7. The electric field is normal to the inner cylindrical surface and is also confined to the space' between the cylinders. The flux of electric displacement vector, D, through the bottom and top surfaces of this Gaussian cylinder ABCD is zero as D is parallel to these faces. The flux of D is only through the curved surface of ABCD and as D is normal to this at all points, the flux through this closed Gaussian surface is given by

Now D =. 6, e E for isotropic'uniformly polarise. dielectrics. Using Gauss' law we get

Zx rD = (2w) a e , E = 3;. ' (6.18)

where X is the free charge'enclosed by the Gaussian surface. Thus

To find the capacitance, we require the potential differeice between the two cylinders. In Unit 3, you have studied that the expression for potential difference is given by

Four our case, Eq. (6.20) becomes

a

4r - $b = - (Edr) . b

h ' = -7 'dr/r

xr €31 Er

- (b/a) . ' 2 n r a &

X * $a- 4 b = *=- In ' " (6.22)

As the. outer cylinder is grounded $b = 0.

Now, ca6acitance per unit length, C is given by Capncltor

Note: In the expression for the capacitance per unit length of a cyli'ndrical capacitor, Eq. (6.23), we find that the capacitancg depends on the ratio of the radii and on their absolute values.

SAQ 6

Two cylindrical capacitors are of equal length and have the same dielectric. In one of them a radii of the inner and outer cylinders are 8 and 10 cm, respectively and in h e other they are 4 and 5 cm. Find the ratio of their capacitances.

6.6 CAPACITORS IN SERIES AND PARALLEL

In Section 6.5, we have seen the method of finding the capacitance per unit length of a cylindrical capacitor. We multiply the capacitance per unit length by the length for cylindrical capacitors and get its capacitance. Now we can consider a cylindrical capacitor of length 2 units as consisting of two cylindrical capacitors of unit length joined end to end so that the inner cylinders are connected together and the outer cylinders also get connected similarly. This is shown in Fig. 6.8,

+ luau

+- - -C

Fig. 6.8: A long cylindricnl capacitor seen us n particular combinntion of unlt cyllndrloP cnpacitor.

We find immediately that in such a combination Lhe charge on the capacitor is doubled and so the capacitance is also doubled since Lhe potenlial difference remains constah. Two capacitors connected in paralldl (symbolic representation) are shown in Fig. 6.8a.

Fig. 6.80: Two cnpncitors cot~ncctcd in parallel.

In this combinzltion, we find lhat I

the potenlial difference between the plates remains the same; I I I s the charge on each capacitor adds up (more area is available for storing charges). I

'We can find an equivalent capacitor chat Hblds the same charge when kept at the same polcnlial difscrcncc as thc combinations of thc capacitors. Thc capaciwncc of thnl

Electrostatics In Medlum capacitor is known as the Effective Capacitance of the combination. Before we proceed further, we note that capacitors can be grouped or combined in another way too. Here alternate plates of the capacitors are connected to the succeeding capacitor so that they form a series. Fig. 6.9 shows the combination, it is known as combination of capacitors in series.

Fig. 6.9: Capacitors In Serles.

If a voltage source is connected across the two end plates of the first and last capacitors of the series, equal charges will be induced in each capacitor whereas the potential difference across each capacitor will depend upon its capacitance.

We shall find the mathematical formulas for the equivalent capacitance of the combination of capacitances in paralleI and in series.

6.6.1 Combination of Capacitors in Parallel

Fig. 6.10 shows the combination of three capacitors in parallel. .

Fig. 6.10: eapncltd~9 in parrrllcl.

Here C1, C2 and C3 are the capacitances of the individual capacitors, Ql, Q2 and Q3 are respective the charges on them and $ is the potential difference between the pIates of each capacitor. We take C to be the effective capacitance of the combination. The total charge Q of the p d l q l combination is equal to

Q = QI + Q2 + Q3 (6.24) ,

Since @ is same for this equivaIent C of the parallel combination

Now C = Q/4 = Qi + Q2 + Q3

4 el+&+@ =- 4 4 $

C = CI+ Cz + C3 (6.24)

Thus the effective capacitance of the parallel combination of capacitor is equal to the sum of the individual capacitances.

6.6.2 Combination of Capacitors in Series

Fig. 6.11, shows the combination of three capacitors in series.

34

Capacitor

Fig. 6.11: Capacitors In serlcs and thc cqulvalent capacitors.

Here C1, C2 and C3 are the capacitances of the individual capacitors. The application of a voltage will place a charge eQ on one plate which induces .a charge -Q to the other plate. The intermediate plates acquire equal and opposite charges, because of electrostatic induction. The potential drop across each will be inversely proportional to its capacitance. (Since C = Q/$ gives t) = Q/C. Since Q is fix+'+ = 1M. Thus $ 1 ,

$2 and $3 the potential drop across the capacitors are such that $1 = 1/C1, $2 = 1/C2 and $3 = 1/C3. Now we replace the capacitors by a single capacitor df capacitance C that holds the charge Q when subjected to a potential difference + = ($1 e $2 + $3)

This capacitance C is known as Ihe effective capacitance of the combination. We now write C = Q/Q, or 1/C = $/Q. But $I = @ I + $2 + 413. Therefore,

Thus for capacitors connected in series the reciprocals of the capacitances add to give the reciprocal of the effective capacitance.

SAQ 7

Determine the equivalent capacitance of the netwrok shows in Fig. 6.12 and the voltage drop across each of the capacitor of the serics of capacitors.

SAQ 8

Calculate the effective capacitance of three capacitors arranged in such a way that two of them C1 and C2 are in series and the third C3 is in parallel wiih this series combination.

6.7 STORED ENERGY IN A DIELECTRIC MEDIUM

In Section 6.31, we have studied that the energy stored in a parallel platc capacitor is given by

We know that

and

Pitting these values in the above Eq. we get

=' Q (Ad) . E2 2

or U 1 - =--a@ (Ad = v) v 2

This is energy per unit volume.

When a dielectric of relative permittivity er fills the space between the plate of the capacitor, then the effective capacitance is given by

The energy stoied in a capacitor wik a dielectric material is given by

In case of a parallel plate condensor, the energy stored per unit volume is 112 eo E? which become 1D eo Er @ = 112 E . D with the dielectric material. Where D is the electric displacement in the dielectric. We have considered here the case of a linear dielectric where E and D are in the same direction. However, there are dielectrics in which E and D are not in the same direction. Thus the energy stored per unit volume in a dielectric medium is given by

1/2 E . D Joules/m2 (6.26) i! I

6.8 PRACTICAL CAPACITORS i

We shall now study some of he capacitors that are commonly in use. Capacitors may be broadly classified into two groups i.e., fixed and variable capacitors. They rhay be further classified according to their consauction and use. Following are the classiIications of the capacitor.

Types PP Capacltor I

4 I

-l. Fixed Capecitot Variable Ctipacitor

c r--7--=, Paper Mica Ceramic Electrolxtic Oauge capacitor Gauge type Trimmer

Capacitor capacitor capacitor capacitor of tube type capacitor of capacitor receivers transistor

receivers '

Now, we will discuss each type of the capacitor one by one.

6.8.1 Fixed Capacitors

These have fixed capacitance. These are essentially parallel plate capacitors, but compact enough to occupy less space. In their make they consist of two very thin layers of metal coated on the surface of mica or paper having a uniform coating of paraffin. The mica or paper having a uniform coating of paraffin. The mica or paper forms the dielectric between the conductors. They are shown in Fig. 6.13.

Fig. 6.13: Fkod capacitors

This arrangement is rolleh up to the compact form. UsuaIly, they are piled "p in parallel to give a large capacitance. Though paraffin-waxed paper capacitors are cheaper, they absorb a good amount of power. For this reason these capacitors are used in alternating current circuits, radio-sets, etd.

6.8.2 Ceramic Capacitors

These are low loss capacitors at all frequencies. Ceramic materials can be made to have very high relative permittivity. For example, teflon has E = 8 but by the addition of titanim the value of & becomes 100 and on adding barium titanate the value of s. may be increased to 5,000. Each piece of such dielectric is coated with silver on the two sides to form a capacitor of large capacitance. Yet another advantage with these ceramic dielectrics is that they have negative temperature coefficient. Ceramic capacitors arc widely used in transistor circuits.

6.8.3 Electrolytic Capacitors

An electrolytic capacitor consists of two electrodes of aluminium, called the positive and the negative plates. The positive plate is electrolytically coated with a thin layer of aluminium oxide. This coating serves as the dielectric. The two electrodes are in contact through the electrolyte which is a solution of glycerine and sodium (or a paste of borates, for example, amonium borate), There are two types of electrolytic capacitors-the wet type and the dry type.

In the wet type the positive plate (A) is in the form of cylinder to present a large surface area. This is immersed in the electrolyte (E) contained in a metal can (M), This can act as a negative plate. It is shown in Fig. 6-14.

Flg.6.14: Wet type capacitor (olcctrolytlc).

In the dry type both plates are in the form of long skips of aluminium foils. Aluminium oxide is deposited electrically on one (A) of Lhe foils. This is kept separated

from the other (B) by cotton gauze (C) soaked in the electrolyte. It is then rolled up t?~ a cylindrical form. The oxide films on aluminium offer a low resistance to current in one direction and a very high resistance in the other direction. Hence an electrolytic capacitor must be placed in a DC circuit such that the potential of the oxide plate is always positive relative to the other plate. It is shown in Fig. 6.15 *.

-1

t+ Fig. 6.15: Dry type electrolytic capncitor

6.8.4 Variable Air CapacitodGang Capacitor

A very common capacitor whose capacitance can be varied continuously is used for tuning in a radio.station. The capacity of this capacitor can be uniformly varied by rotating a knob. (different forms of such a type of capacitor is shown in Fig. (6.16)).

Flg. 6.16: Vnrlnble nlr rnpncltor

The capacitor consists of two sets of semicircular aluminium plates. One set of plates is fixed and the other set of plates can be rotated bjl the knob. As it is rotated, the moving set of plates yr,aduatly gets into (or comes out of )the interspace between the fixed set. The area of overlap between the two sets of plates can thus be uniformly varicd. This changes the capacitance of the capacitor. The air between the plates ac's as the dielectric. Usually it consists of two condensers attached to the same knob (ganged). When the knob is rotated the variation of C in both the plate takes place simultaneously. This is widely used in wireless sets and electronic circuits. See Table 6.1 for a comparative range of voltages for different types of. condensers.

SAQ 9

What'is a variable capacitor? Give an example of a variable capacitor'with a solid dielecrric.

6.8.5 Guard Ring Capacitor

In Section 6.2 we calculated lhc cpacitance of a parallel plate condenser. We neglected the nonuniformity of electric ficld at the edges. It is possible to get over thc prgblem of edge effccts by using a guard ring capacitor. In his capacitor a ring R is uscd around thc upper plates of the parallel plate capacitor. This is shown in Fig. 6.17.

6.17: Guard rhtg cnpocltor.

The inner diameter of the ring R is slightly larger than :he diameter of the capacitor plate A. The diameter of the other capacitor plate B is equd to the outher diameter of the ring. Now the edge effects are absent as far as the plates A and B are concerned. In estimating the capacitance of the guard ring capacitor, we take the effective area of the plates as equal to the sum of the area of the plate A and half the area of the gap between A and R.

In Table 6.1, the capacity range, max. rating voltage and use of different types of capacitors are shown.

6.9 SUMMARY

o Any device which can store charges is a capacitor. The capacity of capacitor is given by

Type of Dielec&c

pap~r

Mica

Ceramic

Electrolytic ( A l d u m Oxide) .

Where the symbols have their usual meaning.

Capacitance Range

. 250PF-10 /.LF

25PF.25 p.F

0.5PF-0.01 @?

1p.F-1000 pF

Max. Rating Voltage

150 KV

2 KV

500 KV

600 V at small capacitance

0 The energy stored in a capacitor is given by

- Remarks

Cheap, used in circuits where losses are not important.

HIgll qulity. used in low circuit

Higb quality used in low loss precision circuit where miniaturisation is important.

Used where large capacitance is needed.

1 W = - C$2 = Q2/2C Joules 2

The symbols have their usual meanings.

If you introduce an insulator of thickness 't' between the two plates of a capacitor, then the resultant capacity is given by

o The maximum' safe voltage is called'rating voltage of a capacitor.

The capacitance of a cylindrical capacitor, per unit length is given by

0 If two capacitors CI and Cz are connected in series, then the resultant capacity is given by

0 The resultant capacity of two capacitors C1 and Cz, when connected in parallel is given by

Electrostarlcs in Medtum e The energy stored in a dielectric medium is given by

e Practical capacitors are made in different ways, to suit the particular applications. Layers of coriducting foil and paper rolled up give a cheap form of capacitor. Mica and metal foil stands high electric field but are more expensive. Electrolytic capacitors, in which the dielectric is a very thin oxide film deposited electrolytically, give very large capacitance. Ceramic capacitors are useful in transistor circuits where voltages are low but small size and compactness are very desirable.

Terminal Questions

1) A capacitor has n similar plates at equal spacing, with the alternate plates connected together. Show that its capacitance is equal to (n - 1) er AID.

2) What potential would be necessary between the parallel plates of a capacitor separated by a distance of 0.5cm in order that the gravitational force on a proton would be balanced by the electric field? Mass of proton = 1.67 x kg.

3) A capacitor is made of two hollow concentric metal spheres of radil a and b @>a). The outer sphere is earthed. See Fig. 6.18. Find the capacity.

4) In the arrangement shown in Fig. 6.19, find the values of the capacitances suchrhat when a voltage is applied between the terminals A and B no voltage difference is set up berween terminals C and D.

Pig. 6.19

5) Two capacitors one charged and the other uncharged are joined in parallel. Show that the final energy is less -than the initial energy and derive the formula for the loss of energy in terms of thk initial charges and the capacitances of the two capacitors.

Answer's of SAQ's ,

1) The potcntid difference (V) between the plates is not changed. But the electric field between the plates is Vl(d/2)= 2(Vld) = twice the value of the electric field E. The doubling of the eleclric field doubles the charge on each plate. Therefore, C = (Q/v) also doubles. Thus if we halve the distance of separation between the

40 plates, he capacitance doubles.

2) We know that Capacitor

and V = 24 V

Q = CV = .OOl x 24 C o d = .O24 Coul.

3) The energy stored in a capacitor is

It can be written

1 w = ? C $ . $

We know that

Q = c$

Using Eq. (ii) in Eq. (i), we get

Hence prove, the result.

Here

Therefore

'. Here C is h e charge that raises the potenlid by unity or the charge holding capacity. I , :

5) We have

Capacitance with the dielectric Er =

capacitance with free space

. Here a, = 3. Thus the capacitance of ihe capacitor will get tr~bled when the dielectric (E, = 3) is filled up in all thc air space.

Now a dielectric malerial is introduced. Let its thickness bc I. Thc capacity of the capacitance is

3 Here t = - d andl a, = 3 4

Therefore,

Tha& is, the capacitance will get doubled.

7) When the capacitor are con~ected in series, the equilent b given by

8) The arrangement is shown in Fig. 6.20. Let C4 be the effective capacitance of Ci and C2. Usinj~ series law of capacitors 1

This capacitance C4 then adds to C3 to give the total capacitance C of the combination i.e..

C = c4 + c3

Flg. 620

Answer's of TO'S

1) As seen from Figure 6.21, n plates provide (n-1) capacitors connected in parallel. The effective capacitance of (n-1) capacitors, of equal capacitance in parallel is

= sum of the individual capacitance

= (n-1) * capacitance of a single unit

= (n-1) er 6x1 Alii.

For example in Fig. 6.21 the first 3 plates A, B, C give two capacitors AB & CB and so on.

2) Let the required potential be equal to 4. Then E = $Id = $15 x lP3 volt,/m.

Electrostatic force on proton = qE

= 2 x lo2 x 1.6 x $ Newtons.

gravitational force = 1.67 x x 9.8 Newtons. Equating the two we get

$ = 5 x 10-lo volts.

3) If a charge q is placed on the inner spherc of radius 'a' an equal and opposite amount of charge appears on the inner side of the outer sphere, The eleclric field gets confined to the space petween the concentric spheres. To evaluate E consider a Gaussian surface. The symmetry of h e problem suggests a concentric sphere of

. radius r as Gaussian surface. The electric field, E, is normal to this surface and so the flux of E is given by

4 4 d - E = - &P

The potential of the inner sphere with reference to lhe outer sphere at zero potential is equal to

Eleclrostntics In Medlu~n

4 4, = - 4 n c p (f - i ) since c = 0

Hence the capacitanci is given by

, = 4 - 4nEoba . @ (b - a)

4) The potentials of the two plates connected to the poht C are the same. Hence if a charge ql is placed on one of these plates the other plate will have an equal and opposite charge. When a voltage is applied between A and B let a charge qi accumulate on C1 and a charge 42 on C2. Then the potentid difference (p.d) between the plates of various condensers are given by

Now

41 41 92 42 - + - = - + - = p.d across AB Cl C2 C3 c4

It the p.d between C and D is equal to zero

41 92 - = - '41 - 42 C2 C4 and - ct - - (A

which on elimination gives

is the required condition for zero potential difference between C and D.

5) Let the initial charge on the capacitor of capacitance C1 equal to q. When this capacitor is joined to the uncharged capacitor of capacitance C2 then the charge distributes in such a way that the potentials are equal as the combination is a parallel one. Let a charge qz flow from the charged capacitor to the unqharged one. The charge which remains on the initially charged capacitor as a result of sharing of charges is then equal to q - qz. As the potentials are equal

Initial energy E of he charged capacitor (before sharing of charges ) is

Final energy Ef of the 'two capacitors is given by . '

Capacitor

Substituting for 42 from above

q2 i 1 Hence the loss in energy = - - - - - - q2C2 2 Cl Cl+C2 2Cl(Cl+C,)

UNIT 7 MICROSCOPIC PROPERTIES OF DIELECTRICS

Structure

7.1 Introduction

Objectives

7.2 Microscopic Picture of a Dielectric in a Uniform Electric Field Review

Definition of Local Field

7.3 Determination of Local Field: Electric Field in Cavities of a Dielectric

, 7.4 Clausius-Mossotti Formula

Polarisation in a Gas

Relation between Polarisability and Relative Permittivity

7.5 Relation between the Polarisability and Refractive Index

7.6 Behaviour of Dielectric in Changing or Alternating Fields

7.7 Role of Dielectric in Practical Life

7.8 Summary

7.9 Terminal Questions

Sohtion//lnswers

SAQ's

TQ's

7.1 INTRODUCTION

In Unit-5, we have studied the macroscopic (average) behaviour of a dielectric in an . electric field. We also foun& that the field is altered within the body of the dielectric, 'Thi's can be accounted for by the charges appearing on the surface of the dielectric i n the case of an isotropic material. In Unit 6, the macroscopic study of the dielectric behaviour was used to study the increase of capacitance in a condenser when a dielectric is placed between the plates of the condenser.

In the present unit, we will describe microscopic picture of a dieiecnic in which We will define the local field (El,), arid the average macmscopip field inside the dielectric (Ei). Further, we will derive the relationship between the local field and the macroscopic field. We will also study the effects of polarisation in nonpolar and polar molecules and derive the famows Clausius-Mossotti formula for polarisation of . these molecules. Then we will derive Clausius-Mossotti equation for a gas. We will also study the relationship between polarisability arrd relative permittivity. After that, we will derive the relationship between polarisability and Refractive index. As you know that capacitors are used in alternating fields. So we will also study the effect of alternating field on a dielectric. In the last section of this unit we will, study the role of dielectrics in our daily life.

In the next block, we will study the electric current produced by moving charges.

Objectives

After going through ihis unit, you will be able to:

0 define the local field and relate it with polarisation,

find the macroscopic field within the dielectric and relate it to polqirisatim. /

e relate the macroscopic elecvic field, the local f ~ l d and the

microscopic field within the dielectric,

, write Clausius-MossoUi equation foi a liquid and a gas,

r establish a relationship between polarisability and Refractive index,

e discuss thr role of dielectrics in daily life,

MICROSCOPIC PICTURE OF A DIELECTRIC IN A UNIFORM ELECTRIC FIELD-REVIEW

F Unit 5 you have studied the average (macorscopic) behaviour of dielectrics. In this section, we will study the microscopic piqture of a delectric in a ul~iform electric field. Let us consider a diclecaic in a uniform electric field as shown in Fig. 7.1.

In an electric field, the elecms and atomic nuclei of the dielectric materid experience forces in opposite directions. We know that the electrons in a dielectric cannot move freely as in a conductor. Hence each atom becomes a tiny dipole with the positive and' negative charge centres slightly separated. Taking the charge separation as a, the charge as q the dipole moment p in the direction of field assbciated with tlie atom or molecule

EQ (7.1) gives the dipole moment induced in the atom/molaule by the field. Hence we cal l it as induced dipole moment. If there are n such dipoles in an element of volume V of the ma@rial, we can define the polatisation vector P as the (dielectric) dl@ moment per unit volume as

Within the dielectric the charges neutcalise each other, the negative charge of one atomlmolecule is ~eubalised by the positive charge of its neighbur, Tlrus within the bulk of the material,the electric field produces on charge density but only a dipole moment density. However, at the surface this charge cancellation is not complete, and a polarisation charge densities of apposite signs a p p at the two surfaces

, perpendicular to the fjeld. Now what is the consequence of the appearance of polarisation charges?

*.The consequence of this is that the elecfrk field inside the dielectric is less than the ! electric field causing the polarisation, The polarisation charges give rise to an electric field in the opposite direction. This field opposes the'electric field causing pdarisation, It is shown' in Fig. 7.2 - I

Electroslntlcs In Medium - Hence we conclude that inside the dielectric, the average electric field is less than the elecbic field causing polarisation. However, the macroscopic or average field is not a satisfactory measure of the focal field responsible for the polarisation of each atom.

, Let us denote the field at the site or location of the atom or molecule as the local field. In next section, we will calculate the local field inside a dielectric.

7.2.1 Definition of Local field 0

In this section we will define the local field in a dielectric material. This is the field on a unit positive charge kept at a location or site from which an atom or molecule has been removed provided the other charges remain unaffected. Fig.7.3 shows a site in a uniformly polarised medium from which a moleculelatom is removed when all other charges are kept intact at their positions.

Flg. 73: A sfta in a uniformly polnrlscd medium.

The extent of the cahrge separation depends on the magnitude of the local field. Hence we conclude that the induced dipole moment, p, is directly proportional to the local field, El,. Thus we have,

P = a El,

Where a is the constant of proportionality and is known as atomiclmolecular polarisability and El, the'local field.

To use Eq. (7.3) we require the value of El,

7.3 DETERMINATION OF LOCAL FIELD: ELECTRIC FIELDS IN CAVITIES OF A DIELECTRIC

The polarisation of dense materials such as liquids and many solids changes the electric fieldinside the material. The field experienced by an individual atom/ molecule depends on the polarisation of atoms in its immediate vicinity. The actual value of the field varies rapidly from point to point. Very close to the nucleus it is very high and it is relatively small in between the atomslmolecules. By taking the tnean of the fields over a space containing a very large number of atoms one gets the average value of the field.

SAQl

Show that the field at the centre of a spherical cavity (filled with air) is zero.

The field experienced by an individual atom/molecule may be called the local field which is different form the average field. The local field is the one which causes the ; polarisation of the atom. The average,field can be expressed as Vld where V is the

potential difference between two points of a dielectric, distant d apart (just as one obtains the field between the plates of a parallel plate condenser). The estimation of local field is not so easy. Let us consider three different cavities to find the local field in a dense dielectric which has been uniformly polarised. See Fig. 7.4.

. ..

Mleroscoplc Propcrtics of Dlclcctricu

ti) (dl

Fig. 7.4: The fleld In a slot cut in n dlclcctrlc depends on thc shape and oricntatlon of the slot. E shown is thc nvcragc flcld.

The directions of electric (average) field E and P are shown in Fig. 7.4. Suppose we cut a rectangular slot ABCDEFGH as i n (a) of Fig. (7.4). The field E and the polarisation P are parallel to the faces ABCD, EFGH. The field inside this slot can be found out by evaluating the line inmgral of E arourrd the curve C shown in Fig. 7.4(b). Since E.dl has to be zero for the closed curve C the field inside this slot has to be the same as the field outside the slot. Therefore the field inside a thin slot cut parallel to the field is equal to the average field E.

Now consider a thin rectangular slot with faces perpendicular to the average field E cut from the dielectric as shown in (c) A'B'C'D'E'F'G'H' of Fig. 7.4. To 'find the field inside this slot we use the Gauss' flux theorem on a surface S with one face outside the slot and one face inside the slot. See Fig,7.4(d). The flux of E through faces parallel to E is zero. Instead of the flux of E let us consider the flux of electric displacement D. Let El,, be the field inside the slot; then Dl inside the slot is ~o Eloc. The D vector outside the slot is ~a E + P. NOW, as the flux of D through the closed surface S has to be zero (no free or exemal charges inside the Gaussian surface). we must have

The field inside the slot in this case is different from the field outside by P/&o because of the surface polarisation charges appearing on A'B'C'D' shown in Fig. 7.4(c).

Another possible slot is a spherical hole which is the most likely way an atom finds itself in most liquids and solids. We would expect that an atom finds itself, on the average, surrounded by other atoms in what would be a good approximation to a spherical hole. What is the local field in a spherical hole? Suppose we cutla spherical hole after "freezing" the state of polarisation from a uniformly polarised material. If we cdl El, as the field inside the spherical hole at its cantre and E p as the field produced by the uniformly polarised bielectric spherical plug at ils centre, then by adding El, and Ep, we should get the average field E inside the dielectric, See Fig. 7.5. This should be m e bWdause of the superposition principle. Thus

Flg. 7.5: The field a t any polnt A In' a dielectric can be considered as a sum of the fleld in a spherical hole plus tho flcld due to the spherical plug.

and the required field

Elw = E - EP (7.6)

One can calculate EP (the field produced by the uniform polarised dielectric ) as I follows: I

The field EP arises from bound charges of density = a. n = P cose. Hence the field '

due to the charges over an area dS is given

where r is the unit vector from &he surface to the centres of the sphere where the field 'is to be calculated.

Resolving dEp into components parallel and perpendiculw to P, it is clear from Ule syrnmeay of the situation that only the components parallel to the direction of P will contribute to the t o d field EP Thus

Ep = &P . cos0

It should be noted that the derection of Ep is parallel to that of P, we hen have

Now & = r2* sineded

and the limits of 0 are from 0 to x and that of from 0 to 21e.

Then the field experienced by an atom in a spherical hole is '

P El, = E - -

3e.O

To determine the field Ep at an arbitrary point r inside the dielectric sphere, we consider the polarised sphere as a superposition of slightly displaced spheres of positive aid negative charges. See Fig. 7.6. Further note that the field at point r, is erltirely determined by the charge contained in the sphere of radius r, interior to point r.

(rl 01

Fig. 7.6: Superposltlon of ellgh~g displaced sphere of podtlve and negatlve charges. I

The sphere of positive charge C a n be regarded as a point charge at its centre and'it P . I

. is the volume charge density then the positive charged sphee is equivalent to charge 4 x

I 50'. at its centre equal to. r3. Similarly the negative charged sphere is equivalent to a' ; j

-- -

4 % Mlrrursroplc ~roperU~s of Dldecerics point charge at its cenm. The magnitude of this point charge is same as - r 3. If 3 'a' is the .separation of the positive and negative charges in an charges in an atom

4 % then the uniformly polatised dielectric is equivalent to a dipole of moment - r3 a. 3 If there are n dipoles per unit volume, q is the charge on each dipole then <rp = qn. [The number of positive or negative charges per unit volume is also equal to n in the spheres considered above]. Then the dipole moment of the sphere is given by

4Yt and the polarised sphere is equivalent to a dipole of moment - r 3P kept at its 3 centre. The potential due to this dipole at the point r on the surface is given by

4 x r3P C ~ S e - P cos 8 r 3 4xeo r2 =3eo

where p, r, 0 are as shown in the Fig. 7.7.

Flg. 7.7: Fldd outslde a uniformly polarlsed sphere.

The polarisation is in the direction of E and if we rake this to be the z-direction with the origin at the centre then the potential at T is

Thii shows that the potential at a point depends only on its z coordinate. Hence the electric field is along z direction and is given by

This shows that the electric field inside the dielectric sphere is uniform and in the direction of the polarisation vector. Hence the f ~ l d experienced by an atom in a spherical hole is

The field in a spherical hole is greater than the average field by P / 3 ~ p .

SAQ 2

Show that the field inside a uniform spherically symmetric charge distribution with as r

charge density is equal to - 3 EO

where r is the position vector of the point with origin

at the centre.

Electrostntica In Medium 7.4 CLAUSNS-MOSSOTTI EQUATION

In a liquid we would expect an individual atom to be polarised by a field obtained in a spherical cavity rather than by the average (macroscopic) field. Thus using Eq. 7.8 and Eq. 7.3 we have

This can rewritten as

The susceptibility x was defined in Unit 5 by the equation

P = E o X E

, Hence . .

Eq. 7.11 gives the relation between susceptibility and atomic/molecular polaiisibility. This is one form of Clausius-Mossotti Equation.

7.4.1 Polarisation in a Gas.

Unlike the atoms/molecules of a liquid or solid it is possible to consider the atoms/ molecules of a gas as far ap&t and independent. We can neglect the field due to the dipoles on the immediate neighbourhood of an individual molecule. Hence the local field causing polarisation is the average or macroscopic field E,Therefore we can write

P = a ~ E = n p

where n is the number of molecules per unit volume. If we consider only an individual atom/molecule and write the dipole moment p as

where a is known as atomic polarisability. Therefore u has the dimensions of volume and roughly equals the volume of an atom.

We can relate a or x to the natural frequency of oscillation of electrons in the atom/ molecule. If the atom is placed in an oscillating field E the centre of charge of electrons obeys the equation

where m is the mass of electron of charge q, mwo2x is the restoring force term and q E 'the force from outside field-this equation is the same-as the equation of forced oscillation. If the electric field varies with angular frequency w then

4

For our purposes in the lectrostatic case w = 0 which means that 52

and the dipole moment p is

From Eq. (7.12) we can write the atomic polarisability as

and

- = ~ ~ = q ( E O - l ) = a n a E

For hydrogen gas we can get a rough estimate of oo.' The energy needed to ionise

the hydrogen atom is equal to 13.6 eV. Equaling this a where h is Planck'n 2it

constant we get

Substituting this in the equation 7.13 (a) we get

The experimentally observed value is E, = 1.00026.

7.4.2 Relation between Polarisability and Relative Permittivity

In Unit 5 , you have studied that one can write P as

P = €4 ( ~ r - 1) E

where E~ is the relative permittivity.

Using Eq. 7.14 in Eq. 7.8 we get

Using Eqs. 7.14 and 7.15 one can rewrite Eq. 7.9 as

which yields

Eq; 7.16 gives us the relation between atomiJmolecular polarisability and the relative permittivity. Eq. (7.16) is another fomi of the Clausius-Mossoti.equation.

EJectrostaties in Medlum SAQ 3

Obtain Eq. 7.15 from Eq. 7.14.

Proofs of thia rclntion will be given in the unit on propgation of electron magnetic wavea.

7. 5 RELATION BETWEEN THE POLARISABILITY AND REFRACTIVE INDEX

For a dielectric, the refractive index p defined as the ratio of the speed of light in vacuum to the speed in the dielectric medium, can be shown to be equal to 6.

Using Eq. 7.15 in Eq. 7.14 we get ,

.Eq. 7.17 gives the relation between polarisability and refractive index. This relation is i known. as the Lorentz-Lorenz formula.

In a l l the equations discussed above n represents the number density of atoms or molecules which is equal to NA dlW where NA is the Avogadro number, d the mass density and W the molecular weight. For gases, we have the gas equation relating pressure, P volume V and absolute temperature T given by

P'V = ET = NAH

where q is the mole number.

and P' = qNA kT / V = nkT

Therefore, n = p' / kT, I

Thus if we determine ei at different pressures for a gas. we can calculate the atomic1 I 1

molecular polarisability of gas. For this we write Eq. 7.16 as I

I

3eokT (Er - 1) a = - p' m 1

1

or i

Eq. 7.18 represents the linear relation between ( E , - 1) / ( E , + 2) and (p' / T). If now I I

a graph is drawn with (e, - 1) / (E, + 2) on the y-axis and (p' / T) on the x-axis, we get a straightline the slope of which gives (a/ 3 ~ 4 .

t 7.6 BEHAVIOUR OF DIELECTRIC IN CHANGING

OR A1,TERNATING FPELDS

So far we have considered only electrostatic fields in matter. Now we would like to look at the effects of electric fields that vary with time, like the field in the dielectric of a capacibr used in an alternating current circuit.

-. Will the changes in polarisation keep up with the changes in the field? Will the polarisability, the ratio of P to E, at any instant be a e same as in a static electric field?. .

For very slow changes or small frequencies we do not expect any difference. However, for high frequencies or faster process we have to look at the response time far the polarisation. We have to separately consider two polarisation processes viz., I. induced polarisation and the orientation of permanent dipoles..we know that the

- - -- - - -.

induced polarisation .occurs by the distortion of the electronic structure. In the Mlcroscoplc Properties of Dlclectrla

distortion mass involved is that of electron and the distortion is very small, which means the structure is very stiff. From our knowledge of oscillatory motion (see the. course on oscillations and waves), its natural frequences of vibration are extremely high. Alternatively, the motions of electrons in atoms and molecules are characterised by periods of the order of the period of a visible light wave (10-l6 seconds). Thus the

, readjustment of the electronic structure i.e. the polarisation response is very rapid, occurring at the time scale of 10-l4 sec. For this reason we find that nonpolar substances behave the same way from dc upto frequencies close to those of visible Ijght.

We shall examine the situation in the light of Eq. 7.15, where we have expressed the Clausius-Mossotti formula in terms of the refractive index. We know that the refractive index is dependent on the wavelength or frequency. Thus, in a way 7.13 implies the variation of the polarisability with frequency.

Experimentally, d.c. values of % can be found. The refractive index of the same substance can be determined by optical methods, using a spectrometer. A fairly good agreement is found between the refractive index and values for'non-polar substances. However for polar substances, EI varies with frequency; it decreases with increase in frequency. The drop in the value of er at high frequencies is due to the fact that the permanent dispoles are not able to follow the rapid alternation of the field. In other words the polarisation response of polar molecules is much slower. However, in the frequency range of visible light the refractive index and E* values shows a fairly good agreement as indicated by nonpolar substances.

7.7 ROLE OF DIELECTRIC CAPA~ITQR IN OUR PRACTICAL LIFE

Dielectrics have several applications. Dielectrics are used very widely in capacitors. Although the actual requirements vary depending on the application, there are certain characteristics which are desirable for their use in capacitors, A capacitor should be small, have high resistance, be capable of being used at high temperatures and have long life. From a commercial point of view it should also be cheap. Specially prepared thin haft paper, free from holes and conductiag particles, is used in power capacitors where withstanding high voltage stresses is more important than incurring dielectric losses. In addition, the kraft paper is impregnated with a suitable liquid such as chlorinated diphenyl. This increases the dielectric constant and thus reduces the size of the capacitor. In addition the breakdown strength is increased.

In addition to paper capacitors for general purpose, other types of capacitors are used. In the film capacitors, thin film of teflon, mylar or polythene are used. These not only reduce the size of the capacitor but also have high resistivity. Teflon is used at high frequencies as it has low loss. In decuic capacitors, an electrolyte is deposited on the impregnating paper. The size of such a capacitor is small as the film is very thin. Polarity and the maximum operating voltage are important specifications for these capacitors,

Some ceramics can be used as temperature compensators in electronic circuits. High dielectric constant materials, where small variations in dielectric constant with temperature can be tolerated, help miniaturise capacitors. Barium titanate and its .modifications are the best examples of such materials.

7.8 SUMMARY

Inside a dielectric the average elecmc field is lcss than the electric field which causes the polarisation. .

' In a dielecaic material, the induced dipole moment p, is direclly proporlional to the local field and mathematically given by

Electrostatics in Medium where the symbols have their usual meanings.

B The field inside a spherical hole is given by

which shows that the field in a spheric& hole is greater than the average field.

B The relation between susceptibiliby and atomic/molecular polarisability is given by

7.9 TERMINAL QUESTIONS

i) A sphere of Linear dielectric material is placed in a uniform electric field Eo (see Fig. TQ1). Find the field inside the sphere and polarisation in terms of external field Elw.

Fig. TQ1 : A linear dielectric material placed in a uhiform map[netic field.

2) The electric field inside a polarised sphere is uniform and equal to -P/3&o. Prove this by superposing the internal fields of two spheres of charge whose centres are separated.

3) Show that I% times the force on a unit charge placed in a disc shaped cavity will measure the electric displacement (D) in A solid dielectric.

4) A dielectric consists of a cubical array of atoms (or molecules) with spacing d between each atom along the (x,y,z,) axis. It is influenced by a field El,, applied along the direction of z-axis. evaluate the average field produced by all the dipoles.

SAQ's

1) We identify pairs of dipoles equidistant from the centre. from Unit 3, we know that the dipole field falls off with distance as 113. Since equidistant pairs have directions of p opposite to one another, the overall field at' the centre due to the pair is zero. This is the case for every other pair. Hence &, = 0.

2) According to integral form of Gauss's law

E . n d s = - dV E 0

Therefore,

or

1 E = - r 3 ~ o

in the vectror form

E(r) = (r)r I 313

Terminal Answers

1) The resultant field is no longer uniform in the neighbowhood of the sphere , because of the polarisation of the sphere. Let P be the dipole moment per unit volume inside the sphere. We would expect P to be uniform as the dielectric material is linear. Then P is proportional ta the electric field, L, inside the sphere. If x is the susceptibility of the material then we can write P as

This polarisation P produces a field inside the sphere which is given by - Pl313.

The electric field inside the sphere, Ei,,, can be regarded as a superposition of the uniform field EO and the field due to polarised (dipoles) charges. Thus.

Substituting for P in terms of E;,

Thus

El, = 3 Eo - 3 (3 + X) - (E, +o

and P = EO (E, - 1) Ek

The assumption of uniform polarisation is now seen to be self-consistent.

(a) (b) Ic) Flg. TQ2 : A sphere of lined-up molecuiar dipoles

The polarise4 dielectric sphere (a) of Fig. TQ2 can be regarded as a superposition of two spheres charged uniformly one with positive charge as in (b) and another with negative charge as in (c) of Fig. TQ2. The two spheres have centres at C1 and C2 which are separated by a distance a, say along the z-axis. This means that P is along the z direction. The field at R due to the positively charged sphere is given by

where r = 61 R and p is the charge density. Similarly the field at R due to the negatively charged sphere is

where r' = C2 R. Now p is along the z-direction and C2 GI = a is also along the

z-direction. Adding E+ and E vectoridly (See Fig. TQ) we get

P a E+ + E, = - ( r - r') = - - 3 ~ o 3 ~ o

as a is equal to the no. of either positive or negative charges per unit volume.

3) We assume that the radius of the cavity to be greater than its thickness, measured parallel to the field . By this assumption, ti.? fields near A and B is the same as E. the field Eo at the centre of the cavity is therefore parallel to EA, the field near A, This is in accordance with the boundary condition for fie normal component of D(= ea E), viz.,

4 EA = EC = (Dl" I Here the normal component measures the cornplek. vector D. Therefore we can conclude that eo times the force on unit charge, viz., E, placed in a disc shaped cavity measures the electric displacement @).

4) The scalar potential due to a dipole of moment P at (x8,y', 2 ' ) is / , I

since (z - 2') = R cos eand

The z-component of the field

We assume the dipole to be present at the origin. To find the average value of the field, we inteaate the field. The integration is taken over by an Octant of the unit celL The unit volume of the unit cell is d3. Thus

[Kindly see PE: -04 & PHE-05 Course]

Structure

8.1 Introduction Objectives

8.2 Electric Current and Current Density

8.3 Conduction Mechanism Drift Velocity and Ohm's Law Temperature Dependence of Resistivity Breakdown of Ohm's Law

8.4 Current-voltage Relationship for Diode (Non-ohmic Conductor)

8.5 Summary

8.6 Terminal Questions

8.7 Solutions and Answers

8.1 INTRODUCTION

Water molecules flowing down a river constitute a water current. Analogously, electric charges flowing in a wire constitute an electric current. All the electrical appliances we use, such as the radio, electric heater and torch light, depend on the flow of electric charge. Such motion of charge usually occurs in coilductors which contain free electrons; in the ionized gases of fluorescent lamps which contain charge carriers of both signs; and also in an evacuated region, for example, electrons in a TV p i a r e tube.

In the last two Blocks we deait with electrostatics, in which charges are at resl. With this background, we now begin our study of electric charge that moves or flows from one point to another. You will find that an electric current results from charge motion due to an applied electric field whenever the charges are free to move. One of the reasons as to why we are studying electric current is that it forms a background for the trealment of electromagnetism in Blocks 3 and 4.

In Block 2 of this course, we discussed certain aspects of the behaviour of a substance under an applied electric field in terms of electric susceptibility of the material. In this unit, we discuss another important of the material called electric conduction, which is also a response to an applied external electric field. The difference between the two cases is that in the former the charges are bound so that they undergo only small

.displacements, while in the latter case the charges are free and under the action of the field they flow and result in a current.

The concepts of electric current will be of much use to us. In the next unit, we will find that a new force field viz. magnetic field arises because of the motion of charge. We will also discuss the experimental relation between current and magnetic field and establish basic laws of magnetostatics, viz., Gauss's law for magnetism and Ampere's law.

Objectives

After studying this unit you should be able to:

explain the concept of electric cumnt and obtain the expression for current density in terms of the driftvelocity,

explain the conduaion mechanism microscopically,

distinguish between ohmic and non-ohmic behaviour,

use the continuity equation to discuss the behaviour of current in a diode.

Ng. 81 t The lastanhnewm rwnent dong e win is defined as Lhc net ante rat which the ehorge passes through an ans perpcndiculnrto the ads of the wire.

8.2 ELECTMC CURWlENT AND CURRENT DENSITY I 'i i

In Block 1, you leanrt that when a charge is placed in an electric field it is acted on by a 1 force and moves in the direction of lines of force. If the ends of a conductor, say, a copper wire are connected to a battery, an electric field E wil1,be set up at every point within the conductor. Due to the presence of the field, th? electrons present in the wire will move in the direction opposite to that of the field and give rise to an electric current in the wirc. An electric current is caused whenever the charges move. (11 the case of a copper wire the flow of electrons constitutes an electric current.) It is defined as the .

I amount of the charge moving across a given cross-section of the wire per unit time. In Fig. 8.1, for a wire, it is defined as the rate at which charge passes through a plane pe~pendicular to the axis of the wire. For example, if charge q crosses the shown

I cross-section in Fig. 8.1 in time t then the average current I is given by I

1 net charge transferred q I - n

time taken t

When the current is not constant, i.e., the current varies with time, we define an instantaneous value of the current I ( t ). If a net charge of Aqcrosses the shaded area of Fig.8.1 in a time At, the instantaneous current is given by

Eq. (8.1) or (8.2) shows that the unit of current is Coulombs per second ( Cs' ' ). In the SI system of units it has been given the name ampere (abbreviated A). In Unit 1 of Block 1 we have stated the definition of ampere.

Current is a scalar quantity, because both q and t are scalars. It is not a vector quantity as it does not obey the vector laws. Often, a current in a wire is represented by an arrow. Such arrows are not vectors, they only show a direction (or sense) of flow of charges along a conductor, 11ot a direction in space.

SAQ 1

Name few other physical quantities, like current, that are scalars having a sense represented by an arrow in a diagram.

An electric current may consist of only one sign of charge in motion, or it may involve both positive and negative chhrges. By convention, the direction of current is defined as that direction in which tlie positive charge flows. If the moving charge is negative, as with electrons in a metal, then the current is opposite to the flow of the actual charges. When the current is due to both positive and negative charges it is determined by the net charge mothl, that is, by the algebraic sum of the currents associated with both kinds of charges. For example, when salt (NaCI) is dissolved in water, it splits up into Nat ions and C1- ions. .The sodium ion is positively charged and the chlorine ion is negatively charged. Under the influence of the electric field established between the two electrodes, these ions inove through the liquid in opposite direction. Thus the inotioiiof both positive and negative ions coi~tributes to the current in the same direction.

i As defined earlier, curreut is the total charge passing through the wire per unit time I

across any cross-section. Therefore, the current is determined by the total charge that 1

flows through the wire, whether or not the charge passing through every element of the cross-section of the wi;k is the same. It is for this reason that current is a macroscopic quantity. If the charge passing through various elements of the cross-section of the wirc is not the same, it is necessary to define a quantity at every point of the copductor. 'lliis is called the current density which is a microscopic quantity and denoted by J. It is defined as the charge flowing per unit tide per unit area normal to the surface, and has a direction in which the positive charge moves. 1 Let us consider a simple system in which particles, each of chargeb, are moving to the right as shown in Fig. 8.2. !

Fb. 8 2 I C a k u L ~ deamnt io t a m s ddrin vdodty.

Imagine a small area dS around 6 i n t P so that all the particles crossing this area may be assumed to have the same speed v. Let us further imagine a cylinder of length vdt as shown in Fig. 8.2. Then all the particles within this cylinder of volume dSvdt would cross the area dS in time dt. If r2 is the number of charged particle per unit volume, tlleii the number of charged particle found in such a volume h d v d t . Therefore, the average rate at which the charge is passing through (LT that is, the current through dS is given by

Since cumnt density is defined as the curreht per unit area held normal to the velocity of the current camers, we have

Since the direction of Jis the direction of the actual flow of charges at that point, the above equation can be written in vector form as

J = nqv 18.5)

Thus J is a vector quantity. In SI system of u ~ t s J is expressed in amperes per square meter. When the current carriers are electrons, q - - e and Eq. (8.5) takes the form

The product rtq in Eq. (8.5) represents the volume charge density p of the current camers. Hence, in terms of p the current density is expressed as follow

If current density is uniform over the cross section S of the wire, we can find the total current by multiplying the current density by the cross section of the wire. If the cumnt density is not at right angles to the cross-sectional area, we consider only that component of J which is perpendicular to it; If we define a vector S the magnitude of which is the cross-sectional area S and the direction of which is along the perpendicular to the area, then a uniform current density J gives rise to a total current I = J.S (Fig.8.3). When the current density andlor surface orientation vary with position; we can do the same thing for many small areas dS, and then sum the results to get the total current (Fig, 8.4). The current through a small area dS is J.dS, so that the total current, I, through the entire surface is

where the limits of the integral are chosen to cover the entire surface. Eq. 8.8 should remind you of the def i~ t ion of the electric flux in Unit 2 of Block 1. (Compare Eq. 8.8 with Eq. 2.4). Indeed, the electric current through a surface is the flux of the current density through that surface. Eq. 8.8 again shows that the current is a scalar because the integral J.dS, is a scalar.

In Fig. 8.4, we have taken the surface S to be open surface, In such situation the vector dS; is taken to be positive in that direction along which the curient throughS is

Fig.8.31 TlcNrnntth&gha s u l i m a c of area S is given byJScas8,wJ.S where 0 is (he .ogle betaten (he vectors S and J.

Electric C u ~ n t and Magnetic Field

Fig. 8.5: The inteZra~J~ . d~

over n dosed +ace is LIK rate or change of the tolal charge q inside.

Fig. 8.4: When the current density a~udlor surface orientation v a r y with pobition, the total current mny be written as1 - J J . ~ S.

required. WhenS is a closed surface, as shown in Fig. 8.5, the directioll of every vector dS is taken along the outward normal to the surface. For such surfaces, the integral of J over S gives the rate at which the charge is going out of the voluine enclosed by S. Now one of the basic laws of Physics is that an electric charge is indestructible; it is never lost or created. Electric charges can move from place to place but never appear from nowhere. We say that the charge is conserved. Hence, i f there is a net current out of a closed surface, it must be equal to the rate at which the total charge within the volume is depleting. The electric current I flowing out of the closed surface S enclosing the volume V, is given by

We can, therefore, write the law of the conservation of charge as

The charge within the volume can be written as a volume integral of the charge density p as follows

ginside l p d v v

where V is the volume enclosed by surface S. Using Eq. (8.11) in Eq. (8.10) we get

d 1 (J.& - -- spdv (8.12) dt v

Since we are dealing with a fixed volume the time derivative opemtes oilly on the function p. Since p is a fuilction of spatiai,coordinates as well as time, the time derivative of p is written rn the partial derivative with respect to time when it is moved inside the integral, Hence

5 J.& - -JT $ d; (8.1 3) S Y

The surface integral on the left hand side of the Eq. (8.13) can be coilverted into a volutne integral through thb divergence theorem (see Unit 2 of Block 1);leading to

8 I

1 - i

But V is completely arbitrary and Eq. (8.14) will hold for an arbitrary volu~ne ele~nellt only when the integrand is zero. Thus

This differential equation is kiio'wn as the continuity equation. It expresses the conservation of charge in a differential form. Its meaning is clearer in Eq. (8.14), . according to which the change in the quantity of charge in any arbitrary volume must be accompanied by a net flow of charge inwards or outwards across its surface. When steady currents are involved we have

This is because a steady current is one for which J is constant in time at every point. In other words, equal charges flow in and flow out of a section and, hence, there cannot be any accumulation of charge at any point of the system. Hence, in this case tlie continuity equation becomes

V.J = 0 (8.16)

SAQ 2

Give an example of steady current system, and using the Eq. (8.16) list its featurc. The equation of continuity can be used to discuss the current distribution in a diode valve. But before discussing it let us find out why do the metals conduct electricity a ~ ~ d what are the factors which influence conductivity of the metal.

8.3 CONDUCTION MECHANISM I In this Unit, we are coilcentrating on the currents flowilig in metal wires. In a metal, the metal ions are fixed in aaregular array, known as lattice, making them relatively immobile. The metal i o y are positively charged because the atoms forming the metal lose one or more electrons which become free in the sense that these electrons wander through the ion lattice as shown in Fig. 8.6. It is the motion of these negatively chargcd electrons that gives metals their conducting properties.

0 0 0 0 0 \ 1(- 8 Mobile negatively charged free electron

It \

0 0 0 0 0

% /

\ r3

0 0 x 0 0 0

1 f

0 4 0 0 0- Immobile posj~ivcly charged ion

? f I \

0 0 0 0 0

Fig. 8.62 A schematic view d the aysh1 mtructutr ot a metal The positive metal ioas exist on a dgid Iat&c Each alom, on terming an ion, gives up m e or mqp dstronq which are Lben f h e (o wander through Lhe crystal,

9

When a battery is connected between the ends of a metallic wire MN as shown in Fig. 8.7,

we find that the current flows through it from M to N (current flowing in the wire can be detected by putting an ammeter inseries). Let us find out why and how the current starts flowing in a particular direction by taking a microscopic view of the situation.

8.3.1 Drift Velocity and Ohm's Law I When the metallic conductor is not connected to the battery, the free electrons present in the metal are in constant motion because of their thermal energy, the motion being random in velocity as shown in Fig. 8.8(a).

b .

F b 88: Motion of some hrc eketrow (a) In the ahstnce dmn ex tend 6cld a d (b) in (be prrscncc d an external field (as shown by the solid Lim).vt rrprrseata chcrmd v d a i ? y , v ~ is the velodly oaly in Ute presence of electric Bcld and v ia (bc net velodly.

In this state, the free electrons encounter frequent collisions with positive ions and impurity atoms (if any). At each collision, the velocity changes both in magnitude and direction; Since the motion is completely random, at any instant, the average velocity along any direction in the bulk of the conductor is zero. Hence no current. But remember that average speed of these free electrons at any instant is not zero. Its value is of the order of ld ms-l. When a battery is connected between the ends of the metallic win!; it maintains a uniform electric field E at each point in the wire. The electrons experience a force in a direction opposite to that of the applied electric field. Due to this force, besides having a thermal velocity v*, an electron also experiences a constant acceleration a = eE/m,, where me is the mass of the electron. You may now wonder and ask if the electron velocity, VE, which is in the presence of the electric field, increases continuously as it moves in the wire. Experiments show that this does not happen. As an electron picks up speed under the action of the field, it collides with the oscillating ion or the impurity atom within the metal. The result of this collision is that the electron loses all its .velocity acquired due to acceleration in the field. In other words, at each collision the velocity of the electron is randornised, and it begins at fresh acceleration in the direction of the field. If u i s the velocity of an electmn just after a collision, its velocity VE just before the next collision will be

eE VE = U f t

me

where t is the time of travel between the two collisions. The avenge ofthe velocities of all electrons before collision can be written as

where the sign c > denotes the average value of the parameter. Since the effect of each collision is to reduce the velocity to zero and to restore the random thermal motion, we can write ( o ) as ( vt ) which is zero, as explained earlier. If ( t ) is represented by z, then we get

For this reason VE does not increase continuously with time, but will rather have an average value ( VE ) as given by Eq. (8.19). Here .;G denotes the average time between successive collisions,i.e., the time over which the electrvn accelerated freely under the action of the electric field. This is called mean free time. The thermal motion of the free electrons is, therefore, modified as shown in Fig. 8.8b. It is clear from the figure that at any instant, the resultant velocity is vt + VE and for each electron it is different. The average ~ u l t a n t velocity of all the electrons can be expressed as

As already stated, vr is zero, but VE is iiot zero because of the fact that the VE for all the free electrons is in the same direction. Therefore, ( v ) = ( VE ). Hence, the free electrons in a metallic wire have an average velocity which is caused only by the applied electric field. This velocity is called the driR vellmity of the electrons denoted by ( vd ). That is,

It is this velocity that appeared in Eq. (8.5). Thus, the current density in a conductor can be written as

In most substances and over a wide range of electric field strengths, it has been experimentally found that the current density is proportional to the'strength of the electric field that causes it. The relation may be written as

I - G Z 1 (8.22)

where a is the proportionality constant and is known as the conductivlty of the material. Eq. (8.22) is a statement of Ohm's law. It is an empirical law, a generalization derived from experiment for some materials under certain conditions. It is not a theorein that must be universally obeyed. The value of a is very large for metallic conductors and extremely small for good insulators. It may also depend on the physical state of the material, for instance, on its temperature, about which you will study in next section. But for many common conductors, for given conditions, it does not depend on the magnitude of E. Such materials are called ohmic or linear and for such materials Q. (8.22) implies that the direction of J is always the s z z c as the direction of E. Eq. (8.22) shows that the units of conductivity are ( Am-2 )/( Vm-I ) or A\:' mbl. But one VAm'is given the name ohm ( symbol SZ ). Therefore, the SI unit of conductivity is ( Q m )-" Instead of the conductivity we can use its reciprocal, called resistivity p in stating the relation between current density and electric field as follows:

The units of resistivity are Q m. Since both E and J are microscopic parameters p also defines a microscopic property of the conductor. In Fig. 8.7, the electric field along the wire is in the direction MN and its value is E - V/L everywhee . Here V is an applied. potential difference between the ends of the wire. Thus

a SV and the total current is I = JS = - L '

where S is the cross-sectional area of the wire. This gives, .

V L p L - P-

I aSms

You must be knowing that a frecly falling body h vacuum has a velocity v 0 gt which increases continuously with lime, but iffhe body blls.through avisaus fluid, the moteion becomes uniiortn wilh a constant limiting velocity. By analogy, the effect of the crystal lattice can be represented by a viscous brce, acting on the conduction electrons when their natural motion is disturbed by the applied electric field.

Eq. 8.22 hot& only for isotropic mrlcrtls-lhosernaterials in whlcb h e electric propalles are the same in ell direclions.

It is customary lo use p as tho symbol for resistivity and a as the symbol for wnductivity inspite of their use in someofour other units for volume charge density and Gurhce charge density respcct ivcly. In Ihc rest of the units, p wit1 denote resistivity and a conduciivity if not stated otherwim.

~kct r ic current md The ratio V/I is called the resistaiiceR of the wire and the Eq. (8.24) is written in the Magnetic Field f0m-I

p] (8.25)

with R = p L /S

This gives another inore familiar expression of Ohm's law. It implies that resistance R of the conductor is independent of the applied potential difference V. Therefore, for the linear conductors (those conductors which obey Ohn's law) a graph between V- I is a straight line as shown in Fig. 8.9.

The resistivity p depends on the nature of the conducting material whereas the resistance depends not only on the nature of the medium but also on its physical dimensions. Resistivity p is of basic significance to those who wish to study the behaviour of the conductor from the atomic viewpoint. If we apply Eq. (8.20) to Eq. (8.21), we find

ne22 J = nevd E - E (Here we have replaced q by e) me

By comparison with Eq. (8.22) we tind.the expression for conductivity as follows:

and for the resistivity

This equation shows that the resistivity of a metal depends on the density of the free electrons, their mass and charge and on mean free time. The cause of the dependence of resistivity on temperature is the variation o f t with temperature. In the next sub section, we shall explain how resistivity depends upon temperature.

SAQ 3

Eq. (8.22) and Eq. (8.25) are mathematical expressions of 0hm"s law. Derive Eq. (8.22) from Eq. (8.25).

8.3.2 Temperature Dependence of Resistivity I i

I Till now we have been considering the simple classical model in which electrons are considered as a "gas" of charged particles getting accelerated under the influence of an

1 external electric field. On this basis, we arrived at theexpression of resistivity. But if we wish to go further in order to understand the dependence of p on temperature or, in other words, to undentand the dependence o f t on tempemture.we have to look for 1

other model. Fortunately, this can be ex~jlained on the basis of quantum mechanical model. In this model, we should not now think of the electron as a tiny charged particle. But we should think electron to be behaving moie like a wave interacting with a larger region of metal. If the temperature of metal is very low, say zero, then all the ions are rigidly fixed at. their regular lattice positions. This makes the classical collision between the electron -nd ion unlikely to occur. It means that, in such situations, the time between collisions

1 i I

is very large or infinite. The hindrance which interrupts the progress of an electron wave is not the regular array of ions but an irregularity in the array. On increasing the temperature, the ions vibrate, and it causes the solid to look less regularly spaced then it would if the ions were at rest. The effect is that the time between collisions is shortened. Hence the mean free time t decreases with increase of temperature. This lea& to increase of resistivity with temperature.

8.3.3 Breakdown of Ohm's Law

You may think that the metals always behave as linear conductors. But it is not so. Under certain conditions metals do not behave as linear conductors. In this analysis, we have been making an unstated assux.nption that the electric field applied to the metal is so small that it does not disturb the electron velocity pattern in the metal in a major

.

way. Ixnmediately, you would be tempted to ask, what will happen if the field is increased to high vahes. Let us first see what is the time between collisions when a small field is applied. In such situation, if the average distance travelled by an electroil before it encounters the next collision is denoted by h called the mean Free path of the electron, the average time t between two collisiorls is given by

where ( I v, + vd I ) is the average speed and not the average velocity between two collisions. For the electric fields normally used in the laboratory, vd ( u 10- cm s- ' ) is very small compared to vf ( - lo8 cm s- ' ). Therefore, Eq. (8.29) can be written as .

When we apply a very large field, the drift speed of the electrons, i.e., vd becomes comparable to vt. Then the time between collisions will be shorter than it was before the field was applied. This is an effect, which is not included in our theory. In this case, the expression for p would also contain parameter vd, which is strongly field-dependent. Therefore, metals under these circumstances would not obey Ohm's law.

At very large fields, yet another thing happens, the free electrons are accelerated so much that they gain sufficient energy. With these energies, they can strike the atom hard enough to knock another clectron out of its grip. Thus, extra electrons are freed and get accelerated. These accelerated, extra electrons release more charges when they collide with other atoms. This process, thus, causes an avalanche of free charge carriers. Under these conditions, it pbd;ces a rapidly increasing cumnt and unless tile avalanche is limited in some way, the process may destroy the material. This is a complete breakdown of Ohm's law.

We end ,this unit with a short discussion of a vacuum tube. We will apply the current electricity ideas, i.e., whatever you have learnt just now to vacuum tube and discover that the vacuum tube does not obey Ohm's law.

8.4 CURRENT -VOLTAGE RELAWONSHIP FOR DIODE (NON-OHMIC CONDUCTOR)

There are many conducters which do not obey Ohm's law. Such conductors are called non-linear conductors. The V- I characteristic of such~conductors is not a straight line as shown in Fig. 8.9. You must be knowing that examples of such non-linear conductors a&: vacuum tube diodes and electrolytes. For vacuum diode, the V- I characteristic have the form shown in Fig. 8.10. In this section we will find out the relation between current and voltage for a vacuum diode using the continuity equation.

In its simplest form, a diode can be assumed to wnsist of two electrodes. One electrode, ca!led cathode, is coated with a material that emits electrons copiously when heated. The other electrode, known as anode, is simply a metal plate. By means of a battery the anode is maintained at a positive potential with respect to the cathode. Both these electrodes are enclosed usually in a glass tube the inside of which is evacuated to a very low pressure ( of the order of lo-' cm of Hg ). Electrons emerge from this hot cathode

4 v -. -_-. Fig. 8.10: Currunt-voltagc

characteristic of a vacuum diode.

Electric Cnrrent an11 with very low velocities and then, being negatively charged, are accelerated towards !he I

Magnetic FicId positive anode due to the presence of electric field between the cathode aiid anode. 111 the space between the cathode and anode the electric current consists of these m o v i ~ ~ g I

electrons. j

In this diode, the local charge density p (here and in this sectioi~ p dciiotes charge density), is simply equal to nq, where ti is the local density of the electrons. Accordii~g to Eq. (8.4), the local current density J is pv, where v is the velocity of eleclrons in the I

region coilcemed, (This is because in this case the drift velocity is the actual velocity.)

x 1 0 x - d ' X If it is assumed that J has no y orz cornponeilts as show11 in Fig. 8.11 and if condiliolls 1

v = o a ~ , are steady, then div J = 0. i.e., - = 0. This iilealls if we have a steady stream of

1

ax Fig. 8.11: A vocuunl diode with electrons moving in the x direction only, the same llumber per second have to cross ally

planc"pnlmue'cathocle and anodc. intermediate plane between cathode and anode. We conclude that pv is constant. But observe that v is not coilstant; it varies withx, because the electrolls are accelerated by the field. Hence p is not col~stant either. Instead, the negative charge density is high near the cathode, low near the anode, just as the density of vehicles OII a highway is I high near a traffic signal and low where traffic is moving at high speed. This is because

I the electrons coming out of the cathode form a space charge which prevents further emission of electrons for low anode voltages (low acceleratioi~ of electrons between the i two electrodes). I

Using this conclusion let us find the diode current. To make the illathelllatics simple, we assume that

i) the potential at the cathode is zero while the polential at the anode is V,;

ii) the distance between the cathode and anode is 'a', which is small so that the field can be assumed uniform and normal to the surfaces of the electrodes;

I

iii) the velocity ( v ) of tlie electrons soon after their e~nissioii at the cathode is zero and anywhere in between the cathode and anode, say at a distance x from the cathode, the velocity is denoted by v, and potential by V ( x ).

Differential form of Gauss's law is given The potential V ( x ) at a distance x from the cathode is given by: bv

P div E - - En

where p is thecharge density.

I n rectangular co-ordinates i t can be where p is the charge density at the point x. - written as ag aE, aE, p -+-+-I - ax ay az EO

Since E, - - a, etc, RS shown in since J = pv and v = v, at x. ax

Unit3, by substitution we f i d Assume the initial velocity of the electron as zero. Iden its velocity v at any point is related with potential Vthrough which the electron has traversed by the following

a 2 v ( x ) + a 2 V ( y ) + a 2 v ( z ) - - - - -2 relation : ax ay dz EQ

This is called Poiseon's equation. -mv; 1 = ~ V ( X ) (8.33) L the given example of diode since we 2 have assumed that the electric Eield acts I

only in thex-direction so that J has noy- Usil%! Eq. in 4% we get or z-component, and E, - E, - 0. 1/2 I Therefore, potential V ( x )at a distance x from the cathode must satisfy the following equation

(8.31) Multiplying both sides by 2 - dv(x ) we get G!x

Integrating above with respect tb x' we get

where C1 is constant of integration. At x = 0, V( x ) and - d V ( x ) arc both zero, so dt.

On integrating again with respect tox, we have

4 3/4 ?V ( x ) = 215(;) j x + C 2

Again constant of integration C2 = 0 as at x = 0 , V ( x ) = 0. On squaring the above equation we get

I From Eq. (8.34), at the anode, x = a, we get

This is the famous Child - Larzgmuir or three fzalvespower law, It suggests that the current density J or current I in a diode is proportional to the three-hakes power of the applied potential difference between the anode and cathode.

I SAQ 4

1 From Iiq. (8.31) obtain the following relati011 :

Eq. (8.37) gives variation of charge density with distance from the cathode. This variation is represented in Fig. 8.12.

Fig. 8.121 Varintion of c h q t dcn~ity Ln a diode as a ItncUon d Lhc d&(mncc from utbode.

SAQ 5

You know that the electric field inside a conductor is zero. Hence, if a charge is placed inside a conductor, it will move to the surface and distribute itself in such a way that the zero field exists within the surface. How fast this happens is of importance,'and

Electic Currcul

Electric Cut~cn t and hfalpthjic liield

irlterestiligly the continuity equation helps in evaluating this time. Deternline the characteristic time for the decay of charge inside a conductor. (Assume that po is the initial charge density.)

8.5 SUMMARY

Current is the flow of charge. The unit of current is the ampere. Current is defined as the amount of charge per unit time passing a given pint.

e Current density J is a vector specifying the current per unit area. The direction of J at any point is that in which a positive charge-carrier would move if placed at that point. '

J = nq vd

e The total current through a surface is the flux of the current density over that surface:

where dS is an element of area and the integral is taken over the surface.

e The total charge crossing a surface S inunit time i s J ~ . d ~ . If S is a closed surface enclosing a volume V, the rate of loss of charge through S must be the same as the rate of depletion of charge contained in V, i.e.

The differential form of the above equation is the continuity ecluation :

Both these statements express the law of conservation of charge.

r Conductivity a is a property of a material which is equal to the ratio of current density to electric field in the material

Resistivity p is the inverse of conductivity.

m Resistance is a property of a particular pie& of material. It is defined as the ratio of voltage Vacross tZle material to the current1 through the material:

The resistance of a piece of material depends on its'resi~tivit~ and physical dimensions. For a material of length I, uniform cross-sectional area S and resistivity p, the resista~ice is

The conduction i n metals is due to the presence of free electmns. In metals, the , combined effects of acceleration of free electrons in an applied electric field and

collisions between electrons and metal ions and impurities result in a drift velocity. On this basis, the expression for electrical resistivity is

me P m - ne2-c

Thus, resistivity depends on the number of free electrons per unit volume, their mass and charge, and mean free time.

I 16

INN, QUESTIONS Electric Curreut

1) TV set shoots out a beam of electrons. The beam current is 10 p4. How many electrons strikes the TV screen each second? How much charge strike the screen in a minute?

2) In the ~ o h r Model, the electron of a hydrogen atom moves in a circular orbit of radius 5.3 x 10- l1 m with a speed of 2.2 x lo6 ms- '. Determine its frequency f and the current I in the orbit.

3) A current of 2.00 A flows in a copper wire of 1.00 inm cross section. What is the drift velocity of electrons in that wire? How long does it take an electrbii to travel 10.0 cm (about the length of an incandescent bulb filament) in this wire under these circumstances? Assume that the number of conduction electron per cubic meter is 8.43 x 10

4) A potential difference Vis applied to a copper wire of diameter d and lengthL. What is the effect on the electron drift speed on (a) doubling V, @) doublingl, and (c) doubling d ?

5) i) See Fig. 8.13. What is the electric field in a copper conductor of resistivity p 5 1.72 x i 0-' ohm meter having a current density J = 2.54 x 1 o6 arnp m-2?

Fig. 8.13

ii) What is the potential difference between two points of a copper wire 100 m apart?

6) As shown in Fig. 8.14, a metal rod of radius rl is concentric wih a metal cylindrical shell of radius rz and lengthl. The space between the rod and cylinder is tightly packed with a high-resistance material of resistivity p. A battery, having a terminal voltage V,, is connected as shown. Neglecting resistailces of the rod and the cylinder, derive expressions for (a) the total current I, @) the current density J and the electric field E at ally point P between the rod and the cylinder, and (c) the

-resistance R between rod and cylinder.

Electric Curnnl and Magnetic Field 8.7 SOLUTIONS AND ANSWERS --

SAQs

1) The other physical quantities like ,current are

i) Thermal current in heat conduction,

ii) Flow current of water or any incompressible liquid in stream line motion.

A wurceor sinkof a current is one by 2) A wire carrying a constant currelit is an example of steady current. As we have which the current may be injected into noted a steadv current svste~n is one for which J is a co~lsta~lt in time at everv or withdrawn from a conducting region.For example, an electrode. point. In such a system, there cannot be any accumulatio~i of charge at any point.

This means that in any region of current flow there is no source or sink of curreilt.

3) Fmm Eq. 8.25, we have = R I

Multiplying both sides by ( A/L )we get

Rearranging the terms, we write

I Since

1 1 :. E . - = - or J = UE which is Eq. (8.22) J a

4) We have = -- d 2 V ( x ) 'Eq.(8.31) d w 2 EO

From above equation we have already derived Eq. (8.34), i.e.,

and also Eq. (8.35), i.e.,

Diyiding one by the other and rearranging we get

F 2 ( x ) = [?)2 . , -

4/3

v ( x ) - va(;)

d V ( x ) Va 4 113" = - a ! ~ a 413 3X

d Z v ( x ) = - Va 4 -m _ P ( x ) 4/3 g --

& a ' Eo

4 EO Va - ~ 3 :. p ( x ) = -- 9 px q .

3) From continuity equation we have

- V . J = - ik dt

18 I

within a conductor J = u E where u is the conductivity. According to Gauss's law V.E = p / ~ where E is the permittivity of the cotlductor. Thus

On rearranging we get

P u Integrating, J LfL = -4 ;dt

Po P'

u Then p = pee-Ti

where po is the initial charge density and p is the charge density after timer. E

The quantity - = t is called the characteristic time. (I

For extrelnely good co~lducto~s t will be very small and for insulators t will be large. ( z is usually referred to as the Relaxalion time.)

Terminal Questions

1) Let n be the number of electrons per second.

lo Cs -' = 6.3 x 1013 electmns per secondlhe charges Q Then tz - - =

e ~ . ~ x I o - ~ ~ c striking the screen is given by

Q = It = ( 1 0 ~ C k - ~ ) ( 6 0 ~ ) a 600pC.

Since the charges are electrons, the actual charge is

Q = - 600 PC.

Each time the electroll goes around the orbit, it canies a charge q around the loop. The charge passing a point on the loop cach second, i.e., currelit is given as follows :

Current I - ef = ( 1.6 x lo-19 c ) ( 6.6 x lo-'' s -' >

Note that the current flows in the direction opposilc to thc electrcui, which is negatively charged.

3) Solving Eq. (8.21) for vd, we obtain

I vd = - ( because I = JS )

nqs

= - 1.48 x ms-l.

(The negative sign appears because the electronic charge is negative, and vd is . therefore directed opposite to I .)

The time required to traverse 10.0 cm at this speed is

t = 0.100 m 1.48 x ms

= 0.674 x 103s = 11 min 14 s

Electnc Cerrcnl

That is a long,time. Yet we know that as soon as we close the proper switch, the charge flows through a circuit and lamps light up. We need not wait several minutes, not even seconds, to witness the effect of the cumnt in a circuit, and . there appears to be no observable dependence on the distance between the wall switch and the light fixture, a distance generally considerably greater than 10 cm.

The point is that one does not have to wait until a particular el&n at the battery t& reaches the lamp for the Lamp filament to respond to the current. When the switch is closed, the entire cbarge distribution within the condudor is set in motion almast irstan-y, muchas water starts to flow in a long pipe as soon as we open a tap.

4) a) drift velocity will be doubled,

b) drift velocity will be halved,

' c) drift velocity will remain unchanged.

5) i) By definition, E, the electric field, is related to the current density J, through the relationship

J E m- u

1 But a = -, and therefo'ore,

P

ii) E is related to V by b

vb-va m $ dl (I) a

From the Fig. 8.13, E is parallel to the axis of the cylindrical wire. If we evaluate (1) along a line in the direction of E and parallel to the cylinder axis, we obtain

Vb-Va = E ( u - b )

Vb-V# - ( 4 . 3 7 ~ l ~ ~ ~ o l t ) ( 0 - 1 0 0 r n )

Vb - Va - 4.37 volts.

Therefore, Vb is at a lower potential than V,.

6) a) Assuming radial flow of charge between rod and cylinder, we have at P

i J - - ~ J E rL and E - p J - L, 2JrrL where p is resistivity, with both J and

E in the direction of r. Then, by definition of the potential,

and so, noting the polarity of V,, we get .

Structure

9.1 Introduction Objectives

9.2 Magnetic Field Source of Magnetic Field Definition of Magnetic Field

9.3 Gauss's Law for Magnetism

9.4 Biot and Savart Law

9.5 Force between Two Parallel Conductors (Definition of Ampere)

9.6 Ampere's Law Applications of Ampere's Law Differential Form of Ampere's Law

9.7 Torque on a Current Loop

9.8 Summary

9.9 Terminal Questions

9.10 Solutions and Answers

9.1 INTRODUCTION

In Block 1 of this course, you were introduced to the concept of an electric charge and studied some properties of charges at rest. You learnt that a static distribution of charge produces a static electric field. Similarly, steady flow of charge (i.e., a &ady current) produces a static magnetic field, which is, infact, the topic of this unit. However, there . are some major differences between the two fields which you will discover in this unit.

In the science laboratory, during your school days, you must have been fascinated with magnets. Recall, when you tried to push two magnets together in a way they didn't want to go, you felt a mysterious force! In fact, magnetic fields or the effect of such fields have been known since ancient times when the effect of the naturally occurring permanent magnet (FesO4) was first observed. The north and south seeking properties of such materials played a large role in early navigation and exploration. Except for this application, magnetism was a little known phenomenon until the 19th century, when Oersted discovered that an electric current in a wire deflects a compass needle. This discovery showed that electric current has something to do with the magnetic field because a compass needle gels deflected and finally points inathe notth-south direction only when placed in a magnetic field.

In this Unit, we shall consider in detail the production of the magnetic fields due to steady currents, and the forces they exert on circuits carrying steady currents and on isolated moving charge.

A good way of gaining a better understanding of the nature of fields is to know how they affect the charged particles on which they act. Hence, in the next unit, you will study the behaviour of charged particles in both electric and magnetic fields.

Objectives

Afier studying this unit you should be able to :

o understand what is meant by the magnetic field, the right hand rule, Biot-Savart . law, right hand method, Ampere's law,

o define the magnetic field at a point in t e r n of the f o m on a steady cutrent element and also on a moving charged particle,

o use the formula for the force on a steady current element-or on charged particle due to a magnetic field to calculate the force on a certain simple current drrying .

, circuits, and solve simple phblems,

TIE point is that om does not have to wait until a particular electron at the battery termiaal reaches the lamp for the lamp filament to mpond to the <xurent. When the switch is closed, the entire charge distribution within the condudor is set in motion almast instantaneously, much as water starts to flow in a long pipe ss soon as we open a tap.

4) a) drift velocity will be doubled,

b) drift velocity will be halved, i

c) drift velocity will remain unchanged.

5) i) By definition, E, the electric field, is related to the current density J, through the relationship

1 But a = - and therefore, P'

ii) E is related to V by b

Vb- Va m J E.d (1) a

Fmrn the Fig. 8.13, E is parallel to the axis of the cylindrical wire. If we evaluate (1) along a line in the direction of E and parallel to the cylinder axis, we obtain

Therefore, Vb is at a lower potential than Va.

6) a) Assuming radial flow of charge between rod and cylinder, we have at P

That is a 1ong.time. Yet we know that as soon as we close the proper switch, the charge flows through a circuit and lamps light up. We need not wait several minutes, not even seconds, to witness the effect of the current in a circuit, and .

there appears to be no observable dependence on the distance between the wall switch and the light fixture, a distance generally considerably greater than 10 cm.

J = - I andE - p.7 = & , where p is resistivity, with both J and 2rc rL

E in the direction of r. Then, by definition of the potential,

and so, noting the polarity of V,, we get .

solving for I, ,

I = ZLVr

p ln ( rz /n )

I b) From (a), J = - I ' and E - Vt 2zrL prln(rz/tr) P J m r l n ( d r 1 )

, . Vr p l n ( d r 1 ) c) Fmm Ohm's law, R - - I I 2 d '

20

UNIT 9 TIC FIELD

Structure

9.1 Introduction Objectives

9.2 Magnetic Field Source of Magnetic Field Definition o f Magnetic Field

9.3 Gauss's Law for Magnetism

9.4 Biot and Savart Law

9.5 Force between Two Parallel Conductors (Definition of Ampere)

9.6 Ampere's Law Applications of Ampere's Law Differential Form of Ampere's Law

9.7 Torque on a Current Loop

9.8 Summary

9.9 Terminal Questions

9.10 Solutions and Answers

In Block 1 of this course, you were introduced to the concept of an electric charge and studied some properties of charges at rest. You learnt that a static distribution of charge produces a static electric field. Similarly, steady flow of charge (i.e., a steady current) produces a static magnetic field, which is, infact, the topic of this unit. However, there . are some major differences between the two fields which you will discover in this unit.

In the science laboratory, during your school days, you must have been fascinated with magnets. Recall, when you tried to push two magnets together in a way they didn't want to go, you felt a mysterious force! In fact, magnetic fields or the effect of such fields have been known since ancient times when the effect of the naturally occurring permanent magnet (Fe304) was first observed. The north and south seeking properties of such materials played a large role in early navigation and exploration. Except for this application, magnetism was a little known phenomenoi~ until the 19th century, when Oersted discovered that an electric current in a wire deflects a compass needle. This discovery showed that electric current has something to do with the magnetic field because a compass needle gets deflected and finally points inathe north-south direction only when placed in a magnetic field.

In this Unit, we shall consider in detail the production of the magnetic fields due to steady currents, and the forces they exert on circuits carrying steady currents and on isolated moving charge. A good way of gaining a better understanding of the nature of fields is to know how they affect the charged particles on which they act. Hence, in the next unit, you will study the behaviour of charged particles in both electric and magnetic fields.

Objectives After studying this unit you should be able to :

r understand what is meant by the magnetic field, the right hand rule, Biot-Savart +

law, right hand method, Ampere's law,

r define the magnetic field at a point in tenns of the force on a steady current element and also on a moving charged particle,

r use the formula for the force on a steady current eleclent-or on charged particle due to a magnetic field to calculate the force on a certain simple current cirrrying '

circuits, and solve simple phblems,

BecQric Current and Magnetic Field

Fig. 9.1: ?When the magmet is Precly suspended, rr p d c u h r end of it points no& lWs end o t Uac magolet is def i i d PO tbe north p B c

(b) Fig. 9.2: a) A mmpm needle

poinls In the direction d the magnetic field. b) Magnetic fie119 lhes of a magnet drawn using the Pact that a compass needle should line up along the Pield lines

show that the divergence of B vanishes,

+ use Biot-Savart law to describe and compute the magnetic field generated by a simple current flow,

s identify the nature of force (attraction or repulsion) on a given length of a long, straight cunent-carrying wire that is laid parallel to a similar currentcarrying wire,

e use Atnpere's law to calculate the magnetic field Erom steady current distributions having simple geometries,

e relate Ampere's law to its differential form via Stokes theorem,

compute the torque exerted by a steady magnetic field up011 closed current loops,

+ appreciate how the forces on current-carrying conductors, placed in a magnetic field, are used to understand the working of galvanometers and motors.

9.2 MAGNETIC FIELD

Wheneverwe speak of magnetic field, we speak in terms of bar magnets since this is the way the fields were first studied. You are already awareof the basic features of the magnetic field from your school days. For example, you know that the poles of a bar magnet experience forces when placed in a magnetic field. If a bar magnet is suspended by a delicate fibre as shown in Fig. 9.1, a particular end of the magnet will always point towards north. This end of the magnet is called the north pole of the magnet. The other end is the south pole. Do you recall that this arrangement is a simple compass? The north poles of two magnets repel each other. The south pole of a magnet is always attracted by the north pole of another magnet. If one tries to break off the north or south pole from a simple bar magnet, then this exercise proves to be futile. The broken magnet becomes two new bar magilets each having a north and a south pole. This shows that an isolated magnetic pole does not exist. a

In order to plot the magnetic field due to a bar magnet, we need only a compass needle. The direction in which the compass needle points is taken to be direction of the magnetic field. In class XII, you must have used this fact to plot the magnetic field in the vicinity of the bar magnet as shown in Fig. 9.2a. The magnetic field lines are drawn in such a way that a compass needle placed on the line will align itself tangentially to the line. Fig. 9.2b shows the typical magnetic field for the bar magnet. Notice that the field lines emerge from the north pole and enter the south pole. These are some qualitative features with which we are all familiar.

92.1 Source of Magnetic Field

As you know, the space near a ~ubbed glass rod (rubbed either by rubber or rabbit's fur) is characterised by an electric field which is denoted by E. Similarly, a magnetic field around a magnet may be represented by the symbol B. In electrostatics, the relation between the electric field E and the electric charge is represented as follows:

electric charge s E electric charge (9.1)

Tbat is, tlie electric charges set up an electric field and the field, in turn, exerts a force (electric in nature) on another electric charge that may be placed in that field. Now, by analogy, can you set up a similar relation for magnetism. Yes, the relation will be as follows:

magnetic charge 2 ]B ;;z magnetic charge (9.2)

You knowthat the two poles, i.e., north and south always, occur together. A single isolated pole is not known to exist. This means that there are no magnetic charges (also called magneticmonopoles). How does, then, the magnetic field arise? The answer to this question you will find in the following lines. Let us consider two wires, running parallel to one another, as shown in Fig. 9.3 a. As soon as the circuit is closed, the current in the two wires, flows in the same direction, and the wires are found to attract. If the direction of one of the currents is reversed, the wires repel each other. Thus the two sections of the wires in Fig.9.3 b, which are part of same circuit, tend to move apaft. If a sheet of metal is put between the two wires, the force with which wires attract or repel is not at all affected (Fig.9.3 c). How do. you explai~m this? Does electrostatic force account for the attraction of parallel ones? No, the

force acting is not an electrostatic or Coulomb force. This is because (i) there is no net charge on the conductor (the charge density of conduclion electrons just compensates for the positive charge on the lattice ions); (ii) the force is reversed in sign by revesir~g the direction of either current; (iii) the force ceases as soon as the circuit is broken; (iv) the force is not affected in the presence of a simple medium; (v) the attraction and repulsion of the electric currents is contrary to the attraction or repulsion of the electric charges.

Mngnelc Field

(0'

Fig. 9.3 : n) Parallel wires carrying currtnts in (be same direction w pulled logether. b) Pnralld wires carrying curnnls in opposik directions nrr pushed apart c) A sheet of metal bclween Lhe two wires docs not nlTect these foms.

'The experiments of Fig. 9.3 show that there is an additional force associated with a moving charge, which is different from the electrostatic force. This new force that comes into play when charges are moving is called the magnetic force. A charge sets up an electric field whether the charge is at rest or is moving. However, a charge sets up a magnetic field only if it is moving. You may ask a simple question. A bar magnet sets up a magnetic field in its vicinity, but where are the moving electric charges in a bar magnet? Actually the spinning and circulating electrons in the iron atoms of the magnetic material are responsible for its magnetism. You will learn more about it in Units 11 & 12 of this block. Hence. in magnetism, we can think in terms of the'following relation :

.PI

moving electric charge S B 3 moving electric charge (9.3)

As the moving charges constitute an electric current in a wire, Eq. (9.3) can be written Bs

electric current * B B electric current (9.4)

Eq. (9.3) or (9.4) tells us that (i) a moving charge or a current sets up a magnetic field and also (ii) if we place a moving charge or a wire carrying a current in a magnetic field. a force will' act on it. Now, let us define the magnetic field. But before doing this g try to answer the follotving SAQ.

SAQ 1

L

Fig. 9.41 A sLrrIgh1 wire CAI y h g cumnl and placed in n mnguetic field experiences n force. You have probably studied about an electric motor in your school, and you may be

knowing the principle on which it works. Briefly explain how an electric motor illustrates Eq. (9.4).

9.2.2 Definition o f Magnetic Field

. In Block 1, we defined the electric field E at a point in terms of the electric force FE that acted on a test charge q at rest at that point as follows :

FE = q E (9.5)

As suggested by Eq. (9.3), we can define the magnetic field in terms of the magnetic force exerted on a moving electric charge. It can also be defined in terms of the force on

23

Ekechic C u m t cmd Magnetic Field

a current. Since current is a flow of electric charge, the two definitions are related. First, let us state the definition in terms of force on a current-carrying wire.

a) Force on currents

Experiments show that a wire, carrying o current placed in a magnetic field, experiences a force. Fig. 9.4 shows a wire carrying a current I in a magnetic field produced by a magnet. Since the field lines come out of the north pole and enter the south pole, the field is directed from right to left. It is found that the wire experiences a force, which is proportional to both the current and the strength of the magnetic field. When the wire is placed parallel (or antiparallel) to the field lines, it experiences no force. But when the wire is placed perpendicular to the field, the force on the wire is the maximum. These two cases are shown in Fig. 9.5. This shows that the force on a wire is due entirely to the component of the field that is perpendicular to the wire. In other words, the force also depends on the relative orientation of the wire and the field lines. in Fig. 9.6, suppose, the angle between the field lines (represented by B) and the current carrying wire is 8. As said above, the force F on the wire of length L is dae entirely to the cornponelit of B that is perpendicular to the wire. This component represented as EL is given by (See Fig. 9.6) . .. .- . .

B-L = B sin 0,

I f there is surface area A placed perpdiculy to a uniform magnetic EieldB then the prodact o f magnctic field B and the surface area A is an important ph ysicaI quantity which i s called magnetic flux through the surface &A It i s dedotcd by 4 asfouows

ThouPit oEmagaetic flux is weber. Hence the unit of B is also weber per square metre. It is alao called Tesla 0.

F = maximum (inlo pagc)

Fm. 9.5 : The ronr oo a wire b due e d d y to (hc compownt &(be Bdd &at is pcrpendiculu to thc w i n

Fig. 9.6; The f o m on (he varks &s B sin 8, Lbat is, In proprtiw to 'B~.

Further, the force on the wire of IengthL depends onL itself and the current I in the wire. We conclude that the force P on a length L of the wire is given by

F =. U B sin 0 (9.6)

Recall that the vector cross product A x B gives rise to a vector of magnitude AB sin 8 which is perpendiciilar to the plane containing A and B. Using this in Eq. (9.6) we get

F = I ( L x B ) (9.7)

Here L is a vector the magnitude of which is the length L of the wire and its direction is along the ahrent. I&. (9.6) or (9.7) shows that the B has the SI units of;N A- ' m- '. This unit is also given the name weber per square meter o r tesla (abbreviated as I?). One tesla h a strong magnetic field, so that a smaller unit called the gauss (G) is often used.

, I ta la = la4 gauss (9.8)

Since gauss is not an SI unit, we should aiways convert it to tesIas before using it in equations. The quantity B has, several names. Its correct name is.magnetic induction. It is also designated as the magnetic field intensity, although another quantity, which you will study in Unit 12 of this Block, is also given this name.

The direction of the force on the wire is always perpendicular to the plane defined by B t Magnetic Field

and I. To find the direction of the force, we use right hand rule as shown in Fig. 9.7. According to right hand rule: if one's right hand is held flat with the fingers pointing in the direction of the field lines and the thumb pointing in the direction of the current, then the palm of the band will push in the direction of the force. I

Let us now apply Eq. (9.6) or (9.7) to a simple situation so that we can better see its f i meaning. - F

Fi. 9.7: Rigbl b n d rule.

Example 1

A horizontal wire is canying a current from east to west. What is the direction of the force on this current if we assume that at this location the magnetic field of eaPth points due north? If the wire carries a current of 20A, find the force per unit length on it due to the earth's magnetic field, which is about 1.0 G .

Solution

Using right hand rule, we find that when the thumb of the right hand points west and the fingers point north, the palm faces down, hence the force on this wire will be down (into the page).

Earth's magnetic field lin& are in a direction perpendicular to the wire. We hove

F - 'ILB sin 90"

Suppose the wire carrying current is not straight so that, at each point, its orientation relative to the field changes, or suppose the field changes in magnitude/direction over -

the length of the conductor, we can still useEq. (9.7) to find the force. For this purpose we imagine the wire to be broken up into small segments so that it is straight, and the field is essentially conitant over its length. See Fig. 9.8. Now Eq. (9.7) can be applied to each segment.

If the length of the small segment is dl then we can write for the small force dF on the segment as

(a) (b)

Fig. 9.8:(n) A curved wire in a nonunif~cm magnetic field (b) A s m d enough sqpment of tbe wire can be treated as rn stmight wire in a uniform Bdd

We can obtain the total force on this arbitrarily' shaped long current-carrying wire placed in a non-unifonn magnetic field B, by summing the expression for dF in Eq. (9.9)~verthe whole wire giving

Electric Current and Magnetic Field

If we let the length dl approach zero, this sum becomes an integral, and we call write the above expression as

F - I J ~ X B (9.1 0)

where, on the right hand side of the Eq. (9.10), we have the line integral taken over the length of the wire and I, being a constant, is taken out of the integral. In particular, if the magnetic field is uniform, which means that B is constant both in magnitude and direction at all points of the wire, then we can write Eq. (9.10) as

In this expressionJdl is the vector joining the initial point of thd wire to its final point. I

Further, if the current carrying wire is straight and of length L, the11 we have

This is the sanie as Eq. (9.7).

So far we considered the force on the current in a wire. An electric current is simply a group of charged particle sharing a colnnlon motion, so we should expect a charge to experience a force in the magnetic field. This gives another way of defining the magnetic fie'ld.

b) Force on a moving charge

The force which a magnetic field exerts on a moving positive charge can be obtained from Eq. (9.7). Recall from the discussion of Unit 8 that the velocity v of the charge q in a wire of cross-sectionA is related to I by Eq. (8.3) as follows :

I = qrdv

wser rz is the number of charge per unit volume. Substituting this expression for I into Eq. (9.9) gives

dF = ( d C ) A n q v x B (9.1 1)

Here (&)A represents the volume of the wire segment of length (dL). So (&)An is the number of moving charges in that portio~i of the wire for which we are writing the force. Hence the force F on a single charge is given dF/( dL )Ail, i.e.,

k

F = q v x B (9.12) I

The magnitude of the force is given by qvB sin 8. The direction of the force'on the moving charge can be obtained by right hand rule (Fig. 9.7 with I replaced by v ). Note, if the particle is negatively charged, the direction of F will be reversed.

SAQ 2 '

Of the three vectors in the equation F = q v x B, which pairs are alGays at right angles? Which may have ally angle between them. Read the following example so that you can understand how the force on a charged particle is calculated. It also illustrates the use of the right hand rule lor determining the

. direction of force.

Example 2

In a certain region a magnetic field of 0.10 T points vertically upward. Three protons enter the region, two horizontally and one vertically, as shown in Fig. 9.9. All the three are moving at 2.0 x lo3 ms-'. What is the force on each proton?

2.

F& 9.9 : Example 2 Solution

P ~ ~ t o n 2 is moving vertically, i.e., parallel to the field ( sin 8 = Q ) ia Eq. (9.12). Therefore, it experiences no force. Protons 1 and 3 are moving at right angles to the field, so sin 8 =, 1 in Eq. 9.12, and thus the forces on these two protons have the same magnitude given by

26

Since the protons carry a positive charge, the direction of the force is the direciion of the vectorv x 18. For proton 1, tnoving to the right, v x 81 is out of the page. For proton 3, moving to the left, the force is illto tlic page. This exi1111ple shows clearly thai the magnetic field alone does not detemiine the force. lde~~tical particles in the sanic field may experience different forccs, if their velocities are not identical. If ihc pallicles were electrons, the negative sign of the electron charge would have i~idicatcd a force opposite to tlie direction of v x B. Eq. (9.12) is equivalent to Eq. (9.7) so that either of them can be taken as the defining equation for B. In practice, we define B from Eq. (9.7), because it is much easier to measure the force acting on a wire than 011 a single ~noving charge.

In this section, you have learnt that a movink charge gives rise to a ~nagnetic field. Now, suppose there is a current carrying wire, and you are asked to calculate the magnetic field produced due to i t at any point of spacc, then ccrtai~~ly you would like to have laws analogous to Coulomb's a~id Gauss's law. Let us first find out Gauss's law for magnetism.

9.3 GAUSS'S LAW FOR MAGNETISM

Suppose magnetic charges-monopoles-exist, then they would give rise to ~nag~ietic ,fields like the electric fields due to electric point charges. Then these fields-and those of magnetic charge distributions-would be described by laws analogous to Gauss's law. hi particular, we would find tliat the flux of the inng~ietic field through any closed surface would depend only on the enclosed ~nag~ietic charge. We may write Gauss's law for nag net is in as

- where the integral on the lefi is the flux of I3 over a closed surface enclosing the magnetic charge or monopoles denoted by g and po is some constant. But the very existence of the magnetic ~no~iopoles is uncertain. And even if they do exist, they seeiii to play no significant role in our world. hi the absence of the lnag~ictic monopoles, we must put g = 0 and then tlie magnetic flux through ally closed surlacc must be zero. We state this mathematically as Gauss's law for magnetism and write it as follows

Pig.9.LO; In Lhe absence of '

magnetic nionopolcs, the magnetic flux through a clnscd surface must be zero.(h) Thi! means there call be no point where magnetic field lhes begin or end, for a cl&ed surface

A consequence of Gauss's law for nlag~ietis~n is that iiiagnetic field lincs can never sumoundiug s id l a begin or end (Fig. 9.10). Unlike tlie electric field lilies, the ~izag~ietic field lines liavc to point would have non- forin closed loops. If we convert the above surface integral of Eq. (9.14) illto a volu~ne zero net flux (b) .

integral using the Divergence Theorem, we obtai~i Instend, n~egnctic licltl liues generally for111

[ v . B ~ v = 0 (9.15) closed lrrolls . .r

The integratio~i is ove'r the volu~ne etlcloscd by thc closcd surface of Eq. (9.14). Since the Eq. (9.16) holds for ai!y arbitrary volume of inlegmtion, we must have

This equation is true, even if B varies witli time hnd is, in fact, one of Maxwell's equations. Eq, (9.14) or Eq. (9.16) says that if the magnetic field exists, they l o ~ k different fonn the fields of the point charges. I11 the next section, we will find out the law analogous to Coulomb's law.

9.4 BIOT- AND SAVART LAW ' - Can we calculate the magnetic fields produced by a current? Can we show that a current loop has the magnetic field of a dipole? Interest in questions like these led the Fre~icli

Uecttic Cumnt and Magnetic Field

scientists Jean Baptiste Biot a~id Felix Savart to experimentally detennine the fornl of the magnetic field arising from a steady current. Known as Biot-Savart law, its result gives the magnetic field-at a point due to a small element of current.

In Unit 1, we learnt to calculate the electric field that a given distributio~~ of charges set up in the surrou~~ding space. Our approach was to divide the charge distribution into

Fig+.p.ll: (a) A ch.rge'daneot dq eshblirbts a di ienntid decCric field dement dE at point P. (b) A curnut clement Id cshblIshes a diertntial magnetic IPekl W at point P. ' Ihc symbol x (the tail of an arrow) shows h a t the drmeut dB points into (he page along tbc ntgativt z - axis,

charge elements dqas in Fig. 9.1 la. We then calculated the field dE set up by a given charge elelllent at an arbitrary poi111 P. Finally, we calculated E at point P by integrating dE over the entire charge distribution.

The magnitude of dE is given as follows:

in which r is the distance from the charge element to the point P.

In the magnetic case, our approach will be the same. Fi$,.9.1 l b shows a wire of arbitrary shape carryinga current I. What is the magnetic fiel'd B at an arbitrary point P

1 near this wire? We first break up the wire into differential current elements Idl, 1 1 corresponding to the charge elements dq of Fig. 9.1 la. Here the vector dl is a

differential element of length, pointing along the tangent to the wire in'the direction of the current. Note that the differential charge element dq is a scalar, but the differential

- - current element Id is a vector.

Then the Biot-Savart law says that the magnitude of the magnetic h l d contributioii set ' up by a given current element at point P is give11 as follows :

po Zdlsin 8 a=--, 4~ r Z , .

Here pa is a constant, called the penneability co~lstaiit of free space. Its value is 4n; x TmA-I. This collstaiit plays a role in magnetic problems, lnGh like the role that the pennittivity constant ep plays in electrostatic problems.

The full expression for dB in vector form is

p, id lx ; a=- - ( Biot - Savart law ) ' ! 4~ r Z

Hew ; is a unit vector pointidg from dl towards P. Eq. (9.18) i s the analogof Coulomb's law, and is called the law of Biot and Savart. TAe direction.of dB in Fig. 9.11b is that of the vector dl 'x ;, where i? is a unit vector that points from the current element to the point P at which yob Wish to know the field. The symbol x (representing the tail of an arrow) in Fig. 9.1 l b shows that dB at point P is directed into the plane of the page at right angles.

Coulomb's law gives the electric tield of a point c h a ~ e in tenns of the charge and the distance from the charge to the Beld point.The electric field varies as the invetse

square of the distance, and its direction lies along the line joining the charge with the '

field point. Analogously, the Biot-Savart law gives the magnetic field at a given point in terms of the current element (its source) and the distance to the field point. Like the electric field of a point charge, the magnetic field of an isolated current element varies . as the inverse square of the distance. But here the similarity ends. Unlike the electric charge in Coulomb's law, the current element Id1 has associated with it a directio~l as well as a magnitude. Hence the magnetic field of the current element is not symmetric about the element, but depends on the position of the field point relative to the directioii of the current element. This directional character is expressed by the cross product in Eq. (9.18). As shown in Fig. 9.1lb, the magnetic field is at right angles to both the ,121 Magactic ~ d d b c s

current element and the vector from the current element to the field point. Also, you geocnUy meircle r

have learnt in the previous section that the magnetic lines have no sources unlike the - current.

electric field lines which end or originate on electric charges, but are continuous and join back on themselves.

h t us see how Eq. (9.18) and Fig. 9.11b show this to be true. Let the point P move around the cumnt axis at a constant distance from the axis. From Eq. (9.18), the magnitude of dB is constant along this path, and at each point it has a direction tangent to the path. These are just the requirements for the lines to be concentric circles around the current. Hence, the magnetic field lines encircle the current as shown in Fig. 9.12. The direction in which the circular field lines point depends on the direction in which the current flows. If the direction of the current flow is reversed, the direction of the field line is also reversed as shown in Fig. 9.13.

( 4 (b) rig. 9.13 m e direction in which (he fidd lines pdat is debnnhcd by the direction in which tbe current

flows (a) When the curtent flows i n t ~ Lbe page, tbedidd lines t o m dockwise circles. (b) Whm the cumnt flows out , d ( h e page, the circles s n mnticlo~kwisc,

However, there is an easy way to remembe~ these directions. Just close the palm of your right hand and point your thumb in the direction of the current as shown in Fig. 9.13. In either case you will f h d that your fingers will naturally curl around in the directioti of the magnetic field lines. This simple tecliiiique for remembering the direction of the magnetic field is illustrated in Fig. 9.14, and is referred to asthe right-hand method. Do not confuse it with the right band rule for vectc cross products shown in Fig. 9.7.

Fig, 9.14 By using your right hand to 'gripp a cunmt- umyhg conduebr, you can llad out the dlrrdioa d a tbc magnetic fld& When your (humb points in the d i d m of current now, your Engem indicate

the dimdon of the mag&lic fId& -

4 Vig. 9-15 Cafculmting the mmgnetic Deld set up by s current tin e loug strnigbt wire. The f idd dB associated with $be cumcut element PiA poi& out of the pap, ns shown,

A horizontal vrlire crrrries a current from east to west. What is the direction of the magnetic field due to ellis current directly above and below the wire? Refer again to Fig. 9.1 1b. Since the inagnetic field obeys the superposition principle, the nct field at P clue to entire circuit, of which the wire is a part, will be the vector sum or line integral of the lields of individual current elements:

wlnere the integratiul~ is taken over the entire path through which the currellt I flows. Eel us apply Biot-Savart law to calculate a magnetic field for sinlple situation.

a) Ca8cmll;ltioaa oQB doe 80 a Ismg straight wise carrying la current

As show11 ita Fig. 9.15, suppose P is a poilie at which we are to calculate the inagnetic lield due to a long straight wire carryi~~g a current I. 7% distance between the point P and the wire is r. The: differential inagi~etic field set up at point P by the current eleiile~~t Id1 is given in ~lxigniiudc b y Eq. (9.191, i.e.,

116) Id1 sin 0 d~ = - (9.17) 4 7 ~ 1.2

The direction of d l 3 is given by the right hand nncthod. Here it points out of page. This is true irrespective of the positioi~ of (4 along the wire, so that at point P all the dB' s from all tlie current ellcme~lls Id! poillt in the same direction. Thus lo find the ~nagliitude of total lnagnetic ficld B at point P we integrate Eq. (9.17) obtainilig

n . . J~B = 4x

where 8 is the angle between and 1~61.

111 order to sulnup the co11trib~ttio11~ from all elelnents of [he long straight wire, we change Lhe variables fi-o1n8 and r to $ (See figure). Now, sill 0 = sin ( JX - ) = cos 41. (Frorn figure ) (9.20)

AC rd0 = cos q, I>rawAC J-PB, then - = - AB dl

[Angle 11etweenAC and AB beillg equal to angle between PA and PO j

. . . ril sin 0 dicosc) - Z -

I' 2 (From Eq. 9.20) I' -

B = - [ sin 4~ + sin $1 ] 43cR

For an infir~itely long straight wire

I The magnitude of B, thus, falls off invers~ly as the first power of the distance from an infinitely long wire. Eq. (9.22) shows that the lines of the magnetic field f o ~ m concentric circles around the wire. This expression for B is,analogous to the expressiot~

for E due to a long charged wire = - 42Ep ( ?), and thus shows its electrostatic

equivalent nature. -

30

A

b) Calculation of B along the axis oCa current loop Mawetic Field

Let us coiisider a circular loop of radius a aiid carrying a current I. As shown in Fig. 9.16, we choose a point P on its axis at a distance b froin its centre. The magnetic field dB at P due to a current element of length dl is give11 by

For all eleil~eiits around the loop, r is perpeildicular to I d , hence, the value of sin 8 in

Fig. 9.16 : Magnetic field along the axis d a current loop.

the cross-product liere is 1 and so I

p4lI dl 1

dB e--

4~ r 2

Since r 1- dl lleilce dB is always perpendicular to the plane coilsistiiig of r and dl. Thus I

dB is perpeiidicular to r at the point P as show11 in Fig. 9.16. It can be resolvcd illto two compoilents, one dB sill cp along the axis aiid the other dB cos $ at right angles to the axis. Here 4 is the angle between rand the axis of the loop. You will iiolice that the compoilents of dB perpendicular to the axis will cancel, as inay easily bc seen by coilsidering the field due to an element opposite to dl. Therefore, the resultant I3 is in I the direction of the axis and will be given by summing only the components dS sin +. 1 Thus, B is given by i 1

B = S dB sill $

= !&?!dlsin$ - 41t r z I

F I *pl - - - 4~ r 2

As we are around the current loop, both + and rare coilstants, so they are taken outside the integral. The length integrated around the loop is b, so that

w,l sill $ 2na B =

4nr

If sin 4 and r are expressed in terms of the constants a and 0, we get

WI B e - a 2 2 3/2

(9.23) 2 ( a 2 + b )

When we choose the point P far from the loop ( b >> u ) , the expression for B is written as

B I -- 4~ r 3

(9.24)

Here we have writtenA = m2, the Area of the loop. This relation shows that the current loop generates its own magnetic field. Notice that the inagnetic field of the loop at large

distances is like the electric field of an electric dipole [ E - &(f ) ]on i tsax is

(See Sec. 3.6 of Unit 3.) This shows that the term (LA corresponds to electric dipole moment p of the electric dipole, Therefwe, the termL4 i s called the magnetic dipole ~ i g . 9.171 he current-canyiug . moment of the loop and is represented by p . This shows there is a similarity between a bar magnet which is.magnetic dipole, and a much like that ofthc

current loop. The similarity can also be seen by plotting the magnetic field around Ule short bar magnet shown m (c), current loop. When a compass is used to plot,the magnetic field, the result show11 in

Elecllic Current and Mngnelic Field

Fig. 9.17a and b is obtained. You should collvi~lce yourself that this is reaso~~able by applying the right halid method to a portio~i of the loop. Observe Fig. 9 . 1 7 ~ and note that rlle magnetic field of loop is much like that of a bar magnet. The current loop call be considered to have the north and south poles. We shall see in a later section that this is one aspect o la very i~llportant similarity between bar magnets and current loops.

Alter going through this section I hope you can tell why the two wires, shown in Fig. 9.3, are attl-acted in one case while they are repelled in another case. If not, read the fol lowi~~g section. It will also help you in defining the unit ampere which we have been using so far without defining it precisely.

But before def ini~~g ampere do the following SAQ.

SAQ 4

Write otie analogy and one differe~~ce between Coulomb's law and Biot-Savart law.

9.5 FORCE BETWEEN TWO P LLEL CONDUCTORS @EFINITI[ON OF AMPERE)

In this section, we will find low much force docs one of the wires in Fig. 9.3 exert on the other. We assulile that the wires are Iiuear, parallel and very long. Here, one of the wires experie~ices a force, because it is in the magnetic field caused by the current in other wire.

Fig. 9.18 shows two long, parallel wires separated by a distance d and carryi~ig currents 11 and I2 in the same direction. The current in wire 2 prnduces a magnetic field Bz at all points around the wire. From Eq. (9.22) the magnitude of Bz at the site of wire 1 is given by

cro I2 Bz = - 2Rd

(9.25)

12

Fig. 9.1&Two parallel wims carrying currents in the same diitiw altrnct cach other.

The right hand method tells us that the direction of Bz at any p o i ~ ~ t on wire 1, is out of the page, as shown in F!g. 9.18.

Now, wire 1, which is carrying a current 11, finds itself immersed in an external magnetic lield B2. If L is the length of this wire, it will experience a lorce given by Eq. (9.7), whose magnitude is

What is the direction of this force? The right-hand rule says that FI points towards the wire 2. This means that wire 1 is attri. ted towards wire 2.

Similarly, for currents in the opposite direction, you should be able to sliow that the wires repel each other. The rule is that parallel current attract and antiparallel current repel.

The force between current-carrying conductors forms the basis for the definition of the ampere. The ampere is that constant current which, if maintaillea in two straight parallel

I conductors of infinite length, of negligible circular cross-section, and placed one meter apart in vacuum, would produce on each of these conductors a force equal to 2 x newtons per meter of length.

In other words, suppose we have two straight pamllel coilductors of infinite length, of negligible circular cross- section, and placed one meter apart in vacuum. When constant

current is made to flow in both the conductors, it is observed that each of these conductors experiences a force. The constant current which produces force equaI ttr 2 x newtons per meter of length of the conductor is known as ampere.

9.6 PEWE'S LAW

In electrostatics, we used Coulomb's law to calculate the electric field due to an arbitrary charge distributions. Do you remember that in Block 1, Unit 2, we used Gauss's law to solve electric field problems of appropriately high symmetry with ease and elegance. There we made simple observatio~a about the number of lines of force emerging from closed'surface and then formulated Gauss's law.

In magnetism, the situation is similar. We can calculate the magnetic field caused by any current distributions using Biot and Savart law. Now the question is can we make a statement analogous to Gauss's law that would help us to calculate a magnetic field with similar ease and elegance? Gauss's law for the electric field relates the amount of charge enclosed by a surface'to the number of lines of force emerging frgm that surface. To quantify "number of lines of force", we introduced the concept of flux. Is there an analogous concept that would prove useful in describing the magnetic field due to a current? Yes, a clue to its nature comes from the fact that the magnetic field lines are closed loops surrounding a current.

Consider the magnetic field of a long, current-carrying wire as shown in Fig. 9.19. The magnitude of this field decreases inversely with distance from the wire as evident from Eq. (9.22). Draw an arbitrary closed loop around the wire. The quantity that will prove useful is the length'of this loop, weighted'at each point by that component of the magnetic field, which is in the direction of the loop. We call this quantity the circulation around the loop. Let us calculate the circulation for a circular loop labelled 1 that coincides with a field line as shown in Fig. 9.19. At all points 011 the circle labelled 1, the magnitude of the magnetic field B1 is given by Eq. 9.22, i.e.,

where I is the current in the wire and r, is the radius of this circle. Since the circular loop is always in the same direction as the field (because it coincides with a field line), the circulatioll becomes simply the loop circumference multiplied by the field strength.

#

This shows that circulation does not depend on the loop radius r l . Let us calculate circulation around another circular loop of radius r2 labelld 2. This loop also coincides with a field line. Therefore,

This shows that the circulation aqund iny field line is the same. It also shows that the circulation around any loop coi~iciding with a field line is proportio~~al to the current in the wire which is encircled by that field line.

What will be the circulation around the closed loop that does not coincide with a field line? Fig. 9.20a shows the two field lines (shown thin). Let us consider, the closed loop (shown by bold line ig.Fig. 9.20b) that encircles the wire but does hot coincide'with a

. , - ..o. 0

(8) @)

Fig. 9.20: (a) M-dc field hes due to currentvrrying wire) (b)A dosed loop that docs noi coincide with a single afcld lint, The clrmlatioo around this loop is Lbe same as the drculntion around a field he.

Fig. 9.19: Magnetic field lines ' surrounding a

current -carrying wire.

Electric Current and Magnetic Field

Fig. 9.212 Ampen's l a w is applied to an arbitrary AmpcrInn loop that encloses two long straight wires but excludes a thM wire. Note the directions of the currents.

single field line. To calculate the circulation around this loop, we move around the loop, taking the product of the distance moved with the field compol~ent in tlie direction of motion. When we move along the straight portion cu anid bd of the loop, it contributes nothing to the ci~culation around the loop because when we move at right angles to the field there is no field component in the direction of motion.

What about tlie contribution to the circulation from the arc ab that lies on the inner field line? It will be the same as the contribution we would get if we moved along the arc cd which is not the part of our loop. This is because (1) the circulation aiound the inner and outer field lines is the same as Eq. 9.27 or Eq. 9.28 shows and (2) the arc ab occupies the same fraction of the inner field line as the arc cd does of the outer field line. Thus the circulation around our closed loop is just what wewould get while going around the outer field line. Therefore, according to Eq. (9.210, circulation has the value pal. This observation shows that the circulation around any closed loop eneinrling s steady straight cmrl-ent is proportiannl to the current I encircled by that loop and Is given by p d .

This statement is a simplified version ofAmperefs law. This law is true for any type of current and ally closed loop, as long as the encircled current is steady (never chonging in time). If the current is not in a single wire, but in a number of wires, we simply add all the currents to obtain the net current encircled by our loop. If there are currents flowing in opposite directions, then we give the opposite signs to opposite directions of the current. The algebraic sum of currents encircled by the loop is the net current that determines the circulation around the loop. For a counter-clockwise traversal of the loop, currents pointing out of the loop are take11 as positive, those pointing inward being negative. To remember this convention the right-hand method will help you. If the fingers of your right hand (the curly elenlent) represent the direction of traversal around the loop, then your extended right thumb (the straight element) represents the positive direction for currents ellcircled by the loop.

SAQ 5

Fig. 9.21 shows thecross-sections of three loiig straight wires that pierce the plarie of the page at right angles to it. The wires carry culTerlts i ~ , i2 and i s in the directions shown. Find the net current encircled by the Amperian loop.

In talking about circulation, we must specify the sense-clockwise or counter-clockwise -in which we traverse the loop. We adopt the convention that dirculation is positive if, when we curl the fingers of our right hand around the loop, our right thumb points in the general direction of the net current encircled by the loop. Let us evaluate the circulation around an irregular loop L in a magnetic field as shown in Fig. 9.22. 7'

(a (b)

Fig. 9.22: (a) Ao Lmgulnr loop in a magnetic tiel$ (b) a magnified view shows that the cmtribullon to the eireulnlion h an inhnitesimd segment d of the loop is just B.d.

Let us examine a small part of the loop so that it is essentially linear and the magnetic field is essentially constant in magnitude and direction over it. This small segment of the loop is represented by dl whose magnitude dl is the length of the segment and whose direction is the local direction of the loop. Then the contribuiion dC from d to the circulation C around the loop is the length of dl weighted by the componerlt of the magnetic field in the direction of dl, i.e.,

where 8 is the angle between the field and the direction of the Ioop segment dl. The circulation around the loop is just the sum bf contributio~ls from all the segrneuts dl, i.e., C = 2 dC. As the segments get arbitrarily small, this sum becomes an integral for the total circulation, i.e.,

You must have encountered line integrals similar to this one in Block 1 of this course when electric potential was defined. Here line integral means just a sum of many dot products of the field with segments dl of the loop. The circle on the integral sign reminds that you are dealing with a closed loop.

Therefore, mathematically, Ampere's law can be expressed as follows

This statement is true for any arbitrary loop provided the current I is steady and it is the net current encircled by the loop.

SAQ 6

Apply Ampere's law qualitatively to the three paths shown in Fig.9.23.

I11 electrostatics, we deterlnined the strength of the electric field due to various charge distributions using Gauss's Law. However, we could use Gauss's law only for certain symmetrical charge distributions by constructing suitable closed surfaces in the electric field. We consider Ampere's law as playing the same role in steady state magnetism as Gauss's law played in electrostatics. We shall see that Ampere's law can be used to determine the magnetic fields only due to symnletric current distributio~ls. For this purpose, we have to co~lstmct suitable closed loops called Amperinn loops in the

magnetic field over which the line integral$l . dl is to be evaluated. We shall illustrate

this by few examples.

9.6.1 Applications of Ampere's Law

1) Magnetic field due to a long strsight.current-carrying wire

We shall determine the magnetic field at a distance r. from a long straight wire carrying a current I as shown in Fig. 9.24. The wire is cylindrically sym~ilctric, so that the magnitude of the magnetic field cannot depend on angular position around the wire. We also know that since there are no magnetic monopoles on the wire, the magnetic field lines cannot begin or end on the wire and go radially outwad. Hence the magnetic field lines must be closed loops. The only field lines that are both closed and exhibt cyclindrical symmetry are circles concentric with the wire. It is useful to &member the right hand method for determining the direction of B (for your benefit wc are giving it again in the margin remark).

Now we shall use Ampere's law to determine the magnitude of B at a distance r metres from the axis of a long straight wire of radius R ( r > R ) metres and carrying a current I amperes. Here we assume that r is small in compariso~~ with the length of the wire so that the wire can be considered to be infinitely long.

To evaluate the line'integral in Ampere's law, welmust find an amperian loop. Here field lines are themselves appropriate loops. We construct a circular path of radius r with its centre on the axis of the wire. From symmetry, the magnitude of the field on this circular path (or line) is constant. Here amperian loop coincides with a field line. So the field is everywhere in the same direction as the loop. Hence

so that B . k -

II I Fig. 9.23 :

Fig. 9.24 I Magnetic nelcl due to long straigl~t wire,

Right Hnnd methodif we grasp the w h with the right haud, tllc thumb pointing in the direction of Lhe current, thou the fingers will curl round the wire in the direction of D,

EZPdric Current and Mmgaetic Field

Example 3 A long cylindrical wire of radius R carries a steady current I which is uniformly distributed over its cross-sectional area. Determine the magnetic field at a distance r ( < R ) from the axis of the wire.

Solution

We first notice that the point at a distance r < R is inside the wire. In this case also, we assert that, by symmetry, B has a constant magnitude at all points on a path which is a circle of radius r with its centre on the axis of the wire and with its plane perpendicular to the axis of the wire. The direction of B at every point along this circle is along the tangent to the circle at that point. We choose this circle as the path of integration for the line integral in Ampere's law. Hence,

However, the current enclosed by this path is not I but the part of the current which passes through the m-section of area zr '. The mmnt = JL r x cumnt per unit area of ms-section

Hence,

B = - IQJ r 2 . n ~ ~

SAQ 7

Plot B as a function of r from the axis of the wire (of radius R ) to some distance outside it.

2) Magnetic field due to a solenoid

In Block 2, we have found that we can produce a uniform electric field between the two closely spaced, charged conducting plates of a capacitor. Is there an analogous device that will produce a uniform magnetic field? Yes, the device is solenoid. Let us see how it produces a uniform magnetic field. You know that when a current flows in a circular loop, the magnetic field is found to be directed as shown in Fig. 9.17. The field lines have been drawn using the right hand method. You observe that field lines circle the win. A solenoid can be thought of as a cylindrical stack of current-caving loops.

'I

Fig. 9.25; A ioosely wound,cail of win The magnetic lldd arising ltam c m n t io Ule rrirc b stroagcet witbin the coil. Ibe field is shown only in tbe phw of the pnge) &b am pdob when current emerges from that plane, rrcrsses where eumnt g a s into plane ofpage.

Fig. 9.25 shows a solenoid of four turns. Here the turn are loosely wound compared to the common solenoids. Close to any part of the wire are magnetic field lines encircling the wire. We show these field lines at the lop and bottom of the coil, where the wires cross the plane of the page. The net field anywhere is the vector sum of the fields of the

individual parts of the loop. You can see that inside the coil, the fields from elements of wire at the top and bottom have a component to the right, and so tend to reinfore. Above the top of the coil, the fields arising from elements at the top all have a component to the left, while fields from elements at the bottom have a component to the right, thereby weakening the net field. 'A similar weakening of the field occurs below the bottom of the coil. Hence the net field is strong and points to the right within the coil, and is weaker and points to the left outside the coil, as shown in the Fig. 9.25.

Suppose the coil is tightly wound and its length is longer than its diameter, as suggested in Fig. 9.26. In such situation the field is still stroiig illside the coil of the solenoid, and as the individual turns get arbitrarily close, the irregularities in the field disappear, giving straight field lines inside the solenoid.

What about the field lines outside the solenoid? The exterior field lii~es must connect the field lines emerging from the right of the solenoid to those going into the left because field lines cannot begin or end. The field lines close to the solenoid axis bend very gradually, and spre'ad far from the solenoid before they return to the other end.

To find the value of B inside a solenoid by use of Ampere's law we must note two points: (i) the magnetic field is directed lengthwise along the axis of a tightly wound, long solenoid; (ii) if the solenoid is long, the field lines emerging from the end of the solenoid will fan out widely as they come back around to enter the other end. This indicates that the magnetic field outside the solenoid is many times weaker than it is inside. Consequently, we approximate the situation and consider the field outside thc solenoid to be negligibly small. We will apply Ampere's law to calculate the field within the solenoid. Consider a closed linear path PQRS as shown in Fig. 9.27. For this path,

The integrals over QR and SP are zero as for part of these paths (outside the solenoid)

Fie. 9.272 Cross stclloo of ? long eolenold, showGg s rectangular rmpcdau loop. . B = 0 and for another part inside the solenoid B is perpendicular to 4. The integral over RS is zero as B - 0 outside the solenoid along this path. The only integral that is different from zero is overPQ. Hence

Q For this path, B is along the direction of the path, which means $ B . dl = B . d

so that

v BL

where L is the length of the path PQ. 1f thi's path encloses N turns of wire of the

Magnetic Ficlld

Fig. 9.26 : A longer ceutrd section of n long, more tightly wound mlenoid.

Elechic Current nnd Mngnetic Field

solenoid each carrying a current I, then the right hand side of Ampere's law is l-4~ NI. Finally, we have

where FZ is the number of turns per unit le~~gtli of the solenoid.

The fornlula given by the Eq. (9.34) derived for an infinitely loug solenoid, holds quite well for actual solenoids, for points well inside the sole~ioid away from its ends. Note that ]B does not depend upon the positio~~ of the point within the solenoid as long as we are far away from the ends of the solenoid. Therefore, we coilclude that B is uniform over the cross-sectio~i of the solenoid. This proves to be a practical way to set up a known uniform rnagletic field for experimeiital purposes.

3) Magnetic field Inside a toroid

If a solenoid is bent illto the form of a circle so as to join its two'ends, one obtains a toroid as shown in Fig. 9.28. Tlie field lines of B inside the toroid, by symmetry, are circular and concentric wit11 the center of the toroid. Also the magnitude of field is coilstant along any field line. Choosing our alnperian loop to be a circle of radius r that coincides with a field line, we call readily calculate the circulation around this loop:

~ i g ; 9.28 : Symmetry requires that the field lines be circular. Also shown is an amperia Imp (bold) for use in calculathg the Ileld

Here we could evaluate the line integral because B is constarit on the a~nperian loop and the loop coincides with a field line. As a result the integral is just the field strength times the circumference 21dr of the loop. How much current is encircled by the loop ? If the toroid consists ofN tunis, and carries a current I, the11 a11 alnperiail loop inside the tomid coil encircles a total current NI. This is because each tun1 carries current in the same direction through the path we have chosen. Usi11g Ampere's law to relate this current to the circulation, gives

~ I ' B = p&TI,

so that

This result holds when our amperiaii loop is within the toroid itself. On the other hand, if our amperian loop is inside the inner edge of the toroidal coils, there is no current encircled, and the magnetic field is zero. If the amperian loop is outside the outer edge of the coils, it encircles equal but opposite currents, again giving zero field. From Eq. (9.35) we see that B is not constant over the cross-section of the toroid unlike the stmiglit solenoid.

Magnetic Field

There is another way to express Ampere's law. Eq. (9.31) gives the integral form of this law. Zn this section, we will transform the integral form into the differential form viz:

GurlB = w j (9.36)

where j is the current density.

Let us consider the line integral$^. d around a closed pitlie, as shown in Fig. 9.29a. Here we have replaced the magnetic field, B by a general vector field F. The closed path C call he visualized as the hou~idary of some surface S which spans it. In the integrand, dl is the element of path which is an infinitesimal vector locally tangent to C. - - Now the closed path C is divided into two, thus making two loops C1 and C2, (Fig.9.29b). Take the line integral around each of these in the same directional sense. The sum of the circulatioils around C1 and Cz will be the same as original circulation around C. The reason is that the extra co~itribution from the bridge dividing the original F surface area into two parts cancels out since the two line integrals over the bridge have (a)

the same but opposite signs. The same reasoning is applied, no matter how many sub-divisions are made. Further sub-division into many loops C1, Cz, . . ., Ci, . . .. (Fig. 9.29~) leaves the sum unchanged, i.e.,

$ P . ~ - $ F . d l ~ t $ F . & + . . . C c1 cz

where each tenn in the sum of circulations around subareas of the original area is indicated .by a particular value of the subscript i. If we conti~iue to subdivide the whole loop indefinitely, then in the limit, we amve at a quantity, characterislic of the field F in a local neighbourhood. When we sub-divide the loops, we not'only make loops with s~naller circulation, but also with smaller area. So il is-natural to consider the ratio of the loop circulation to the loop area, just as we considered ratio of flux to volume in Unit 2 of Block 1. The area ai of a bit of surface that spans a small loop Ci is a vector, hence, the surface has an orientatioii in space. As we make loops smaller and smaller in some neighbourhood, we get a loop oriented in any direction we choose. Let us choo%e some particular orientation for the loop as it is sub-divided finally. If the unit vector n denotes the normal to the loop then it has to remain fixed in direction even if the loop su~ounding a particularpoint P shri~iks down towards the zero size. The direction of n, according to the right hand screw method, is as shown in Fig. 9.30. The limit of tlie ratio of circulatio~l to loop area will be written as follows:

SGF . mi lim - (9.38)

Oi-. 0 ai

Fig. 9.29: a) The circulation or F around the curve C is . the Line iutegral d R , the tangential component of F, b) The circulation around Ihe whole loop i s lhe sun1 ut the drculaiions around the two loops. c) When the whole loop b divided Into a number d snlall loops, fhe cirmlntion around the whde Iwp is the sum of (he clrcul~Uons around the little loops.

The limit obtained in the Eq. (9.38) is a scalar quantity which is associated with the point P in the vector field F and with direction I?. If the directions are towards 2, y^ and then we get three different numbers. It turns out that these numbers can be coilsidered as components of a vector called curl F. That is, the number we get for the limit with d in a particular direction is the component, in that direction, of the vector curl F.

( curl F ) . n" = lim - (9.39) oi - 0 ai

FIE. 9.301 Right-hand-smw Let us write again the expression for the circulation around the original path: relalion between the

N N surface normal md tbe

$ , ~ . r n - ZJ F e d a 1 . i (9.40) direction drculption lo line which Antegrad the

i - 1 '' i - 1 Is Wrca

III the last step we merely multiplied and divided by a;, In the.right hand side suppose N is made enormous and all the a'i's shrink. Then according to Eq. (9.40) the quantity in parentheses becomes ( curl F).$i where fii is the unit vector normal to the ith.100~. So, we have on the right hand the sum of the product "loop area times the normal

39

connporlent of curl F ". The sum is over all loops that make up the entire surface S ~~,aer~ l ing C. This is equal to the surface integral of the vector curl F over S, Thus

F . d i

.i ai ] ai ( curl F . i i i- 1 i - 1

We thus find that

where dS is the area enclosed by the closed path. Now using Eq. (9.42) we can write Eq. (9.31) as follows:

Since I = SJ . dS we write Eq. (9.43) as

Since d§ is nonzero, the quantity within brackets must be zero. INrlB;] (9.44)

which is the differential form of Ampere's' law. The relation given by Eq. (9.42) is known as Stokes Theorem. I

We now use the definition of the curl (Eq. 9.39) to find an expression for the curl in terms of cartesian co-ordinates. We would learn how to calculate curl F when the vector function F ( x, y, z ) is explicitly given. Since the curl of a vector isitself a vector we

will find its expression by finding its three mutually perpendicular components. Let us start by a calculation of the circulation about a path of very simple shape that encloses a rectangular patch of surface in the xy-plane as shown in Fig. 9.31. That is, we am taking n̂ 3 2. m e patch is considered to be ve j small so that vector F. does not change much along any one side of the rectangle. In Fig. 9.32 we look down on to the rectangular patch from above. To calculate the circulation we start at the point ( x, y ) - t h e lower left corner of the figure--we go around in the direction indicated by arrows. Along the first side which is marked (I), the tangential component is Fx (1) and the distance is Ax. Hence the first part of the integral is Fx ( 1 ) A x. Along the second side, marked (2), we get Fy ( 2 ) A y , Along the third we get - Fx (. 3 ) A x and along the f~ur th - Fy ( 4 ) A y. Qe minus sign is introduced because the tangential component has to be in the direction oftravel. The wlqle line integral is then

Fig. 9.32: Looking down on the patch in Fig. 9.31.

Now let us consider the first and third terms. Together they are

As we have assumed the patch to be infinitesimally small, the difference is zero. This is true to the first approximation. But we can be more accurate by taking into account the rate of change of F,. When we do so, we write

(See also Unit 2 of Block 1 .)

If we had considered the next approximation, it would have involved te rm in ( A y ) also. But since we will ultimately think of the limit as A y - 0, such terms can be neglected, Using Eq. (9.47) in (9.46) we find

Similarly, for the other two terms in the circulation, we can write

Thus, the line integral around the whole rectangle is

Now Ax A y is the magnitude of the area of the enclosed rectangle wlrich we have represented by a vector in the z - direction.

Hence the quantity ( ---Y- at - - z) is the limit of the ratio :

line iitegral arouhd patch as the patch shrinks to zero si72.

Area of patch

If the rectangular patch had been oriented with its nonnal in the positive y-direction like shown in Fig. 9.33, we would have found the expression

for the limit of the corresponding ratio.

If the patch had been oriented with its normal in the positivex-direction as shown i n Fig. (9.33), we would have obtained

Thus curl F is given by the following expressions:

Magnetic Field

Flectric Cumat and Magnetic Field

FiS. 9.33: For each aritntatlon, (he Urnil of tbc nth dcircdatioa/m gives a componzent of F d ahnt point. H a e the patches hnve been shown sepamted. In arturl dl the patches should dwter a m w ~ d the polnt where curl F is king detenui~~ed

Now that we have finished our discussio~l of the magnetic field generated by a steady current, we turn again to the study of forces on conductors carrying currents. The force on a currentcarrying wire is the basis for many practical devices, including electric

- motors that start automobiles and run refrigerators. In this section, you will know that the working of galvanometer, which is the most important current measuring

.instrument, depends on the action of the magnetic field in exerting a torque on a current loop. Many electromechanical devices make use of the fact that a current-carrying coil of wire is caused to rotate by a magnetic field. When a current loop is placed in a uniform magnetic field as shown in Fig. 9.34, it is acted upon by equal and opposite forces and having same line of action. Therefore, the total force on the current loop is zero. But, a toque does exist on such a coil which can make it rotate. This $ easily seeh in Fig. 9.34.

If you apply the right-hand rule to the wires of the loop shown in Fig. 9.34 you will notice @e following: The forces on the upper and lower sides of the loop (not shown in Figure) are parallel to the axis of rotation and are equal and opposite. They cannot cause any rotationeither. However, the forces hat act on the sides of the loop can indeed cause it to turn. The turning effect zero when the coil is in the position shown in Fig. 9.34a, a torque is seen to exist for the position shown in Fig. 9.34b. Even though the magnitude of forces FI and Fi are same in the two casesF the lever arm from the axis is zero in (a) while non-zero in @). Let us now find the quantitative relation for the torque.

Consider the rectangular loop PQRS carrying a current I and immersed in a uniform magnetic field B as shown in Fig. 9.35a. Let PQ RT = 1 and QR - SP - b. The vertical sides PQ and RS of the loop are perpendicular to the magnetic field. Therefore, the magnitude of the force on these sides is given by

Fi = F2 = 113.

These two forces are equal, parallel and oppositely directed, and hence they form a couple.

42

(a) Parapoaivc view (b) Top 6~

Fi~9.35: Torque on the coil. i

Suppose, at any instant, the axis of the loop NN' (axis normal to the plane of the loop) makes an angle 0 with the magnetic field as shown in Fig. 9,35b. Then at that instant the magnitude of the torque z' due to forces F1 and W2 is given by

z1 .. F1 ( - Fz. ) x perpendicular distance

But 1 x b = A (area ofthe loop)

:. z' - L4B sill 0

This toque acts on every turn of the coil. If there are N turns, the total toque-c is

The quantities in parentheses are grouped together because they are all properties of the coil viz., its number of turns, its area and the current it carries. Eq. (9.53) tells us that a current carrying coil placed in a magnetic field will tend to rotate. We can express the torque in vector notation if A is defined as a vector such that its magnitude is the area ab of the loop and its direction is along the perpendicular to the plane of the loop. The direction of A is given by the right hand rule: Wrap your fingers around the loop in the direction of the current; your thumb then points in the direction of A. 'Ihen we can write

Eq. (9.54) should remind you of Eq. (3.40) for the torque on an electric dipole in an electric field. Eq. (3.40) is

with p the electric dipole moment and E the electric field. Comparison yith Eq. (9.54) suggests that a current loop in a magnetic field behaves analogously to all electric dipole in an electric field. The quantity NIA is called the magnetic dipole moment of the current loop. That is

p = NIA (9.55)

Magnetic dipole moment is a vector quantity and for a current loop its direction is along the direction of A. Using Eq. (9.59, the torque on a current loop can be written as

t = V X B (9.56)

The torque tends to align the magnetic moment with the field.

SAQ 8

A circular loop of radius 5.0 cm consists of 10 turns of wire, A current of 3.0 A flows in the wire. What is the magnitude of the loop's magnetic moment? Suppose initially the

Electric Current and Magnetic Field

magnetic moment is aligned with a uniform magnetic field of 100 Gauss. Now the loop is turned 90" from its original orientation. How much torque is required to hold the loop in its new orientation?

Eq. (9.56) shows that if a current carrying loop is placed in a fixed magnetic field then the torque experienced by it will depend on the current flowing in the loop. This is the principle of the galvanometer.

9.8 SUMMARY

e A long straight wire carrying a currentZ through a uiliforln lnag~ietic field B experiences a force due to the action of the field on the moving electric charges that constitute the current. The force on a section of the wire of length 1 is given by

F = Z l x B

where I is a vector of magnitude 1, pointing in the direction in wliicll the current flows along the wire.

e A magnetic field B is said to exist in any region in which a rlloving charge experiences a force that depends on its charge, its velocity v and the magnetic field. If B and v make an angle 8, the force on the moving charge is given by

F = q v x B

or F = qvB sin 0

e The direction of force on currents and moving charges is given by the right hand rule as shown in Fig. 9.7. It states that if one's right hand is held flat with the fingers pointing in the direction of the field lines and the thumb pointing in the direction of the current, then the palrn of the hand will push in the direction of the force.

Q Gauss's law for magnetism relates the number of magnetic field lines emerging from a closed surface to the net magnetic monopole enclosed. Siilce magiletic monopoles do not exist, Gauss's law for magnetism says that the magnetic flux through any closed surface is zero

This shows that magnetic lines have no beginning or end rather they formclosed loops.

r A current gives rise to a magnetic field. The magnetic field set up by a current carryiiig conductor can be found from Biot-Savart law:

where dS is the contribution to the field from a current I flowing along an infinitesimal vector dl . is a constant whose value is 4n x 10- NA-~ and f is a unit vector from the current element Id1 towards the poini where the field is being calculated.

The direction of the magnetic field is given by means of right' hand method as illustrated in Fig. 9.13. 8

e The magnetic field at a point which is at a distance of r from a long straight wire carrying a cumrit I is given by

e The magnetic field at a point along the axis of a circular loop carrying current is given by

Magnetic Field

where a is the radius of the circular loop carrying current I and bis the distance of the point (along the axis of the loop) from the centre of the loop.

When the point is far from the loop then

2 where A = rn . This shows that current loop behaves like a magnetic dipole.

Two parallel wires carrying currents in the same (or opposite) d i~ct ion attract (or repel) each other. If these two wires are separated by a distance d in a vacuum, then the force (q of attraction (or repulsion) on a segment of length I of either wire is given by

where I1 and I 2 are the currents flowing in the two wires. The force between two current carrying wires is used to define the ampere which is the basis of the whole system of electrical units.

Another way of finding the magnetic fieId in symmetrical cases is by use of Amperes law. This law relates the circulation (or line integral) of the magnetic field around an a&itraly closed loop to the cumnt encircled by the loop, i.e.,

It is used to calculate the magnetic field in situ,ations with high symmetry.

r A solenoid is a long cylindrical coil having many turns of wire, Inside the solenoid, there is a uniform field given by

B = W J where n is the number of turns per unit length.

r The field inside a toroidal coil is given by

MI B e - 2m

where r is the distance from the aytre of the toroid andN is the total number of turns wound on the toroid.

Differential form of ampere's law is

curl B = p~ J

where J is the cumnt density at a given point.

Stoke's theorem states that :J curl F . dS = SF. dl. S

o A closed current loop in a magnetic field behaves like a magnetic dipole with magnetic dipole moment ( p )

p = NIA

whereN is the number of turns in the loop, I the loop current and Avector perpendicular to the plane of the loop with magnitude equal to the loop area. The torque (z) experienced by such current loop is given by

z = p x B

This torque is the operating principle in ammeters, voltmeters and electric motors.

1) In Madras, ti= horizontal component of the earth's field is 3.6 x ~ b m - ~ . ~ f a vertical wire carries a current of 30A upward there, what is the magnitude and direction of the force on lm of the wire?

2) Find the force on each segment of the wire shown in Fig. 9.36, if B = 0.15 T. Assume that the current in the wire is 15A. ( sin 65" = 0.9063 )'

C

P

P i

B P

F

3) Fivevery long, straight, insulated wires are closely bound together to form a small cable. Currents carried by the wires are ill 5 204 I2 = - 6 4 13 = 1 2 4 14 = - 7 4 Is = 18A (negative currents are opposite in direction to the positive). Find B at a distance of 10 cm from the cable.

4) In the Bohr model of the hydrogen ato~n the electron follows a circular path centeredon the nucleus. Its speed is v and the radius of the orbit is r.

a) Show that the effective current in the orbit is e v / a r.

b) Show that p = - ( e/2m ) L , where L = rnr v is the angular momentum of the electron in its orbit.

9,10 SOLUTIONS AND MSWERS

SAQs

1) In most motors current in a wire sets up a magnetic field, and tlie magnetic field, in turn, exerts a force on a second current-carrying wire causing the shaft to rotate.

2) The pairF and v, and F and Bare always at right angles. Vectors v and B may have any angle between them.

3) Refer to Fig. 9.37, applying the right hand method shows that directly above the I wire B points north; and directly below it points south.

4) Both are inverse square laws. In Coulomb's law electrical force acts along r on Fig. 9.37 stationary charge. In Biot-Sava~t law, magnetic force acts perpendicular to r .

5) Net current i = il - i2. Current i3 lies outside the loop and is not encircled by it helice it is not included in calculating i.

6) ~ o r ~ a t h ~ o . 2 $ B . d = 0s-ntcunedbzem.~or~~~o.1and3$~.d - mi. k

7) Inside the wire, the field increases linearly with distance from the w i k axis. O m L

- Inside Ouuidc

, r --,

b W 8 r ' I h c m r g n d k W d M ~ d ~ b a t e . ' I b e d m B F k l B n t t b c b d t b e w i P .

46 1

we reach the surface bf the wire the field begins to decrease inversely with distance. Fig. 9.38 shows the rough plots of the field strengths both inside and outside a wire.

8) As described by Eq. (9.55), the magnetic moment p, is given by

The magnitude of the torque needed to hold the new orientation is given by Eq. (9.56) as

= 2.4 x Nm.

Terminal Questions

1) The vertical component of B is parallel to the current and does not contribute to the force; therefore, - F - EBH = ( 3 0 A ) ( 1 m ) (3.6 x 10-5 ~ b m - ~ ) - 10.8 x IO -~N , west.

2) For each straight segment F = I L x B, where E is the directed line segment. In sections AB and DE, L and IB are parallel, so sin 0 3 0, and F = 0. In section BC, F = ILB - ( 5 A ) ( 0.16 m ) ( 0.15 T ) - 0.12 N, into page, In section CD,F = (5A) (0 .20M)(O. l5T)s in65" = 0.136N,outofpage.

3) By superposition the field is just the sum of the fields due to the individual currents. At r = 10 cm all the fields are either parallel or antiparallel as the currents are parallel or antiparallel. Then

4) a) Since charge e passes a point once every revolution, I e/T, whereT = (2nr) /v,sothat I = (ev)/(2nr).(b)In magnitude, the dipole moment is

and because the electron is negatively charged, p is antiparallel to L.

Magnetic Field

UMT 10 MOTION OF C ELECTNC FIELD

Structure

10.1 Introduction Objectives

10.2 Motion in an gectric Field Initial Vclocity in the Direction of the Ficld Initial Velocity in any Direction

10.3 Motion in a Magnetic Field Initial Velocity Perpendicular to the Field Initial Velocity in any Direction

10.4 Cathode Ray Oscilloscope (CRO) ,

Electrostatic Deflectian Magnetic Deflection

10.5 Lorentz Force and its Applications Velocity Selector Cyclotron

10.6 Summary

10.7 Temlii~al Questions

10.8 Solutions and Answers

- 10.1 INTRODUCTION

By now you are familiar with three kinds of force, namely, gravitational, electric and magnetic forces. These forces are best described in terms of fields, All forces have a property by virtue of which they act on a suitable kind of particle located in the region occupied by the field. If once you know exactly how the fields affect the particles 011 which they act, you are in a position to understand the nature of the field and, hence, the nature of the force.

You are all aware of the way in which objects move in a gravitatio~ial field. In such fields, when an object is thrown upward though the air, it follows a parabolic path. What kind of path is followed by a particle in an electric field or n~agnetic field? What is the behaviour of a particle in electric and magnetic fields? This Unit attempts to answer these questions. Later, in this Unit, you will find liow the behaviour of a charged particle in electric and inagnetic field is put to several applications.

With this Unit we complete our study of magnetostatics -the magnetic field associated with steady currents, and its effect on other currents and 011 isolated moviiig charges. So far, we considered the magnetic field in vacuum. In the next two Units, we turn our attention Lo the study of the magnetic fields when matter is present.

Objectives

After studying this unit you should be able to:

e carry out simple calculations involving the motion of charged particles in a uniform electric field,

e describe the main features of the motion of charge in a magnetic field, and define the term cyclotron frequency,

e explaiii the helical trajectory of a,charged particle moving in a uniform magnetic field,

e explain the working principle of Cathode Ray Oscilloscope,

a appreciate the applications of the combined electric and magnetic fields acting perpendicular to each other.

48

10.2 MOmON IN AW EILIECTHBPG FIELD

In Block 1, we defined the electric field at a point as the force per unit charge at that point. This definition reminds us that electric fields are important because of the effect they have on charged particles. In this section, we consider the problem: how do charges respond, when placed in a known electric field? We will investigate the motion of a particle moving through a uniform electric field.

You are already acquainted with one important experiment that involved the motion of' a charged particle in a uniform electric field. This experiment is Millikan's oil drop experiment, about which you have learnt in your school physics course, and which has been also mentioned in Unit 1. In that experiment, the electric force, due to the uniform field beyeen two charged metal plates, is used to prevent a charged oil drop from falling under the influence of gravity. Let us now consider how such a drop or any other charged particle would behave if there were no gravitational force, and if initially the charged particle was moving in the direction of field.

10.2.1 Initial Velocity in the Direction of the Field

Suppose a uniform electric field E iS set up between the two charged plates as showii in Fig. 10.1 Let us consider the particle with charge q moving in the direction of the field with velocity v. From Eq. (1.7) of Unit 1, the force acting on this particle will be given by ,

P = qE (10.1)

Eq. (10.1) shows that the force is independent of both the velocity and position of the particle. This constaiit force gives the particle a constant acceleration. From Newton's second law (F = ma), this constant acceleration is given by

F @ 8 1 - P (10.2)

m m where m is the mass of the particle. Eq. (10.2) says that the acceleratioii is in the same direction as the electric field. This equation also shows that it is the ratio of charge to -s that determines a particle's acceleration in a given electric field. This explains why electrons much less massive (about 2000 times) than protons but carrying the same charge, are readily accelerated in electric~fields. Many practical devices, like electron microscope.and W tubes makeuse of the high accelerations possible with electrons even in electric field of modest strength. You know how to solve problems iiivolving constant acceleration from your mechanics course studied in school days. Such problems arose while discussing the motion of an object in the uniform gravitational field near the earth's surface. In such problems the coilstant acceleration is the acceleration due to gravity (g); the velocity acquired and the distance travelled in a given time intelval is calculated by using one or inore of th:r,c simple equations known as constant (or uniform) acceleraiioa equations. These equations are

v - u+at (10.3)

In a given electric field, Eq. (10.2) shows that the particle undergoes constant acceleration. Hence Eqs. (1 0.3) to (10.5) can be used to study the motion of particles in the electric field. After solving the following SAQs you will realise that there is a subtle difference between the constaiit acceleration caused by gravity and that of a uniform electric field.

&

SAQ 1

Suppose the electric field shown in Fig. 10.1 is of the strength 2.0 NG'. An electron is released from rest in this field. How far and in what direction does it move in 1.0 p?

SAQ 2

Tick ( 4 ) the correct answer and give reasons. In the problem given above, if a proton (instead of an electron) is released, the distance travelled by it would have been

m, laic The force ~n n (psitivcly) charged pattielo in an elec(lic field. aPIe velocity of the particle has no Muencc on the force,

Electric Current and a) more than distance travelled by an electron Magnetic Field

b) less than distance travelled by an electron

c) equal to the distance trave,lled by an electron

d) proton would not be affected.

In what direction would the protoil move?

In an electric field the acceleratioll varies fro111 object to object depending on the ratio of charge to lilass whereas for gnvity the accelcratioii is the salne for all ob-jccts, no matter what their mass, This is a useful pheno~nena, for it allows us to separate charged objccts (ions) according to their charge- to-mass ratios.

10.2.2 Initial Velocity in any Direction

Till'now, we llave studied the effect of an elcctric field on those charged particles which are at rest or whose direction of iilotion is along the electric field. Let us liow co~lsider the case in which the charged particle is moving in the clectric field with a velocity in the direction shown in Fig. 10.2.

You already know that when a particle (or a projectile) moves with colcjtalll Fig. 10.2:The ditrclionoCthe initial acceleration under the earth's gnvitatio~ml field, it follows a parabolic path. A similar

I -

ve'ocityofthepositive'y thing happens hcre (Fig. 10.3). In such cases, the velocity v is regarded ns n sum of two charged particle is not in the dir~ct ion of the Geld. other velocities; one pl~rallcl to the field de~ioted by vll, and the other pe~pendicular to it

I denoted by VJ-. So tlic total velocily v can be written as follows:

I The horizo~ifal position of the charged particle at any time r is given by

The vertical position of the cliargcd particle is given by

Substituting the valucof r from Eq. (10.6) illto (10.7) we get

This is the equation of the particle in electric field. Because v, q, E and m are

conslants, Eq. (10,8) is of the f o m y = ax + bx 2, in which a and b are constants. This ' is the equation of a parabola. Suppose a particle that sets out from poi11tA with velocity

LI VA describes a parabolic pathABCD as shown in Fig. 10.3. To u~lderstalid the motion of such particle we sllould be able to calculate the velocity of the particle at ally point on the parabolic path. Lct us find out the velocity vn at the point B. Here again, VB is the

i s 6

sum of the two other vectors: v , ~ and vv. Since VJ has been defined in such a way that it is perpendicular to tllc acceleration, it relnaills unaffected by the electric field.

v u = VIA (10.9) , Fig. 10.3: P n r a b d c t r ~ e d o r y oCn

(negativeIy) charged This is because, as the particle inoves v,--component does not change, as there is no pariic~e in a unifonn acceleratiol~ in that directio~~. This is true for any other point on the path. electric field.

In other words, perpendicular colnponent of the velocity remains colistalit in both magnitude and direction throughout the motio~l of the particle. As vp is along the direction of electric field i.e. along the constant acceleration, so it will be affected by thc electric field. The magnitude of vlgt call be found out by applyi~lg ally of the Eqs. (10.3) to (10.5). If the time taken to travel fronlA to B in Fig. 10.3 is r, and if s, is tlle distan2c fromA to B ~ncasured parallel lo E the11 ,

! (1 0.10)

I 50

. I --

Motion of Charges in Electtic and Magnetic Field

(10.11)

Usilig Eqs. (20.9) to (10.12) you call solve wide range of problcliis colicenii~ig the ~iiotio~i of a charged particle in a u~iiforni clcctric ficld. You will co~iic across such proble~lis while sludyi~ig the Cathode ray oscilloscope in Scc. 10.4, where the clcclro~~s arc allowed lo pass through the electric field region alid then the electron bcalii strikes the fluorescelit screen. However, to make ccrlaiu that whatever siiitl in this scctio~i is clear lo you try all the parts of the followi~ig SAQ.

SAQ 3 - 20 an ------3

All electroll is ~i lovi~ig horizontally lo the right at a speed of 4.0 x 10' ~i is-~. It e~iters :a Fig. 10..l:'ille deflecled elcdnrr1 li-olr is ils

region of length 3.0 CI~ I i l l which there is an electric field of2.0 x 10" NC-I pointing straight-line pu111 dow~iward as show11 in Fig. 10.4. Answer Ihe following ques~ions: by a nnifona

cleclric field.

i) How long Goes i t spe~id lravellilig InmA to B?

ii) By how ~nucli and in what dircctio~i is tlic electron dcllcctcd, wheii i t leaves the electric field'? Describe the motion of the particle in the clcctric ficld and d n w rough sketch of the path.

iii) What is the ~iiag~iitudc o l the vertical co~iiponc~it of tlic velocity ollhc clcctron, when i t leaves.thc electric Geld ?

iv) What is the speed o l t'hc electm~i wlie~i i t leaves the clcctric field ?

v) Through what anglc has the elcctro~i bee11 dcllcclcd wlic~i i l Icavcs thc cleclric lie Id '!

vi) Describe and draw the rough sketch of its subscquc~~t ~iiotion.

vii) I11 the si~ililar situation, what will happen lo a positively charged particle of same chargc and Inass as that of an electron.

viii) For wliat purpose call such an arrangemeel be used?

10.3 MOTION IN A MAGNETIC FIELD

We havc dealt with ~no t i o~ i in a u~iifonn electric field, let us now tun1 lo the prohle~n of motion in a u~iiforln maglietic field. If a cliargcd particle is moving along tlie maglietic field, i t will nlove as i t is, because it will experie~ice no lorce. Let us now co~isider thc case in which the particle is i~iitially movi~ig in a plane ~ionnal to tlie ~nag~ictic field.

10.3.1 Initial Velocity Perpendicular to the Field

I11 the lasl unit, you havc learnt that a particle having charge q and ~ilovi~ig with a velocity v in a ~ilag~ietic field 0 experic~iccs a maglietic force given by

Fro111 Eq. (10.13) it follows that lilng~ietic force always acts perpe~idicular lo the directio~i of motion. This means that the ~ilag~ielic ficld call do no work on a charged particle. Because no work is dolie, the ki~lelic ellergy of the particle canliot change -

both the speed v and kinetic energy (i mv ' rc~nr i~ i constant. r17ierefore, the rnag~ietic \ / force chaligcs o~i ly the direction of particle's lnotio~i hut ~iot its specd.

To u~iderstnlid how the'directisn of particle's motion is changed let us colisider the casc of a particle of charge q ~nov i~ ig at right angles to a unifonn ~ilag~ietic field as show11 in Fig. 10.5. Suppose at some i~ista~lt at the pointA, thc velocity v poilits lo the right, so with the ficld bcing out of the page, the cross product v x I? points dow~iward accordi~ig to right halid rule (see Unit 9). If the particle is positive it will experience a dowiiward force. This force changes Ilic dircctio~i of the particle's motion, hut not its speed. A lilllc while later, the particle is ~iioving dowliward alid to tlie right. Now the force poi~ils

Fig. 10.5; A chagfd yhrlicle ninvi~~g at right angles to a ur~il'onal magnetic Bcld describes a circuli~r path.

Electric C~lrreut and Magnctic Field

downwards and to the left. Since thc specd of the particle is still v and the velocity is still at right angles to the field, so the n~agnitude of the force remains the same. Thus, the particle describes a path in which thc force always has the same ~nagnitude and is always at right angles to its niotion. Each time, under thc i~lflucl~cc of the force the particle is deflected fro111 the rectililiear pat11 resulting in the simples1 possible curved path -a circle. Now in ally circular path, thc particle experie~~ce n cc~~tripctal forcc Fc directed towards tlic centre of thc circle. It is given by

where r. is the radius of the circular orhit and v the tailgc~~tial specd of the particle. Therefore, in the present case, t l~e centripetal force being the ~nagaetic force wc can write

so that

The larger the particle's mo~~ient~un rirv, the larger the n d h s of the orbit. On the other hand, if [he field or charge is m;dc larger, the orbif beco~nes smaller. Therefore, observatio~~ of a charged particlc's tr;~jectory in a ~nagnetic field is the standard tech~ique for lneasuril~g lnolneiltuln of the particle. A charged particle can tfilverse a circular path either in clockwise direction or al~ticlockwise direction. In Fig. 10.5 thc particle is describing a circular path in clockwise direction. Solvc thc followil~g SAQ. You will understand that a circular path traversed in the al~ticlockwise direct io~~ is also possible.

SAQ 4

In Fig. 10.5, if the particle is llegatively clurged, what will tl~e circular orbit look like'? Draw with pencil, the orbit oLtl~e negatively charged particle on Fig. 10.5.

Thus, it is possible to identify the sigl~ of the charge on the particle 21s well as its lno~nclltu~n by observilig the particle's trajectory.

Since the circu~~lferelice of the orbit is 2 nr, the time taken by the particlc'to complete olie fill1 orbit is

2Jcr T = - (10.16) v

Using Eq. (10.12) for the radius r. gives

The frequency of rotati011 of a ~novillg charge is give11 by

This qualitity is called cyclotron frequency. It is so called because it is the f r e q ~ ~ e ~ ~ c y at which the charged particles circulate in a cyclotro~l particle acceleratbr. Using the known value of charge-to-mass ratio ( e/m ) of an el~ciron, the n:agnetic field strel~gth can be determined by measuring the cyclotron frequency of the c!ertron. YQII will learn more about cyclotron later in this unit. Now let us find out what path, a charged particle will have, if ii~itially its velocity is neither perpendicular nor parallel to the field,

10.3.2 Initial Velocity in any Direction

In this case the velocity call be resolved hito two vectors: v l perpel~clicular to tlie field and vl along the field. Then Eq.(10.13) becomes :

52

I , a,

Motion of Cl~arges UI Electric and Magnetic Field F = q ( v j - + v ~ ~ ) X B = q v l x B + q v l l x B

Sillce the second tenn on the right hand side of this equation is the cross product of two parallel vectors, it is zero. Therefore,

The force F is clearly perpendicular to B, i.e., there is no acceleration in the direction parallel to B. This nlea~ls

Eq. (10.19) shows that the force is perpelldicularto lhc field i.e., it iliflucnces tlie particle's motion in a plane perpendicular to the ficld.

But we know that the particle's inolioll perpclldicular lo the rnagnctic field is circular. Eq. (10.19) further shows that no force acts ;ilong thc magnetic field. Therefore, the co~npoiiellt of the vclocity which is along the field relnains ullaffected by the ficld. Thus tlie particle mrivcs with a uniform vclocity vll alolig the mag~ictic field, evcli as il exccutcs a circular motion with velocity VL perpendicular lo the field. The resulting path is a helix shown in Fig. 10.6. The radius of the helix is given by Eq. (10.15) if we replace v by v ~ . Thc motion of tlie chargcd particle can be visua1i;lxd like beads strulig on a wire, with the wire being thc magnetic field as show11 in Fig. 10.7.

Fik 10.6: Motion of a l~nrliclc in a unirorni niagnctic ficld.

Fig. 10.7: (a) Charged particlci uuclergoing liclical ~l~otion about nmg~letic ficld lines rlre like (b) bends that are frcv to move along a wirc but uot at right angles to i l

SAQ 5

An electroll with a velocity of lo7 ms-I enters a lnagl~elic ficld of strength 1,5 x Wb I I I -~ at an angle of 30" with it. Calculate tlie radius of the helical path a~ id the time taken by the electroll for one revolulion? Take e / n ~ = 17.6 x 10" C kg1.

In the next section wc illustrate the working principle of Cathode Ray Oscilloscope (CRO) so that you understand the rolc played by tlie electric and ~nngnetic fields.

10.4 CATHODE RAY OSCILLOSCOPE (CRO)

The Cathode Ray Oscilloscope (CRO) is a very useful and versatile laboratory instrument used for display, measurement and analysis of wavcforna and other pheiiomena in electrical and electronic circuits. It is used for a nuinber of purposes, such as, ineasuretiieilt of current, voltage, observation of waveforms of alternating voltages; recreation of television images; as indicator in radar for visual presentatioli of target data such as distance, height, etc. It is based on the followi~lg two principles:

I) When fast lliovi~ig electrons strike the glass screen coated with zinc sulphide, they cause fluorescence.

2) Since the inass of electroils is very sinall, they are easily deflected by the electric and ~nagnetic fields and follows their variation with ppctically no time lag.

Electric Current and Mngnetic Field

The CROs are infact very fast X - Yplotters, displaying an input signal versus another signal (or versus time). The plotter is a luminous spot which moves over the display area in response to. an input voltage. The luminous spot is produced by a beam of electrons striking a fluorescent screen. The extremely low Inass of the electroll enables the beam of electrons to follow the changes of the rapidly varying voltages.

Norrnally CRO uses a horizontal input voltage, which moves the lulninous spot periodically in a horizolltal direction left to right on the screen. The vertical input voltage inoves the luminous spot up and down. The lulnillous spot thus traces the waveform. When the input voltage repeats itself at a fast rate, the trace or display on the scree11 appears stationary on the screen. The CRO, thus, provides a means of visualizing voltage waveforms.

A Cathode Ray Oscilloscope consists of a cathode ray tube, which is the heart of the device. In that tube, as shown in Fig. 10.8, the electrolls produced by the cathode are accelerated towards the final anode of the electron gun under the influence of the anode potential VA. By proper focussing, the electrons are compressed into narrow high velocity beam.

e * -. Fial anode

.

Fig. 10.8 : Electrostatic deflection.

The electrons leave tlie final anode with a velocity along the x - axis. The velocity is given by

If the beam next passes through a field free space, the velocity of the electrons remains constant. In that case, the electrolls go straight and strike the screen at R as shown in Fig. 10.8. But if the beam of electrons is made to pass through an electric field or magnetic field then the electrons will get deflected' from their rectilinear path. The amount of deflection will depend upon the strength of the field. Let us see how the beam of electrons is deflected by an electric field.

I 10.4.1 Electrostatic Deflection

If the electron beam passes between two plane parallel charged plates M and N, which have a uniform electric field at right angles to the direction of lnatioli of the electron

I

beam, the electrons will experience an acceleration along the y-axis. The beamof I

I electrons will get deflected vertically and, hence, strike the screen at point S. Since there is no force and, hence, no acceleration along the x-axis, the x-compoaent of the velocity I

I ~

of electrons remains constant.

If V is the potential difference between two deflectiug plates, d their separatian, then the acceleration along the vertical direction in this region is give11 by

I 4E q v a,=- - - - (10.21) 1

rn m d 1

If 1 is the length of the plates, the11 the time for which each electmrl remains in'the region between the two is

., 4 54

-- .

while leaving the field, the final velocity alongy-axis is given by

v v, = 0 + a Y r = 4--r Using Eq. (10.10)

m d

While leaving the field, the electrons get deflected by the distance given by

Substituting tlie value of ay from Eq. (10.21), we get

Using Eq. (10.12)

It shows that the electron moves along a paribolic path in the region between the two plates. This vertical coinponent of velocity remains constant after the electron move out of tlie electric field. From the11 onwards, the electron travel in a straight line, because the space iS field free.

Outside the field, the electron velocity has two components, i.e., ux and v,. Hence, the resultant electron velocity is

v = G. When the electron leaves the electric field region it will travel in straight line towards ' the screen. This straight line wlien produced backwards meets thex-axis at point 0'. If. the line inakes an angle of 9 with thex-axis, then

using Eq. (10.23)

''I or act = A!- = Since tan 8 = - 2 m d u : 1 P -

OC ' tan8 q V 1 2' m d u:

Point 0' is at the centre of the deflecting plates. The vertical deflection D of the beam on a screeii which is at a distance L from the point 0' is given by

D = L tan 8

Substituting the value of u, from Eq. (10.20), we get

From Eq. (10.25) we conclude that for a given accelerating voltage Vp and for particular diqeiisions of the cathode ray tube, the deflection of the electron beam is directly proportional to the deflecting voltage. The deflecting voltage may be a time varying quantity and thus the image on the screeii follows the variations of the deflecting voltage in a linear manner,

Motion of Charges io Eleclric and Magnetic Field

Electric Current and Magnetic Field

Magnetic deflection is used where a wide angle of deflection is required, as in television tubes. The magnetic field is produced in such a way that it is perpendicular to the electron beam as shown in Fig. 10.9.

. . . . . C Fig.lO.9 : Magnetic deflection

In the absence of the magnetic field, the electron beam goes straight and strikes the fluorescent screen at a point R. When the magnetic field is set up in the region of length 1 the beam in this region moves along a circular path of radius R. After leaving the region of the magnetic field at the point B it continues along a straight path and strikes the screen at point S. The vertical deflection traced out on the screen is D.

Let the circular pathAB subtends an angle 8 at the point C. When SB is produced backwards it cuts thex-axis at point 0 which can be taken as the mid-point of the magnetic field for small deflection. Since triangles SOR and ACE are similar, we have

Now

If the electroll has been accelerated by the final anode to cathode potei~tial Vo, then

I I Thus, the deflection on the screen is proportional to B and inversely proportional to

square root of the accelerating potential V,.

In the next section we will study the effect of combined electric and magnetic fields on I

the motion of charged particles. The general equation for force containing both electtic and magnetic fields is called Lorentz equation, We will also discuss a few important \ applications of the Lorentz force law. Before moving to the next section, try the

I following SAQ. I ) I i SAQ 6 ,

I

A beam of protons is deflected by an electric field and also by a magnetic field. If either could be responsible, how would you be able to tell which was present?

10.5 LORENTZ FORCE AND ITS APPLICATIONS I i J I

Suppose a particle having charge q is moving with velocity v through a space, in which I

, both magnetic and electric fields exist simultaneously, then the force exerted on such a ' ! particle is given by

1 I

i

. 56 1 ' I

I

Eq. (10.27) brings together Eqs. (10.1) and (10.13). Eq. (10.27) is called the Lorentz force equation and F as Lorentz force. It is the vector sun1 of the electric force q E and the inagiletic force q v x B.

Let us now discuss ixnportailt applications of the col~lbined electric and 111ag11etic fields, acting perpendicular to each other, in devices, such as, the velocity selector and cyclotron.

10.5.1 Velocity Selector

In a large class of experiments, in which the motioi~ of charged particles or ions or electrolls is to be studied, it is iinportant to have a source of particles, all having the same velocity. Since most sources of electroi~s or ioils elnit particles with a widc range of velocities, a velocity selector is oftcn essential. Let us understal~d the action of one such selector which uses both electric and ~nagl~etic force. Here, a capacitor like arrangement provides a uniforln electric field downward in thc plane of paper and magnetic field is provided perpendicular to the paper pointing into it, as shown in Fig. 10.10.

Fig. 10.10 : Crussed electric and magnetic fields act as a velocity selector. Only electrons with v, - E/B pass through the field region undeflected.

Suppose a narrow beam of identical charged particles (for example, a beam of electrons), travelling in a vacuunl enters a region which co i~ ta i~~s a ui~ifor~n electric field E and a unifornl magnetic field B, with E and 13 perpendicular to each other. The particles in the bean1 have a spectrunl of velocities, but they all enter the field region perpendicular to both Geld vectors, as indicated in Fig. 10.10. The electric field produces an upward force qE on the electrolls in the beam, whercas the magnetic field produces a downward force qvB (check this by using the right hand method). There is a unique velocity v, for which the electric and lnagrietic forccs exactly canccl. The value of v, is obtained by inaking these two forccs equal. We find

Electroils that have velocities less thall v, are deflected upward and strike the exit wall at points such as a and b. Electrons that have velocities greater than v, are deflected downward and strike the wall at poillts sucli as c, Electrons that have the velocity vo (and only these electrons) are ulldeviated and pass through the exit slit S. The arrangeineilt of crossed electric and magnetic fields is called a velocity selector.

10.5.2 Cyclotron

Cyclotron is the most familiar of all the machines for accelerating charged particles and ions to a high velocity, It is based on the fact that electric and magnetic fields exert force on the ions. A sketch of cyclotroi~ is show11 in Fig. 10.11.

Motion of Charges iu Electric and Magnetic Field

Electric Current and Magnetic Field

High- frequency ac voltaga

lource

Fig,lO.ll: Cyclotron. Top view oFcyclolron electrodes placed in an evacuated cl~a~nber betweeu the poles of an electromagnet Positive ions en~itted by the sources travel in circular orbits inside the hollow electrodes perpendicular to the n~agnetic field. B c h time the ious trnversc the gap, between the electrodes, they are accelerated by a potentinl dinerencc due to nu applied nlternrting voltage synchronized with the ion nlotion. As the ions gaiu energy, the radi~~s'of their path i~~creases. Finally, they are brought out of the nlngnetic field region by n negatively charged deflector plate.

The charged particles starting from a central source are caused to move in circular paths by a magnetic field perpendicular to their motion as showii in Fig. 10.11. They travel inside the two hollow electrodes, between which an alterliatiiig voltage is applied. This alternaling voltage is syilchroliized with the ion motion so that the ions that start out at the right tiine feel an accelerating electric field each time they pass from one electrode to the other. Such ions make larger and larger orbits as they gain kinetic energy.

:$$;;I,T;E;~ However, they continue to stay in step with the alteniating voltage, since their angular I- frequency is constant giver1 by Eq. (10.18). Finally, tliey are brought out of the

nlagiletic field region by a negatively charged deflector plate..

The essential eleinents of a cyclotron are sliown in Fig. 10.12. It consists of two electrodes Dl and D2 whichare hollow metallic semi cylindrical chambers shaped like

p.dck w m a pill -box cut in half along a diameter. These are called the "Dees" because of their resemblence to the letterD in shape. The two Dees are conliected to a high frequency

Fig.lO.12 : Essntial elelnents oscillator capable of generating voltages of 10,000-100,000 V. Thus potelitial difference of cycIolron. appears across the narrow gap between the two Dees and, thus, in this region a strong

electric field is established which reverses its direction at regular intervals. The Dees are enclosed within a vacuunl cllainber (not shown in Figure), which is tlien placed between the poles of a powerful electromagnet. At the center of the machine tliere is an' ion source S, which releases the charged particles or ions which iire to be accelerated. Suppose the strength of the magnetic field has been so set that the cyclotroil frequency, given by Eq. (10.18), of the ion emitted by the source S just matches the frequency of the high frequency oscillator connected to the Dees. At ally illstant, if Dl is at a peak negative potential, then a positive ion released from the source will get accelerated towards Dl by the electric field in !he gap. While inside the Dee, the ion describes a circular orbit under the illfluei~ce of the magnetic field and returns to the gap at the end of a half circle. (Withh the Dees there is no electric field because the Dees are metallic). By then, the potential produced by the oscillator has also completed half a cycle so that now Dl is at the positive potential wliile D: dt the negative potential. Hence, ions get filrther accelerated by the electric field in the gap and enter&. There they follow a circular path of increased radius and again arrive at the gap when D l is once Inore negative.

Each time the ion crosses the gap, it gains energy qVo , where q is the charge on the ion and Vo is the maximum potential difference generated by the oscillator betweeii the two Dees. As it gains energy, its speed increases and the radius of circular orbit also increases. After ions have reached their outermost orbit, they are deflected out of Dee's by means of a deflecting electrode.

W e are providing a solved example, which will help you in desigili~lg thc cyclotron.

I Example 1

I

The pole faces of a cyclotron magnet are 120 cm in diameter; the field between the pole I

faces is 0.80 T. The cyclotron is used to accelerate protons. Calculate the kinetic energy, in eV, aiid the speed of a proton as it emerges from the cyclotron. Determine the frequency of the alteniating voltage that must be applied to the Dees of this accelerator.

( lev = 1.6 x 10-l9 J ); inass of the protoil = ( 1.67 x kg )

58

Solution

The kinetic energy of the particle is given by

and froin Eq. (10.15)

v = w m

For protons, q = e - 1.6 x 10-l9 C and m = m,, = 1.67 x kg. Thus

B r e ( 0.80 T )2 ( 0.60 m )2 ( 1.6 x lo-'' C ) KEp =

&P 2 ( 1.67 x kg)

and

The frequency of the voltage applied to the Dees is tlie cyclotron frequency given by Eq. (10.18). For protons

= 1.22 x lo7 Hz = 12.2 MHz.

Before summiilg up what you have learnt, answer the following SAQ.

SAQ 7

What is the primary functioil of electric Geld and magnetic field in the cyclotron? Let us now sum up what we have learnt in this unit.

10.6 SUMMARY

Motion of Charges in Elccttic and Magnetic Fiem

e The force on a charged particle in an electric field is simply the product of the charge and the electric field

F = q E

When no other f o m act oil the particle, the resulting acceleration, given by Newton's law, is

a A charged particle moving in a uniforin electric field follows a parabolic pat11 because it is subjected to a constant acceleration. Given the inilial velocity of the particle, tlie velocity a1 any other point on the parabolic path can be computed using Eqs. (10.9) to (10.12).

e A chargkd particle, velocity of which is v in a plane perpendicular to a magnetic field B, describes a circular trajectory. The radius rof this circular pafh is given by

where m is the inass of the charged particle.

0 The number of revolutions made by this particle per second is known as the cyclotron frequency and is given by

Electric Current and Magnetic Field s When the direction of motion of the charged particle is neither parallel nor

perpendicular to the direction of magnetic field it describes a helical trajectory.

o In CRO, the electron beam can be deflected either by an electric field or by a magnetic field. In both cases, the deflection of the electron bean1 is proportio~ial to the applied electric (or magnetic) field.

The motion of a charged particle, ll~ovilig through a combi~latioll of the electric and magnetic fields, is described by the Lorentz force

10.7 IFEMHNAL QUESTIONS

1 ) A particle with charge q and Inass m is shot with kinetic energy K into the region between two plates as shown in Fig. 10.13. If the magnetic field between the plates is B and directed as shown, how large must B be, if the particle is to miss collision with the opposite plate?

Fig. 10.13

2) hl Fig. 10.14, a proton ( q = +e, rn = 1.67 x lo-" kg ) is shot with speed 8 x lo6 ms- at an angle of 30" to anx-directed field B = 0.15 T. Describe the path followed by the proton (including the radius, pitch etc.)

Fig. 10.14

3) As shown in Fig. 10.15, a beam of particles of charge q enters a region where an electric field is unifonn and directed downward. Its value is 80 kVm- I . Perpendicular to E and directed into the page is a magnetic field B = 0.4 T. If the speed of the particles is properly chosen, the particles will not be deflected by these crossed electric and magnetic fields. What speed is selected in this case? (This device is called a velocity selector.)

X X X X X

x x x . x x

+9 B=0.4 T (into page)

X X X X X

X X X X X

I

I

4) A beam of electrons passes undeflected through two mutually perpendicular Motion of Charges iu Eleclnc I

electric and magnetic fields, the electric field is cut off and the same magnetic field and Magnetic Field I

is maintained, the electrons move in the magnetic field in a circular path of radius 1.14 cm. Determine the ratio of the electronic charge to inass if E = 81 Vm-' the

I magnetic field has flux density 2 x T.

5) A cyclotron is being used to accelerate protons to a kinetic energy of 5.0 MeV. If the magnetic field in the cyclotron is 2.0 T, what must be the radius of the . cyclotron and the frequency at which the Dee voltage is alternated?

--

10.8 SOLUTIONS AND ANSWERS

SAQs

1) Using Eq. (10.4) we have

1 2 \,'I

y = 0+- at (distance is measured along vertical direction) I i 2 / i

with the acceleratioil'given by Eq. (10.2), we have

The inillus sign indicates that motion is downward, opposite to the field direction. It is expected because electron carries a negative charge.

2) b) A proton in the same situation would not move as far because its acceleration is much less due to its much higher mass. The proton wilrmove in the upward direction.

3) i) The electric field acts along the y-axis. The horizontal cotnponent of velocity v, of electron remains unaffected by the electric field. Thus, the 3 time spent travelling fromA to B is

k

ii) During this time the electron experiences acceleration in vertical Z

m I

direction and undergoes a vertical deflection (y) given by !

Thus the path of the charged particle within the electric field is a parabola in thexy plane. Now

= 2.2mm.

The positive value ofy means that electmn is deflected upward. In the field region, it follows an ~pward~arv ing parabola, as shown in Fig. 10.16.

61

EkctrlcCumt m d Mamelic Field

iii) At the time of entrance there was no component of v along the y-axis, hence, using Eq. (10.10) we get

iv) The electron leaves the region with a speed v given by

v) The angle of deflection at B is 0 = tan-' (4 = tan-

1 8.8 x lo5 ms-' 4.4 x lo6 ms-'

= 12" above the horizontal.

vi) Once it leaves the field region, the electron will again move in a straight line along the tangent to the parabola at the point where the electron leaves the field region as show11 in Fig. 10.16.

vii) Positively chirged particle is also deflected by 12", but because of their sign they are deflecled downwards.

viii) Such an arrangement can be used for separating positive arid negative particles in a beam.

4) When the sign of the charge is negative, the right hand rule shows that the force experienced by it will be in upward direction. The Fig.lO.17 shows what your answer should look like.

X X X X X X X X B

X X X X X X X X

5) = lo7 m~ - ', vll = v cos 30°, V_L = v sin 30"

nt V_L R 5 - - - - lo7 sin 30" - 1 -- e B 1.5 x lo-' x 1.76 x loll 52.8 I

n a

6 ) 011 reversing the direction of the flow of protons, if the protons are deflected in tlie same direction, the deflcctioli is due to electric field, if the protons are deflected in the opposite direction, tlre deflection is due to magnetic field.

7) In a cyclotron, the purpose of the electric field is to energize the beam and that of magnetic field to give it the circular motion.

Terminal Questions

1) To just miss the opposite plate, the particle must move in a circular path with radius r so from Bqr = mv; and using K = ( mu2 )/2, we have B = ( 2mK ) ' /( qr ).

2) Weresolve the particle velocity into co~nponents parallel to and perpendicular to the ~ilagnetic field. The ~nagnetic force due to vll is zeta ( sin 0 - 0 ) ; the force due to vl,has no x compo~lent. Therefore, tliex notion is u~~ifonn, at speed yl = ( 0.86 ) ( 8 x 10 ms" ) = 6.88 x lo6 mns-', while the transverse ~notio~r is circular with radius

The proton will spiral along thex-axis; the mdius of the spiral (or helix) will be 28 cm. To find the pitch of the helix (thex distance travelled during one revolution), we note that the time taken to complete one circle is

23tr 23t(0,281n) period = - = - 4.4 lo-'

V (0 .5 ) (8~10~111s- ' )

During that time, the proto11 will travel x distanceof pitch given by ( vr ) ( period ) = ( 6.88 x lo6 ms-' ) ( 4.4 x 10-~s ) - 3.0 m

3) The electric field causes a downward force Eq on the charge if it is positive. 'Ilic right-hand rule tells that the magnetic force, qvB sill 90" , is upward if q is positive. If these two forces are to balance so that the particle does not deflect, then

When q is negative, both forces are reversed, so the resdlt v - E/B still holds.

4) If tlie beam is u~ideflected, wlien the crossed fields are on, we have evB - eE and v = E/B. Whe~r the elcctric field is cut off, the electrons move in a circle with: e/m - v/RB = E/( RE ) - ( 8 x lo3 )/[ ( 0.0114 ) ( 2 x )2 ] - 1.75 x 10" C kg-'

5) The cyclotron icquency is given by

This is the frequency required to accelerate pmto~ls at each crowing of the Dee gap. An cnergy of 5 0 MeV is equal to

( 5 . 0 ~ 1 0 ~ e V ) 1 , 6 x 10"'9J(eV)-' 8 . 0 ~ 10-13J.

Motion of' Charges io Electric and Magnetic Field

Electric Current md Magnetic Field ,

so the protoll kinetic energy is

Solvilig for the speed v, gives

The radius needed to acco~nlnodate 5-MeV protolls is given by

T 11 GNETISM OF TE S-I

Structure 11.1 Introduction

Objectives

11.2 Response of Various Substance to a Magnetic Field

11.3 Magnetic Moment and Angular Momentum of an Atom

11.4 Diamagnetism and Pammagnetism Diaeniaenetism -Effect of Maenet ic Field on Atomic Orbits - - - - Paramagnetism -Torque on Magnetic Dipoles

11.5 The Interaction of an Atom with Magnetic Field - Lannor Precession

11.6 Magnetisation of Paramagnets

11.8 Tenninal Questions

11.9 Solutions and Answers

11.1 INTRODUCTION

In the last two Units, we have discussed the magnetic fields produced by moving , charges or cui~ents in conductors. There, the moving charges and conductors were

considered to be placed in vacuum (i.e., in air). In Units 11 and 12, we learn how the magnetic field affects materials and how some materials produce magnetic field. You must have learnt in your school Physics Course that in equipmnent such as generator and motor, iron or iron alloy is used in their structure for the purpose of enhancing the magnetic flux and for confining it to a desired region. Therefore, we will study the magnetic properties of iron and a few other materials called ferromagnets, which have * similar properties as iron. We shall also learn that all the materials are affected by the magnetic field to some extent, though the effect in some cases is weak.

When we speak of magnetism in everyday conversation, we almost certaii~ly have in mind an image of a bar magnet. You may have observed that a magnet can be used to lift nails, tacks, safety pins, and needles (Fig. 1 l.la) while, on the other hand, you canllot use a magnet to pick up a piece of wood or paper (Fig. 11.1 b).

I

,

TACKS MADE OP MAGNBT

Fi&11.1: a) M r k h b that are attracted to a magnet are ulkd mngoccic matetids b) MaLuLls tbai do not read tor magnet are called nonmagnetic materids

Materials such as nails, needles etc., which are influenced by a magnet are called magnetic materials whereas other materials, like wood or paper, are called non-magnetic materials. However, this does not mean that there is no effcct of magnetic field on non-magnetic materials. The difference between the behaviour of .

, Such materials and iron like magnetic materials is that the effect of magnetic field OJI

non-magnetic material is very weak. There are two types of non-magnetic materials: diamagnetic and paramagnetic. Unit 11 '

deals with diamagnetic and paramagnetic effects. The ideas, concepts and various terms

Electric Current and Magnetic Field .

so the proto11 kinetic energy is

Solvi~~g for the speed v, gives

The radius needed to accoln~nodate 5-MeV protots is give11 by

Structure 11.1 Introduction

Objec'tives

11.2 Response of Various Substance to a Magnetic Field

11.3 Magnetic Moment and Angular Momellturn of an Atom

11.4 Diamagnetism and Paramagnetism Diagmagnetism -Effect of Magnetic Field on Atomic Orbits Paramagnetism -Torque on Magnetic Dipoles

11.5 The Interaction of an Atom with Magnetic Field - Larmor Precession

11.6 Magnetisation of Paramagnets

11.7 Suminary

11.8 Tenninal Questions

11.9 Solutioils and Answers

11.1 INTRODUCTION

In the last two Units, we have discussed the magnetic fields produced by moving , charges or cuirents in conductors. There, the moving charges and conductors were

considered to be placed in vacuum (i.e., in air). In Units 11 and 12, we learn how the magnetic field affects materials and how some materials produce magnetic field. You lnust have learnt in your school Physics Course that in equipinei~t such as generator and motor, iron or iron alloy is used in their structure for the purpose of enhancing the magnetic flux and for confining it to a desired region. Therefore, we will study the magnetic properties of iron and a few other materials called ferromagnets, which have similar properties as iron. We sliall also learn that all the materials are affected by the magnetic field to some extent, though the effect in some cases is weak.

When wespeak of magnetism in everyday conversation, we almost certaiiily have in mind an image of a bar magnet. You may have observed that a magnet can be used to lift nails, tacks, safety pins, and needles (Fig. 1l.la) while, on the other hand, you cannot use a magnet to pick up a piece of wood or paper (Fig. 1l.lb).

TACKS MADE OP M A G r n C MATERIAL

w BLOCK OP WOOD

Fi~11.1: a) Materials that are a t h d a l to a magnet are uUedmagnctic mabdds b) Mabrlrls Lbal do not read to 8 magnet arc called nonmagnetic mabhls.

Materials such as mils, needles etc., which are i~lfluellced by a magnet are called magnetic materials whereas other materials, like wood or paper, are called non-magnetic materials, However, this does not mean that there is no effect of magnetic field on non-magnetic materials. The difference between the behaviour of .

, such materials and iron like magnetic materials is that the effect of magnetic field on non-magnetic material is very weak.

There are two types of non-magnetic materials: diamagnetic and paramagnetic. Unit 11 deals with diamagnetic and paramagnetic effects. The ideas, concepts and various tenns

Electric Current and Mnguelic Field

that you become familiar with in this Unit would help you in the study of fenornagnetism in the next Unit. In this unit, we present a simple classical account of the magnetism, based on notion of classical physics. But you must keep in mind that it is not possible to understand the magnetic effects of materials from the point of view of classical physics. The magnetic effects are a completely quantum mechanical phenomena. Only modem quantum physics is capable of giving a detailed explanation of the magnetic properties of matter because the study requires the introduction and utilization of quantum mechanical properties of atom. For a co~n~ le te explanation, one must take recourse to quantum mechanics; however, a lot, though i~icomnplete, of information about matter call be extracted by combining classical and quantum concepts.

Basically, in this unit, we will try to understand, in a general way, the atomic origin of the various magnetic effects. The next unit is an extension of this unit. There, we wiIl try to develop a treatment of nlagnetised matter based on some observed relations between the magnetic field and the parameters whichcharacterise the material. Finally, we consider the analysis of the magnetic circuit, which is of particular importance in the design of the electromagnets.

Objectives

After studying this unit you should be able to:

r understand and explain: gyromagnetic ratio, paramagnetism, diamagnetism, Larmor frequency,

e reIate the magnetic dipole moment of an atomic magnet with its angular inomeiitum,

c expIain the phenomena of diamagnetism in terms of Faraday inductio~l and Lelu's principle,

o explain panmagnetism in terms of the torque on magnetic dipoles,

e find the precessional frequency of an atomic dipole in a magnetic field,

o appreciate that a lot of information about magnetism of matter can be obtained from the classical ideas of ato~nic magnetism.

11.2 RESPONSE OF VARIOUS SUBSTANCE TO A GNETIC FIELD

To show how the magnetic materials respond to a magnetic field, consider a strong electromagnet, which has one sharply pointed pole piece and one flat pole piece as shown in Fig. 11.2.

Small piece of malerial

Fig.ll.21 A mall cylinder of bismuth is weakly repelled bithe sharp pde; a piece of aluminium is attracted.

The magnetic field is much stronger in the region near the pointed pole whereas near the flat pole the field is weaker. This is because the lines must concentrate on the pointed pole. When the current is passed through the electromagnet (i.e., when the magnet is turned on), the hanging material is slightly displaced due to the small force

66

acting on it. Some materials get displaced in the direction of increasing field, i.e., towards the pointed pole. Such materials are paralnagiletic materials. Examples of such material are aluminium and liquid oxygen. On the other hand, there are materials like .

bismuth, which are attracted in the direction qf the decreasing field, i.e., it gets repelled from the pointed pole. Such materials are called diamag~letic. Finally, there is a small class of materials which feel a considerable stronger force ( 10 - 10 times ) towards the pointed pole. Such substances are called ferromagnetic materials. Examples are iron and magnetite.

. HOW does a substance experience a force in a magnetic field? And why does the force act in a particular direction for some substance while in opposite direction for other substance? If we can answer these questions, we will understand the mechanisms of pramagnetism, diamagnetism and ferromagnetism. In Unit 9, you have already leanit that the magnetic fields are due to electric charges in motion. In fact, if you could examine a piece of'material on an atomic scale, you would visualize tiny current loops due to (i) electrons orbiting around nuclei and (ii) electrons spinning on their axes. For, macroscopic purposes, these current loops are so small that they are regarded as the magnetic dipoles (see Section 9.2 of Unit 9) having magnetic moment. It is this magnetic moment, via which the a t o m at a substance iiiteract with the external field, and give rise to diamagnetic and paramagiletic effects. In this unit, you will understand the origin of paramagnetism and diamagnetism. Fernmagnetism has been left to be explained in the next unit. Let us first find out the value of the magnetic moment and see how it is related to the angular momentum of the atom.

11.3 GNETIC MOMENT AND ANGULAR ILTOMEN'IVM OF AN ATOM

Study comment: You may find it useful to look back at Unit 9, Section 9.2.2, in which the idea of magnetic dipoles has been introduced. Electrons in an atom are in coilstant motion arouild the nucleus. To describe their motion, one needs quantum mechailics, however, in this unit we shall use only classical arguments to obtain our results, though we repeat here that our description of the physical world is incomplete as we shall be leaving out quantum mechanics.

We consider an electron in the atom to be moving, for simplicity, in a circular orbit around the nucleus under the influence of a central force, known as the electrostatic force, as show11 in Fig. 11.3(a). As a result of this motion, the electron will have an angular momentum L about the nucleus.

Fbll.31 n) C l d c a l model of an atom in wbich an eltclron moves nl speed v in n drrular orbit. b) The average elcchic cumnl is the same ns if the charge - e were divided iato small bils, fonning a rotating ring d charge. c) The orbital angular momentum vector and the magnetic mommt vector both point ia opposite dindona

The magnitude of this angular momentum is given by the product of the mass m of the electron, its speed v and the radius r of the circular path (see Fig. 11.3), i.e.,

L = mvr

Its direction is perpendicular to the plane of the orbit. As you have already read Unit 9, and worked out the terminal questions given at the end of that unit, the fact that orbital

-

Elcclric C u m t and Magnetic Field

motion of the electron constitutes an electric current will immediately strike your mind. The average electric current is the same as if charge on electron were distributed in small bits, forming a rotating ring of charge, as shown in Fig. 11.3@). The magnitude of this current is the charge times the frequency as this would equal to the charge per unit time passing through any point on its orbit. The frequency of rotati011 is the reciprocal of the period of rotation &/v, hence the frequency of rotation has the value v/%. The current is then

The magnetic moment due to this current is the product of the current and the area of which the electron path is the boundary, that is, p = I JW. 2. Hence we have

evr p = -- 2

It is also directed perpendicular to the plane of the orbit. Using Eq. (11.1) in Eq. (1 1.3) we get as follows:

The negative sign above indicates that p and L are in opposite directions, as shown in Fig. 11.3(c). Note that L is the orbital angular momentum of the electron. The ratio of the magnetic moment and the angular momentum is called the gyro- magnetic ratio. It is independent of the velocity and the radius of the orbit.

I

Acmrding to quantum mechanics, L = 7i V I ( I + I ) , where kis a positive integer and I1 ti = - Ii beingplanck's constant. However, in some physical cases the applicability

2 n' of classical models is close to reality, therefore, we will go ahead with the classical ideas. Further, the early work on the nature of magnetic materials was based on classical ideas which gave intelligent guesses at the behaviour of these materials.

SAQ 1

a) Show that the magnetic dipole moment can be expressed in units of JT - ' ( Joule per Tesla ).

b) In the Bohr hydrogen atom, the orbital angular momentuln of the electroil is

quantized in units of f i, where Ir = 6.626 x JS is Pla~lck's constant. Calculate the smallest allowed magnitude of the atomic dipole moment in JT - '. (This quantity is known as Bohr magneton.) Mass of the electroit is 9.109 x 10 -31kg.

!r. addition to its orbital motion, you know that, the electron in an atom behaves as if it were mtating around an axis of its own as shown in Fig. 11.4.

t Angular momentum

Flg.11.4r Tbe spin and che sssochtcdmognclc momcot d & e dedroa

This property is'called spin. Though strictly it is not possible to visualise the spin of a point particle like electron, for many purposes it helps to,regard the electron as a ball of

negative charge spinning around its axis. Then you can say that it is a current loop. Spin is entirely a quantum mechanical idea. Nevertheless, the spin of the electron has associated with it an angular momentum and a magnetic moment. For purely quantum mechanical reasons with no classical explanation, we have

where S is the spin angular momentum and p is the spin magnetic moment. The gyromagnetic ratio in this case is twice that in the orbital case.

In general, an atom has several electrons. The orbital and spin angular momenta of these electrons can be combined in a certain way, the rules of which are given by quantum mechanics, to give the total angular momentum J and a resulting total magnetic moment. It so happens that the direction of the magnetic moment is opposite to that of the angular momentum in this case as well, so that we have

where g is a numerical factor known as Lande g-factor which is a characteristic of the state of the atom. The rules of quantum mechanics enable us to calculate the g-factor f ~ r any particular atomic state. g ;. 1 for the pure orbital case and g = 2 for tlie pure spin case. The atoms and molecules interact with the external magnetic field due to its magnetic moment. But there is another way in which atomic currents and hence moments are affected by the field. In this case the magnetic moment is induced by the field. This effect leads to diamagnetism which we study in the uext section. But before moving to the next section, try the following SAQ.

SAQ 2

a) Compare Eq. (1 1.6) with (11.4) and (1 1.5), to find the value of g for (i) pure orbital case and for (ii) pure spin case.

b) The experimentally measured electron spin magnetic moment is 9.27 x ~ m ~ , Show that this value is consistent with the formula given by the Eq. (11.5).

A h (Hint : According to Bohr's theory S = - Here ii = - h being Planck's constant.) 2' 27C'

11.4 DI GNETISM AND NETISM

In many substances, atoms have no permanent magnetic dipole moments because the magnetic moments of various electrons in the atoms of these substances tend to cancel out, leaving no net magnetic moment in the atom. The orbital and spin magnetic moments exactly balance out. These materials exhibit diamagnetism. If a material of this type is placed in a magnetic field, little extra currents are induced in their atoms, according to the laws of electromagnetic induction (to be discussed in detail in Unit 13), in such a direction as to oppose the magnetic-field already present, Hcnce, in such a substance, the magnetic moments (on account of induced currents) are induced in a direction opposite to that of the external magnetic field. This effect is diamagnetism. It is a weaker effect. However, this effect is universal.

There are other substances of which the atoms have permanent magnetic dipole moments. This is due to the fact that the magnetic moments due to orbital motion and spins of their electrons do not cancel out, but have a net value. When such a substance is placed in a magnetic field, besides possessing diamagnetism, which is always present, the dipoles of such a material tend to line up along the direction of the magnetic field. This is paramagnetism and the material is called paramagnetic In a paramagnetic substance, the paramagnetismusually masks the ever present property of diamagnetism in every substance. Diamagnetism involvd a change in the magnitude of the magnetic moment of an atom whereas paramagnetism involves change in the orientation of the magnetic moment of an atom. Let us see how.

Electric Current and Magnetic Field

11.4.1 Diamagnetism - Et'fect of Magnetic Field on Atomic Orbits

We collsider an atom, which has no intrinsic magnetic dipole moment, and imagine that a magnetic field is slowly turned on in the space occupied by the atom. The act of switching the magnetic field introduces change in the magnetic field which, in turn, generates an electric field given by Faraday's law of induction (to be discussed in detail in Unit 13). It states that the line integral of E around any closed path equals the rate of change of the magnetic flux through the surface enclosed by the path.

Fig.ll.5: An electron moving in circular orbit in a uniform magnetic field tbnl is nonnal lo the orbit.

For simplicity, we choose a circular path along which the electron in the atom is moving (see Fig. 11.5). The electric field around this path is given by Faraday's law as

where I-,- is the radius of the circular path perpendicular to B. The above equation gives the circulating electric field whose strength is

This electric field exerts a torque t = - e ErL on the orbiting eleclron which must be dL equal to the rate of change of its angular momentum - , that is, dt

The change in angular momentum, AL due to turning on the field is obtained by integrating Eq. (11.9) with respect to time from zero field as follows:

Thus Eq. (11.10) shows that a build up of a magnetic field B muses a change in the angular momentum of the electron, U and hence a change in the magnetic moment governed by Eq, (1 1.4) as follows:

The direction of the induced magnetic moment is opposite to that of B, which produces it as can be seen from the negative sign in the Eq. (1 1.11). In this equation, we have the

70

term r t which is the square of the radius of the particular electron orbit whose axis is Magnetisn~ of Material$ -I

along B. If B is along the z-axis, we put 6 = x ' + y '. Thus, the average ( r:) would I

be 2( x ' ), since ( x ' ) = ( y ' ) = ( z ' ) due to spherical symmetry. Further 7 1 , 1

( x 2 ) = 0 1 ' ) = ( 2 - ) = - ( x - + ~ ~ ' + z ' ) 3 = - ( ~ ~ ) ~ i v f i ( d - ) 3 = $ ( r ' ) .

Hellce the Eq. (11.11), which we shall write as

We find that the induced ~nagllctic tllonlellt ill a diamagnetic atom is proportional to B and opposing it. This is diamagnetism of matter. I f each ~nolecule has rz electrons each with an orbit of radius r, the11 the change in the 111ag11etic ~nolnellt of the atom is

There is all alternate way of u~lderstanding the origin of diamagnetism which is based on the fact that electroll either speeds up or slows down depending on the orientatio~~ of

u the lnagnetic field. Let us see how. As show11 in Fig. 11.6, in the abse~ice of the Fig. 11.6: There is no extenla1

mv ' magnetic Field. magnetic field, centripetal forcc - is balanced by the electrical force as follows: Centripebl force is

I' balanced by the electrical force.

I ~ " N I V ' - (11.13) 4n EO I' 2 r

Let us find out what hi~ppens to one of the orbits when an external magnetic field is applied as shown in Fig. 11.7.

Fiy.ll.7: Magndic Field is perpendic~~lar lo llrc plane of the orbit.

In the presence of thc ~nagnetic field there is an additional term e ( v x B ) and under these conditions speed of the electron changes. Suppose the new speed is vl, then

If we assume that the change A v = vl - v is small, we get

A change in orbital speed means a change in the dipole inornent given by Eq. (11.3) as follows:

EI&c Current and Magnetic Field

This shows that change in p is opposite to the direction of B. In the absence of all external magnetic field, the electron orbits are randomly oriented and the orbital dipole moments cancel out. But in the presence of a magnetic field, the dipole lnolnent of each atom changes and all get aligned antiparallel to the external field. This is the ~ n e c h ~ ~ ~ i ~ ~ ~ responsible for diamagnetism. This property of magnetic material is observed in all atoms. But as it is much weaker than para~nagnetism it is observed only iii those material where paramagentism is absent.

11.4.2 Paramagne tism-Torque on Magnetic Dipoles

Paramagnetism is exhibited by those atoms which do not have the magnetic dipole moment. The magnetic moment of an atom is due to moment produced by the orbital currents of electrons and their "unpaired spins". In Unit 9 you have leanit that a current loop having p as its magnetic dipole moment when placed in a rliiifom field experiences a torque 'G which is given by Eq.(9.16), i.e.,

The torque tends to align the dipoles so that the magnetic ~no~ne~ i t is lined up parallel to the field (in the way the permanent dipoles of dielectric are lined up with electric field). It is this torque which accounts for paramagnetism. You might expect every material to be paralnagnetic since every spinning electron constitutes a magnetic dipole. But it is not so, as various electron of the atom are found in pairs with opposing spills. The magnetic moment of such a pair of electrons is cancelled out. Thus parainagnetism is exhibited by those atoms or molecules in which the spin magnetic moment is not cancelled. That is why the word "unpaired spins" is written above. Paramagnetism is generally weak because the lining up forces are relatively small compared with the forces from the thermal motion which try to destroy the order. At low temperatures, there is more lining up aiid hence stronger the effect of paramagnetism.

Of the following materials, which would you expect to be paralnag~~etic a ~ ~ d which diamagnetic?

Copper, Bismuth, Aluminium, Sodium, Silver

Would it be possible to prepare an alloy of, say, a diamagnetic ~ilatcrial like copper and a paramagnetic material like aluminium so that the alloy will neither be paramagnetic nor diamagnetic?

11.5 THE INTERACTION OF AN ATOM WITH MAGNETIC FIELD -LARMOR PRECESSION

In the last subsection, while explaining paramagnetism we, coilsidered an at0111 as a magnet with the magnetic moment p. When placed in a unifonil lnagnetic field B, it is acted upon by a toque 'G = p x B, which tends to line it up along the direction of the magnetic field. But it is not so for the atomic magnet, because it has an angular momentum J like a spinning top. We already know that a rapidly spinniiig top or a gyroscope in the gravitational field is acted upon by a torque, the result of which is that it precesses about the direction of the field. (To know more about precessio~i you call read Unit 9 of the coulse 'Elementary Mechanics', PHE- 01). Similarly, instead of lining up with the direction of the magnetic field, the atomic mag~~et precesses about the field direction. The angular momentum and with it the magnetic niolnent precess about the magnetic field, as shown in Fig. 11.8a.

Due to the presence of the magnetic field, the atom will feel a toque z whose magnitude is given by

z = pBs in8 (11.16)

where 8 is the angle which p makes with B. The direction of the torque is perpendicular to t he direction of magnetic field and also of ~r, as shown in Fig. 11.8b,

72

m.11.8: a) 'Ibe angular momtutum associated with rtamk m.gnct p m c s s s about ma+ tkld (b) The prrscnce of magntUc ictld mulls h the torquer. It Is at right angles to (be anplnr mommtum, c) 'Ibe torque changes the dimtion oftbe angular momentum vector, causing prrassioa

Notice that the torque is perpendicular to the vector J. Now according to Newton's second law

For small changes, we can write it as

In other words, the torque will produce a change in the angular momentum with time. Suppose that A J is the change in the angular momentum in an interval of time A t. This A J will be in the direction of z. This will result in the tip of J moving in a circle about Bas the axis. This is, in fact, a precession of J (so also of p) about the direction of B. The magnitude of A J can be written by using Eq. (11.16) in Eq. (11.18)as follows:

Although the toque r, being at right aAgles to J, cannot change the magnitude of J, it can change its direction. Fig. 11.8~ shows how the vector AJ adds vectonally onto the vector J to bring this about. If cop is the angular velocity of the precession and A 9 is angle of precession in time A t, then

From Fig. 11 .8~ we see that

AJ I

( p B s i n 0 ) A t A + n ~ x Jsin 8

Dividing above by A t, approaching the differential limit and putting o, = 9, we get

Substituting for p N from the Eq. (11.6), we get

(1 1.22)

as the angular speed of precession of an atomic magnet about the direction of B. If in e. (11.22) g 1, then up is called the Lamor hquency, and is proportional to 11. It should be borne in mind that this is the classical picture.

NOW you may wonder if the atomic magnets (dipoles) precess about magnetic field, how many of these dipoles get aligned along the direction of magnetic field. We know that the potential energy of a dipole in the applied field is given by - Cr . B - - p B cos 0. Therefore, an unaligned dipole has a greater potential energy than an aligned one. If the energy of the dipole is conserved then it cannot change its .

'

Electric Currcot m d Magnetic Field

direction with respect to the field, i.e. the value of angle 0 remains constant. So it keeps precessing about the field. However, by losing energy the atomic dipole gets aligned with the field. In a solid, the dipole can lose energy in various ways as its energy is transferred to other degrees of freedom and so it gets aligned with the field depending upon the temperature of the solid. To change the orientation of the dipole, the maximum energy required is 2ylB. If y. is about ~m-' and a large field, say, 5T is applied then the potential energy will be of the order of lo-= joules. This is comparable to the thermal energy kTat room temperature. Thus only a small fractio~i of the dipoles will be aligned paralIel to B. In the next section it will be shown, using statistical mechanics, what fraction of dipoles is aligned along B. In the presence of the magnetic field, when the tiny magnetic dipoles present in the material get aligned along aparticulardirection we say that material becomes magnetized or magnetically polarized. The state of magnetic polarization of a material - is described by the vectur quantity called magnetisation, denoted by M. It is defined as the magnetic dipole mornerit per unit volume. It plays a role analogous to the polarization P in electrostatics. In the next section we will also find the expression of magnetisation for paramagnets. But before proceeding do the followil~g SAQ.

SAQ 4

Water has all the electron spins exactly balanced so that their iiet inagnetic lnoinent is zero, but the water molecules still have a tiny magnetic momelit of the hydrogen nuclei. In the magnetic field of l.O.wb m-* protons (in the form of W- nuclei of water) have the precession frequency of 42 MHz. Calculate the g - factor of the proton.

I 11.6 GNETS

In the presence of an external magnetic field, the magnetic moment tends to align along the direction of the magnetic field. But the thermal energy of the n~olecules in a macroscopic piece of magnetic material tends to randomize the direction of nlolecular dipole moments. Therefore, the degree of alignment depends both on the strcngth of the field and on the temperature. Let us derive the degree of alignment of the molecular dipoles, quantitatively, using statistical methods.

, A~~~~~ to B ~ ~ ~ ~ ~ ~ - ~ taw Suppose there are N magnetic molecules per unit volume each of magnetic moment p, the probability offinding at a temperature T. Classically, the magnetic dipole can make any arbitrary angle with molecules in a given state varies exponentially with the

the field direction (Fig. 11.9). In the absence of an external field, the probability that the

negativeof the potential energy dipoles will be between angles 8 and 0 + d0 is proportional to 3 sin 8 d0 , which is the ofthat sate divided b y m . In solid angle dS.2 subtended by this range of angle. This probability leads to a zero tbiscasethecnagy depends average of the dipoles. When a magnetic field k is applied in the r - direction, the upon the angle 9 that the moment makes with the probability beco~nes also proportional to the Boltzmam distribulioi~ e-'"kT. He,, magocric field. SO probability U = - p . B = - p B cost) is the magnetic energy of the dipole when it is making an is proportional to angle 0 with the magnetic field, k is the Boltzmann constant and T is the ahsolute exp(-u( e)/m-3. temperature.

I

r t tB 1

I

Fill.% Cddatiw oftbe pvtmyactk prgmlics dml tc r t l s in erttr0.l rnmpdic 64d. ,

74 .

L

Hence, the number of atoms (or molecules) dN per unit volume for which p makes '' angles between 8 and 8 + d0 with B, is given by

dN = ~ e + @ ~ / ~ ~ s i n 8 d 8 (11.23)

where K is a constant . Calling p B/kT as a, the total number of dipoles per unit volume of the specimen is

Magnetism dhlatrrinls-1

Putting cos8 = x, we have + 1

N 2 x ~ J e ~ " d r

The magnetic dipole, making an angle 8 with B, makes a contribution p cos8 to the intensity of magnetization M of the specimen. Hence, the magnetization of the specimen obtained by summing the contributions of all the dipoles in the unit volume is given by

- 1

where, again, w e b v e put cos 8 - x and p B/kT a. Evaluating the above integral, we obtain

Substituting for b K from the Eq, (11.24), we get

where Ma pN is the saturation magnetization of the specimen when all the dipoles 1

align with the magnetic field. The expression cotha - - is called the Langevin function a

which is denoted by L ( a ).

We now consider two cases : (i) when @ is very large. Tbis would happeq if the kT

temperature were very low andlor B very large. For this case,

(i) is b a d on c h i d akulrtion with no m t r k t h o a ( h e direction d&pale

fi1b-m quantum md.nical ca lda th with rrstrktioo on (he d i rdoo d dipole.

Hence M = M, These would be saturation.

(ii) ~ k n @is small which means that T h large and / or B is small. In this case kT

The complete dependence of M on B is shown in Pig. 11.10. For your comparison, the dependence of M on B based on quantum mechanical calculation is also shown.

W c cumat and SAQ 5

+ 1

Evaluate the integralle " x dr . - 1

SAQ 6

MS a Show that when a Q @/kT is small, M

Let us now sum up what we have learnt in this unit.

e All materials are, in some sense, magnetic and respond to the presence of a magnetic field. Materials can be classified into mainly three groups: diamagnetic, paramagnetic and ferromagnetic. Diamagnetism is displayed by those materials in which the atoms have no permanent magnetic dipole moments. Paramagnetism and fernmagnetism occurs in those materials in which the atoms have permanent magnetic dipoles.

e The orbital motion of the electron is associated with a magnetic moment p, which is proportional to its orbital angular momentum J. We write this as

where e is the charge on electron, m the mass of electron and g is Lande g - fac~or which has a value -1 for orbital case and 2 for spin case.

9 The ratio of the magnetic dipole moment to the angular momenhim is called the gymmagnetic ratio.

9 The magnetic dipoles in the magnetic materials are due to atomic currents of electrons in their orbits and due to their intrinsic spins.

9 Change in the magnitude of the magnetic moment of atoms is responsible for diamagnetism whereas change in the orientatioh of the magnetic moment accounts for paramagnetism.

e Because the magnetic moment is associated with angular momentum, in the presence of a magnetic field, the atom does not simply turn along the magnetic field but precesses around it with a frequency op = g ( e/2m ) B. This is called the Lamor precession.

e When a dirnagnetic atom is placed in an external magnetic field normal to its orbit,

I the field induces a magnetic moment opposing the field itself (Lenz's law) as

I where r and m are tbe radius of tbe orbit and mass of the electron.

1 e When atoms of magnetic moment p are glaczd in a magnetic field B, then the > i Magnetisat ion M is given by

I I M = M,(cotha -l/a)

i I ?

1 I

where a - @and M, - ph'k the saturation maystisatioz whelk 311 t)wc dipoles a= 1

kT 1 I I aligned in the direction of field.

1 I

11.8 TERMINAL QUESTIONS

i 1. A uniformly charged disc having the charge q and radius r is rotating with I I

constant angular velocity of magnitude w. Show that the magnetic dipole moment I 1

, . hastbemsgnitudez(grn r 2 ) I i ? , 76 t ' ; - L

( Hint : Divide the sphere into namw rings of rotating charge; find the current to MalpleUsm d Matcri~h-l which each ring is equivalent, its dipole moment and then integrate over all rings.)

2. Compare the precession frequency and the cyclotron frequency of the proton for the same value of the magnetic field B.

ON§ AND ANSWERS

SAQs

1) a) Potential energy U of the magnetic dipole is given by the relation : U = yr . B. where. p is the dipole moment and B is the magnetic field.

Since U is expressed in Joules and B in Tesla, the above relation gives the unit of magnetic dipole moment as JT- '

nh L I - (because angular momentum of electron is quantized) 2n

where n is an integer.

Hence minimum allowed magnitude of dipole moment is given by putting n = 1, as follows:

or V ~ n = 9.27 x 10- 24 C JS k g

= 9.27 x 1rZ4 J T-'

eh ;. the Bohr magneton is given by - = 9.27 x J T 43vn

2) a) ( i ) g - l ( i i ) g - 2

e b, Eq. (11.5) is p - -S m

hence

so that

A But the spin angular momentum S is - , themfore 2

Eleciric Current and Magnetic Field

which is indeed the value of Planck's constant.

a) Copper is slightly diamagnetic. Bismuth, Silver - diamagnetic, Aluminium & Sodium - paramagnetic

b) No. Since the diamagnetic material is characterised by the absence of intrinsic magnetic dipoles and paramagnetic substances have magnetic dipoles, the alloy of these materials will be the material with intrillsic magnetic dipoles. Such a material will exhibit the property of paramagnetism which masks the diamagnetism of both components of the alloy.

We have the formula

but 2nfp = cop

hence

Now

2.m 2 x 1 8 6 0 x 9 . 1 x 1 0 ~ 3 1 ~ g C ~ ~ For proton, - = e 1.6 x 10''~

Using this above we obtain g :. 5.584, which is the proton g - factor.

x e ax

5-r e 'dx = -j> dr integrated by parts

x e m l e a xe" e m = ---- pi ---

a a a a a2

e a + e - a 6) We have coth a = and also that eO-e -O

Hence

so that

1 + -

coth a = a

Therefore,

1 a coth a - - = - and M = Ms a / 3

a 3

Terminal Questions

4 1) The surface charge density is

nr

The disc can be thought of as made up of number of rings. Let us consider a ring of radius R and width dR.

The charge within this ring is given by

The current carried by this ring is its charge divided by the rotation period:

The magnetic moment contributed by this ring has magnitude

dp = ad1

where a is the area of the ring.

Therefore,

I

Taking into accoullt all the rings (radius varying from 0 to r ), we get the magnitude of the magnetic moment as follows:

2) Precession frequency op of a proton in a magndtic field is given by

P q , (because 7 - - where q is the charge and mp is mass of the proton) h p

ELtctric Currtnt urd Magattic Field

Cyclotroll frequency W L is

UNIT 12 GNETISM OF TE

Structure Introduction Objectives

Ferroinagne tism

Magnetic Field Due to a ~ h ~ i l e t i s e d Material

The Auxiliary Field H (Magnetic Intensity)

Relationship between B and H for Magnetic Material

Magnetic Circuits

Suininary

Terminal Questions

Solutioiis and Answers

2.1 INTRODUCTION

In Block 2 of this course you have studied the behaviour of dielectric materials in response to the extenial electric fields. This was done by investigating their properties in terms of electric dipoles, both iiatural and induced, present in these materials and their lining up in the electric field. The inacroscopic properties of these materials were studied usiiig the so-called polarization vector P, the electric dipole inomelit per unit volume. The magnetic properties of materials lias a similar kind of explanation, albeit in a more complicated form, due to the absence of free magnetic inonopoles. The magnetic dipoles in these materials are understood in terms of the so-called Arnperian current loops, first-introduced by Ampere.

All materials are, in soine sense, magnetic and exhibit lnagnetic properties of different kinds and of varying intensities. As you know, all materials, can be divided into three main categories: (i) Diamagnetic; (ii) Paramagnetic and (3) Ferromagnetic materials. In this unit, we s'hall study the inacroscopic behaviour of these materials.

We understood the macroscopic properties of the dielectric materials using the fact that the atoms and molecules of these substances contain electrons, which are mobile and are responsible for the electric dipoles, natural and induced, in these substances. The polarisation of these substances is the gross effect of the alignment of these dipoles. Similarly we describe the magnetic properties of various materials in terms of the magnetic dipoles in these materials.

In Unit 11, we have already explained diamagnetism and paramagnetism in terms of magnetic dipoles. In this unit, first, we will mention the origin of ferromagnetism. Later, we will develop a description of the macroscopic properties of magnetic material.

With Unit 12, we end our study of magnetism. In the next Block we will deal with the situation where both electric and magnetic fields will vary with time. This will'lead,

, ultinlately, to the 'four differential equations known as Maxwell's equations.

0 bj ec t ives After studying this unit you should be able to :

understand and explain Lhe tenns: ferromagnetism, amperiail current, magnetisation, magnetic intensity H, magiletic susceptibility, lnagnetic permeability, relative permeability,

relate magnetisatioll M (which is experimentally measureable) and the atomic currents (which is iiot measureable) within the material,

0 derive and understand the differential and integral equations for M and and apply these to calculate fields for simple situations,

Electric Cumat and Magnetic Field

UNIT 12 GNETISM OF

Structure 12.1 Introduction

Objectives

12.3 Magnetic Field Due to a ~kgnetised Material

12.4 The Auxiliary Field H (Magnetic Intensity)

12.5 Relationship between B and H for MagneticMaterial

12.6 Magnetic Circuits

12.8 Terminal Questions

12.9 Solutions and Answers

12.1 INTRODUCTION

In Block 2 of this course you have studied the behaviour of dielectric materials in response to the external electric fields. This was done by investigating their properties in terms of electric dipoles, both ~latural and induced, present in these materials and their lining up in the electric field. The macroscopic properties of these materials were studied using the so-called polarization vector P, the electric dipole moment per unit volume.

The magnetic properties of materials has a similar kind of explanation, albeit in a more complicated form, due to the absence of free magnetic monopoles. The magnetic dipoles in these materials are understood in term of the so-called Amperian current loops, first*introduced by Ampere.

All materials are, in some selae, magnetic and exhibit magnetic properties of different kinds and of varying intensities. As you know, all materials, can be divided into three main categories: (i) Diamagnetic; (ii) Paramagnetic and (3) Ferromagnetic materials. In this unit, we shall study the macroscopic behaviour of these materials.

We understood the macroscopic properties of the dielectric materials using the fact that the atoms and ~nolecules of these substances contain electrons, which are mobile and are responsible for the electric dipoles, natural and induced, in these substances. The polarisation of these substances is the gross effect of the alignment of these dipoles. Similarly we describe the magnetic properties of various materials in terms of the magnetic dipoles in these ~naterials.

In Unit 11, we have already explained diamagnetism and para~nagnetism in terms of magnetic dipoles. In this unit, first, we will mention the origin of ferromagnetism. Later, we will develop a description of the macroscopic properties of magnetic material.

With Unit 12, we end our study of magnetism, In the next Block we will deal with the situation where both electric and magnetic fields will vary with time. This will lead,

, ultimately, to the 'four differential equations known as Maxwell's equations.

Objectives

After studying this unit you should be able to :

0 understand and explain the terms: ferromnagnelism, amperian current, magnetisation, magnetic intensity HI magnetic susceptibility, magnetic permeability, relative permeability,

0 relate lnagnetisatio~l M (which is experimet~tally measureable) and the atomic currents (which is 11ot measureable) within the material,

0 derive and utldersta~ld the differential and integral equations for M and H and apply these to calculate fields for simple situatioils,

Flectk C u m t and Magnetic Field

Fig. 12.18 Domain

Cansider rwo electrons on atoms that are close to each other. If the electron spins are parallel, they stay away from each ot l~er due to Pauli principle, therehy reducing their caulomh energy of repulsion. On theother hand, i f these spins are anti-parallel. the electrons can come close lo eachother and their wulonlb energy is higher. 'Thus, by making their spins parallel, Ihe electrons can reduce their energy.

Fig. 122 : l l ~ e domains in an unmagnctised bar of iron. m e m w s show the alignmen1 dirtclion or the rnagnecc moment in each domain.

e interreIate B, H, M, w, p a i~d X,

* relate B & H for various magnetic a l~d non-magnetic materials,

* derive all equatior~ in allalogy with Ohtn's law for ;I ~nagnetic circuit.

Ferrolnaglietic materials are those materials, which respoi~d very slrongly to the presence of magnetic fields. In such ~natcrials, the lnaglletic dipole monlenl of the atoms arises due to the spins of ullpaired eleclrolls. These lelld lo liue up parallel lo each other. Such a line-up does not occur over the whole material, but it occurs over a srnall volume, known as 'domain'., as shown ill Fig. 19.1. However, these volulnes are large compared to the atomic or lnolecular dimensions. Such line-ups take place even in Ihe absence of an external magnetic field. You must be wollderil~g about the nature of forces that cause the spin lnagnetic nlonlellts of different atoms to lille up parallel to each other. This can be explailled o111y by using quantum mechal~ical idea of "exchar~ge forces". We will not go into the details of exchallge forces. About this, you will study ill other courses of physics, but we are giving you sollle idea of exchange forces in the margin remark.

In an unmagnctized fernmagnetic material, the lnagnetic momc~lts of different domails are rai~doinly oriented, and the resulting inagnetic,moment of the ~naterial, as a whole is zero, as shown in Fig. 12.9. However, in the presence of an extcrllal ~llagnelic field, tile magnetic moments of the domairls line-up in such a manner a s to give n net magnetic mornent to the material in the direction of the field. The mechanism by which this happens is thatJhe domains with the magnetic ~nornents in the favoured directions increase in size at the expense ofthe other domains, as show11 in Fig. 12.3a.

Fi.123: In a ferromagnetic ninlerial domniu changes, resulting in a net magnetic momenl, occur through (e) domain growth and (b) domnin realignment,

In addition, the magnetic momel~ts of the entire domaills can rotate, as shown in Fig. 12.3b. The material is thus magnetised. If, after this, lhe extenla1 lnagl~ctic field is reduced'to zero, there still remains a considerable amount of lnaglleti7aIio11 in the material. The material gets permanently magnetized. The behaviour ol I'erromagnelic materials, under the action of changing magnetic fields, is quite tolnplicaled and exhibits the phenomenon, of hysteresis which literally means 'lagging behind'. You will study more about this in Sec. 12.5. Above a certain temperature, called 'Curie Temperature', because the forces of thermal agitation domillate 'exchange' forces, the do~naills losc their dipole ~noments. The ferromagnetic material begills to behave like a paramagnetic ~naterial. When cooled, it recovers its ferromagnetic properties.

Finally, we briefly mention two other types of nagn net ism wllich are closely related to ferromagnetisrn. These are anti-ferromagnetism and ferr imagnetism ( also called ferrites). In this course, we will not study the physics of antiferro- and ferrimagnetism. The main reason for mentioning these materials is that they are of techi~ological imporlance, being used in magnetic recording tapes, antenna and in colnputer memory. In antifemmagnetic substances, the 'exchange' forces, as we menlioned earlier, play the role of setting the adjacent atoms into antiparallel alignmen1 of their equal ~nagnetic

moments, that is, adjacent magnetic moments are set in opposite directions, as shown in Fig. 12.4 a.

Magnetism of Makrials-I1

Such substances exhibit little or no evidence of magnetism present in the body. However, i f these substances are heated above the temperature known as Neel temperature, the exchange force ceases to act and the substance behaves like any other paramagnetic material.

In ferrimagnetic substances, known generally as femtes, the excha~ige coupling locks the magnetic moments of the atoms i n the material into a pattern, as show11 in Fig. 12.4b. The external effects of such an alignment is intcnncdiate betwee11 ferromagnetisin and a~ltiferromagnetism. Again, here the exchange coupliiig disappers above a certain temperature.

Thus, we find that the magnetization of tlie materials is due to permanent (and induced) magnetic dipoles in these ~nalerials. The magnetic dipole moments in these materials are due to the circulalillg electric currents, k~iowii as ampcrian currcnts at the atomic and molecular levels. You arc expccted lo understand the correct relationship between magnetization in a material and the amperian currents, togetherwith the basic difference (and somelimes similarities) between tlie behaviour of the magnetic materials in ~nagnelic fields, and dielectrics (and conductors) in electric fields.

Though physics of paramagn~tic and ferrolnagnetic materials have analogues in the electric case, dian~agnetism is peculiar to ~naglietism. The student is advised to read the matter in this unit and find the analogies and appreciate the differences, i f any, by referring back to the units on dielectrics. In the next section, we will find out the relationship between the macroscropic quantity M, which is experilnentally measurable and the atomic currents (a microscopic quantity) within the material which is not measurable. With the help of this relationship, we can find out the magnetic field that lnagnetised matter itself produces.

12.3 MAGNETIC FIELD DUE TO A MAGNETISED MATERIAL

In Unit 5, we have.described the macroscopic propeflies of dielectric materials in tcrnls of the polarization vector I', the origin of which is in the dipole lnolnents of its natural or iliduced electric dipoles. We shall adopt a similar procedure in the study of magnetic materials. You would be tempted to say that we should carry over all the equatio~ls in the study of diclectrics to magnelic materials. One way of doing this would be to replace the electric field vcctor E by D, then replace P by an allalogous quantily which we,shall call lnagnetizatio~i vector M which is the lnagiietic dipole moment per unit volume. Further, we replace the polarization charge density p, by magnetic 'charge' density p, whatever that means, by writing V . M - - p, just as we had V . I' = p,. In fact, people did something Iike this, and they believed that magnetic charges or monopoles exist. They have built a whole theory of electromagrielis~n on this assumption. However, we know that magnetic 'charges' or monopoles have not yet been detected in any experiment so far, despite a long search for them. Now, we know that the magrletization of matter is due lo circulating currenls within the atoms of the materials, This was origirially suggested by Ampere, and we call these circulating currents as 'amperian' current loops. These currents arise due to either the orbilal motion of electrons in the atolns or their spila. These currents, obviously, do not involve large scale charge trailsport in the magnetic materials as in the case of conduction currents. These currents are also known as mag~ietization currents, and we shall relate these currents to the lnagnetizatioli vector M. Let us consider a slab of uniformly magnetised material, as shown in Fig. 12.5a. It contains a large number of atomic magnetic dipoles (evenly distributed throughout its volume) all pointing in the same direction. If y is the magnetic moment of each dipole then the magnetisation M will be the product of y and the number of oriented dipoles per unit volume. You know that the dipoles can be indicated by tivy current loops. Suppose the slab collsists of many tiny loops, as shown in Fig. 12.5b. Let us consider any tiny loop of area a, as shown in Fig. 12.5~. In terms of magnetisation M , the magnitude of dipole moment p is written as follows:

p = Madz (12.1)

where dz is the thickness of the slab.

Fig. 124 :Relative onentalion ul' electron spins in (n) antiferromagetic niatcrinl and (b) ferrite.

Electric C u m t lad Magnetic Field

(dl Fig.125: (a) A thin slab duni lomly magnetized material, with Lhe dipoles indicated by (h) and (c) tiny

current loops is cquivdent to (d) n ribbon ofcurrent I flowing around the boundnry.

If the tiny loop has a circulating current I, the11 dipole lnoinent of the tiny loop is given by

y l a (1 2.2)

Equating (12.1) and (12.2) we get

I M = - or I 5 Mdz.

dz

Here we have assumed that the current loops correspondi~ig to magnetic dipoles are 1 large enoughso that magnetisation does not vary appreciably rrom one loop to the next, so Eq. 12.3 shows that the current is the same in all current loops o l Fig. 12.5b. Notice

I

that within the slab, currents flowing in the various loops cancel, because everytime if I there is one going in one particular direction, then a co~i t i~~uous one is going io the exactly opposite direction. At the boundary of the slab, there is no adjacent loop to do

I

the cancelling. Hence the whole thing is equivalent to the single loop of cur redI I

flowing around the boundary, as shown in Fig. 12.5d. Therefore, the thin slab of lnagnetised inaterial is equivalent to a single loop carrying the current Mdz . Hence, the

I magnetic field at any point extenla1 to the slab, is the same as !hat of the currelit Mdz . ! I In case there is non-uniform magnetization in the material, the atr_l!nic currents iu the

I I (arnperian) circulating current loops do not have the same inagtlitude at all points inside

the material and, obviously, they do not callcel each other out illside such a material. I Still we will find that magnetised matter is equivalent to a curre~lt distributio~i I

, J = curl M. Let us see how we have anived at this relation.

In the non-uniformly magnetised material consider two little blocks of the volurne 'Ax A y A z, cubical in shape adjacent t~ each other along y-axis (see Fig. 12.6a). Let us call these blocks '1' and '2' respectively. Let the z-compo~ienl d l M i n these blocks be M z ( y ) and Mz ( y + Ay ) respectively.

I ;

. I / 84 i >

, L

~ e t the ampenan currents circulating round the block '1' be I ( 1 ) and round the block '2' be 1 ( 2 ).Using Eq. (12.3) and refemng to Fig. 12.6a we write,

I x ( 1 ) = Mz(y )Az

and 1,(2) = M, (y+Ay )Az

Fi.116: Two adjacent chuncksolnmgnetised malerial, with a larger arrow on the w e to the right in (a) and above in (b), suggesting greater magnetisation rt that phi On the surface where they join them i s r net cumn t in the x-direction

At the interface of the two blocks, there will be two contributions to the total current: I( 1 ) flowing in the negativex-direction, produced due to block 1, and1 ( 2 ) flowing in the positivex-direction produced due to Block 2. The total current in the positive x-direction is the sum :

Eq. (12.4) gives the net rnagnetii.ation current in the material at a point in the x-direction in t e r n of the z-component of M. The current per unit area, i.e., current density Jm flowing in thex-direction is give11 as follows :

where Ay & is the area of cross-section of one such block for the current A Ix. Hence

In these equations, we have put suffixes x to the currents to indicate that, at the interface of the blocks, the current is along the x-axis.

There is another way of obtaining the current flowing inx-direction by considering these two tiny blocks, one above the other, along the z-axis, as shown in Fig. 12.6b. We obtain the relation as

By superimposition of these two situations, we get

I f the magnetisation in the first block is M ( x. y. : ). the magnetisation in the second block is

The tcomponent of magnetisation ofthe first block in termsoil. ( 1 ) is wrillen as

M,Az - I , ( l )

Similarly, the:-component o l magnet isation oithe second block neglecting high -order ternis which vanish in the limit where cach'block becomes very sn~all. isgiven by

Magnetism oCMateri&-I1

Eq. (12.7) is obviously the x-component of a vector equation relating Jm and the curl of M. Combining this withy and r components, we obtain

Eq. (12.8) is a more general expression representing the relationship between the magnetisation and the equivalent current. We see from Eq. (12.8) that inside a . uniformly magnetized material in which case M = constant; we have Jm = 0. This is true. See Eq. (12.8), the current is only at the surface of the material where the

Electric C u m t nnd Magnetic Field

magnetization has a discontinuity (dmpi'ng from a finite M to zero). Inside a non-uniformly magnetized material, J, is nonzero.

We shall see in the next section that J,, which is introduced to explain the origin of magnetisation in a material, is made to make its exit from the equation, and only the conductioncurrent density indicating the actual charge transport and which is experimentally measurable remains.

12.4 THE AUXILIARY FIELD H GNETIC INTENSITY) '

So far we have been considering that magnetisation is due to current associated with atomic magnetic moments and spin of the electron. Such currents are known as bound currents or magnetisation amperian current. The current density J, in Eq. (12.8) is the bound current set up within the material. Suppose you have a piece of magnetised material. What field does this object produce? The answer is that the field produced by this object is just the field produced by the bound currents established in it. Suppose we wind a coil around this magnetic material and send through this coil a certain current I. Then the field produced will be the sumof the field due to bound currents and the field due to current I. The current I i s known as the free current because I t is flowing through the coil and we can measure it by connecting an ammeter in series with the coil. (In case the magnetic material happens to be conductor, the free current will be the current flowing through the material itself.) Remember that free curre~~ts are those caused by external voltage sources, while the internal currents arise due to the motion of the electrons in the atoms. The current is free, because someone has plugged a wire into a battery and it can be started and stopped with a switch. Therefore, the total current density J can be written as

where Jfrepresents the free current density.

Let us use Ampere's law to find the field. In differential form, it is written as (See Unit 9)

V x B = W J (9.46)

Using Eq. (12.9), Ampere's law would then take the form as follows:

As mentioned earlier, we have no way to measure J, experimentally, but we have a way to express it in terms of a measurable quantity, the magnetization vector M through the Eq. (12.8). We then have

! I

Eq. (12.10) is the differential equation for the field in terms of its source Jfi

the free current density. This vector is given a new symbol H, i.e.,

The vector H is called the magnetic 'intensity' vector, a name that rightly belongs to B, f but, for historical reasons, has been given to H. Using Eq. (12.11). Eq. (12.10) becomes

V x H = 31 (12.12)

In other words, H is related to the free cumnt in the way B is related to the total current, bound plus free. This surely has made you think over the purpose of introducing the new vector field A. For practical reasons the vector H is very useful as

I

! it can be calculated from the knowledge of external current only; whereas B is related to the total current which is not known. Eq. (1312) can also be written in the integral form as

I 86 ' ' I

Magnetism oCMnttrinls-11 (11.13)

where Ijis the conduction current through the surface bounded by the path of the line integral on the left. Here the line integral of H is around the closed path which may or may not pass through the material.This equation can be used to calculate W, even in the presence of the magnetic material.

SAQ I

Fig. 12.7 shows a piece of imn wound by a coil canying a current of 5A. Find the value of JH . dl around the path (I), (2) and (3). Also state for which path(s) B = B and B ;. N.

Fmm Eq. (12.3), we see that the units in which M is measured is amperes per meter. Eq. (12.11) shows that the vector W has the units asM, hence W is also measured in amperes per metre. The electrical engineers working with electromagnets, transformers, etc., call the unit o f H as ampere turns per metre. Since 'turns', which is supposed to imply the number of turns in the coil canying a current, is dimensionless, it need not confuse you.

Magnetic properties of substance Fig. 127 : SAQl

In paramagnetic and diamagnetic materials, the magnetisation is maintained by the field. When the field is removed, M disappears. In fact, it is found that M is .

proportional to B, provided that the field is not too strong. Thus

It is conventional to express Eq. (12.14) in terms of H instead of B. Thus we have

M = x m H (12.15)

The constant of proportionality -A, is called the magnetic susceptibility of the material. It is a dimensionless quantity, which varies fmrn one substance to another. We can cl~aracterise the magnetic properties of a substance by %.It is negative for diamagnetic substances and positive for paramagnetic materials. Its magnitude is very small compared io unity, that is ( X, I < < 1. For vacuum X, is zero, sirice M can only exist in. magnetised matter. We give below a short table giving the values of X, for diamagnetic and paramagnetic substances at room temperature.

Paramagnetic

Paramagnetic

Paramagnetic

Paramagnetic

Diamagnetic

Diamagnetic

Diamagnetic

Diamagnetic

Aluminium

Sodium

Tungsten

Oxygen

Bismuth

Copper

Silver

Gold

We have not given a table for the susceptibilities of fernmagnetic subs&nces as f i, depends not only on H but also on the previous mangetic history of the material. Using Eq. (12.1 1) in the fonn

B = w ( W + M )

we have

Electric Cumnt and Mmgnttic Field . . (12.17)

where c~ = w K , = clo( I + % )

p is called the permeability of the medium and Km is called the 'relative' permeability. We see that p has the same dimensions as po and Km is dimensionless. In vaccum xm = 0 and p = po. Relative permeability Km differs from unity by a very small amount as Km = ( 1 + X, ). K,,, for para- and ferromagnetic materials are greater than unity and for diamagnetic material it is less than unity.

The magnetic properties of a material are completely specified if any one of the three quantities, magnetic susceptibility, )(F, relative permeability K, or permeability p is known.

Example 1

A toroid of aluminium of, length lm, is closely wound by 100 turns of wire canying a steady current of 1 A. The magnetic field B in the toroid is found to be 1.2567 x wbm-'. Find (i) H, (ii) %, and Km (iii) M in the toroid and (iv) equivalent surface magnetization arrent I,.

Solution

i) According to Eq. (12.1 3)

To evaluate H produced by the current, we consider a circular integration path along the toroid. H is constant everywhere along this path of length lm. The number of cumnt turn threading this integration path is 100 x 1A . Since H is everywhere parallel to the circular integration path, we get

H x l m = 100xlA

(ii) From Eq. (12.16)

(iii) From Eq. (1 2.15)

M = x m H

(iv) I , = ML

In this solution, we have assumed B, H and M to be uniform over the cross-section of the toroid and along the axis of the toroid.

Try to do the following SAQ.

Magnetism of Mr~nls-11 SAQ 2 ~n air-core solenoid wound with 20 turns per centimetre canies a current of 0.18 A. Find N and B at the center of the solenoid. If an iron core of absolute permeability 6 x lom3 H m-I is inserted in the solenoid, find the value of H and B? ( w = 4xx 1 o d 7 ~ m - l )

12.5 RELATIONSHIP BETWEEN B AND H FOR GNETIC

The specific dependence of M on B will be taken up in this section. The relationship between M and B or equivalently a relationship between B and H depend on the nature of the magnetic material, and are usually obtained from experiment. A convenient experimental arrangement is a toroid with any magnetic material in its interior. Around the tomid, two coils (primary and secondary) are wound, as shown in Fig. 12.8.

Fh.128 i Anangcmcnt for invatiptlng Lbe rebtion between B and M, or I3 and H, in r rnagoetic rnalrria~'

If we consider the radius of the cross-section of the toroidal windings to be small in comparison with the radius of tbe toroid itself, the magnetic field within the toroid can be considered to be approximately uniform. A current passing through the primary coil establishes H. The establishment of the current in the primary coil induces an electromotive force (emf). By measuring the induced voltage, we can determine changes in flux and hence, in B inside the magnetic material. If we take H as the independent variable, and if we keep the track of the changes in B starting from B - 0, we can always know what B is for a particular value of H. In this way, we can obtain a B-H curve for different types of magnetic material. The experiment described above can be carried out for diamagnetic and paramagnetic materials by commencing with I = 0 and slowly increasing the value of I to obtain a series of values of B and H . A plot of B against H for these substances is shown in the Fig. 12.9(a). We see that the graph is a straight line as expected from the relation

B = w ( l + X r n ) H (12.16)

hl291 latema1 mmp& hld (B) versus applied magnetic &Id (H) for diBemml lypa of rnqnetlc rnatrrtk (a) In d l rnnpd ic and prunrgnclic mrlrrhb, Lbc trhhwhlp Is bear. @) In fermmgaetic mrlnlab, Lbt dolrt imdp depends m Ihc s l r c ~ q t h d t b c r ppW lkkl and OD h e p u t history of (be materid, In (b), lbt Lkkl rtrmgthr .low tbt v d a l axis u e much gruLcr lbrn along Lbe hmimn(.l .xis, Arrows Indlutc Lhc d W o n in wbieb Lbc Oclbr ut changed

Electric Currenlmd Magnetic Field

where pa and -h are conslants. The slope of the graph is given by ( 1 + -h ) from which call be determined using the following relation:

For diamagnetic substances, slope < making X, < 0. For paranlag~letic materials slope > po so that y, > 0.

I f in the experiment givenabove we use ferromagnetic materials like iron, we obtain a lypical B - Hcurve as showri in Fig. 13.9(b).

i) At I = 0 , i.e., when H = 0, B is zero. When I is increased, B and H are detennined for increasing values of 1 . At first, B increases with H along the curve 'a'. At some high value ofH, the curve (shown by the dashed line in the figure) becomes linear, indicatillg that M ceases to increase, as the material has reached saturation with all the domain dipole molilents in the saliie direction.

ii) If, afterreaching saturation, we decrease the current in the coil to bring H back to zero, the B-Hcurve falls along the curve 'b ' . When H reaches zero, there is still some B left implying that even when/ = 0, there is still some magnetizatio~~ or M left in the specimen. The li~aterial is pennaiieiitly magnetized. The value of B for H = 0 is called remanence.

iii) If the current is reversed in the primary coil and made to increase its value, the B-H curve rulls along the curve 'b ' until B becomes zero at a certain value of H. This value o fH is called the coercive force. If we co~~tinue to increase the value of the current i ~ i the negative direction, the curve col~tinues along '6 ' until the saturatioli is reached again.

iv) The current is now decreased until it becomes zero once again. This corresponds to H = 0 , bul B is not zero and has magnetization in the opposite direction. Here we reverse the current again, so that the current in the coil is once Illore along the positive direction. With the increasing current in this direction, the curve continues along the curve 'c ' to meet the curve 'b ' at saturation.

If we alternate the current between large positive and negative values, the B-Hcurve goes back and forth along ' b ' and ' c ' in a cycle. This cycle curve is called hysteresis curve. It shows that B is 1101 a single valued function of H , but depends on the previous treatment of the material.

The shape of the hysteresis loop varies very widely from one substance to another. Those substances, like steel, alnico, etc., from which pennanent magnets are made, have a very wide hysteresis loop with a large value of the coercive force (see Fig. 12.10). However, those substances, like soft iron, permalloy, etc., from which electromagnets (temporary magnet) are made, should have large remaliance but very small coercive force. Those ferromagnetic materials, which are used in the cores o f transformers, like iron-silicon (0.8-4.8%) alloys, have very liarnjw hysteresis loop.

Fig.1210: h r hysteresis curves for a few materials. ~ w e s ' ( a ) aod @) are respectively for specimen of a d ' iron and steel mnteriels,

12.6 MAGNETIC CIRCUITS

A magnetic circuit is the closed path take11 by the lnaglietic flux set up i11 an electric machine or apparatus by a magnetising force. (The maglietising force may be due to a currelit coil or a pennanent magnet.)

'90

_-ei

In order to study the resernblarice between a ~iiagnetic circuit and an electric circuit, we will develop a relation corresponding to Ohii's law for a magnetic circuit. Let us consider the case of an iron rillg (Fig. 11.1 1) ~ilagl~etised by a current flowing through a coil wound closely over it. Suppose,

I = current flowing in the coil

N = number of tunls in the coil

I = length of the magnetic circuit (mean circulnference of the ring)

A = area of cross-section of the ring

p - pernleability of iron.

In this case, all the magnetic flux produced is confined to the iron ring with very little leakage (we shall see the reason for this later). We have seen earlier that H illside the

'I4 Fig. 12.1 1 : Magnetic circuit

J H . ~ = NI (from Ampere's law) I

where the path of integration is along the axis of the ring. As the line integral of electric field E over a circuital path is the electromotive force (e.m.f), by analogy, the line integral of H is termed.as magnetomotive force (M.M.F.)

... M.M.F. = JH . d l = NI.

At every point along fhis path in the ring, we write

Further if Q is the magnetic flux given by @ = BA, then H - a/@, hence

where we have taken @ outside the integral as it is constant at all cross-sections of the ring. Eq. (12.19) rerni~~ds us of a similar equation for an electric circuit colltaining a source of E.M.F., namely, .

e m f . - current x resistance = IJ

The Eqs. (12.19) and (12.20) suggest that:

i) The magnetomotive force ( J H . ~ ) is analogous wit11 e.1n.f. (JE . ti^ ).

ii) The magnetic flux is analogous with current I in Ohm's law.

iii) The magnetic ~s is ta l lce known as reluctance is analogous with electric

resistance (J y) . .

:. M.M.F. = flux x relucta~ice

- . - - M.M.F. NI

Electric Current and Magnetic Field

If we take p to be constant throughout the ring then

where L is the letigth o f the ring. However, we must recogrlise the significant difference betweeti all electric circuit and a inag~letic circuit:

i ) Energy is colltinuously being dissipated in the resistallce of the electric circuit, whereas no energy is lost in the relucta~lce of the ~nagnetic circuit.

ii) The electric current is a true flow of the eleclro~ls bul there is no flow of such paflicle in a magnetic flux.

iii) At a given teinperature, the resistivity p is iildepeildcilt of current, while the 1

correspondillg quantity - i n reluctailce varies with inag~ietic flux a. CI

Reluctances in Series : Let us assume that the toroid is made of no re thaii one ferromag~ietic ~naterial, each of which is of the same cross-sectional area A, but with different pern~eabilities pi, p ~ , . . . . .

Fig.12.12 : (a) A magnetic circuit conlpmed of several ~naterials: Reluctances in series. (I)) 1 G e t i c circuii consisting of two loops: Reluctances in parallel.

Then, (see Fig. 12.12) as before, we have

NI = J H A

where the itltegnls on the right are taken over axial paths in the materials 1,2, .... There fore,

so that the total relucta~lce of the given magnetic circuit is given by % = 81 +&+ . . . . . (1 1.23)

Reluctances in Parallel : We shall next illustrate the case of a magnetic circuit i11 which the reluctances are in parallel.Fig. 12.12b shows such a magnetic circuit. The current canying coils have N tun6 each, canying a current [amperes. The magnetic flux threading the coil splits into two paths with fluxes @I and 4): as shorv~i in the figure. dhviously,

= + 02. We assume that the area of cross-sectionA is coilsta~it everywkre in the circuit.

I Let the lengths of the paths ABCD, DA and DEFA shown in the figure be L, L1, L2 I I respectively. For the pathABCDA , we have

~irnilarly for the closed path ADEFA , we haye

Notice that we have used CLI and y: for the paths AD and DEFA, As a ' s being different for these paths, H s would be different. This inakes ps different in these paths. Using @ = a 1 + 0. and Eq. (12.25), we write

\ /

Substituting the value of @I from the above equation in the Eq. (12.24), we have

This shows that the relu'ctaiices of th6 paths DA alld DEFA are in parallel as the magnetic flux 4) splits into $1 and $2 along these paths respectively. The combined reluctance 'Si of these paths is given, in tenns of the reluctances ?Ill and 8: of these paths, as follows

Notice that the Eq. (12.24), (12.25) and = 01 + are the statements of Kirchoff's laws for the inagnetic circuits. Now we see why the magnetic flux does not leak through the air. Air forms a parallel path for the flux, for air, p = po and for a ferromagiletic material p L. lo4 pc, hence the aiipath is a very high reluctaiice path compared to that through the ferromagnetic material. The ~nagiietic flux will follow the path of least reluctance, a situation similar to that in the electric circuit.

The magnhtic circuit fonnulae are used by the electrical engineers in calculations relating electromagnets, motors and dynamos. The problem is usually to find the number of tunsand the current in the winding of a coil, which is required to produce a certain flux density in the air gap of a11 electromagnet. Knowing the reluctance of the circuit, M.M,F. is calculated froin the relation:

M.M.F. = flux x reluctance

Since M.M.F. is also NI (see Eq. (13.19)), the magnitude of aippere turns can be calculated. Let us illustrate it by studying the magnetic circuit of an electromagnet.

Magnetic Circuit of an Electromagnet

The magnetic circuit o f an electromagnet coi~sists of the yoke which fo& the base of the magnet, the limbs on which the coil is wound, the pole pieces and the air gap. See Fig. 1113. Let 11 be the effective length and a the area of cross-section of the yoke. If

11 1 1 is the permeability o f its iron, then -is the teluctance of the yoke. Similarly the

I11 a1 1: . 13 reluctance of each limb is -and the reluctance of each pole piece is - while w a? p3 03'

14 the reluctance of the air gap is -(because pair I \ ~ ) . Hence the total reluctance ~f ~b a4

Ydro

Fig. 12.13 : Mqpclic circuit d an dcclromagnd

the magnetic circuit is

11 2 2 a3 14 - L A - + -

If the magnetic circuit carries one and the same flux cP across all its parts, then according to Eq. (12.19), the number of ampere turns is:

Let us take another example of calculating the magnetic field B in the air gap of a toroid of Fig. 12.14. Here the toroid is of a ferromagnetic materiaL (soft iron) with a small air gap OF width 'd ' which is small compared to the length L of the toroid. For this case, we have

NI = @ [v + --$I O being the flux through this magnetic circuit.

B - - [ u ( L - d ) + l u l ] w

NI ,PI B = (I 2.29)

or c l d , + ( v - w ) d

I

This is the value of the magnetic field in the air gap. Read the following example which Air @ shows bow the air gap effectively increases the length of the toroid.

I& U L I ~ ~ q p a e t k w ~ &. l rw~ Example 2

Compare the examples of a complete toroid of length L wound with a coil of N turn each carrying a current I amperes and of a toroid of lenth( L - d ) with an air gap of

I length d ( d << L ) . Show that the air gap effectively increases the length of the toroid I by ( Km - 1 ) d, where Km is relative permeability.

Solution

In the case of a complete toroid without the air gap, we have B = NI/(:). In the

event of an air gap of length d, we have fmm Eq. (12.29) :

B - NI ,Po Pd+(P--lJ4l)d

Dividing both the numerator and the denominator by ppo, we get

NI , .. NI - 1 - so that

- [ ( L - d ) + E d ] CL w

B = NI

1 - [ L + ( K m - l ) d ] P'

I If we compaE this formula with that for the complete toroid, we see that L is effectively I increased by ( K,,, - 1 ) d. 1

i I

I Before ending this unit solve the following SAQ. I >

I I

I SAQ 3 ' I A soft iron ring with a 1.0 cm air gap is wound with a coil of 500 turns and carrying a

I I cumnt of 2 A. The mean length of iron ring is 50 cm, its cross-section is 6 cq2, its

, permeability is 2500 pa Calculate the magnetic induction in the air gap. Eind also B and H in the iron ring. I

' I , , I

i j 94

i 1 C

Let us now sun1 up what we have learnt in this unit.

The behaviour of the ferromagnetic materials is complicated on account of the permanent magnetization and the phenomenon of hysteresis. This behaviour is explained by the presence of the domains in these materials. In each domain the dipole moments are locked to remain parallel due to 'exchange' force. However, in the unmagnetised state, the magnetisation directions of different domains are random, resulting in a zero net magnetisation. There also exist two other kinds of magnetic materials : antiferromagnetic and femmagnetic.

For non-uniform magnetisation, magnetised matter is equivalent to a current distribution J = curl M, where M is magnetisation or magnetic moment per unit volume.

The magnetic field produced by the magnetised material is obtained by Ampere's law as follows:

V x B - J f + J ,

where J f is the h e current density which flows through the material and J , is the bound current density which is associated with magnetisation. This gives

B when B ( - - M ) is a new field vector.

e For paramagnetisin and diamagnetism B, M aiid H are linearly related to each , other except for ferromagnetic materials which exhibit hysteresis, a non-linear

behaviour.

e The study of the electromagnets, motors and dynamos involves the problem of current carrying coils containing ferromagnetic materials, i.e., it involves the study of magnetic circuits. We speak of the magnetic circuits when all the magnetic flux present is confined to a rather well-defined path or paths.

e Magnetic circuit formula is:

magnetomotive force ( M.M.F. ). = flux x reluctance

e M.M.F. is also equal to NI where N is the number of turns of the coil wound over . the magnetic material and I the current flowing through each coil.

1 e Reluctance 3 = -

Cur where I, a and p are the length, area of cross-section and permeability of the material. Additions of reluctances obey the same rules as additions of resistances.

I

12.8, TEWINAL QUESTIONS

1) Find the magnetizing field H and the magnetic flux density B at (a) a point of 105 mm from a long straight wire carrying a current of 15 A and (b) the center of a 2000-turn solenoid which is 0.24m long and bears a currefit of 1.6A. ( - 4n x lom7 H m" )

2) A toroid of mean circumference 0.5 m has 500 turns: each carrying a current'of 0.15 A. (a) Find H and B if the tordid has an air core. (b) Find B and the

'

magnetization M if the core is filled with imn of relative permeability 5000.

3) A toroid with 1500 turns is wound on an iron ring 360 mm2 in cross-sectional area, of 0.75-m mean circumference and of 1500 relative permeability. If the windings carry 0.24A, find (a) the magnetizing field H (b) the m.m.f,, (c) the magnetic induction B, (d) the magnetic flux, and (e) the refuctance of the circuit.

12.9 SOLUTIONS AND ANSWERS -- -

SAQs 1) Path (1) encloses I = 5A

:. IH. d - I = 5A

For Path (2), $H .dl - 71 = 35A

ForPath(3), J H . ~ = 2f = 1OA

B = H forPath(1)

B t H for Paths (2) a~ld (3) because these paths pass through iron.

2) H E n l - (2000m-' ) (0.18~) - 360~rn- '

If an iron core of absolute permeability 6 x H m-' is inserted in the solenoid then H remains unchanged i.e.,

H = 360 Am-' ( unchanged )

andB = pH = ( 6 x lom3 ~ m - l ) . ( 3 6 0 ~ m - ' ) - 2.16T

3) The expression for the magnetic induction in the air gap is

B - N I P [ L + ( K m - l ) d ]

Substituting the values, given in question, we get

B in the iron ring has tbe same value as in air, but H in iron is given by

Tenninal Questions

B w I 1 I 1) i (a) H I - - - x - - -I 15A - 22.7 A me' clo 2m 1.l0 2nr (b) (O. l05rn)

2) For a toroid H = nl, a d we & B - (&/ lo7) ( K J I ) - pH. Thus

(b) B = 5000 ( 0.188 mT) = 0.94T

Using B/w = H + M

d) O = BA = ( 0.905 Wb mb2) = 3.26 x Wb. 106

m.rn.f. (e) Reluctance = - = 360

3 . ~ ~ 1 0 - ~ = 1:l x 106~''

Magnetism of Materials-11

Structure

13.1 Introduction Objectives

13.2 Induced Currents Faraday's Law of Electromagnetic Induction Induction and Conservation of Energy: Lenz's Law Motional Electromotive Force

13.3 Inductance Self-Inductance Mutual Inductance The Transformer

13.4 Energy Stored in a Magnetic Field Energy Stored in an Inductor Magnetic Field Energy

13.5 Summary 13.6 Terminal Questions

13.7 Solutions and Answers

13.1 INTRODUCTION

When you step inside a dark room and illuminate it by a mere flick of a switch, do you ever pause and wonder for a minute: What fnakes this possible? To find the answer you could follow the copper wires (connected to the switches) which bring . electric power to your homes and are spread out across the countryside. You w most certainly find an electric generator which would most likely be situated hydel or thermal power plant. Somewhere between your homes and these generators, there *ill be several transformers. Both these devices are essential for large-scale generation and distribution of electrical energy today. Both are based on momentous discoveries made independently by Michael Faraday and Joseph Henry more than 250 years ago. Indeed, it can be said that their discoveries form the basis of modem electrical technology.

In 1831, Faraday and Henry discovered that electric currents were induced in circuits subjected to changing magnetic fields. k t us, in this unit, study these phenomena associated with changing magnetic fields, explored by Faraday in' his experiments and understand the principles underlying them. We shall also study some applications of these principles, particularly the ones that form the cornerstone of electrical technology, namely, the generator and the transformer.

0 bjectives

After studying this unit you should be able to

6 apply Faraday's law of electromagnetic induction p ~ d Lenz's law

6 compute the self-inductance and the back emf of an inductor possessing a simple geometry

. . 6 compute the mutual inductance of circuits h simple configurations

Elecbromngnetism 4 determine the energy stored in any given magnetic field.

'13.2 INDUCED C

Consider the simple circuit of Fig. 13.1, connecting a loop of wire to an ammeter. Is there a way by which you can induce an electric current in this circuit? Now study Fig. 13.2, in which we show two simple experiments similar to Faraday's which demonstrate the existence of induced currents.

~k.13.lt Cm cumntbc Aide ta now in this cirmit?

(a) (b) Fig.13.2 The amme(+rrrgisieru m nvrent both when (0) the circuit is moved in (he magnetic fidd ot 0

magnet, and (b) (he circuit b hdd Bud .ad (he magnet is moved An low M both the chmit and (hb mpr a n held e(lll, &ere is no a m n l 'Ihls cumeht, termed induced nurrnt, ibws d y whea the m q p d and L e loop move relative Lo each other.

There is another way of inducing an electrical current in a circuit. This experiment is shown in Fig. 13.3. Study the figure before proceeding further.

I What do all these experiments reveal? Can you explain these observations on the basis of what you have studied so far? In fact, it is possible to explain the results of the first experiment (Fig.13.2.a) on the basis of Lorentz force. Would you like to try 1

giving the explanation? 1

f-"--- I . , li

5 min Explain qualitatively how an induced current results in the circuit of Fig.13.2a? iTr ""I I I What about the other two experiments shown in Fig. 13.2b and Fig. 13.31 In both these cases, the charges are stationary (ignoring the thermal and drift effects) and no Eaa Lorentz force acts on them. So how do we explain the origin of induced currenls? The answer is: we are observing a new phenomenon and we have to look for new

i explanations. Now, did you notice that there was a common element in all these' experiments? The magnetic field was changing in all cases. An induced current

2 appeared in a circuit subjected to this changing magnetic field. This leads us to introduce a new basic physical principle: i

alf.31 Whtntherr*lLihS& lafltutdra~It(l)hdoscd, (he rmmekLthtstcaaddrmlt[Z)

A changing magnetic field produces an electric field. -

8h0w9 This new principle explains the origin of the induced current. It is the electric field rh+r a B b o ~ Urns tbs cun?nt in t b e r r e o l l d - t m b Z s r a produced by the changing magnetic field that sets the charges in the wire of the Therch.prln r brlctrespcmw & circuits into motion, the drrull.2 whtn the MvlLch in tbe * .

dndtlhoponsdagaiu,Tbb su- b t the a d dopcning

This explanation is, of course, qualitative. Wershould also be able to give a .nd wg &,,"* quantitative relationship betwetxi the induced current and the changing magnetic to indud c a m ~ ~ k ~ , field: This is precisely what Faraday's law #of electromagnetic induction does.

13.2.1 Faradsay's Law of Electromagnetic Induction

What is it that gives rise to a current in a circuit? As you may know, we need a source of electromotive force (emf) a device, like a battery that supplies energy to the circuit. similarly, when an induced current flows in a circuit, an induced emf must be present. It is along these lines that Faraday developed a general theoretical fonpulation of his experimental results. He formulated the law that the emf induced in aHy cir,cuit depends only on the time rate of change of flux of the magnetic field going through the circuit. Mathematically, this law can be expressed as follows:

d@ E I -- fit

where E is the induced emf and c2, the magnetic flux linked by that circuit. We can rewrite Faraday's law givin by Eq. (13.1) so that we can omit any reference to circuits. The induced emf is simply the work per unit charge done on a test charge that is moved around a circuit. Since work is the line integral of force over distance, and electric field is the force per unit charge, we can write the emf as the line integral of the electric field along the circuit C.

I I

In Unit 9 you have already come across the integral representation of magnetic flux

@ - $ D . ~ s S

where S is any surface whose boundary is the circuit C. Faraday's law can then be expressed as

In this form of Faraday's law, there is no need to have a physical circuit. C can just represent a loop in space and S a surface bounded b;r C. It simply describes induced electric fields, which arise whenever there are changing magnetic fields. If electric circuits are present, induced currents arise as well. Using Stokes' theorem (see Unit 9), we can also express Eq, (13.2a) h the differential form:

Since the surface S is arbitrary, we must have

dB curl E I - dt

dB dB SinceB may depend on position as well as time, we write - instead of - to

at dt account for only the time variation of B. We thus have two entirely equivalent statem'ents of Faraday's law:

d $ c ~ . d = - ; j i & ~ . d ~ (integralform)

aB curl, E - -- at

( differential form )

Remember that in explaining the experiments discussed in the beginning of the section, we have used two different explanations in terms of Faraday's law and Lorentz force law. When we move the circuit, it is just the Lorentz force that giv&

Spend 3 min

rise to induced current. But when we move the magnet, the changing magnetic field induces an electric field and it is the electrical force due to this field that induces the current, However, the law that the induced emf in a circuit is equal to the rate of change of the magnetic flux through the circuit applies to both the cases. Viewed in this light, does it not seem surprising that the two processes yield the same emfs? In fact, consideration of this symmetry between.the two cases led Einstein to his special theory of relativity. But, that is a different story altogether.

So you have found that electric fields are produced by sptic charges as well as changing magnetic fields. But is the nature of these fields the same? Try to find out yourself!

SAQ 2

What is the basic difference in the nature of the electric fields produced by static charges and those induced by changing magnetic fields?

So far we have not said anything about the negative sign in Faraday's law. The negative sign has an important purpose in the law: it tells us the direction of the ' induced current; and it is nothing but t4e consequence of conservation of energy. Let us explore this connection a bit further. In doing so, we will arrive at Lenz's law.

13.2.2 Induction and Conservation of Energy: Lenz's Law When an induced current flows through a coil, electrical energy is dissipated as

. heat, because the coil possesses some resistance. Where does this energy come . from? The principle of conservation of energy tells us that energy cannot be created

from nothing. Now, the induced current is caused by a changing magnetic fiux. So the agent that changes the flux must do work to supply the energy.

(a) @) Fig, 13.4: (a) As the bar magnet moves Lowad Ibe loop, the changing magnetic flux Inducts an anf that

i M v e o a cumnt in (be loop. Conselvation of energy r q u h s tbnt Lbc magnetk field due to the induced nurrnt oppose (bc motion of b e magnet, lhus the agent movhg tbc magnet must do work ha t coda up as heat In the cmdnctlng loopj (b) when we pull away &e mqnet &am the loop, the d W o n d L b c i n d u d arnmt L stub that the m.gn& force due to the loop opposg the wiihdrawrl of the magnet, BM is Lhc magnetic h l d doc to the mrlplet md BL ia due to ihe loop

Let us consider a simple example of a bar magnet which is moved near a loop of wire. As we move the magnet towards the loop, a current is induced in the loop (Fig. 13.4a). Like any. other current this induced current gives rise to a magnetic field. The magnet experiences a force in the field. This force must be repulsive so that we have to do work to push the magnet toward the coil rather than having work done on us. Suppose we are moving the north pole of the magnet towards the loop. With its induced current, the loop behaves like a magnetic dipole givinig rise to a field similar to the bar magnet's. For a repulsive force on the magnet, the north pole of the loop'dipole must be toward the approaching magnet. Applying the right-hand rule to the current loop, we see that 'the induced current must flow in the direction shown in Fig. 13.4(a).

What happens when we pull the magnet away from the loop? Conservation of energy requires that we work against a force to do this. The loop must then present

a south pole to the magnet. This would result in an attractive force which opposes the withdrawal of the magnet. As a result, the loop current is in the opposite direction (Fig. 13.4b).

Both these results stemming from the principle of conservation of energy are presented in the form of Lenz's law:

The direction of the induced current (emf) is such as to oppose the change giving rise to it.

Lenz's law is reflected mathematically in the minus sign on the right-hand side of Faraday's law given by Eq. (13.2).

To concretise this idea, think of what would happen if the direction of the induced current aided the change in the flux. For instance, when you pushed the magnet toward the loop in Fig. 13.41, a south pole appeared towards the magnet. Then you would need to push the magnet only slightly to get it moving and the action would carry on forever. The magnet would accelerate toward the loop, gaining kinetic energy in the process. At the same time thermal energy would appear in the loop. Thus we would have created energy from practically nothing. Needless to say, this violates energy conseivation and does not happen. While applying Lenz's law you should keep in mind that the magnetic field of the induced current does not oppose the mslq~le~c f ~ l d , but the change in this field. For example, if the magnetic flux through a loop decreases, the induced current flows so that its magnetic field adds to the original flux; if the flux is increasing, the current will flow in the opposite direction. This is a sort of an "inertial" pheiiomenon: A conducting loop 'likes' to keep a constant flux through it; if we try to change the flux, the loop responds by sending a cvrrent in such a direction as to counter our efforts.

You may now like to apply Lenz's law to certain simple situations and determine the direction of the induced current.

SAQ 3

a) In Fig. 13,5(a) what is the direction of the induced current in the loop when the . area of the loop is decreased by pulling on it with thc forces labelled F? B is directed into the page and perpendicular to it.

b) What is thc direction of the induced current in the smaller loop of Fig. 13.5(b) when a clockwise current as seen from the left is suddenly established in the larger loop, by a battery not shown?

Let us now consider an application of these laws, which is, perhaps, the 'most important technological application in use today.

Example 1: The ac generator

Do you know'that at present the world uses electrical energy at the rate of about 1013 watts daily, and virtually all of this power comes froin electric generators? A generator is nothing but a system of conductors in a magnetic field. Fig.13.6 shows a simple version of the generator. Mechanical energy is supplied to rotate the coil placed between the-gales of a magnet. In the power stations the source of mechanical energy is either falling water (hydroelectric power plants), or sbam from burning fossil fuels (therr~lal power p1aiits)- or h m nuclear Cission (nuclear power plyts). The rotation causes a change in the magnetic flux through the coil which induces an emf ant1 a current flows through the coil. Let us determine the magnitudes of the induced emf and the induced current, Let S

' be the area of the coil and 8 the angle between the magnetic field and the normal to the plane of the coil: The flux through the coil is

Q1 - BS cos 8 (13.3a)

-13.6: A simple dimgram &an rc e l d c genemtw. As the loop rotates in the nugnetic Rekl, an induced emf is prpdwced in it by the changing magnetic llux. Cu-t flows through (he rotaling contacts and ' 1 slntiooary brushes to en eleetrica1 load

If the coil is rotating with a uniform angular speed w, 0 varies with time as 0 = o a. The emf in the coil is then I

If we bring the wires of the coil to a point P, quite some distance from the rotating coil, where the magnetic field (due to the magnet) does not vary with time, then from Eq. (13.2b) the curl of E in this region will be zero, i.e., E is conservative. Then we can define an electric potential associated with this field. Let the two ends

'

of the coil be at a potential difference V. If no current is being drawn from the generator, the potential difference between the two wires will be equal to the emf in the rotating coil, i.e.,

V = BS w sin wt I. Vo sin u t (13.4a) '

where V, 9 BS w, is the peak output voltage of the generator. As given by Eq. 1 (13.4a), V is an alternating voltage. If we now attach a load R to these wires, we &n

generate an alternating current given by AQ

Spend I0 mirt

Would you now like to work out a numerical problem on the design of an ac generator?

SAQ 4 I

An electric generator like the one in Fig.13.6 consists of a 10 turn square wire loop of side 50 cm. The loop is turned at 50 revolutions per second, to produce the standard 50 Hz ac produced in our country. How strong must the magnetic field be for the peak output voltage of the generator to be 300V? I Recall that for the first experiment shown in Fig.13.2, you had solved SAQ 1 to show qualitatively that the emf was induced in a constant magnetic field due to the Lorentz force acting on a moving wire. Later while formulating Faraday's law we had said that Eq. (13.2) was valid for this case as well. The emf generated in this manner is termed the motional emf. Let us determine the motional emf quantitatively.

6.ia~137'Thaindmcedcmli.ibe 13.23 Motional Electromotive Force Loup cm be srpktrrd b tekma d thdorrcs on~ecbvgcs incho Let us consider the simple situation shown in Fig. 13,7. A wire CD of length L is moving wiw. pulled to the right with constant velocity v. It is in contact with the wire GABH at

points C and D so as to form a wire loop ABCD. We measure the position of the wire by the distance x shown in the figure. The loop is immersed in a constant magnetic field B which is directed perpendicular to the page and into it. Each charge in the wire experiences a force

Here we have ignored thethermal velocities of the electrons in the wire. As you how, thermal velocities of electrons in the wire are very large. The reason we @ore them here is that they are randomly oriented and if we add up the magnetic force on all the eletrons of the wire CD, the net contribution of thermal velocities to Lorentz force is zero. Now since v I B, P = q v B and F points in the direction shown in Fig. 13.7. To calculate the effective emf around the loop, we must find the ngt work per unit charge; which is

where dl is an infinitesimal element of lerigth directed dowi~ward along CD. Thus, the induced emf is

, , !!! , l$ padl. (~ereI?:dl - - P d l s i & ~ 4 4 c and d are in the opposite

directions, and F - qvB. ) - - v B Jdl

But the quantity vLB is just the change in the flux through the loop, since the magnetic flux is - BLx, where Lx is the area of the loop

and

Therefore, comparing Eqs. (13.5b) and (13.5~) we once again find that the net motional emf resulting from the motion of a lrop in a constant magnetic field is given by

This is in perfect agreement with Faraday's law. Thus Faraday's law is a new principle which is consistent with the law of magnetic force on a moving charge. In summary, we can say that the physics of induction is governed by the following two basic laws:

F - q ( E + v x B ) Lorentz force law

Faraday's Law

So far, you have studied Faraday'd law which relates the changing magnetic flux through a circuit to the emf induced in it. Let us now consider a situation when the changing flux through an elemical circuit is itself mused by a changing current in it or in a circuit nearby. We then speak of inductance of the circuit. Let us now study this phenomenon.

.13.3 INDUCTANCE m

When a current changes in a circuit, there is a consequent changing magnetic field around it. If a part of this field passes through the circuit itself then an emf is '

induced in it. If there is another circuit in the neighbourhood, then the magnetic flux

through that circuit changes, causing an induced emf in that circuit. Thus, induction in circuits can occur in two Ways:

i) a coil of wire can induce an emf in itself,

ii) for a pair of coils situated near enough, SO that the flux associated with one a i l passes through the other, a changing current in one coil induces an emf in the other.

in the first case we associate a property called self-inductance with the coil, while in the second we speak of mutual inductance. Let us consider these effects separately.

63.3.1 Self-Inductance

Consider a circular loop carrying a current 1 (Fig. 13.8). A magnetic field is set up by the current in this loop, so there is a magnetic flux through it. As long as the current is steady, the magnetic flux does not change and there is no induced current. But if we change the current in the loop, the flux changes and an emf is induced. The direction of the induced current is determined by Lenz's law, i.e., it opposes the change in the loop current. The more rapidly we change the current in the loop, the

4 greater is the rate of change of flux, and the induced emf which opposes the change in current.

B To discuss this property quantitatively, we associate a quantity L, called self-inductance, with every circuit. The self-inductance is defined as follows:

!6g, 13.8 1 selr-tnda-ccQI s d Q L = - (13.6a) I

where @ is the flux passing through the circuit when a current I flows in the circuit. The unit of self-inductance is the henry (H), named after Joseph Henry. It is defined as

2 1 henry = 1H - 1 tesla - m /ampere

For a coil having N turns

All loops whether in the form of straight wires or coiled ones, possess self-inductance. However, the effect of self-inductance is important only when the magnetic flux through the circuit is large or wher! current changes very rapidly. For

I example, a 1 cm length of straight wire has an inductance of about 5 x lo-' H and it exhibits very little opposition to current changes in the 50 Hz ac. But in TV sets, high speed computers or in high frequency communications, such as satellite communication, current changes on time scales of the order of s. Then the self-inductance of the wires themselves must be taken into account. There are devices, called inductors, designed specifically to exhibit self-inductance.

From Faraday's law, the emf induced in an inductor is

This induced emf is also called the back emf. Eq. (13.7) tells u s that the back emf in an inductor depends on the rate of change of the inducwr current and acrs to oppose that change in current. Since an infinite emf is impossible, so from Eq. (13.7), an instantaneous change in the inductor current cannot occur. Thus, we can say that

( The current through an induciar cannot change instantaneously ] You have just studied that the self-inductance of an inductor is a measure of the opposition to the change in current arough it. How do we determine the magnitude of the self- inductance of an inductor?

n e inductance of an inducfor depnwm iB geo-. principle, we can alm]ate -qgwtkhd* the self-inducm~e of any ckcuit, but in practice it is difficult unless ggemeky is shele:A solenoid is a device with a simple geometry and is widely used in eiectrical drcuits. k t US, therefore, dekrmine the self-inductanw of n solenoid.

The self-inductance of a solenoid

consider a long solenoid of cross-sectional areaA and length L, which consists of^ turns of wire. To find its inductance we must relate the current in the solenoid to the

flux through it. In Unit 9, You have used AmpQe's law to determine the field of a long solenoid, which is given as

B - p o n I

where n is the number of turns per unit length ahdl is the current through a e solenoid. For our problem n = N/I, which gives

The total flux through the N turns of the solenoid is

Pa 3 . a - N J B . I S - N B J d s - N B A - -- 1 turn 1 turn 1

Here we have written ]B . d S - BdS, since the magnetic field of the solenoid is uniform and perpendicular to the cross-section of the individual turns. The self-inductance of the solenoid is

Given this information you may like to determine the self-inductance and the back emf for a typical solenoid to get an idea of their magnitudes.

SAQ 5 Spend I 0 min

A solenoid 1m long and 20 crn in diameter contains 10,000 turns of wire. A current of 2.5 A flowing in it is reduced steadily to zero in 1.0 ms. What is the magnitude of the back emf of the inductor while the current is being switched off? Take P~ = 1.26 x ~ m - '

As we have said earlier, the back emf in an inductor opposes the change in current and its magnitude depends on how rapfdly the current changes. If we try to stop

current in a very short time, dl is very large and a very large back emf appears. This dt

is why switching off inductive devices, such as solenoids, can result in the destruction of delicate electronic devices by induced currents. Having worked out SAQ 5 YOU would realise that you have to be extremely cautious in cloghg switches in circuits containing large inductors. Even in your day-to-day experience, you may have seen that you often draw a spark when you unplug an iron. Why does this happen? This is due to electromagnetic induction which tries to keep the current going, even if it has to jump the gap in the circuit. Now, you may wonder what happens when you plug jn the iron and put on the switch? This brings us to the role of inductors in circuits, Let us consider the example of an LIP circuit to understand this role.

Example 2: LR circuit

Consider the circuit in Fig, 13.9a which has a battery (a source of mns(ant emf EO)

Wnnectedsto a resistanceR and an inductance&. What current flows in the circuit When we close the switch?

. ,

through that circuit changes, causing an induced emf in that circuit. Thus, induction in circuit. can occur in two ways:

i) a coil of wire can induce an emf in itself,

ii) for a pair of coils situated near enough, so that the flux associated with one coil passes through the other, a changing current in one coil induces an emf in the other.

In the first case we associate a property called self-inductance with the coil, while in the second we speak of mutual inductance. Let us consider these effects separately.

13.3.1 Self-~nductnnre

Consider a circular loop carrying a current I (Fig. 13.8). A magnetic field is set up 1 by 'the current in this loop, so there is a magnetic flux through it. As long as the current is steady, the magnetic flux does not change and there is no induced current. But if we change the current in the loop, the flux changes and an emf is induced. 1 The direction of the induced current is determined by Eenr's law, i.e., it opposes the change in the loop current. The more rapidly we change the current in the loop, the + greater is the rate of change of flux, and the induced emf which opposes the change in current.

To discuss this property quantitatively, we associate a quantity L, called self-inductance, with every circuit. The self-inductance is defined as follows:

Fib 138 r SrU-fnducbnceda coil @ L w - (13.6a) I

where 0 is the flux passing through the circuit when a current I flows in the circuit. The unit of self-inductance is the henry (H), named after Joseph Henry. It is defined as

1 henry - 1H - 1 tesla - mZ/ampere

For a coil having N turns

All loops whether in the form of straight wires or coiled ones, possess self-inductance, However, Lhe effect of self-inductance is important only when the magnetic flux through the circuit is large or when current changes very rapidly. For example, a 1 crn length of straight wire has an inductance of about 5 x lom9 H and it exhibits very little opposition to current changes in the 50 Hz ac. But in TV sets, high speed computers or in high frequency communications, such as satellite communication, current changes on time scales of the order of s. Then the self-inductance of the wires themselves must be taken into account. There are devices, called inductors, designed specifically to exhibit self-inductance.

From Faraday's law, the emf induced in an inductor is i

This induced emf is also called the backemf. Eq. (13.7) tells us that the back emf in an inductor depends on the rate of change of the inductor current and acffi to oppose I

that change in current. Since an infinite emf is impossible, so from Eq. (13.7), an instantaneous change in the inductor current cannot occur. Thus, we can say that

The current through an inductor cannot change instantaneously.

You have just studied that the self-inductance of an inductor is a measure of the opposition to the change in current through it. How do we determine the magnitude of the self- inductance of an inductor?

The inductance of an induktor depemen its geomCay.. hi principle, we can calculate H&mnpetlc hduction

the self-inductance of any circuit, but in practice it is difficult unless the geometry is petty sim~le:A solenoid is a device with a simple geometry and is widely used in electrical circuits. Let US, therefore, determine the self-inductance of a solenoid.

The self-inducbnce of a solenoid

Consider a long solenoid of cross-sectional areaA and length L, which consists of N turns of wire. To find its inductance we must relate the current in the solenoid to the magnetic flux through it. In Unit 9, you have used Ampere's law to determine the magnetic field of a long solenoid, which is given as

where n is the number of turns per unit length &d l is the current through the solenoid. For our problem n - N/l, which gives

The total flux through the N turns of the solenoid is

PO Q - N J B . ~ s - N B J ~ S - N B A - - 1 turn 1 turn

1

Here we have written B . d S - BdS, since the magnetic field of the solenoid is uniform and perpendicular to the cross-section of the individual turns. The self-inductance of the solenoid is

Given this information you may like to determine the self-inductance and the back emf for a typical solenoid to get an idea of their magnitudes.

SAQ 5 spend 10 min'

A solenoid lm long and 20 cm in diameter contains 10,000 turns of wire. A current of 2.5 A flowing in it is reduced steadily to zero in i.0 ms. What is the magnitude of the back emf of the inductor while the current is being switched off? Take CL, = 1.26 x 1 0 - ~ ~ r n - l

As we have said earlier, the back emf in an inductor opposes the change in current and its magnitude depends on how rapitlly the current changes. If we try to stop

d l current in a very short time, - is very large and a very large back emf appears. This dt

is why switching off inductive devices, such as solenoids, can result in the destruction of delicate electronic devices by induced currents. Having worked out SAQ 5 you would realise that you have to be extremely cautious in c1o"sing switches in circuits containing large inductors. Even in your day-to-day experience, you may have seen that you often draw a spark when you unplug an iron. Why does this happen? This is due to electromagnetic induction which tries to keep the current going, even if it has to jump the gap in the circuit. Now, you may wonder what happens when you plug in the iron and put on the switch? This brings us to the role of inductors in circuits, Let us consider the example of an LR circuit to understand this role.

Example 2: LR circuit

Consider the circuit in Fig. 13.9a which has a battery (a source of constant emf go) . connected.to a resistance R and an inductance L. What current flows in the circuit when we close the switch?

I - "

Elech.arnagee(ism Solution

Let us analyse this current quantitatively. The total emf in the circuit is the sum of the emf provided by the battery and the back emf of the inductor. Therefore, Ohm's

E L law gives

I We can solve ttiis equation to: obtain I(t) :

@> Fig, 13,9: (a) LR dreuit ; @) current in an LR circuit

Of

This gives

The constant C can be determined from initial conditions.

Before we close the switch the current in the circuit is zero, giving us the condition that at t - 0, i - 0. This initial condition yields

Had there been no inductance in the circuit, the current would have jumped immediately to EdR. With an inductance in the circuit, the current rises gradually and reaches a steady state value of E ~ R as t oc. The time it takes the current to reach about two-thirds of its steady state value is given by LIR, which is called the inductive time constant of the circuit. Significant changes in current in an LR circuit cannot occur on time scales much shortei- than LIR. The plot of the current with time is shown in Fig. 13.9b.

You can see that the greater the L is, the larger is the back emf, and the longer it takes the current to build up. Thus, the role of an inductance in electric circuits is somewhat similar to that of mas. in mechanical systems. You know that the larger the mass of an object is, the harder it is to change its velocity. In the same way, the greater L is in a circuit, the harder it is to change the current in the circuit.

You may like to get an idea of the time scales involved in actual circuits. Work out the following SAQ.

Spend 10 min

SAQ 6

A 15-V battery, 2000 S2 resistor and 10 mH inductor are connected in series. Wha$ is the steady state current in the circuit after a sufficiently long time interval *I

(4100 L/R) ? How long does it take' the current to reach half of its steady state vaIue?

Let us now consider the second situation wherein the changing current in a circuit induces a current in an adjacent circuit, i.e., the phenomenon of mutual induction.

13.3.2 Mutual Inductance Consider two circuits situated close together and at rest. If we pass a current Il through circuit 1, it will produce a magnetic field B1 (Fig. 13.10).

' Let a2 be the flux of B1, through 2. If we change 11, Q2 will vary and an induced emf will appear in the circuit 2. This induced emf will drive an induced current in 2. Thus, every time we vary the current in circuit 1, an induced current will flow in circuit 2. Since from Biot-Savart's law (Unit 9), the magnetic field B1 is proportional to the current I,, the flux of B1 through loop 2 is also proportional to I,:

and

Thus, we can write

From Faraday's law, the induced emf in coil 2 is

The proportionality constant M is known as the mutual Inductance of the two circuits. It is a purely geometrical quantity which depends on the sizes, shapes and the relative arrangement of the circuits. In arriving at Eq.(13.12b) we could as well have changed the current in the second coil to induce an emf in the first. We would have got a similar result fox the induced emf in coil 1:

The unit of mutual inductance is also henry (H). Mutual inductances found in common electronic circuits range from microhenrys ( pH ) to several henrys.

1 Let us consider an eximple to determine the mutual inductance of coupled solenoids, Such an arrangement is also used in the ignition coil of vehicles,

Example 3: Mutual inductance of coupled solenoids

A short solenoid of length L2 is wound around a much longer solenoid of length L1 and areaA, as shown in Fig.13.11. Both solenoids have N turns per unit length. What is the mutual inductance of this arrangement?

Solution

The magnetic field produced by the long solenoid is

B - p,-,NI

This field is constant, and at right angles to the plane of the solenoid coils. It is co,nfined to the interiors of the longer solenoid. So the flux through a single loop of the short solenoid is the product of the field strength and the area of the longer solenoid. Since the total number of turns in the shorter solenoid is NL2, the total flux through it is

14 - N L ~ $ B.dS - NL2BA S

Fig. 13.10 I Mutual lnductsaee d two drcuib.

Big. 13,111 Mutunl iudufloncs d coupled solend&

and the mutual inductance is

The reverse calculation, i.e., computing the flux from the shorter one that was Iinked by the longer one, is pretty difficult, because of the diverging field lines at the ends of the short solenoid. However, such a calculation done by a computer gives the same result.

You may like to apply these ideas to a practical situation.

Spend SAQ 7 10 min

Petrol in a vehicle's engine is ignited when a high voltage applied to a spark plug causes a spark to jump between two conductors of the plug. 'Fhis high voltage is provided by an ignition coil, which is an arrangement of two coils wound tightly one on top of the other. Current from the vehicle's battery flows through the coil with fewer turns. This current is interrupted periodically by a switch. The sudden change in current induces a large emf in the coil with more turns, and this emf drives the spark. A typical ignition coil draws a current of 3.0 A and supplies an emf of 24 kV to the spark plugs. If the current in the coil is interrupted every 0.10 ms, what is the mutual inductance of the ignition coil?,

An extiemely important application of the phenomenon of mutual inductance is found in the transformer. Let u s study it in some detail.

13.3.3 The Transformer

You have just studied that a changing current in one coil induces an emf in another coil. And the emf induced in the second coil is given by the same law: that it is equal to the rate of change of the magnetic flux through the coil. Suppose we take two coils and connect one of them to an ac generator. The continuously changhg current produces a changing magnetic flux in the second coil. This varying flux generates an alternating emf in the second coil, which has the same frequency as the generating current in the first coil. The induced emf in the second coil can, for example, produce enough power to light a bulb (Fig.13.12).

Fig, 13.12: T w o coh al low a gentrator C with no dirret connatioa, to light a bulb 8- an appliuLiaa d mutual induction.

r Now, the induced emf in the second coil can be made much larger than that in the , first coil if we increase the number of turns in the second coil. This is because, in a given magnetic field, the flux through the coil is proportional to the number of turns. In the same way, the emf in the second coil can be made much smaller, if the number of turns in it is much less than the first coil.

Let us compute the magnitude of the voltage in the second coil (also known as the secondary coil) vis-a- is the voltage in the first coil (known as the pflnrary coil).

Voltages In the primary and secondary coils Electmmagnetic ~ndocd iw

Fig. (13.13b) shows a schematic diagram of a transformer shown in Fig. (13.13a). It has a primary coil (P) with N, turns. When the switch S, is closed, a current starts flowing in the primary coil. As the current increases, it generates an increasing magnetic flux in the circuit, which induces a back emf in the primary. The back emf exactly balances the applied voltage E if the resistance of the coil can be neglected. According to Faraday's law (Eq.13.1) the magnitude of the back emf is given as

Now the changing flux in the primary cuts the secondary S and sp generates an emf E, whose magnitude is given by

d@ E2 ' N 2 ; i ~ (13.15b)

d@ whereN2 is the number of turns in the secondary. Eliminating - from Eqs. (13.15

dt a and b) we get

P S (a) E l Nl - = - (13.16) E2 N2

Thus, the instantaneous voltages in the two coils are in the ratio of the number of turns on the coils. By settinglhe ratio of turns in the primary and secondary coils, we call make a step-up or step-down transformer that transforms a given ac voltage to any level we want. For example, if we want a step-down transformer to convert the high grid voltage of 22,000 V, to the low lriains voltage of 220 V we must have

Thus for every turn on the secondary there must be 1000 turns on the primary. In practice, the high grid voltage is reduced to the low mains voltage via a series of ~ig.13.131 (a) AII actual ,

0~)

substations rather than via a single transformer. As you can sce from Fig. 13.14, trandormeq (b) schematic dtprPm oP r translonner. The transformers are used throughout the power distribution network, to transform high boa aecilruits

voltages to low voltages. becrusc d(b&being identical geomchkauy.

Substation Substation Electric wlee

Fig. 13.14 : TmasPormers in a power distribution netwark

So far you have studied the phenomenon of electromagnetic induction and learnt about Faraday's Law, LRnz's law, self- inductance, mutual inductance and some of their applications. We will now turn our attention to another important aspect associated with this phenomenon, viz. the storage of energy in magnetic field.

13.4 ENERGY' STOlRED IN A GNETIC HELD

While discussing self-inductance, we had encountered the concept of back emf of an inductor. You know that we must do work against the back-emf to get the current going in a circuit. So it takes a certain amount of energy to start a current flowing in the circuit. This energy can be regarded as energy stored in the magnetic field of the current. In this section of the unit, we will determine the magnitude of the energy stored in a current-carrying circuit, and then the energy stored in a magnetic field.

Let us first determine the energy in a loop of inductance L. This is equal to the work required to build up a current I in it.

13.4.1 Energy Stored in an Inductor

When the current is built up in a circuit there is an induced back emf which opposes the flow of current. Suppose the back emf at some instant is &. Then the v:ork done on a unit charge againt the back emf E, in one trip around the circuit is - &. If the current at that instant is I, the charge passing through the wire in a small interval of time dt is I dt. Thus the work dW done in the interval dt is

Remember that here we have used dW - - E l d t = L - I d t (7

So, the total work done in building the current from a zero value to a value I. is

This was a specific example of storage of energy in an inductor. We can generalise the equation (13.17) to surface and volume currents, and show how this energy can be regarded as being the energy of the magnetic field produced by the steady current.

13.4.2 Magnetic Field Energy

You know that the flux through a single loop, is equal to L I where L is its inductance and I the current through the loop:

a? = L I (13.18a)

You also know that

Since the divergence of B is zero (Eq. 9.20 of Unit 9), we can use the vector identity V . ( V x A ) - 0 , where A is a vector field, to express B, in terms of A:

Here A is termed the vector potential associated with the magnetic field B.

Therefore, from Eq. (13.18b)

@ = J ( v x A ) . ~ s (13.18d) S

18

Using Stokes' theorem we get

Thus, from Eqs. (13.18a and 13.18e) we get

Therefore, the energy'of this loop is

Now, to generalise this expression, let us suppose that we do not have a current circuit defined by a wire. Instead, let the 'circuit' be a closed path that follows a line of current density. Then U given by Eq. (13.20) can approximate this situation very closely if we replace

and

where V is the volume occupied by the current.

Hence we can write

1 u - Z J v ~ . ~ ~

Using AmpCre's law, ( V x B = & J ) we get

We now use the following relation

V . ( A x B ) - B . ( V x A ) - A . ( V x B )

to write

A . ( V x B ) = B . ( V x A ) - V . ( A x B ) = B . B - V . ( A x B )

As a result we get

where we have used Gauss' divergence theorem in the second term !nd S is the surface that bounds V. The integration is to be taken over the entire volume occupied by the current. However, we can even choose a larger region for integration without altering the result, since J will be zero beyond the volume occupied by the current. Let us extend the volume integral to include all space. In such an event, the contribution from the surface integral goes to zero, since the farther the surface is from the current, the smaller B and A are. Thus, we are left with

Spend 10 min

In view of this result we say that energy of the current- carrying circuits can be regardeti as stored in the magnetic field produced by these currents, in the amount ~ ~ / 2 ~ ~ per unit volume. Thus there are two ways to think about the energy stored in

1 2 circuits which are entirely equivalent: i.e., either- ( A . J ) or B /2po of energy

2CLo per unit volume.

Does it appear strange to you that it takes work to set up a magnetic field? The point is that setting up a magnetic field where there was none r dires a changing . .

magnetic field. And, as you know, a changing magnetic fi 7 Id induces an electric field. The electric field can do work. Thus, in the beginning a@ at the end there is no electric field. But, in between, when the magnetic field is building up, there is an electric field, and it is against this that the work is done. This work done appears as the energy stored in the magnetic field.

You may like to evaluate the magnitude of the energy thus stored for a specific situation.

SAQ 8

A long coaxial cable carries current /which flows down the surface of the inner cylinder of radius a and back along'the outer cylinder of radius b (Fig. 13.15). Find the energy stored in a section of length 1 of the cable. It is given that the magnitude

i of the magnetic field between the cylinder is

B = - 2 c r

and zero elsewhere. - I Hence find the self-inductance per unit length of the cable.

1215 We will now summarise what you have studied in this unit.

13.5 SUM Y

In this unit we have introduced you to two important phenomena: elktromagnetic induction due to a changing magnetic field and motional induction.

Motional induction can be explained by the Lorentz force. However, the explanation of electromagnetic induction requires the introduction of a new fundamental principle: A changing magneticfield gives rise to an induced electric field.

C

. Electromagnetic induction is described by Faraday's law which gives the, emf induced in a circuit when the flux through the circuit is changing with time as

Faraday's law applies to either of the two.kinds of induction.

Faraday's law can be rewritten in both integral and differential forms, to relate the induce4 electric field and the changing magnetic field

__-- 2 0

aB V x E = -- at

The inducd electric field k nonconservative unlike the conservative electrostatic field of a stationary charge. Thus it can do work pn charges as they move around a closed loop.

0 Tlie direction of an induced current is specified by Lenz's law: The direction in which induced ciwrentflows is such as to oppose the change that produced it. This law is reflected mathematically in the minus sign on the right hand side of Faraday's law. Lenz's law is a consequence of the conservation of energy principle.

o A changing current in a coil or circuit gives rise to a changing magnetic flux through the same circuit, which induces a back emf in it. The back emf opposes the original change in the current. This property of the circuit or the coil is called its self-indnctance. Special devices that exhibit the property of self- inductance are called inductors. The self-inductance L of an inductor is the ratio of the magnetic flux to the current through it:

An inductor opposes insta~~taneous change in current. Faraday 's law relates the emf in an inductor to the rate of change of current:

0 When a pair of coils or conductors is placed so that the magnetic flux of one coil links the other, a changing current in one coil induces an emf in the other. This electromagnetic interaction of coils is'called mutual induction. The mutual inductance of a pair of coils is defined as the ratio of the total flux in

,. thesecond coil to the current in the first:

Faraday's law relates the emf in the second coil to the rate of change of current in the first

The same mutual inductance 1U describes the emf induced in the first coil as a result of changing current in the second coil.

0 Work needs to be done to build up current and, therefore, magnetic field in an inductor. This work ends up as stored energy in the inductor, given by

where L is the self-inductance of the inductor carrying current I.

This energy can also be regarded as the energy stored in the magnetic field produced by the current I and can be written as

This expression is very general and applies to a single inductor, coupled inductors, and surface and volume distributions of currents.

Spend 45 min

1. A wire loop of radius 20 cm having a resistance of 5.0 SZ is immersed in a uniform magnetic field B at right angles to it (Fig. 13.16). The field strength is

M . . . ~ increasing at the rate of 0.'10 tesb per second. Find the magnitude and direction of the induced current in the loop.

Fb 13161nem.ploeticfkHB 2.a) Ametal ring placed on top of a solenoid jumps when current through the points into tbc pgc solenoid is switched on (Fig. 13.17a). Why?

Fig. 13.17

Two coils are arranged as shown in Fig.13.17b. If the resistance of the variable resistor is baing increased, what is the direction of the induced current in the fixed resistor R?

Determine the self-inductance of a toroidal coil of rectangular cross-section, having N, turns and inner radius a, outer radius b and height h. If the current through the coil is i,, what is the total magnetic energy stored in the coil?

Suppose a coil C of N2 turns is wound over the toroidal coil of part (a) as shown in Fig. 13.18. Show that the mutual inductance for this arrangement is

. .

4. The rim of a horizontally suspended wheel of radius R carries charge q. The wheel (with wooden spokes) is free to rotate. In the central region of the wheel upto a radius a, there exists a uniform magnetic field B pointing up (Fig.13.19). What happens when the magnetic field is turned off? How much angular momentum will be added to the wheel?

5 .a) A sheet of mpper is placed in a magnetic field as shown in Fig. 13.20. If we try to pull it out of the field or push it in, a resisting force appears; Explain its origin.

b) A superconducting solenoid designed for imaging the, human body by nuclear pig. 13.19 magnetic resonance is 0.9 m in diameter and 2.2 m long. The field at its centre

is 0.4 @la. Estimate the energy stored in the fielg of .this coil. I -

13.7 SOLUTIONS mP) ANSWERS

SAQs (Self-assessment Questbns)

1. The circuit wire is made of metal and contains electrons which are relatively free to move. When the wire is moved in the magnetic field, the eleqtrons in it also move with it. When these electrons move in the magnetic field, they experience the force F - - e v x B that tends to push them along the wire. The relatively free electrons are set into motion by the force and form an electric current as they move along the wire. It is this current that is deteded by the ammeter. This is termed the induced current.

Recall from Unit 4 that the curl of the electric field due to st& charges is zero and the force field corresponding to it is conservative. From Eq. (13.2b) you will note that the curl of the electric field induced by a changing magnetic field is non- zero. Hence, the electric field g$en by Eq. (13.2b) is nonconservative. This is the basic difference between th& two kinds of electric fields. The force corresponding to electric field Of Eq. 3.2b) can do work on charges as they move around a closed loop. Moreove , we cannot associate scalar potential with this field.

Y 3.a) As we stretch the loop, its area and hence the magnetic flux through it decreases.

The direction of the induced current is such as to oppose this deaease, i.e., its magnetic field should add to the existing magnetic field. Thus the induced current should flow in the clockwise direction as we view it from top.

b) When a clockwise current is established in the bigger loop, it sets up a magnetic field similar to a bar magnet's with its north' pole facing toward the smaller loop. The induced current in the smaller loop should be such as to dppose this change in the magnetic field, i.e., it should offer a north pole to the

, magnetic field of the bigger loop. This will happen if the current in the smaller loop is in the counterclockwise direction as seen from Ihe left.

4. The induced emf is given by

From Eq. (13.3a), for a uniform magnetic field the flux through one turn of the loop is

Qlg = BScoswt

o = 2 xf, where f is the frequency at which the loop rotates. S is the area of

, the loop given by s ', where s is the lengtb of the side loop. Thus, the induced emf for an N turn coil is

This is the typical field strength near the poles of a strong permanent magnet. -

We will first have to determine the self-inductance of the solenoid which is

For the paramet&s given in the problem

Now since the current changes steadily, the magnitude of its rate of change is

The magnitude of the back emf is

The voltage is high enough to produce a lethal shock. Note that this voltage is unrelated to the voltage of the source supplying the inductor current. We could have a 6V battery and still be electrocuted trying to switch off a circuit rapidly when a large inductance is present. So be careful if you have to handle such circuits.

In the steady state the current is not changing and there is no emf in the inductor. So the current in the circuit is simply given by

When the cunent has half its final value, E ~ R , Eq.(l3.l0b) gives

Taking the natural logarithm of both sides

t = @ 10 x ~ o - ~ H x ( 0.69 ) _ 3,5 irs

200061

Although short, this time would be significant in a TV, computer or elec!rogic communication involving high frequency signals,. .

. - 7. The rate of change of current is

The mutual inductance is

8. The energy per unit volume is

The volume of the cylindrical shell of length 1, radius r, and thickness dr is ( 23~ r d r ) I . Therefore, the energy stored in the cylindrical shell is

Integrating from a to b we have:

1 2 Since U - - LI , we get iin expression for L:

2

clo b The self-inductan?$ per unit length is - In -. 2 ~ c a

Terminal Questions

1. To find the'induced current we must know the induced kmr which is given by Faraday 's law:

Now the flux is

Since the field is uniform in space and the loop is at right angles to it we get

where r is the radius of the loop. Thus, with the loop area being constant

We can obtain the magnitude of the induced current through the loop using Ohm's law:

magnetism

I Fig. 13.21

Since the magnetic field, pointing into the page, is increasing, the direction of the induced current is such as to oppose this increase. Thus, the current's magnetic field should be in the opposite direction, i.e., the current should flow

.counterclockwise in the loop as viewed from top of the page.

2.a) Before the current is switched on, the flux through the ring is zero. When the current is switched, a flux appears upward in the diagram. Due to the change in flux, an emf and a current is induced in the metal ring. The direction of the current is such that its magnetic field is directed opposite to that of the solenoid. Thus the'current in the loop is opposite to the current in the solenoid. You must have studied that the force between two conductors carrying currents in the opposite directions is repulsive. This causes the ring to jump.

b) The current in the coil 1 on the left flows counterclockwise in it, so that its magnetic field points towards the second coil. As the resistance increases, the current in coil 1 decreases, causing a decrease in the magnetic flux linked by coil 2 The induced current opposes this decrease, so its magnetic field should point to the right of the coil 2. Thus, from the right- hand rule the induced current should flow counterclockwise in coil 2, i.e., it should flow in the resistor R from right to left.

3.a) You know that the magnetic field within the toroid is given by

where i, is the current in the toroid windings. We must now find the flux through each turn of the toroid, which is the flux through the toroid (Fig. 13.21).

Let us consider an elementary strip of area h dr. The flux through the strip is

since B is normal to the toroid cross-section. The total flux through the toroid cross-section is found by integrating dQ from r - a to r = b :

'4

The self-inductance of the toroid having N, turns is

The total magnetic energy stored in the coil is

b) The flux linkage through the coil C of Nz turns is due to the flux within the toroidal coil. Therefore, it is given by

Thus, the mutual inductance for this arrangement is

4. The changing magnetic field induces an electric field. Since the magnetic field is decreasing, the E field will be in the counterclockwise direction; to oppose the decrease, The induced electric field exeks a force on the charges on the rim driving them around. This causes the wheel to rotate in the counterclockwise direction, as seen from above. Quantitatively, from Faraday 's law we have ,

, 26

The torque on a small element dl of the rim referred to the centre of the wheel is ( R x F ) or ( Rq E dl ). The magnitude of total torque on the wheel is

The total angular momentum imparted to the wheel is

5.a) As we try to pull the sheet of copper out of the magnetic field, induced current. appear in it. Since the flux through tbe sheet is decreasing, the direction of the current in it is clockwise to oppose the decrease. The magnetic force ( = I d 1 x B ) due to the induced cuirent will be towards the left, i.e., it will oppose the motion of the loop.

Similarly, when we push the sheet in, induced current in the counterclockwise direction appears. The magnetic force due to this current points towards the right opposing the direction of motion. The currents induced in solid conductors due to changing magnetic fields are termed eddy currents. As you have seen here, eddy currents can make it difficult to move a conductor through a magnetic field.

b) The energyU stored in the solenoid is 44 L i '. The magnitude of the magnetic , field of the solenoid of length 1 is

Thus the current through the solenoid is

You know that the inductance of a long solenoid is

Thus

Electromagnetic Induction

UNIT 14 EQUAnONS GNETIC WAVES'

Structure

14.1 Introduction Objectives

14.2 axw well's Equations Asymmetry in the Fundamental Laws of Electromagnetism Generalisation of AmpEre's Law: Displacement Current Putting Maxwell's Equations Together

14.3 Electromagnetic Waves Tl~e Wave Quation Plane Wave Propagation in Empty Space

14.4 Maxwell's Equations and Plane Wave Propagation in Dielectric Media Maxwell's Equations in Dielectric Media Plane Wave Propagation in Dielectrics

14.5 Energy Carried by Electromagnetic Waves: Poynting's Theorem

14.6 Summary

14.7 Terminal Questions

14.8 Solutions and Answers

14.1 INTRODUCTION

At this point of the course, you h o w the four fundamental laws that govern electric and magnetic phenomena, nameIy Gauss' law for electric fields, Gauss' law for magnetic fields, ArnpCre's law and Faraday's law. All these laws together explain the electric and magnetic interactions that make matter act as it does. Recall that in the introduction to this block we had promised to assemble all that you had learnt in this course into a single set of equations, called Maxwell's equations. This is precisely what we will do in this unit.

' MaxweIl's equations govern the behaviour of electric and magnetic fields everywhere and describe all electromagnetic phenomena. For example, they help us explain why a compass needle points north, why light bends when it enterswater, why thunderstorms occur, why we see aurora in the polar regions, and many other natural phenomena. These equations also form the basis for the operation of a large number of devices in use today, e.g., electric motors, television transmitters and receivers, microwave ovens, telephones, computers, radars, cyclotrons, etc. Understanding Maxwell's equations is, indeed, a truly rewarding experience.

Apart from explaining this wealth of phenomena, Maxwell's equations lead us to a fundamental insight into the nature of light and other electromagnetic radiations. You will also share this insight when you study about the nature of electromagnetic waves and their propagation in this unit. In the next unit we shall discuss the reflection and refraction of electromagnetic waves and some of their technological applications.

Study guide

To be able to get the most out of this unit you should be well aware of the materials of Blocks 1 to 3. Further, it would help you to review Block 2 of the cou'w on 'Oscillations and Waves' (PHE-02). It will also be useful to keep the Block 1 of the physiq course Mathematical Methods in Physics-I (PHE-04) at hand for the sake of reference. We have put the complex derivations of some important results in an

to this unit. You may like to study the relevant sections in the Appendix whenever we refer to them in the text. This will lead to a clarity of perception and also show you how the results are based on logical mathematical reasoning. You will not he examined for the ~apnte~al give11 ha the Appendix. This is a big unit, However, you should be able to complete Sec. 14.2 in a couple of hours and devote the remaining 6 h to Secs. 14.3 to 14.5.

Objective

After studying this unit you should be able to

explain the symmetry considerations which led to Maxivell's equations

write Maxwell's equations in charge-free and current-free regions, regions containing charges and currents, and in dielectric media . derive electromagnetic wave equations from Maxwell's equations . explain the nature of electroinagnetic waves

apply Maxwell's equations and the electromagnetic wave equations (with plane wave solutions) in vacuum and dielectric media

compute the Poynting vector and the energy carried by electromagnetic waves .

14.2 WELL'S EQUATIONS

Recall all the laws governing electric and magnctic phenomena that you have studied in this course. Which ones amongst them can be thought of as fundamental? Let us try to list these fundamental laws in a table.

Table 14.1: A tentative list &the fuadnmenbl laws governing dccMc and n~agnetic phenomma

S.No.

1 ~ a u s s ' law for the magnctic field

1 1.

Rraday's law of electromogentic induction

Law

Magnctic field lincs closc on themselves, they do not begin or end at my point. (This in~plies tllnt an isolated

Gauss' law for the electric fiold

d@8 I t& ~ , d , - --.,- Changing magnetic field gives dl

rise to electric field I or

Whnt the law snys Mothen~nticnl statement

Charges give rise to electric field; olcctriic field lines begin and end on charges

You may be wondering why we have not put Coulomb's law and the law of Biot and Savart in the table. As you have already studied in Blocks 1 and 3, these laws may be viewed as fundamental only for stationary or slowly moving charges. In fact, Coulomb's law is equivalent to Gauss' law when the flux is not changing, i.e., dis/at = 0. Similarly, Biot-Savart's law follows from Gauss' law for magnetic field and AmpCre's law. Thus, both Coulomb's law and Biot-Savart's law can be obiained from a combination of two of the four laws listed in Table 14.1. All the other-equations that you have studied in this course apply to special situations and are incorporated in the four equations given in Table 14.1.

4.

M n x w ~ * e Equations .nd agnetic Waves

&E. d S - 2 €0

or

V . E - t o

(14.1)

Amfire's law (for steady currenii only) Electric current gives rise to

magnetic field

qC dl - or

V x B - W J

Let us now examine the four laws assembled in Table 14.1, together. Do you see some similarities in them? Notice that the left hand sides of the first bvo laws (Eqs. , 14.1 and 14.2) and the last two laws (Eqs. 14.3 and 1 4 4 , respectively, are completely identical except for the interchanging of E and B. In the first two laws we have surface integrals respectively, over closed surfaces. Similarly, in the last two laws we have line integrals around closed loops. These pairs of equations differ only in the interchange of E and B on the left-hand side. So we call say that the left sides of the equations in Table 14.1 are symmetrical in E and B, in pairs.

What about the right-hand side of these laws? These do not seem symmetrical at all. What is the asymmetry in these laws? k t ' us find out.

14.2.1 Asymmetry in the Fundameanbl Laws of Els&mmagnetkm We can identify two kinds of asymmetry in the right side of these laws. The right-side of Gauss' law for electric fields bas a charge q enclosed by a surface. But Gaus' law for magnetism has zero on the right side. This asymmetry. arises from the following fact: isolated electric charges exist in nature, but there is no evidence so far that isolated magnetic charges exist. Therefore, the enclosed magnetic charge on the right-side of the second law is zero. If and when magnetic monopoles are discovered the right side of this law would be non-zero for any surface enclosing a net magnetic charge. In the same way, the term yo i ( = yo dq/dt )repaese~ting the flow of electric charges appears on the right side of Eq. (14.4). But no similar term (representing a current of magnetic monopoles) appears on the right side of Eq. (14.3). This is one kind of asymmetry in these laws which would be resolved if we knew for sure that magnetic monopoles existed. Current theories of elementary particles suggesting the existence of magnetic monopoles l~ave prornpted an earnest search for them.

There is another asymmetry in these laws. On the right side of Faraday's law (Eq. 14.3) we have the term - dQ?B/dt. Recall that we interpreted this law by saying that changing magnetic fields produce an electric field. We find no similar term in Am@reYs law. Are we missing something? From symmetry considerations, could we suggest the following?

Changing electric fields produce a magnetic field.

This was the line of thought followed by Maxwell. Showing remarkable insight into the symmetry of electric and magnetic phenomena, be introduced the concept of induced magnetic fields and displacement current. Thus, he generalised Am@rels law to arrive at the symmetrical counterpart of Faraday's law. k t us see how this -was done.

14.22 Generalisation of Amgreys ILnw : Dtplacement Current Let us reconsider Ampere's law for steady cutrents. For mathematical convenience we use its differendal form.

where J is the current density associated with the electric current i. You know that

i - J . dS. Let us see if we can use Eq. (14Ja) for fields that vary with time. If

we take the divergence of both sides of Eq. (145a), we get

v. ( v x p ) = k ( V - J ) (14.5b)

F~! vecwr h l d ~ , itaa be The left-hand side of Eq. (14.5b) is zero (see the margin remark). This gives us that showuthal V . ( V x A ) - 0 V * J - 0

This equation is true only in the special case of steady currents. Recall the continuity equation that follows from the conservation of electric charge. It tells us that

or in integral form Maxwell's Equatiola.. Eledmmagnetic Wu\ -

that is, the divergence or net outflow of J from some region is the rate of decrease of charge contained in the region. It is only for steady currents that V . J = 0. Thus, a term is missing from Eq. (14.5a) for time-varying fields. This term should

be the time derivative of some vector field so that for static fields, the equation would reduce to (14.5a). Perhaps, Maxwell's most important contribution was the determination of this missing term. Maxwell modified Eq. (14.5a) by

aE adding a term po ee -to Eq. (14.5a); which was rewritten in the differential form as

at

The integral form of Eq. (14.6a) can be obtained by integrating both sides of the equation over some open surface S and applying Stokes' theorem. We have given the derivation in the margin alongside. Thus, we obtain the generalisation of Ampdre's law as carried out by Maxwell.

Notice that in writing this equation the minus sign in Faraday's law is replaced by a plus sign. This is dictated by experiment and considerations of symmetry. The factor EO is inserted to express the equation in SI units. Eqs. (14.6a and b) tell us that there are two ways of setting up a magnetic field:

1) by an electric current and

2) by a changing electric field

Eqs. (14.6a and b) give the generalised differential and integral forms of Ampt5re's law. These are also termed the ArnpBre-Maxwell's law. Remember that Maxwell did not derive this law from any empirical considerations. He was motivated by symmetry considerations and he deduced the additional term by requiring Am$re's law to be consistent with the conservation law of electric charge. Since Maxwell's time many experiments including direct measurement of the magnetic field associated with a huge capacitor, have confirmed this remarkable insight of Maxwell. Let us now study Eq. (14.6b) closely and understand its meaning. You can verify that the term EO dCDE/dt has the dimensions of a current. Let us examine this term further.

Displacement current

Although the changing electric flux is not an electric current, it has the same effect as a current in producing magnetic fields. For this reason Maxwell called this term the displacement current and the name has stuck. It is given as

where 4 ( - EO 5) is termed the displawrncnt current density. The word

'displacement' does not have any physical meaning. But the word 'current' is relevant in the sense that the effect of the displacement current cannot be distinguished from that of a real current in producing magnetic fields. So, we can say that a magnetic field can be set up by a conduction current i or by a displacement current id, Thus, we can express Anpire-a ax well's law as

Integrating both sides of &.(14.60) over PO open surface S we get

Q V X B ) . ~ - W & J . ~ S

+ w r o ~ s & ~ . d ~

Applying Stokes' law to the LHS and noting that i = J S ~ . d s

and* - & E , ~ s

we obtain

~ Q E $ l . d - poi+,Ao~-- dl

To better understand the role of the displacement current, let us consider a parallel plate capacitor and detMnhe the displacement cumnt in a cqeuit containing the capacitor.

Example 1: Displacement cunrent In a circuit containhg a parallel-plate capacitor.

Let a parallel plate capacitor be charged by a constant current i as shown in Fig. 14.1. Suppose the plates are large in comparison with their separation.

Fig. 14.1 I

Then, there will be an E field only between the plates and to a good approximation, it will be uniform over most of the area of the plates. Under these conditions we shall use Eq(14.n) to determine id for the parallel plate capacitor. Let us apply Eq. (14.7b) to a circuit around the wire leading to the capacitor. Let us construct two "balloon-liken surfacess, and S2 as shown in Fig. 14.1. Note that the wire and hence the free current i penetrates the surface &and the related contour C encircles the wire. The second surface Sz encloses one of the capacitor plates, No free current penetrates this surface but the contour C again encircles the wire. For the surface S, and the related contour C, the right-side of Eq. (14.7%) gives

The displacement current term is zero since E is constant id a wire carrying a constant current. Hence for S,, Eq(14.n) becomes

For the surface S2 with the same contour C,

J J.dS - 0 4

since no free current penetrates S,. However, as free charge is being stored and removed from each capacitor plate, a time varying field E will develop between the plates. The lines of flux of this field will penetrate Sz, thus

since the contour C is same for both S1 and S2, it should require that i I id. To k I m ~ ' r Equqtions mud alpl la tic waves

SRQW this, we apply Gauss' law to a pill box volume enclosing a surface of area S which surrounds the positive plate:

where q idthechargo on the plate and S is its area ( - n R~ ) . ~hus

and

Hence for the circuit

for either S1 or S2 when both conduction'and displacement currents are taken into account.

Note that the contours bounding the two surfaces are chosen to be the same. If the displacement current term were not present in Eq. (14.7b), we would have had an inconsistency: choosing the two surfaces having the same contour would yield d'fferent results (po i for S1 and zero for S,). Therefore, we can think of the

acement current as 'completing the circuit': where conduction current stops 47 flo ing,'displacemedt current takes over to complete the circuit. This is another way of thinking about continuity equation for charge, i.e,, Eq. (14.5~).

This example of a parallel plate capacitor shows concretely the necessity for the displacement current term in the fourth Maxwell's equation. To get an idea of the importance of displacement current we would like you to work out the bllowing SAQ.

SAQ 1 spend 5 min Obtain the ratio of conduction and displacement current densities

aE J - a E, Jd I w, E = Eo sin a t in copper at a frequency of 1 MHz. Repeat at

for Teflon at 1 MHz. For copper 6 = EO, p = p,-,, a = 5.8 x 10' ~m-'. For Teflon,

e = 2 . 1 ~ 6 , ~ = poando = 3 x 10-~~m- ' .

It was indeed Maxwell's genius to ;ecogniw that Ampere's law should be modified to reflect the symmetry suggested by ~arau!~'s law. To honour Maxwell, the four complete laws of electromagnetism are given the name Maxwell's equations. Maxwell's equations belong to the category of the fundamental laws of nature. As you have seen, they are not derived from any fundamental precepts by logical reasoning and mathematical calculations. Fundamental laws of nature are generalisations of our h o d e d g e and they are discovered, found or ascertained. We will end this section with a brief oveiview of these equations.

14.2.3 Potting Maxwell's Qua tions Together Let us first list these equations.

Table 14.2 : Maxwell's muations

Ampire's law: Electric current and changing electric field give rise to v B - Clo J + Eo - dcPE magnetic field 1 ( l')l$cB.d - w+weo- dl

I Gauss' law: Magnetic field lines close on themselves, no magnetic charges exist

This set of equations, first published by Maxwell in 1864, governs the behaviour of electric and magnetic fields everywhere. These are written for the fields in vacuum, in the presence of electric charges and electric currents. Notice that the lack of symmetry in these equations, with respect to E and B is entirely due to the absence of magnetic charge and its corresponding current. In charge-free regions, the terms containing q and i (or p and 3) are zero and Maxwell's equations take the following form:

Integral from

4 E . a - ' go

dl

Table 14.3: Maxwell's equations in vacuum with no source charges or currents

Differential form

- ' v . E - WI

aB V x E - --

at

S.No.

1.

2.

V . B - 0

What the Equation says

Gauss' law: Charges give rise to electric field, eledric field lines begin and end on charges

Faraday's law: Changing magnetic field gives rise to electric field

4 B . d ~ - 0

Differential Form

You can see that in charge-free regions the symmetry is complete; the electric and magnetic fields appear on an equal footing. The constants eo and appear in Amptre-Maxwell's law due to our choice of units. You could be wondering about the discrepancy in sign. The difference of signs in Eqs. (14.9) and (14.11) or in Eqs. (14.13) and (14.15) is actually due to symmetry: it reflects the complementary way in which electric and magnetic fields give rise to each other.

Integral Form

V . E - 0

aB V x E - -- at

What do Maxwell's equations tell us? In a nutshell, the first two equations (in Table 14.2) tell us that an electric field is set up in two ways: by electric charges and by a variable magnetic field; the last two equations tell us that a magnetic field ,has no sources (there are no magnetic charges) and it is set up by electric currents and a variable electric field. Maxwell's equations also indicate that a variable magnetic field cannot exist without an electric field, and a variable electric field, without a magnetic field. This is why the two fields are not regarded as separate. An electromagaetic field is a single entity. In this manner, Maxwell succeeded in formulating mathematically a unified theory of electricity and magnetism.

The consequences of Maxwell's formulation are legion - all of electrical and radio engineiring is contained in these equations. M e r , the presence of the displacement current term in Eq. (14.15) alongwith Eq. (14.13) implies the existence of . ' electromagnetic waves. This forms the discussion of the next section. But before proceeding further you should apply Maxwell'ls equations to time-varying fields.

J S ~ . d S - 0

dt

(14.12)

(14.13)

--

SAQ 2

Maxwell's Eqwtions end F3cdmmrgnefic Waves

Under what conditions do the following time-varying electric and magnetic fields satisfy Maxwell's equations (Eqs. 14,12 to 14.15) ?

Spetrd A 18 mitt

E = kEos in(y-v t ) A

B = iBosin(y-vt)

where Eo and Bo are constants.

14.3 ELECTROMAGNETIC WAVES

As we have said earlier, one of the great successes of Maxwell's equations was that they predicted the existence of electromagnetic waves. They were subsequently discovered by Hertz in 1887. Moreover, at high enough frequencies, these waves were shown to be light waves. Now we know that radiowaves, infrared, visible, ultraviolet, X-rays and gamma rays are all electromagnetic waves differing only in frequency. In this section we will see liow Maxwell's equations lead to the prediction of electromagnetic waves. From Maxwell's equations, we will derive an equation which is just the wave equation and understand its physical meaning.

14.3.1 The Wave Quzlticpn We shall first derive the wave equation from Maxwell's equations in a region of space where there is no charge or current (Eqs, (14.12 to 14.15) given in Table 14.3). As ybu can see, these equations are coupled, first order, partial differential equations. But we can uncouple these equations. For the sake of mathematical convenience, we will use their differential form. Taking the curl of Eqs. (14.13) and (14.15) we get,

and

We now make use of the following vector identity for any vector field F:

V X ( V X F ) = v (v .F ) -V 'F

Thus, on using Eq. (14.15) alongwith this vector identity, Eq. (14.16a) yields

Since V . E - 0 from Eq. (14.12), we get

Similarly, Eq. (14.16b) alongwith Eq. (14.13) yields

Thus, from Maxwell's equations, we get two uncoupled second order partial differential equations for the time-varying E and B fields in vacuum in the absence of charge or current:

Electromagnetism

You should gain complete familiarity with this kind of a mathematical manipulation. So we are giving an exercise for you. Follow the same procedure and obtain second order uncoupled partial differential equations from Maxwell's equations in the presence of electric current only, (with the free charge in the region being zero).

Spend SAQ 3 10min Show that the uncoupled partial differential equations for the E and B components

of an electromagnetic field in charge-free material media are given by

Here p = 0, J = a E and E ~ , p0 have been replaked by the E, p of the medium, in Maxwell's equations (Eqs. 14.8 to 14.11).

Let us now consider Eqs. (14.17a and b) in detail. Recall the classical wave equation

which describes a wave travelling with speed v. You must havestudied this equation in the Physics Courses PHE-02 and PHE- 05. Comparing this with Eqs. (14.17a) and (14.1%) tells us that the latter are wave equations and imply changing electric and magnetic fields which propagate like waves in space. Recall from Maxwell's equations that a changing electric field gives rise to a magnetic field, which itself may be changing with time. Taken together Eqs. (14.17a) and (14.1%) suggest the possibility of self-sustaining travelling electromagneticfields in which a change in the electric field continually gives rise to the changirtg magnetic field, and vice-versa. Thus, electromagnetic waves can be thought of as structures consisturg of electric and hragneticfields that travelji-eely through empty space. Comparing Eqs. (14.17a and b) with Eq. (14.19) gives us the speed of the electromagnetic waves. It is

This is just the speed of light in vacuum! The implication is extremely exciting: light is an electromagnetic wave. This conclusion would not surprise you today. But imagine what a triumph it was in Maxwell's times. Do you recall where EO and came in the theory in the first place? They appeared as constants in Coulomb's law and Biot-Savart law. We can measure them in exper'iments involving charged pith balls, batteries, and wires-xperirnents which have ;<Thing to do with light. And yet, in Maxwell's theory these two are related in a beautifully simple manner to the speed of light! Notice also the crucial role of the displacement current term in AmpBreMaxwell's law. Without this term, the wave equation would not have emerged. Thus, according to Maxwell's equations, empty space supports the propagation of electromagnetic waves at a speed given by Eq. (14.20). We would like you to understand clearly the nature of these waves.

Nature of Electromagnetic Waves

We have suggested above that an electromagnetic wave is constituted of s time-varying electric and.magnetic fields. How do we visualise such a travelling

electromagnetic wave? To do so, let us consider an electromagnetic wave travelling

through a region in empty space (i.e., charge-free and current-free region). As the Maxwell'e Equations and Elcdmmagnetic Wavcs

wave passes over it, the magnetic flux through the region will change and according to Faraday's law, induced electric fields will appear in that region. These induced electric fields are, in fact, the electric component of the travelling electromagnetic wave. But as the wave moves through the same region, the flux of the induced electric field will also change in time. The changing electric flux will induce a magnetic field as per Ampire-Maxwell's law. This induced magnetic field is simply the magnetic component of the electromagnetic wave. Thus, Faraday's and ~rn~ire-Maxwell 's law describean induced field (E or B field) that arises from the other changing field. The other field, in turn, arises from the change in the first field. So we have a self-perpetuating electromagnetic wave whose E and B fields exist and change without the need for charged matter. In this way, Maxwell's equations teach us that a beam of sunlight is a configuration of changing electric and magnetic fields travelling through space. The same is true for radio waves, microwaves, infrared rays, ultraviolet rays, X-rays and y-rays.

In relation to electromagnetic waves, we would like to stress on one important point. It is not enough that an electromagnetic field satisfies the wave equations (14.17a and b). It must also satisfy Maxwell's equations. Altbough the wave equations are a necessary consequence of Maxwell's equations, the converse is not true. Thus, in solving wave equations, you must take special care to see whether the solutions satisfy Maxwell's equations. Only then would they represent an electromagnetic wave.

As an example, consider the electromag~letic field of SAQ 2. You can verify that it vE0 satisfies Maxwell's equations provided Eo - vBo and 3, = -T, where c is the C

speed of light given by c - I/=. Together these require that v = * c and Bo c - Eo. You can also verify that E and B of SAQ 2 satisfy the wave equations (14.17a and b). This is an example of a plane electromagnetic wave.

fig kt.% 'libc wave described by the E rad B Belds ol SAQ 2 Is shown at two different ht6 Raarbcr thnl w*(hlng varies with x or z; whtcvtr is happening at a point ou Uley.uis b h r p p n b tvtqwberr 00

tb@p*rpmdicular pknc t k ~ u g h the point As time pas- Ulc enlirr pUan slida to the dpbt bEuu~ ~ - ~ t h a ~ - ~ ~ ~ u e a t ~ ~ ~ ~ a ~ d t r ~ t u i t b d a l ~ ~ d t , ~ w ~ * d d ) - Y A L L O ~ U W & W ~ have a plnlme wave h v e l l h g with a w a s b u t spctd v In the J dirrcli01.1.

Electromapetism Did you notice that we have introduced a new term: plane electroinagnetic wave? You may well ask: What is a plane electromagnetic wave? The term plane is meant to indicate that the field vectors E and B at each point in space lie in a plane, with the planes at any two different points being parallel to each other (study Fig. 14.3). Hereafter in our discussion we shall be mainly concerned with plane electromagnetic waves, as these are found to be very useful in various areas of physics, engineering and technology. As you have seen, the electromagnetic field given in SAQ 2 is a specific example of a plane electromagnetic wave. Our interest now is to find the general plane wave solutions of the wave equations (14.17a and b) in empty space, which also satisfy Maxwell's equations.

r' 14.3.2 Plane Wave Propagation in Empty Space z

Let us consider a scalar function of the form A ( k . r - vt ). Now k . r -,d, where d ~ i g . 14.d Plane is a constant, is the equatioqof a plane whose normal is in the dirsction electromagnetic wave. Therefore, the functionA ( k . r - vt )represents a plane wave for k . r = d. We can

show that E and B fields of this form satisfy the wave equations. For this, let us express these fields in terms of their scalar components

and

We can show that these scalar components, and hence the E and B fields satisfy Eqs. (14.17a and 14.1%). The scalar components Em Ey, E,, Bm By, B, are of the form A ( k . r - vt ). Then ths fields given by Eqs. (14.21a and b) represent waves propagating in the direction k at a speed v.

A

Let q = k . r - vt. We can takeA to stand for either of the components Ex, E,, E,, B, B,,, B,. We now use the chain rule to express the differentiation ofA with respect to x,y,z and t in terms of q:

And

a2A a2j4 Similarly, you can detenniiie - and - and show that ay2 a2

Why don't you try proving this result in the following SAQ?

Spend SAQ 4 10 min

Prove Eq. (14.22a). (You will have to use the result that l2 + 2 + n2 - 1, where 1, m, n are the direction cosines of k)

Similarly, we can show that

. . . a2a 2 a ' a - T = V -

(14.22b) at q2

Compating Eqs. (14.2%) and (14.22b) yields the wave equation for A:

We can express this result for the three components of the electric field :

A A A

Since x, y, z are constant vectors, we can write that

In the same way, we can arrive at a similar result for the B field:

A

Thus, wehave shown that fields of the form E = E ( k . r - vt )and B - B ( k . r - vt ) satisfy the wave equation. The speed of the propagation of

1 wave is given by v = ------ - c, where c is the speed of light. G So far we have established that fields of the general form

A

~ ( 2 . r - v t ) a n d ~ ( k , r - v t ' )

represent plane wave solutions of the wave equations in empty space, We must also make sure that they satisfy Maxwell's equations, Let us ascertain th? conditions under which these solutions satisfy Maxwell's equations.This will give us an idea of the properties of the electromagnetic waves. We can show that if we insert these solutions in Maxwell's equations (14.12 to 14.15), we obtain the following conditions:

MaxweU's Equations and. Electromagnetic Waves

with

1 c p - (14.2&) G

If we take the magnitudes of either Eq. (14.24b) or (14.24d), we get the relationship between electric,and magnetic field strengths:

Eleetromgnetism The mathematical procedure for deriving these equations is given in the Appendix. Thus, the plane electromagnetic waves given by Eqs. (14.21a and b) satisfy Maxwell's equations yielding a set of algebraic equations (14.24a to f) in the case where there are no sources. These equations also reveal the properties of plane electromagnetic waves. Eqs. (14.24b and d) together tell us that E and B are perpendicular to each other. You can verify these properties by taking the scalar product E . B which is zero. And Eqs. (14.24a and c) show that E and B are perpendicular to k, i.e., the direction of propagation of the electromagnetic wave (Fig. 14.4). An electromagnetic wave is, thus, a transverse wave. h general E and B can have any functional dependence on r and t. Such a plane electromagnetic wave is shown in Fig. 14.5. To sum up, we have learnt that a plane electromagnetic wave in empty space has the following properties.

Fig.144: Eldmmqndic waves ue trrmversc.

Properties of plane electromagnetic waves in empty space

The field pattern travels with speed c. I 2. At every point in the wave at any instant of time, the electric and magnetic field

strengths are related. by Eq. (14.24b).

13. The electric field and the magnetic field are perpendicular to one another and also to I ille direction of wave propagation, is . , the wave is transverse.

A particularly important type of plane electromagnetic wave solution used in many areas of physics is the sinusoidal wave of a given frequency. Let us briefly consider such waves.

~onochrom'atic sinusoidal electromagnetic plane waves

Of all possible wave forms you have studied, waves of the form

would surely be the most familiar. Such waves are called monochromatic sinusoidal electromagnetic plane waves. (Monochromatic means single colour. Since, frequency corresponds to colour, especially for light, sinusoidal waves of a single frequency are called monochromatic). The argument of the cosine function ( k . r - wt ) represents the phase of the wave. In view of Euler's formula

ill e - cos 8 + i sin 8

the sinusoidal waves given by Eqs. (14.25a and b) can be written as

where Re (A) denotes the real part of A. Suppose we introduce the complex function

F = ~ ~ i ( k . r - m t ' )

Then the actual (real) wave is the real part of the complex wave

If we know g, it is a simple matter to find F. The advantage of using the complex notation is that exponential functions are much easier to manipulate than sine and cosine Wctions. Therefore, from now onwar* we shall express E and B fields using the exponential notation: I I

In general, Eo and Bo are complex vector amplitudes independent of r and t. However, when dealing with actual waves, we shall take the real parts of Eqs. (14.25~ and d). The complexity of this notation should not bother you. For ekample, you can see that the E and B fields of SAQ 2 are special cases of Eqs. (14.25a and

w b) where k is parallel to the y direction, v - -and Eo ( k, w ) I Eo ; a d

A k

B, ( k, w ) I Bo x. The advantage of dealing with waves of thk form is two-fold:

1) Any wave train can be regarded as a linear superposition of waves (for different values of w and k) of the above forms.

2) For monochromatic plane electromagnetic waves given by Eqs. (1'4.25a and b) the operations in Maxwell's equations assume a particularly simple form:

In general, we summarise this property by the equivalence relations I v * [ } - i k * ( ) (14.27a)

The symbol * means any of the operations (grad, div, curl) of V upon either a scalar or a vector quantity. With this equivalence we can rewrite Maxwell's equations for monochromatic plane waves for empty space.

k .Eo = 0 (14.28a)

kxEo - wBo (14.28b)

k .Bo - 0 (14.28~)

As we have already said, it is now well known that radiowaves, microwaves, blackbody radiation, light waves, X-rays and gamma rays are all electromagnetic radiations. These constitute what is called as the electrornngnetic spectrum. .What distinguishes each of these radiations is their frequency or wavelength (see Fig.14.6).

Fig. 1461 The dccttwnrlplclic spcdrum

C

id 10s I@ 101 16 10.' itY 10s I@ 10d 106 '1~7 I@

10 OH^

id, 101 16 id I@ I@ I@ id I@ I@ 100 '1011 ion 1013 IOU 10B loM 1017

1 l l I l l ~ r 1 1 1 1 1

7 - m ~ ~ .- i;

i ~ l i I 1 1 I I ~ ~ .

d i o . Wio . 5

miayavw

e r I I

3 -5 a

t i " h d , r .C

4 I (

!j x

Elcctromngaetlam Can you determine the wavelength of the plane em wave of Eq. 14.25a7 Give it a try.

Spend SAQ 5 5 min Show that the wavelength of the plane monochromatic sinusoidal electromagnetic

wave (Eq. 14.Za) is given by

2~ L = - (14.29a)' k

C and hence the frequency is given by f a h (14.29b)

The quantity k is called the wave number as, there are k/2n wavelengths per unit aisthnce, and k is called the wave vector.

You may now like to stop for a while and recapitulate what you have studied so far in this section: In empty space (i.e., charge-free and current- free region) Maxwell's equations predict electromagnetic waves. Maxwell's equations model a travelling electromagnetic wave as being constituted of time- varying electric and magnetic fields. The electric and magnetic fields are not produced by any external sources, but are mutually induced. Thus, an induced E (B) field arises from the changing B (E) field. The B (E) field itself arises from the changing E (B) 'field resulting in a self- perpetuating electromagnetic wave which travels with the speed of light. The E and B fields in an electromagnetic wave must satisfy the wave equations as well as Maxwell's equations. Plane electromagnetic waves of the form A ( k . r - vt ) are of special interest in physics. Such waves have the property that their E and B field vectors at each point in space lie in a plane with the pkanes at N o different points being parallel to each other. They are perpendicular to each other and also to the direction in which the wave travels. Thus electromagnetic waves are transverse waves. Even amongst plane electromagnetic waves, sinusoidal waves of single frequency [ A ( k, w ) e

- i ( w f - k . r ) ] are used very frequently in physics for many reasons.

We will now illustrate these ideas with an example. - -

Example 2

Consider an electromagnetic wave in empty space whose electric field is given by

Determine the direction of propagation, the wave number, the frequency and the magnetic field of the wave.

solution

We will make use of the properties of E, B and k to determine these details of the wave. To find the direction of propagation of the wave, let us examine the argument of the exponential.

$ompaging 10' t + Bz with ot - k . r we a n see th$ the direction of propagation is k = - z. This direction also satisfies the property k . E = 0. We can find the wave number from the relation

w lo8 The frequency of the wave isf = - = - HZ 1.67 x lo7 Hz. The magnetic 23c 23c field is of a form similar to the E field:

To evaluate Bo, we make use of the properties k . B = 0 and k x E = c B. This gives the direction of B:

A A A A A A

B = k x E = - z x x ,= -y The magnitude of Bo can be found from the relation ( E ) / I B I = c, which yields Bo = 60 /~ . Thus

Be---yexp - i 10 t + - z tesla C O A [ ( ) ]

TO ensure that you have grasped the ideas of this subsection, you must now work out an SAQ.

SAQ 6

a) The electric field given by

L \ I 1

represents the E field of a plane electromagnetic wave in a charge-free and current free region. Calculate the associated magnetic field. Find the wavelength and frequency of the wave.

A

b) Consider two electromagnetic waves travelling in the y and - j directions, respectively, described by

Show that when the two waves are both present in empty space, with the resultant electric and magnctic fields

E - El+E2, B = B1+&

they still satisfy Maxwell's equations.

Now that you have understood plane electromagnetic wave propagation in empty space, we would like to discuss another related aspect: propagation of plane waves in dielectric media. It is important to study this aspect as it helps us understand many phenomena, viz. propagation of light in glass and water; and of X-rays and gamma rays in human body, and so on. But in order to study electromagnetic wave propagation in dielectric media, we must first write Maxwell's equations for the same and derive the wave equation. This is what we shall study in the next section.

14.4 MAXWELL'S EQUATIONS AND PLANE WAVE PROPAGATION IN DIELECTRIC MEDIA

Ma.well's Equations mcl Electromagnetic Waves

Spend 15 min

As listed in Table 14.2, Maxwell's equations are complete and correct. However, when you are working with materials, there is a more pertinent way to write these equations. So let us now write Maxwell's equations inside matter. We are especially interested in dielectrics. At this Stage you will frnd it usefrl to look up the relevant equations given for dielectric nlatt$al,, in the relevant units of Blocks 2 and 3.

Electromagnetism

You know that P is simply the electric dipole moment per unit volume. For the nonstatic case, any change in the electric polarization involves a flow of charge which yields a polarization current. The corresponding polarization current density

aE Jp , which arises because pp changes with time, is given by -, i.e., at

alp Jp = -

at

Similarly, the magnetization M in a dielectric results in a bound current

JM = V X M (14.30~)

Magnetization, as you know, is just the magnetic dipole moment per unit volume. You must note, however, that the polarization current Jp has nothing to do with the bound current J,. The bound current is associated with the magnetization of the material and involves the spin and orbital motion of electrons. By contrast, Jp is the reslilt of linear motion of charge when the electric polarization changes. If, for example, P points to the right and is increasing, then each positive charge moves to the right and each negative charge to the left, resulting in the polarization current. Keeping this discussion in mind, we can write the total charge density a's

P = Pf+Pp p r - V * p (14.31a)

and the total current density as

Here pr is the free charge density and Jf is the free current density. Substituting these expressions for p and J in Maxwell's equations (Eqs. 14.8 to 14.11) we obtain Maxfiell's equations for material media:

. We can recast Eqs. (14.32) to (14.35) in a form similar to Eqs. (14.8) to (14.11). For this we introduce auxiliary fields

D = E,E+P (14.36a)

14.4.1 Maxwel19s Equations in Dielmectris Mdta Recall that dielectric materials are subject to electric polarization and magnetization. In such materials there is ad accumulation of 'bound' charge and current. You have learnt that in a dielectric, for the static case, an electric polarization P results in an accumulation of bound charge

Then we can rewrite Eqs. (14.32) and (14.35) as follows:

In an-isotropic (linear) dielectric medium, P is parallel to E ( P = co X, E ) and M is parallel to W ( M - X, M ). For such materials we can write Eqs. (14.36a and b) as

and

Here

x is the electric susceptibility, and X , is the magnetic susceptibility. In general E

and are frequency dependent. If the medium is hoinogeneous, E and p are constants, i.e., they do not vary from point to point. Given this information you can write the Maxwell's equations inside a dielectric yourself.

Spend SAQ 7 10 min

Show that Maxwell's equations for an isotropic (linear) dielectric media take the following forms:

Differential form

V . E - 5 Integral form

J S ~ . d S f (14.39)

, Eqs. (14.39 to 14.42), given Eqs. (14.36a and b) and Eqs. (14.38a and b) are the fundamental laws of electromagnetism inside dielectrics. We can now consider plane wave propagation in dielectric media which are linear and homogeneous.

14.4.2 Plane Wave Propagation in Dielectrics Let us first consider regions inside matter where there is no free charge or free current. Maxwell's equations for such regions insidk a linear medium become

V , E - 0

You can see that these equations are similar to Eqs. (14.12 to 14.15). Once again you can follow the procedure adopted in Sec. 14.3.2 and derive the wave equation for electromagnetic waves propagating through charge-free and current-free linear homogeneous media.

and

Maxwell's Equations mad Electromagnetic Waves

Electromagnetism

SAQ 8

Spend Prove Eqs. (14:43a and b) 5 min

Thus, we find that in a linear homogeneous medium, electromagnetic waves propagate at a speed

Now a well known result from optics tells us that the speed of light in a transparent medium is reduced by a factor of n:

where n is the index of refraction. It follows that n is related to the electric and magnetic properties of materials by the equation

For a dielectric p = and n = = K where K is the dielectric eonstoat of

the material.

You should realise that to be able to relate the expression of the index of refraction of a material with its electric and magnetic properties was another triumph of Maxwell's equations. After all, if the index of refraction as measured optically could be calculated so easily from the dielectric constant measured electrically, it would be a convincing piece of evidence for Maxwell's identification of light with electromagnetic radiation.

We can once again write plane wave solutions similar to Eqs. (14.21a and b)or (14.25a and b) for the wave equations (14.43a and b) for a linear, homogeneous media with no free charge and no free current. The only difference is that the . waves travel with a speed v given by Eq. (14.44) :

E - ( - v t ) ; B = ~ ( i ( . r , - v t )

with the conditions that

You can verify that monochromatic sinusoidal plafie waves of the form

E ( r , t ) = E o ( k , o ) e x p ( - i w t + i k . r ) (14.46a)

satisfy the wave equation and Maxwell's equations inside charge-free and current-free linear, homogeneous dielectric media, with the conditions that

k .Eo = 0 k.Bo P 0

These equations show that the plane wave solutions in a dielectric having no free charge and no free current resemble the'plane wave solutio~s in vacuum: B is perpendicular to E and the wave travels in the direction of k which is perpendicular to both E and B. What is the difference? The speed at which the wave travels in the dielectric is different from c, the speed of light in vacuum, by a constant factor, termed the index of refraction (in optics).

.Thus, if we compare a wave in a dielectric with a wave of the same frequency in vacuum, the wavelength in the dielectric will be less than the vacuum wavelength by a factor [ l/n ), since frequency wavelength = wave speed.

Now in the last section of this unit we shall consider another interesting aspect of electromagnetic waves. Consider the following situation. When you sit outdoors on a cold winter morning in bright sunshine, you feel warm after a while. Why does this happen? Obviously, it is the energy carried by the sunlight which gets transferred to you and gives you this pleasant sensation. You already know that waves transport energy from one region of the space to another. In this section, we will determine the amount of energy carried by electromagnetic waves across the space. '

14.5 ENERGY CARRIED BY ELECTRO GNETP1IC . WAVES : POYNTING'S THEOREM

Recall that in Unit 4 , we expressed the work necessary to assemble a continuous static charge distribution (against the Coulomb repulsion of like charges) as

We can rewrite this expression in terms of the electric field E. We first use Gauss' law to express p in terms of E:

Thus

Now we use the following vector identity

V . ( E V ) = ( V . E ) V + E . ( V V )

Since E - - VV, we get

Applying Gauss' divergence theorem to the first term, we obtain

Now suppose we enlarge the volume to include all the cha&z-Any extra volume will not contribute to WE, since p - 0 in that voluhe. But as we enlarge the

volume, the integral 012 can only increase, since the integrand is positive. Therefore, the surface integral must decrease so that the sum remains the same. If we integrate over a11 space, the surface integral goes to zero, and we are left with

Maxwell's Eguatiors and Elfftaromagnetic Waves

Electromagnetism where E is the resulting electric field. Likewise, we have in Unit 13 shown that the work required to get currents going (against the back emf) is (Eq. 13.24)

where B is the resulting magnetic field. Evidently, the total energy stored in a current and charge distribution can be expressed in terms of electric and magnetic field produced by this distribution as

We would now like to derive Eq. (14.48~) more generally, keeping in view the energy conservation law.

Suppose some charge and current configuration produces fields E and B at time t. Suppose the charges move around. We would like to know: Wow much work, dW, is done by the electromagnetic forces on these charges in the small time interval dt? According to the Lorentz force law, the work done on an element of charge dq is

Now, dq - p dVand p v = J, so the total power delivered on all the cl~arges i11

some volume V is given by

Let us express Eq. (14.49) in terms of the fields alone, using Amphe-Maxwell's law to eliminate J:

A well known vector identity gives us the result

Combining this result with Faraday's law, V x E = - - at 7 it follows that

We can also write

so that

Putting this into Eq. (14.49) we get

Now applying divergence theorem to the second term on the RHS, we hav.e

where S is the surface bouding the volume

Eq. (14.51) is the mathematical statement of Poynting's theorem; it expresses conservation iof energy in electromagnetism. The first integral on the right represents the total energy stored in the fields, WEB(Eq. 14.48~). The second term represents the rate at which energy is carried out of V, across its boundary surface, by the electromagnetic fields. Poynting's theorem says, then, that the rate of work

- done on the charges by the electromagnetic force is equal to the decrease in energy stored in thefield, minw the energy which flowed out through the surface.

The quantity

is called Boynting's.vectos; it represents the energy flux density - that is, S.da is the energy per unit time transported by the fields across a surface da. We can state Poynting's theorem more compactly in terms of S and WEB:

Of course, the work W. done on the charges will increase their mechanical energy (kinetic, potential, or whatever). If we let U, denote the mechanical energy density, so that

and use UEB for the energy density of the fields,

i then i

i and hence I

This is the differential version of Poynting's theorem. Compare this with the continuity equation expressing conservation of charge:

The charge density is replaced by the total energy density (mechanical plus electromagnetic), while the current density is replaced by the Poynting vector, Thus, the Poynting vector S describes the flow of energy in the same way that J describes the flow of charge.

Let us now.recapitulate what you have studied in this unit.

'\ Maxwell's ]Equations and Electromagnetic Waves

-

m Maxwell's equations constitute the fundamental set of differential equations describing electric and magnetic fields. These equations in their integral and differential forms are tabulated below for different situations.

i

Maxwell's equations In vacuum in charge-he and current-free regions V . E - o

aB V X E - -- at

V . B - 0

aE V x B - P o E o Z

Maxwell's equations in vacuum In the presence of charges and currents

Maxwell's equations allow wave solutionsfor electric and magnetic fields in vacuum and in dielectric media. These equations also model electromagnetic waves, as constituted of time-varying, selfperpetuating, electric and magnetic fielak, which are mutually perpendicular and perpendicular to the direction of propagation. Ehctromagnetic waves ate transverse waves. The electromagnetic wavespropagate in vacuum at the speed of light. For

- i ( u t - k . r ) monochromatic sinusoidal electromagnetic waves of the form Eo e , Bo e-i(u~-k.r) , Maxwell's equations in charge-free and current-free empty space simplify to yield the following set of equations.

k.Eo - 0 k.Bo - 0

' k xEo = oB, 0 kxB0 = -,Eo C

J , E . ~ s - o

d@E % ~ . m = -- dl

J , ~ . d s - 0

d @ ~ $ c ~ . d -

P V . E - - €0

dB V X E m -- at

V . B - o

V X B - p,, ( ~ t ~ ~ $1

Maxwell's equations inside isotmplr, linear dlelectdc media

Such electromagnetic waves are described by their angular freq~ency o), the speed of propagation c (in empty space), the wave number k, the wavelength A.

4 J s ~ . d s .. - €0 -

~ O B $ c ~ . d ~ = -- dt

J s ~ . d s = o

t jCB .d l = p,,,i+weo- ~ Q E dt -

V . E - B e

aB V X E = - - at

V . B - 0

V x B - p ( ~ , t ~ $ )

In a linear dielectric medium, Maxwell's equations allow w.ave solutions for

4 - ; d@E $..dl - -- dt

J'l3.d~ - 0

$ c ~ . d l = p i t p c - d@fi dl

1 ' C electric and mainetic fields which propagate at a! speed v - - - - where fi !n' n is the refractive index of the medium. .. .

.1 The Poynting vector S = - ( E x B ) represents the energy flux density, i.e.,

Po the energy per unit time transported by electromagnetic fields across a surface. Poynting's theorem expresses the conservation of energy in electromagnetism. The work done on the charges by the electromagnetic force is equal to the decrease in energy stored in the field, minus the energy which flowed out through the surface.

IN& QUESTIONS Spend 30 min

1. A plane electromagnetic sinusoidal wave is characte~ised by the following parameters: the wave is travelling in the direction - x ; its frequency is 100

n megahertz (MHz, lo8 cps); the electric field is perpendicular to the z direction. Write down the expressions for the E and B fields that specify this wave.

2. Show that the electromagnetic field described by

E - E, ices kx cos 6 cos wt

B = Bo ( coskr sin ky - ; sin kx cos ky ) sin a t . will satisfy Maxwell's equations in charge-free and current-free empty space if Eo = f i c ~ ~ a n d o = f l c k .

3. The electric field of an electromagnetic wave in vacuum is given by

where E is in volts per meter, t is in seconds, and x is in meters.

i ) Determine the frequency v.

ii) Determine the wavelength h.

iii) Determine the direction of propagation of the wave. .I.

' iv) Determine the direction of the magnetic field. '

14.8 SOLUTIONS AND ANSWERS

Self-Assessment Questions (SAQs)

1. The conduction current density is

J = a E - a ~ ~ s i n o t ~ m - ~

The displacement current density is

where w = 2 x f and f - lo6 Hz. The ratio of the magnitudes of these.current densities is

For copper at 1 MHz

Maxwell's Equations and Eiectromagnetic Waves

For T~,ilon at 1 h1H7

Thus, for copper the conduction current dominates the displacement current. But for Teflon which is a reasonably k ~ o d insulator at 1 MHz, the conduction I

current can be neglected in comparisonbith the displacement current.

2. Substituting the E and B fields in the Maxwell's Eqs. (14.12) to (14.15), we get

p a c d A a A

= ( I - t j - + k - ) . kE ,s i n ( y - v t ) a x a y a t .

a - -[E0sin ( y - v t ) ] az - 9

Thus, Eq. (14.12) is an identity.

a~ A - = - ivBOcos(y-vf) at

Eq. (14.13) gives the condition that

Eo = Bov

a iii) V. B = -Basin ( y- v t )

ax

1 aE iv) V x B = --

C2 at

where

~ a B , s i n ( y - v t ) A V x B = -k = - kBocos (y - v t )

8~

vE0 Eq. (14.15) gives the condition that Bo = C

These two conditions, Eo = Bo v and Bo = ' tugether mquire Lhat 7

1 and Eo Boc l 1

11 3. Maxwell's equations for charge-free matserial media are '1

V x B = p J + E - ( Z ) where J - a E.

Once again taking the curl of the second equation we can write

The LHS is simplified as follows

v x ( V X E ) - v ( v . E ) - $ E

= - V ~ E sinceV.E I 0

Thus, we get

which is Eq. (14.18a).

Similarly, taking the curl of curl B, we get / a V x ( V x B ) - p ( V x u E + e - V x E )

at

The LHS yields

V X ( V X B ) v ( v . B ) - V ~ B .." .? " - -9% [.: V . B - 0 ]

.- Thus, we have

Maxwell's Equations and Eketromrgnctic Waves

which is Eq. (14.18b).

Similarly,

a2A d2A A A 2 7 = - ( z * k ) az a

now a2A a2A a2A

d 2 ~ A A 2 A A 2 A - - ~ [ ( r . k ) + ( y . k ) + z . g ) i

A A A A A A drl

A

But x . k , y . k and z . kare the direction cosines of k Using the result that A A 2 A A 2 A A 2 a2A

( k ) + ( y . k ) + ( z . k ) = l , w e g e t $ ~ = 7whichisEq.(14.22a). %

5. For mathematical convenience, we choose the direction of propagation to be in the x-direction. The expression of Eq. (14.25a) simplifies to

Now the cosine function completes one complete cycle when x advances by the wavelength h. Thus, we must have

This condition .is satisfied only if

Since wave speed - (frequency) x (wavelength),

C *

we have frequency f = -

6.a) The associated magnetic field can be obtained from Eqs. (14.24 a to f). Using Eq. (14.24b) we have

1 A - -kx1000f exp 9 c A A

2 y - z k 2;-l and k = - whence k - - 100 I k l * 6

Thus the associated magnetic field is 58

firirdl'e Equ.tiwr and EkctroM@I& W ~ V &

This is consistent with Eq. (14.240 since

The wavelength of the wave is

c 3x ldms- ' = l$s-l The frequency of the wave is f - - - A 300m

b) The resultant electric and magnetic fields can be expressed as follows:

E - El+&

and B - B1+&

- - 2 illo ms 2*Y sin h

': sinA-sinB = 2cos- 2

sin -

We have to show that the resultant E and B fields satisy Maxwell's equations:

and

T ~ U S , V X E -- dy yields Eo = ow

Similarly, V x B - - - dE yields the following c2 at

dE - 2 n c -z2EO- asin 2"" sin - at A A A

This is the same as Eq. (14.24f).

Thus, the resultant E and B fields of the plane electromagnetic wave. satisfy '

Maxwell's equations. Such a wave is called a standing wave.

7.(i) From Eq. (14.32) we have

V . ( e 0 E + P ) = pf

or V . D = pf

or E V . E = pf (Since D - e E from Eq. (14.38a))

To cast it in ihe integral form we integrate both the sides in a volume V

In the RHS d the equation$ pf dV - q, the charge enclosed by the volume. v

To the LHS we apply Gauss's divergence theorem and obtain

where S bounds the volume V. Eqs. (14.40) and (14.41) are the same as Eqs. (14.33) and.(14.34).

ii) V x B -

aD V x H - dl + C\O JI psing Eqs. (14.36a and b)]

aD V x H - --+ Jp at

Using Eqs. (14.38a and b) we can write

which is Eq, (14.42).

To cast this in the integral form we integrate both the sides on an open surfiw S .

The surface integrals on the RHS are related to the flux and the current

Applying Stokes' theorem to the LHS we get

where the open surface S is bounded b; the tontour C.

8. Once 'again we take the curl of V x E:

Taking the curl of V x B:

. a V x ( V x B ) - p e - ( V x E ) at

Terminal Questions A

1. Since the wave is travelling in the ; direction, E field wilibe normal to x. It is given that E is perpendicular to z. Therefore, E isin the y direction. The expression for the E field is thus

where . w - 2 n f - 2 ~ c x 1 0 ~ ~ z

and 21c k - - - h c

The corresponding B field is given by

2.' Maxwell's equations in charge-free and current-free empty space are

V,E - 0 V . B = 0

aB iii) V x E P -- at

1 dE iv) V x B - -- . C2 at

2 W or. --r - 2c2 k

and W Eo = -Bo = ~ Z C B , k 3. Comparing with the expression E Eo cos ( la - ot )

i) The frequency of the wave is

v - 1o8HZ(~: 0 - 2 3 % ~ )

2n 2n.(3)13m ii) The wavelength is h, = - I k 2 3d

A

iii) The direction of propagation is in the + x direction.

iv) Since E is along + j and k is along + ; the B field will be$long + z direction.

APPENDIX: H LATIONSHIP BETWEEN E,

You have studied in the unit that plane waves of the form ~ ( & . r - c t ) and ~ ( k . r - c t )

satisfy the wave equations in empty space. We will now insert these wave forms into Maxwell's equations for charg~ tee and current-free empty space, and obtain the relationships between E, B and k For this purpose, we will consider each of Maxwell's equations separately.

I. V . E - 0 A A A

Let E - xE,+yE,,+zE,

Thus, we can write Eq. (A.la) as

(A. la)

(A. lb)

(A. lc)

Here Em E , E, are functions of ( & . r - ct ). Let us make the substitution A

q - k.r-ct

Using the chain rule, we can write

aE'%-dE'(r.;) - =- a~ at, a~ (See Sec. 14.3.2)

aE aE aE, aE, A A

~imiluly, - - $ ( &. ; ) and - aY - - ( k . z )

az al Thus,

or

A a A A A V.E E k.-(E,x+E,y+E,z) (.; ~e$~p rod~ i scommuta t i veand q X, y, z are constant vectors. )

Thus, Maxwell's first equation becomes

aE This implies that i; is perpendicular to - We can show that this also implies that k al* A

is perpendicular to E. For this we assume that E iq along k, i.e., E - f ( q ). Then

and

unless f ( q ) is a constant. put for waves f ( q ) is not copstant, Therefore, E can have no component along k. Thus E is perpendicular to k, that is

Using a similar method as in (I), you can show that

Maxwcll'~ Eqartiws nnd Eke(rornalplctk Waves

Structure .

15.1 Introduction Objectives

15.2 Reflection and Refraction at a Boundary Between Two Dielectric Media Boundary Conditions Reflection and Refraction at Normal Incidence Laws of Reflection and Refraction

15.3 Generation of Electromagnetic Waves Radiation from an Oscillatory Electric Dipole Antenna

15.4 Summary , 15.5 Terminal ~ h t i o n s . 15.6 Solutions &swers

1 INTRODUCTION

In Unit 14, you have studied Maxwell's equations and derived equations for electromagnetic waves from them. You have learnt that light and several other forms of radiation, viz. radiowaves, infrared, X-rays and gamma-rays are electromagnetic radiations. You have also obtained plane wave and sinusoidal solutions of the electromagnetic wave equations in empty space devoid of free charges and currents, and in dielectrics. In this unit we come to another interesting question related tb electromagnetic waves in dielectric media. What happens when an electromagnetic wave passes from one dielectric medium to another? For example, you~how what happens when light pass8 ' from air to glass or air to water. You get a reflected wave and a refracted wave, which we also call the-transmitted wave. We shall derive the equations for the reflected and transmitted electromagnetic waves, when such waves are incident perpendicular to the boundary of the media. So far we have not talked about how electromagnetic waves are generated. In the last section of this unit, you will study how an oscillating electric dipole produces electromagnetic waves. Finally we shall briefly discuss the antenna - a device widely used to transmit and receive electromagnetic waves.

0 bjec tives

After studying this unit you should be able to

I solve problem based on reflection and refraction of electromagnetic waves at . the boundaries of dielectric media

e explain qualitatively the generation of electromagnetic waves from an oscillating electric dipole antenna.

15.2 REFLECTION AND IWFRACTION AT A By the term linear, we mean $at D and H are proportional to E

I and B . Also e and p are

BOUNDARY BETWEEN TWO DIELEC MEDIA

I I inaependeot of position and

direction. Thus, we are takiig I about linear, homo~enaous and Let us consider the situation in which uniform plane electromagnetic waves are

isotropic media. incident on a boundary beisveen two linear, dielectric media, e.g., light passing from air

to glass, or from oil to water. Let us assume that there are no free charges or currents in the materials.

Reflection aud k e f r a d m d Electromagnetic Waves

rctkded (Irnsmitted) waves.

Fig. 15.1 shows a plane boundary between two dielectric media having different material properties: E, , p1 for medium 1 and E~ , p2 for medium 2. A uniform plane wave travelling to the right in medium 1 is incident on the interface of the media, normal to the boundary. What happens to the wave? As in the case of waves on a slring, and from our experience, we can expect to get a ref ected wave propagating back' into the medium a11d a transmitted (or refracted) wave travelling in the sedond medium. We would like to determine the expressions for the reflected and refracted waves in terms of the incident wave. We would also like to know what fraction of the incident energy is reflected and what transmitted? In order to do this we would need to know the boundary conditions satisfied by the waves at the interface of the media. These are the conditiotts we obtaitt wtteta we stipulate tltat Maxwell's equations must be satisfied at the boundary betweett the media. So let us first obtain the appropriate boundary conditions.

15.2.1 Boundary Conditions We will derive these boundary conditions from Maxwell's equations in a dielectric fr& of charge and currents. The relevant equations are:

(a) E S E , d S - 0 (15.l.a) S over a closed surface S.

Fb) J l 3 . d ~ = 0 (15.1b) S

by the closed loop C. 1 d

(d) -f ~ . d = r - S E . ~ s (15.16) P c dt s

d (c) # ~ . d - -;iiSs B . d S

C

@) Fb.lJ.2~ The pcmitivc d i d o n for S m d E l in (a) i s kcinn m d u m 2 lowanla 1,

(15.1~)

for any surface S bounded

Let us apply Eq. (15.la) to a tiny, thin Gaussian pill box extending just a little bit (hair-like) on either side of the bougary df the niedia (Fig. 15.2a). Eq. (15.1a) implies

The edge of the wafer contributes nothing in the limit as the thickness goes to zero. Thus, the components of the electric fields perpendicular to the interface satisfy the condition

E ~ E ~ - E ~ E ~ = 0 (15.21)

Using the same process, we obtain from Eq. (15.lb) the following boundary condition for the magnetic fields:

Bu-Bu 0 (15.2b)

We now apply Eq. (15.1~) to a thin Amgrian loop across the surface(Fig. 15.2b) and obtain

d El . I - E 2 . 1 - --$ B,dS dt s

Now in the limit as the width of the loop goes to zero, the flux vanishes and the

contribution of the two ends to$ E . dl is zero. Therefore, C

(E l -EZ) . I 0

which implies that

This means that the coiponents of E parallel to the interface ar2 continuous across the boundary. In the same way, from Eq. (15.ld) we can obtain the condition that

which yields

Thus, we have derived the boundary conditions satisfied by the electric and magnetic fields at the interface of two linear dielectric media where there is no free charge or current. Let us put them together :

Ell s &I (15.2~)

Wl=,A -* cl-c wave ibddent nom.Uy at the boundyy of two m& , -

We shall now use these boundary conditions to study reflection and refraction (transmission) at normal incidence.

15.2.2 Reflection and Refraction at Normal Incidence

Consider Fig. 15.3. Suppose theyzplane forms the boundary between the two linear dielectric media. Let a sinusoidal incident plane wave of frequency w,travelling in the x direction approach the interface from the left. Suppose its electric field is along they direction. The electric and magnetic fields of the incident wave are given by

E I ( x , t ) 5 ~ , f e x p [- i (wt-kIx)] (15.3a)

At the interface of the media, the incident plane wave gives rise to a reflected wave and a refracted (or transmitted) wave. The reflected w a ~ e ' ~ r o ~ a ~ a t e s back into the f i t medium and can be represented by the following E and B fields:

A

E R ( x , t ) = E O R j e x p [ - i ( w t + k l ~ ) ] (15.4a)

Why have we put the minus sign in Eq. (15.4b)? This is because the direction of , propagation is reversed and the fields of the wave must obey the relation

Will you now like to try writing down the E and B fields of the transmitted wave which travels to the right in medium 21 Fill in the following blanks.

These three electric and magnetic field vectors must satisfy the boundary conditions given by Eqs. (15.2) at every point on the plane interface at all times. Thus, at x = 0, the combined field to the left, viz. E, + ER and BI + BR, must join the fields to the right, E, and B, according to the boundary conditions. In this case there are no field components perpendicular to the interface, since neither E nor B field i's in the x direction. Thus, Eq. (15.21 and b) are trivial. ?'he remaining Eqs. (15.2~ and d) require that

where Cl1 Vl Cll n2

P2 "2 P2 n1

You may like to solve Eqs. (15.6a) and (15.6b) to obtain the reflected and transmitted amplitudes in terms of the incident amplitude. Try the Iollowing SAQ.

SAQ 1

Show that

l - a = (%)EN

For most dielectric media, the permittivities are close to their values in vacuum. In v1 such -es a - - and we have "2

R d l d o a and Rt&diw d Electromagnetic Waves

Spend 5 min

I

n, ~:r ! (a) m d k d 1 ,

wave j4 in phase with the inddent wave (b) the rtflcclrd J

wave bout or pburc with (be /"

i d e n t wave.

69

Electromagnetism

Spend 10 mi11

The reflected wave is in phase with the incident wave if v2 > v , and out of phase if C

v2 < v1 (See Fig. 15.4). In terms of the index of refraction n = -, we can write Eqs. v * (15.8) as

when nl< nz, i.e., when the wave passes from a less dense medium to a more dense medium, the reflected wave is 180 "out of phase with the incident wave. This is well known in optics.

What fraction of incident energy is reflected and what fraction transmitted? You may like to work out this result yourself!

SAQ 2

Given that the intensity (average power per unit area) is 1

I = - F v & , and p1 = k2 = pO, show that the ratio of reflected intensity to 2

incident intensity is

and the ratio of the transmitted intensity to the incident intensity is

Show that R + T = 1. (15.10~)

R is called the reflection coeEciemt and T, the transmission coefficient of the surface. They measure the fraction of incident energy that is reflected and transmitted, respectively. You can now explain why most of the light is transmitted when it passes from air ( n, = 1 ) to glass ( n2 = 1.5 ). You only need to calculate R and T to know the answer! Why don't you do so before studying further?

We will end this section by deriving the laws of reflection and refraction in a simple manner for the case of oblique incidence.

15.2.3 Laws of Reflection and Refraction

Consider Fig. 15.5. An incident plane wave in medium 1, at an angle €II, results in a reflected wave in 1 at an angle 9, and a transmitted wave in medium 2.

We represent the waves by the following plane wave forms:

Recall that in Sec. 15.2.2 we had said that the boundary conditiohs (Eqs. 15.2) must hold at every point on the interface at all times. If the boundary conditions hold at one point and at one time, they will hold at all points and all times only if the phases of the three waves are equal, i.e. . '

H~15.51 A p h wave, represenled by propegi&on vector k, encountem m Lntedmcc bt twep two m d i ( sir and 8% pz ) . ?lhe normal to the interface, pointing into medium 2, is the unit vedor n.

The equality of phases at all times requires that the three angular frequencies must be the same:

0 Since k = - n, we have

C

Equality of phases for all points on the interface requires that

k,.r - b . r - kT.r (15.12)

for all r on the interface. These equations yield

( k , - b ) . r = 0 and (k , -kT) . r = 0

Since r * 0, the above equation tell us that either

i) k, = handk, - kT

or

ii) ( k, - k, ) and ( k, - kT ) are perpendicular to r, for all r on the surface.

Case(i) is the trivial case of no reflection and no refraction. So we consider only case (ii). Since r is any vector in the plane interface, condition (ii) will be satisjied only if (*k, - k, ) and ( k, - kT ) are along the normal to the plane interface. If n represen'ts the unit vector normal to the plane, it will be parallel to ( k, - b - ) and ( k, - k,. ). So we can write that

Refledon and Rtnprtioo oP Electromnpetic Wnves

and k(&-k , ) 0

This gives us that A

k x k , - n x k , = C x k , (15.i3) 3

Eq. (15.13) gives us the following informations.

1. ; x k, - ; x k, says that the plane iefined by i, 4 (the phne of incidence) coincides with the plane defined by n, kT (the plane of refraction).

2. The equality of magnitudes

I ; x k, I = 1 ; x k, ( gives the condition that

kI sin 8; P kT sin 8,

3. Similarly the relation

A

tells us that the plane of incidence coincides with the plane formed by n and b, the plane of reflection and

kI sin 0, - k, sin 0, (15.15)

Since k, = kR from Eq. (15.llb), Eq. (15.15) yields

OR = 01 (Law of reflection) (15.16)

Substituting Eq. (15.14) in Eq. (15.11b) we get

sin 8, n2 'P-

sin eT n l (Snell's L a w of refraction) (15.17)

'

Thus, we have arrived at the well known laws of reflection and refraction in optics.

We end this section with an SAQ for you.

spend SAQ3 10 min

(a) A uniform plane wave whose electric field is given by

is incident from a region having = 4 E, pl = pO normal to the plane surface of a material having EZ = 9 E& ~ 4 1 1 4 po.

Write complete expressions for the incident, reflected and transmitted electric and magnetic fields.

(b) A plane electromagnetic wave propagates from one dielectric to another at normal incidence. Find the ratio of the indices of refraction of the two dielectrics for which the reflection and transmission coefficients are both equal to 0.5.

15.3 GENERATION OF ELECTROMAGNETIC WAVES

So far you have studied aboui the existence of electromagnetic waves, their properties in free space and in dielectrics, and their reflection and refraction at the boundaries of dielectric media. But how are these electromagnetic waves , generated? This is the question we are going to answer in the present section. From what you have studied about the nature of electromagnetic waves, you can

72 - -

-

guess that to generate such a wave, we must create a changing electric or magnetic field. Once we do so, the self-regenerating process described by Amp6re- Maxwell's Law and Faraday's law occurs, and the wave propagates on its own. So how do we create a changing electric or magnetic field?

We create changing fields when we aldr the motion of an electric charge. This can be done by accelerating an electric charge. It is also possible to create such fields when an electric dipole oscillates. Let us now study qualitatively how electromagnetic waves are generated from oscillating electric dipoles.

15.3.1 Radintio~l from an Oscillatory Electric Dipole Fig; 15.6 shows an experimental arrangement depicting an oscillating electric dipole. Since an ac source of current is connected to the wires, charges will move back and forth along the vertical wires. This will produce an electric dipole, alternately pointed up and then down. Currents also flow in the wires, producing a magnetic field B around the wires. The B field also reverses direction periodically. Let us now examine the electric and magnetic fields produced near the oscillating dipole. When the charges near the ends of the wire are at a maximum, the current will be zero. Consider a point P near the wire. The E field at P will be a maximum when B is zero, and vice-versa. Thus, near the wire, E and B are 90 " out of phase in time. Hence, the direction of energy flow which is determined by the Poynting .

1 I

vector S = - ( E x B ) alternates between outward and inward directions. So the h

net energy flow at P is zero. Thus, the near field of an oscillating dipole moves energy in and out equally and does not result in a radiated wave. - What happens at distances far away from the oscillating dipole? Recall that time-varying E and B fields can produce B and E fields, respectively. he time- varying electric and magnetic fields generated by the oscillating dipole continue moving outward, because the local time-varying E fields produce in-phase B fields and in-phase E fields are produced by the local time-varying B fields (in accordance with the two Maxwell equations for induced electric and magnetic fields). Thus, the total E and B fields at P are the vector sums of the out-of-phase components generated by the oscillating electric dipole and in- phase fields induced locally by the local time-varying fields. The energy density of out-of-phase

1 components falls off rapidly with distance as The in-phase components provide

r the net transfer of energy'and become the dominant fields at large distances from the dipole. We can see that the time-varying fields near the dipole serve to launch the electromagnetic waves at large distances.

An interesting device based on this simple process of generating electromagnetic waves is the antenna. Can you think of any modem ~ommunication network without an antenna? An antenna is used for the generatjon (transmission) as well as for the reception of electromagnetic waves. In the last subsection of this Unit we are briefly going to discuss the antenna. I

/ 15.3.2 Antenna

i The antenna based on the mechanism of an oscillating electric dipole is termed the oscillating electric dipole antenna. It is, perhaps, the simplest of antenna systems. Such an antenna consists of two conducting rods connected to an ac generator as shown in Fig. 15.6. Fig. 15.7 shows the electromagnetic wave generated by this kind of antenna. The field pattern shown in the figure holds for all radial distances r from the antenna such that r z k, where A is the wavelength of the electromagnetic wave. Remember that certain electromagnetic radiations such as X-rays, gamma rays and ligbt come from atomic and nuclear sources. In this case, we have restricted outselves to the region of the spectrum ( I - lmm to l m ) in which the source of radiation is both macroscopic and of

manageable dimensions. Essentially we are speaking of radio wave aisd microwave generation.

b l L 6 1 An 0dll.ting electric dlpde. 'Zbe E rod B IkW mdting ltom CURYULS and cb~rges on the wlrr are shown dongwlth Ibe Poynting vectors. At the instu~t shown, Ulc PoynUng vector is dirrdtd outward at the point P. At p a t e distance6 them i s a nd llorrof urtrlly, p l a d d by tbe the-vay lng E and B lklda

FTg.15.7: lh t k d r i c m d magndic field lines associated with tbc clcrtnom.gn& wave rsdtttd by la

oscillating dipok rnlcnn~ Ibc doh and crcrsses represent Udd lines anedng f m m rad entering into (be phne of the flgarr The Uekl pattern close to (he antenna is not shown.

Antennas are made of metals as well as dielectrics and come in different shapes. In fact, any metallic or dielectric structure which is designed so as to launch (or radiate) waves efficientIy into space and to focus (or concentrate) these waves in a particular direction is referred to as an antenna. Two common antennas in use today for various purposes are shown in Fig. 15.8.

(a> @) Q.15.8: (a) Monopdc (Made) mttnna d on (b) log-paiodie mtmnr.

Let us now summarise what you have studied in this unit.

e The boundary conditions satisfied by electromagnetic wavei at the interface of two charge-free and current-free dielectric media charaderised by el, p1

'

(medium !) and €2, (medium 2) are given as follows:

RdlcdIau and Reeradim d Ekctnoolagnetic Waves

The boundary conditions are consequences of Maxwell's equations.

o The iacident, reflected and transmitted waves at the interface of two dielectric media can be represented as follows:

E, = E, exp [ - i ( w , t - b . r ]

ER - EOR exp [ - i ( o R t + k , . r ]

ET = EOT exp [ . - i ( o $ - k ~ . t ]

. The amplitudes of the electric fields of the reflected and transmitted waves when the incident wave is normal to the interface of the dielectric media are:

where

The reflection and.transmission coemcients are defined as the ratio of reflected intensity to incident intensity, and that of transmitted intensity to incident intensity, respectively. For the dielectric media for which p1 - ~1;? = pch these are given as

0 The law of reflection and Snell's law of rekaction for oblique incidence are derivable from boundwy conditions in electromagnetic fields.

0 An oscillating elhctric dipole generates electromagnetic waves. The antenar, a device used for radiating. and receiving electromagnetic waves in the radio

. and micro frequency region is based on this principle.

15.5 TEFWINAL QETESTIONS Spend 20 min

1. A uniform plane wave has a wavelength of 3 cm in free space and 2 cm in a dielectric for which p - 4.7 x NA-~. Determine the dielectric constant of the dielectric.

2. A uniform plane wave of 200 MHz travelling in free space strikes a large block of a material having E - 4 E, , p = 9 and a = 0 normal to the surface. If the incident magnetic field vector is given by

write complete expressions for the incident, reflected, and transmitted field vectors.

115.6 SOLUTIONS AND ANSWERS

Self-Assessment Questions (SAQs)

1. Summing Eqs. (15.6a and b) we get

Substituting EoT in Eq. (15.6a)

l-a Em - ( & - I ) & - (=)GI

P1 n2 " 2 where a = - - - in this case,since b1 = p2 P z n1 n1

2 1 2 1 since n: - c/vl , n, - c/v , , vl E - , v2 = - CLlE1 CL2 €2

A

Since k is in the x-direction, BI will be in the y-direction. Its magnitude& given by the relation

lEIl . , - = vl , where v, = 1

IBII Tr It is given that 5 4 E ~ , p1 = p0

The reflected wave travels back in the negativex- direction in medium 1. The E and B fields of the reflected wave are

where EM - ( L : ) E ~ ~ -

1 C where C - 1 V 2 - -- r -ms

6

and

Thus,

loo ER = - c o s ( ' u t + 6 x r ) ~ ~ r n - ' 7

and

,The E and B fields of the transmitted wave are Y

where

and

and

b) It is given that R T - 0.5

Let

Then we have that

or I-p+p2 = o whence fl 1 5.83.

Terminal Questions

E ' 1. The dielectric constant is given by u = -. It is given that the plane wave has €0

wavelength 3 cm in free space and 2 crn in the dielectric. Hence, the speed of the wave in the dielectric is

I

Thus

2. The incident magnetic field is given by

0 2 -1 whereo = 200MHz and (3 = - = -m . c 3

Titus, the wave is travelling in the y-direction. So the electric fiefd is in the x-direction. The magnitude of the electric field can be obtained from

a

The reflected wave travels in the negative x-direction with amplitude

where

. .

and A

B, = - 2 x 1 0 ~ c o s ( o l + ~ ~ ) z t e s 1 a

The E and B fields of the transmitted wave are

where

and *

and

FURTHER ING

1. Electricity and Magnetism - Volume 2; Edward M. Purcell; International Student Edition, McGraw-Hill Book Company; 1985.

2.. Introduction to Electrodynamics. - David J . Griffiths; Prentice.Hal1 of India Private Limited, 1984

3. Fundamentals ofElectricity and Magnetism - Arthur P. Kip; International Student Edition; McGraw-Hill International Book Company; 1984

Table of Constants /

I _

Symbol QusnUty Vnlue I

c speed of light in vacuum 2.998 x 10' m s-'

PO permeability of free space 1.257 x I O ~ N I So permittivity of free space 8.854 x lo-" C ~ N - ' ~ - ~ I c chargeof the proton

--c chargeof the electron - 1 . ~ 2 l0 l9c I Planck's constant

hm

elmtron rest mass

electroncharge to mass ratio - -1.759 x 10 ' '~ kg-'

"'P proton rest mass 1.673 x I mn neutron rest mass 1.675 x l ~ - ~ ~ k ~ I

Bohr radius

Avogadro constant

R Universal gas constant 8.314 J K-' mol" I b a Boltmann constant . 1.381 x 10-"J K-' I G Universal gravitational constant 6.673 x 1 0 - " ~ , r n ~ k ~ - ~ 1 .

W O K W I S E LIST OF STUDY ClNTRW FOR R%r. PR

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1. H Y D m RECION(Andbm PPrrduL) 1 0102 V R College. Nellom-524 00 1, .ban ~adnh 2 " 0103 KBN CoUcgq Kolhapa. Vijsyewoda.520 001, Arrlbn W h 3 Ol l l Aurora's D q m College, Hydcrabd-500 020,

AodblrF'mdUh 2. m m n R E G I O N ( h u m . .4nlorhai P d a b B si-) 4 0401 w w h d ~ n i v ~ l y . Guwahati-781 014, . bum. 5 01lM Bijhora Mahavldydaya. Bqa~gaon-783 280. Cuwahrtl 6 0408 Mcdque Girh Collepc, G u w h - 7 8 1 100. A=rn 7 W09(p) G m . Scicnm College, Jorhat-705 010, Arum 8 Ojl l B q d i College. P d x d b . P&da P.O.. B a r p A a M d - 7 8 1 325,

b m 9 0116(D) Dcbroj Roy CoUtge. GOI~I&P.O., Golagha-785 6,l. h a m 10 041'3 Ldkkmnplrr G i h College, kkr18ti P.O., SdLdEbunplr-7gl031. Asnm 11 2431 S& G m t ' C o U c s Tadcmg Gmgtok-Z7 IO22Sikkim 3. PAT?iA RKGION (Blhar) 12 0501 Vanijyn Mahavidyulay. P u U m ~ l i t y , Prtn.-8W 005, BBn

O'ma Science College, Paina, W a r ) . 13 OStU 13.RS. Bhar Univcnity L i b ~ ~ ~ ~ - g l l 0 1 ) 1 . B i b r (LS L'ellqc.

Mudfnrpw. Wur). I4 0505 Manvnri Collcg~~bI.Bhegalpur Univmity), Bhagdpur-8 12 007. BUlu I5 0508 h a Collcgc, Rtrrun-854 MI. B l h r 16 0509 IbjcndFa CoUege, C b p r % M l 301, B l h r 17 OSl5R Bdika Vidyapztb Lakhiaariu-811 31 1, Blhr. I B 0521 S i d CoUce. P.O. Sindri-818 121 Dhmbad Umr. 19 0522 C.kL Collegc, Kilaghq Darbhanga. Blkar. 20 0524 Bihu N a t i d CbUegc. Palna-BM MW, D l b r 21 0525 hintvla Collcgc. Chaibea P.O. Chnibara-833 201. Dia W ~ I

Singhbhum. BIhar E 0528D St. Columbns College. P.O. Colleee .More, tlaznnha&-8:5 301 23 0.529 Anugrnh Naraym C o l l ~ p , B h g Road Pabra-8lN 013 4. DEUII RECION(1) (South and W e d Rqloe, Curgmn. Farklabd n d M.(hur.) 24 0707 MCRC. J m ~ n Millla trlnrma Jamla N a p , New Dclhl-l I0 025. 2 5 . 071 1 Gnrg CoUeg. S ~ r i Fan Ron4 3ew Dcml- l I0 M9 26 07 15 A c h q a Narcndrn Dor College. Ukaji . New Whl-I10 019. 5. DELHJ REGIONC,) (North and Errl Reglon lnclvcling Mnrut, ,Modln.gmr mm( G h d s b d D M M olUttar P n d 4 ) '7 0728 Ah~aknrochnryn Collcgc ofAppl~cd Scicnm. Vccr SavarkarComplcx P u ~ s a ,

NmDclhl-I IOOl2 28 0729 Knltnd~ Collcgc, En?! Patel Sagu, k w Delhl-l lO Oc)X 29 27-13 Lnjpat Rai (P G ) Collsgu, SalubnLad-lnl 005, 'Cttr Pnderh 6, AIhlF.I),\nAD REGION (Cujmrnl, l)mman R; Diu. Dudm B Nvgnr Hnvcli) 30 090 l L D An? C~~llcge, Nnvfimypura, Mrnednbd-380 CAB, C u j a n t 31 (r)K Genmal Education B d l d ~ n g MS. Unncnity, Yadudun-393 OK, CuJarnt. 32 @)06 JLI 1hncl;c.r Commerce Collcgc.Dhuj-371) 001 ,Guja:at (Lalan Collegc.

I l l~ t~ j , CuJ:~rnl) 33 ( m y ) Kcw F'r11~1c:;sivc I<tlucalion l'rusl, hlcbnn.1.381 M?, GlrJ*ral 3-1 EI?2 (I< ) Sl~rcc Cinw ViJynloy~t.l'lot No 910. GlUC Es~;t~e,XnLlcahtvc,CuJnnt 35 ol)?H(lt) Na t i~n~ ,~ l In*.ti!ulc fijr bI;moycn:unl and InCormalion Tcc!molog~(NL\fll')

Clo Parnfl Ad, 1sns;tlla I'rrxs, Rajko1-5 36 2OUl Govl h t s Collcgc, 011rn11n nnll DIU (l1.T.)-196 310 7, KARNhI. RE(;ION (Ilaryana and Purd;111) 37 1001 hlukanndl~tl Notiondl Colluyc, YumunaNag~r-131011, tlnryuna 38 1005 Cllhl>tu Ryn Collcl~c ol'Educmon. Rohmk-l:4OOI, Hayatu

(tU1 Irulia J d L11cnxs Memarid Collcgc. Roh& H~ lqana j . 33 I008 Gm'c. Collcge (Girls Wing1 Sector-1 4, R~ lwn) ' R o d K.~11nl-13?. 001,

Elmrpnm 40 1007 Gmt. P.C. Colltgc, I.Ii%iar.I3 001, Hmr)xnn. 4 I 1012 hfa~karula Xational Collcgu, Sh~hsbad, Kurulrshc!ra, IGq,ana 42 1013 GwernmuntP.G. College, Jind-126 102, Hrr).mn 43 220 I D A.V. Colltgc, Jdmdhiu-1.W 008. Punjab 8. SIUh11A HECIOiY(l1lmach~l I ' r x l o h and Chandlgrrh) 4 1101 Ciovzmnlcnt Roy3 Callcgc, Smjaul!, Shimla-171 M6,

Il lrmchal P n d c ~ h . 45 I I05 Governmen! College, Dhnramshda-I76 215, & I l m d d P n M .

1113 G a t . P G. Collcgo. B1l=pur-l74 001, Hlmnchal P n d n h 47 1115 Gm?. Dcgrcc Collqy, Rccay Pno, Kinmur D u L , EJlmrrhrl P- '9-LIhDE REClOX(J&K)

' 4s. 1301 ' L'nivcnlty ofJlmmu. Ikput~s'r& of'bfm- Mt% I J~nmuTawi- l80001, JBK

(Gmdhi 4lanorid Science Collqe, Jammu Tam< J k K ) . 49 1206 G a l . DL- College, Mur. JkK M 1207 Gnt. Delyec Collcgc, Rajouri J&K 51 I ?On W. DcprPc CoUeg~. Poanch. JWC. 52 I?-) ~ l D J h r &harid Couege, Clmp b i p . B d a b , Jammu-1 81 123. J&K 10 U G A L O R E REGI0.-h rnd Ga) 53 0 8 E m p c Collgc a f b & Scim~c. P.O. B y 1 2 0 . E P a p 4

C1.403 001 El 1303 J S.S. Cdlesr; D h d - 5 8 0 004, Kamtbb 55 1320 mt S-CoUcg+ Ei- R d I3+-560 001, m h 11. cOClB3 REGlO.yll"m31 .ad L.Mwhmp) -56 1401 ~mtiancof~m%YbBlvLls-pmm69S~3.

r r r ~ ~ p-- CW.W #la, ~ ~ ~ j . 17 1403 J.D.T. kLm. C r l i n t d h 018. Urrrh IS 1-w ~ x t h a t i u t c ~o l l t ge , ~--a 645. i- 9 I s;? S M N u y m Cdcgc. Ljrmla470 007 i. '1 IJI: s t ~ l k r u C o l l q c E1nakuIam~682 018. Lmh.