ELECTRICITY AND MAGNETISM  ELECTROSTATICS

If a rod of ebonite rubbed with fur, or with a coat-sleeve, it gains the power to attract light objects (bodies), such as pieces of papaer or tin-foil or a suspended glass beads. The discovery that a body could be made atttractive by rubbing is attributed to Thales (640-548 B.C) but discovery is made by chinese long before than f.He seems to have been led to it through the Greeks' practice of spinning silk with an amber spindle; the rubbing of a spindle in its bearings caused the silk to adhere to it. The Greek word for amber is elektron, and a body made attractive by rubbing is said to be “electrified”. This branch of Electricity, the earliest discovered, is called Electrostatics. Little progress was made in the study of electrification until the sixteenth century A.D. Then Gilbert, who was physician-in-ordinary to Queen Elizabeth, found that other substances besides amber could be electrified: for example, glass when rubbed with silk. He failed to electrify metals, however, and concluded that to do so was imposibble.  More than 100 years later-in 1734- he was shown to be wrong, by du Fay ; du Fay found that a metal could be electrified by rubbing with fur or wool or silk, but only if it were held in a handle of glass or amber; it could not be electrified if it were held directly in the hand. His experiments followed the discovery, by Gray in 1729, that electric charges could be transmitted through the human body,water,and metals. These are examples of conductors; glass and amber are examples of insulators. Positive and Negative Electricity In the course of his experiments du Fay also discovered that there were two kinds of electrification in nature: he showed that electrified glass and amber tended to oppose one another's attractiveness. To illustrate how he did so, we may use ebonite instead of amber, which has the same electrical properties. We suspend a pith-ball, and attract it with an electrified ebonite rod E (fig.1(i); we then bring an electrified glass rod G towards the ebonite rod, and the pith-ball falls away (fig.1(ii). Benjamin Franklin, a pioneer of electrostatics, gave the name of “positive electricity” to the charge on a glass rod rubbed with silk, and “negative electricity” to that on an ebonite rod rubbed with fur.

 

Fig.1 (i) and (ii)

1

Electrons Electric charge is a fundamental property of matter. Charge comes in two types, arbitrarily called positive and negative. Charge is quantized, with one elementary charge—the

magnitude of the electron or proton charge— equal to 1.60X10−19

C, where the coulomb (C) is the SI unit of charge. Charge is also conserved, in that the algebraic sum of the charges in a closed system never changes

Towards the end ofthe 19 th century J.J.Thomson dicoverd the existence of the electron. It is about 1/1840 th of the mass of the Hydrogen atom-and experiments show that it carries a tiny quantity of negative electricity. Later experiments showed that electrons are present in all atoms. Electrons exist round nucleus which carry positive electricity. Normally atoms are electricly neutral which means the total negative charge on the electrons is equal to the positive charge on the nucleus.

Robert Millikan (1868-1953), in1909 discovered that electric charge is quantized: any positive or negativecharge can be written as q=ne, where n is positive or negative integer,

e is a constant of nature called the elementary charge.( 1.60x10−19C).

The elementary charge e is one of the important constants of nature

q=ne n=±1 ,±2 , ±3 , . . .

Electric charge is quantized.

Electric charge isconserved: the algebraic net charge of any isolated system cannot change

 Gold-leaf Electroscope 

One of the earliest instruments used for testing positive and negative charges consisted of a metal rod A to which gold leaves L were attached (Fig..2). The rod was fitted with a circular disc or cap B, and was insulated with a plug P from metal case C which screened L from outside influences other than those brought near to B.

When B is touched by an ebonite rod rubbed with fur, some of the negative charge on the rod passes to the cap and L; and since like charges repel, the leaves diverge ( Fig.2 (i). If an unknown charge X is now brought near to B, an increased divergence implies that X is negative(fig.2(ii). A poistive charge is tested in a similar way; the electroscope is first given a positive charge and an increased divergence indicates a positive charge.

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Figure 2 Induction and Charging By Induction We shall now show that it is possible to obtain charges, called induced charges, without any contact with another charge. An experiment on electrostatic induction, as the phenomenon is called, is shown in fig.3(i). Two insulated metal spheres A,B are arranged so that they touch one another, and a negatively charged ebonite rod C is brought near to A. The spheres are now separated, and then the rod is taken away. Tests with a charged pit-ball now show that A has a positive charge and B a negative charge fig.3(ii). If the spheres are placed together so that they touch, it is found that they now have no effect on a pith-ball held near. Their charges must therefore have neutralized each other completely thus showing that the induced positive and negative charges are equal. This is explained by the movement of electrons from A to B when the rod is brought near. B has then a negative charge and A an equal positive charge.

3

 Fig.3 An experiment on electrostatic induction Fig 4 Electrostatic induction Fig 3 shows how a conductor can be given a permanent charge by induction, without dividing it in two. We first bring a charged ebonite rod, say, near to the conductor, (i);next we connect the conductor to earth by touching it momentarily (ii); finally we remove the ebonite. We then find that the conductor is left with a positive charge  This phenomenon of induction can again be explained by the movement of electrons. If the inducing charge is negative, then, when we touch the conductor, electrons are repelled from it to earth, as shown in fig 3(ii) and a positive charge is left on the conductor. If the inducing charge is positive, then the electrons are attracted up from the earth to the conductor, which then becomes negatively charged.   Conductors and Insulators

In some materials, such as metals,tap water and the human body,outer electrons of atoms are loosly bound. therefore negative charge can move rather freely. We call such materials conductors. İn other materials, such as glass, chemically pure water and plastic non of the charge can move freely. We call these materials nonconductors,or insulators.

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İf you rub a copper rod with a wool while holding the rod in your hand you will not be able to charge the rod,because both you and the rod, are conductors  Again conductors are material in which a significant number of charged particles (electrons in metals) are free to move. The charged particles in nonconductors,or insulators are not free to move  Law of Force between two Charges The magnitude of the force between two electrically charged bodies was studied by Coulomb in 1875. he showed that, if the bodies were small compared with the distance between them, then the force F was inversely proportional to the square of the distance r, 

Fα

1r 2             (1)

This result is known as the inverse square law, or Coulomb’s law. It is not possible to verify the law accurately by direct measurement of the force between two charged bodies. In 1936 Plimton and Lawton showed, by an indirect method, that the power in the law cannot differ from 2 by more than ±2¿ 10-9. We have no reason to suppose, therefore, that the inverse square law is other than exactly true.

Quantity of Charge

The SI unit of charge is the coulomb (C). The ampere(A), the unit of current, is defined later. The coulomb is defined as that quantity of charge which passes a section of a conductor in one second when the current flowing is one ampere.

By measuring the force F between two charges when their respective magnitudes q1 and q2

are varied, it is found that F is proportional to the product q1 q2

Thus Fq1 q2 (2)

Law of Force

Combining (1) and (2), we have

F

q1 q2

r2F= k

q1 q2

r2 (3)

where k is a constant. For reasons explained later, k is written as 1/4πε o, whereε o is a constant called the permittivity of free space if we suppose the charges are situated in a vacuum. Thus

F=

14 πε0

q1 q2

r2

In this expression, F is measured in newtons (N), q in coulombs (C) and r in metres (m). Now, from (4),

5

εo =

q1 q2

4 π Fr2

Hence the units of ε o are coulomb2 newton -1 metre-2 (C2N-1m-2).

Permittivity

So far we have considered charges in a vacuum. If charges are situated in other media such as water, then the force between the charges is reduced. Equation (4) is true only in a vacuum. In general, we write

F=

14 πε

q1 q2

r2

Where ε is the permittivity of the medium. The permittivity of air at 1.005 times that, ε o, of a vacuum. For most purposes, therefore, we may assume the value of ε o for the permittivity of air. The permittivity of water is about eighty times that of a vacuum. Thus the force between charges situated in water is eighty times less than if they were situated the same distance apart in a vacuum.

The unit of ε o, more widely used, is farad metre-1 ( Fm-1). We shall see later that ε o

has the numerical value of 8.854¿ 10-12, and 1/4πε o then has the value 9¿ 109 approximately. The elementary charge e is one of the important constants of nature and its charge

( 1.60x10−19C).

Particles Charge (C) Mass (kg)

Electron (e) -1,6021917x10-19 9,1095x10-51

Proton (p) +1,6021917x10-19 1,67261x10-27

Neutron (n) 0 1,67492x10-27

Table:1

Fig.3 Vectoral force to show vector form of Coulomb’s Law

Vector form of Coulomb’s Law

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F= k

q1 q2

r2 ûr (4) ûr =

r γ

If there are more than 2 charges resultant charges acted on one charge is

F1= F21+F31+F41 (5)

PROBLEMS

1-a)Calculate the value of two equal charges if they repel one another with a force of 0.1 N

when situated 50 cm apart in a vacuum.

14 πε0 =9¿ 109

b) What would be the size of the charges if they were situated in an insulating liquid whose permitivity was 10 times that of a vacuum?

SOLUTION:

From F=

14 πε0

q1 q2

r2

Since q =q here,

0.1=

9 x109 q2

(0 . 5)2

or q2

=

0 .1 x (0. 5 )2

9 x 109

q=2.7x106

C (coulomb),approx.

=2.7μ C ( microcoulomb).

b) The permitivity of the liquid ε =10ε 0

F=

14 πε0

q1 q2

r2

=

110( 4 πε0 )

q2

r 2

q2=

(0 . 1)×(0 .5 )2×109×10(9 )

q=5 .310−6C=5.3μ C

7

--------------------------------------------------

2) Three charged particles are arranged in aline as shown below. Distance between q1

and q2 0.30m. Total distance between q1 and q3 is 0.50m. k=9.0x109N.m

2/C

2;

๐ ⊕๐

q1 = -8μ C q2 =+3μ C q3 = -4μ CCalculate the each electrostatic force on particle 3 (4μ C) due to other two charges

SOLUTION:

F31 =

(9 . x109 )( 4 x10−6 )(8 x 10−6 ) .

(0 .50)2=1.2N repulsive

F32=

(9 x109 )(4 x 10−6 )(3 x10−6 )( 0. 20 )2

=2.7N attractive

THE ELECTRIC FIELD

Electric Intensity or Field-strengthAn “electric field” can be defined as a region where an electric force is experienced. As in gravitation or magnetism, The force exerted on a charged body in an electric field depends on the charge of the body and on the intensity or strength of the field. If we wish to explore the variation in

intensity of an electric field, then we must place a test charge q0 at the point concerned which is small enough not to upset the field by its introduction. The intensity E of an electrostatic field at any point is defined as the force per unit charge which it exerts at

that point. Its direction is that of the force exerted on a positive charge. q0

From this definition,

E=

Fq0 (6)

E=

limq0→0

Fq0

F= Eq0

8

Since F is measured in newtons and q0 in coulombs, it follows that intensity E has units of newton per coulomb (NC-1). We shall see later that a more practical unit of E is volt metre-1 (Vm-1)jThe electric field at a point charge

E=k

q

r2(7)

The electric field at a point is a vector giving the electric force per unit charge

that would be experienced by a charge at that point:

E=

Fq0

The electric field of a point charge q is therefore

(8)

We can easily find an expression for the strength E of the electric field due to a point charge situated in a vacuum .For the force between two such charges.

F=

14 πε0

Qq0

r 2 ∴E=

Fq0 =

Q

4 πε0 r2

The direction of E is away from the point charge if the charge Q is positive; it is radially inward if the charge Q is negative.

Continious Charge Distributions

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With continuous distributions of charge, the sum over all point charges becomes an

integral, giving

∆E=k e∑

i

Δqi

ri2 ri

(9)

Volume charge density ρ=

QV (10)

Surface charge density σ=Q

A (11)

Line charge density λ= Q

ℓ (12)

For any finite charge distribution with nonzero net charge, the field approaches that of

a point charge at large distances.

LİNES OF FORCE

A convanient way to visualize the electric field produced by a charge or charge distribution is to construct a map of the field lines of the force of the field. By other words electric fields can be mapped out by electrostatic lines of force, which may be defined as a line such that the tangent to it is in the direction of the force on a small positive charge at that point. Arrows on the lines of force show the direction of the force on a positive charge; the force on a negative charge is in the opposite direction.

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Fig.5

Fig.6

3) An electron of charge 1.6¿ 10-19 C is situated in a uniform electric field of intensity 1200 volt cm-1. Find the force on it, its acceleration, and the time it takes to travel 2 cm from rest (mass of electron, m,=9.10¿ 10-31 kg )

Force on electron F= eE

Now E= 1200 volt cm-1= 120 000 volt m-1

F = 1.6¿ 10-19¿1 .2×10 5

=1.92¿ 10-14 N (newton)

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Acceleration, a=

Fm =

1.92 x10−14

9 .1 x10−31= 2.12¿10 16 m s-2 ( metre second-2)

Time for 2 cm travel is given by

s=

12 at2

∴ t= √ 2 sa = √ 2×0 . 02

2 .12×10(16 ) = 1.3¿ 10-9 seconds.-------------------------------------------------

4) Calculate the magnitude and direction of the electric field at a point P which is 30 cm to

the right of a point charge q=-3x10−6C k=9.0x109

Nm2/C

2

←30 cm------→

• ←⋅¿ ¿Pq=-3x10−6

C E

SOLUTION: the magnitude of the electric field due to sıngle poınt charge ıs gıven

E=k

q

r2=

(9 . 0 X 109 )(3 .0 X 10−6 )(0 . 30)2

=3.0X105N/C

Direction of the electric field is toward the charge q ---------------------------------------

5) Determine the magnitude of the electric force on the electron of a hydrogen atom exerted by the proton Asume the distance between electron orbits and proton is

r= 0.53x10−10m., k=9.0x109

N.m2

/C2;

SOLUTION:

From F=

14 πε0

q1 q2

r2

Since q1 =q2 =1.6x10−19

C here

F=

(9 x109 )(1. 6 x10−19 )(1. 6 x10−19 )(0 .53 x 10−10)2

=8.2x10−8

N-----------------------------------------------------------

6) The Earth, which is an electrical conductor,carries a net charge of Q= - 4.3x105

C distributed approximately uniformly over its surface . Find the surface charge density and

caculate the electric field at Earth’s surface. Earth’s radius R=6.37x106

m

12

SOLUTION:

Earth’s surface charge density σ =

QA =

Q

4 πR2=

−4 . 3 x105

4 π (6 . 37 x106 )2=-8.43x10

−10 C/m

2

E.A=

σ . Aε

∴E=σε =

−8 . 43 x10−10

8 . 85x 10−12=-95 N/C

Where the minus(-) sign indicates that the field direction is downward

ELECTRIC FLUX

Before discussing Gauss’s law itself we first discuss the concept of flux

Flux from a Point Charge

We have already shown how electric fields can be described by lines of force. It can be said that the density of the lines increases near the charge where the field intensity is high. The intensity E at a point can thus be represented by the number of lines per unit area through a surface perpendicular to the lines of force at the point considered. Asurface area A through which auniformelectric field E passes

Fig.7

Fig.8

The electric flux , through this surface is defined as the product.

=ExA (13)

"number of field lines crossing a surface." We call this quantity the electric flux, Ф, through the surface

13

If the area is not perpendicular to E but rather makes an angle θ fewer field lines will pass the area. In this case the electric flux through the surface as

=ExA= ExAcos θ (14)

Fig.9

7) +1μC charge in the center of sphere whose radius 1m..Find the flux emerging from

the sphere. k=8.89x109N.m

2 /C2

SOLUTION:

Electric field at the surface of the sphere

E=k

Q

r2=(8.89x109

N.m2 /C2

)

1 x 106 C(1m )2

=8.99x103N/C

ФE =E.A= E(4r2

) A=4 r2

=12.6 m2

ФE =E.A= E(4r2

)=(8.99x103N/C)12.6m2

=1.13x105N.m

2 /C2

8)Three charges -1 µC, 2 µC and 3 µC are placed respectively at the corners A, B, C of an equilateral triangle of side 2 metres. Calculate (a) the potential, (b) the electric field, at a point X which is half-way along BC.

14

=

= 18 x 103 V.

VA = =

= - 5 x 103 V (approx.)

= (18+27-5) x 103 V

= 40 x 103 V.

EB = =

= 18 x 103 V m-1.

EC = 27 x 103 V m-1.

EA = =

= 3 x 103 V m-1.

E2 = EA2 + (EC - EB)2

= (9 + 81) x 106

= 90 x 106

E = 3 x 103 V m-1 = 9.5 x 103 V m-1.

Tan = = =

= 18º 25´.

GAUSS’S LAW

We can in preciple determine theelectric field due to any given distribution of electric charge using Coulmb’s law. The total electric field at any point will be the vector sum(or integral) of contributions from all charges present

E=E1 +E2 +….. E=∫ dE

Except for some simple cases, the sum or integral can be quite complicated to evaluate.

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In some cases , theelectric field due to given charge distribution can be calculated

more easily using Gauss’s law.The major importance of Gauss’s law is that it gives us

additional insight into the nature of electrostatic fields and more general relationship

between charge and field. If a surface is curved, then we divide the surface into many

small patches, each small enough that it's essentially flat and that the field is essentially

uniform over each. If a patch has area dA, then Equation below gives the flux through it:

dФ = E • dA. (15)

where the vector dA is normal to the patch (Fig. 10). The total flux through the surface is

then the sum over all the patches. If we make the patches arbitrarily small that sum

becomes an integral, and the flux is

Fig.10The magnitude of electeric field on the surface of sphere is the same as E = keq/r2

For ΔAi It can be written

Net flux through Gauss surface

In many cases (in particular) we deal with the electric flux through closed surface

=∮E . dA Gauss’s Law (16)Consider a sphere of radius r drawn in space concentric with a point charge. The value of

electric field E. It can be written at this place. ∮

dA.=A= 4πr2

The total flux through the sphere is, =ExA

16

The net flux through the sphere is ke=

14 πε0 therefore

ФE = E(4r2

)=( q

4 πr2 ε0)(4r

2)=

qε0

ФE =

qε0 ФE=

qε =

charg e .inside . spherepermittivity (1)Gauss’s Law

(17)

The electric flux through any closed surface is proportional to the charge enclosed by

that surface.

This demonstrates the important fact that the total flux crossing any sphere drawn outside and concentrically around a point charge is a constant. It does not depend on the distance from the charged sphere. It should be noted that this result is only true if the inverse square law is true.

Field due to Charged Sphere

The flux passing through any closed surface whatever its shape, is always equal to

qε

where q is the total charge enclosed by the surface. This relation, called Gauss’s Law, can be used to find the value of E in other common cases.

(1).Outside a charged sphere

The flux across a spherical surface of radius r, concentric with a small sphere carrying a charge q is given by,

Flux=

qε ∴E×4 π r2 =

qε E=q /4r

2(18)

17

Fig.11

This is the same answer as that for a point charge. This means that outside a charged sphere, the field behaves as if all the charge on the sphere were concentrated at the centre.

Fig.12 Two gaussian surfaces surrounding a spherical charge distribution. Surface 1 lies outside the distribution, and encloses all the charge q. Surface 2 lies inside the distribution, and encloses only part of the charge

(2)Inside a charged empty sphere

Fig.13 Flux and Electric field inside a charged empty sphere

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Suppose a spherical surface A is drawn inside a charge sphere, as shown in fig. 12 Inside this sphere the flux is zero.. Hence. E must be zero everywhere inside a charged sphere.

E=0 (19)

The electric field is zero inside a conductor ( E=0 ) in electrostatic equilibrium. If

a conductor in electrostatic equilibrium carries a net charge, all excess charge resides on

the conductor surface.

(3) Outside a charged Plane Conductor

Now consider a charged plane conductor with a surface charge density of σ coulomb metre-2.

Applying equation

∴E×area=ch arge . inside . surfaceε

Now by symmetry, the intensity in the field must be perpendicular to the surface. Further, the charges which produce this field are those in the projection of the area on the surface . The total charge here is thus .A coulomb

E.A=

σ . Aε

∴E= σε (20)

Field Round Points and The Action of Points,  we saw that the surface-density of charge (charge per unit area) round a point of a conductor is very great. Consequently, the strength of the electric field near the point is very great. The intense electric field breaks down the insulation of the air, and sends a stream of charged molecules away from the point,by other words charge leaks away from a sharp point through the air. This mechanism of the breakdown, is called a ‘corona discharge’. Corona breakdown starts when the electric field strength is about 3 million volt metre-1. the corresponding surface-density is about 2.7¿ 10-5 coulomb metre-2

 

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In all types of high-voltage equipment sharp corners and edges must be avoided, except where points are deliberately used as electrodes. Otherwise, corona discharges may break out from the sharp places. All such places are therefore enlarged by metal globes, these are called stress-distributors.  

POTENTIAL IN FIELDS

When an object is held at a height above the earth it is said to have potential enrgy. A heavy body tends to move under the force of attraction of the earth from a point of great height to one of less, and we say that points in the earth’s gravitational field have potential values depending on their height.

Electric potential is analogous to gravitational potential, but this time we think of points in an electric field. Thus in the field round a positive charge, for example, a positive charge moves from points near the charge to points further away. Points round the charge are said to have an electric potential.

Electric potential in the space is described potantial energy per unit of charge at that point

V=

EP

q (21)

POTENTIAL DIFFERENCE

Let us consider two points A and B in an electrostatic field, and let us suppose that the force on a positive charge q has a component in the direction BA. Then if we move a positively charged body from B to A, we do work against this component of the field E. We define the potential difference between A and B as the work done in moving a unit positive charge from B to A. We denote it by the symbol VAB

VAB =V B-V A =

W AB

q (22)Fig.14 The work required to move a charge q from A to B in a uniform electric field is qEℓ .

We define the potential difference ∆V between two points in an elektrik field as WORK AND ENERGY

The work done will be measured in joules(J). The unit of potential differences is called the volt and may be defined as follows: the potential difference between two points A and B is one volt if the work done in taking one coulomb of positive charge from B to A is one joule.

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From this definition, if a charge of q coulombs is moved through a p.d. of V volt, then the work done W in joules is given by

W= qV (23) V=

Wq

Let us consider two points A and B in an electrostatic field, A being at a higher potential than B. The potential difference between A and B we denote as usual by VAB.If we take a positive charge q from B to A, we do work on it of amount

W=q.VAB : the charge gains this amount of potential energy.

If we now let the charge go back from A to B , it loses that potential energy: work is done on it by the electrostatic force, in the same way as work is done on a falling stone by gravity. This work may become kinetic energy, if the charge moves freely, or external work if the charge is attached to some machine, or a mixture of the two.

The work which we must do in first taking the charge from B to A does not depend on the path along which we carry it, just as the work done in climbing a hill does not depend on the route we take. The fact that the potential differences between two points is a constant, independent of the path chosen between the points, is the most important property of potential in general.

POTENTIAL DIFFERENCE FORMULA

To obtain a formula for potential difference, let us calculate the potential difference between two points in the field of a single point positive charge, q in fig.14. for simplicity we will assume that the points, A and B, lie on a line of force respectively

(24)

Generally, you may see our ΔVA→B written as VAB or VBA or VB — VA. We use the Δ here to show explicitly that we're talking about a change or difference from one point to another, and we use the subscript A→B to make it clear that this the potential difference going from A to B. In the next section we'll show how our notation is equivalent to the

21

commonly used VB — VA, and in subsequent chapters we'll sometimes use just the symbol V for potential difference.

In the special case of a uniform field, Equation above reduces to

(25)

There ℓ is a vector from A to B. Figure 14 shows the special case when the field E

and path ℓ are in opposite directions.

The potential difference can be positive or negative, depending on whether the path

goes against or with the field. Moving a positive charge through a positive potential difference

is like going uphill: We must do work on the charge, and its potential energy increases.

Moving a positive charge through a negative potential difference is like going down hill: We

do negative work or, equivalently, the field does work on the charge, and its potential energy

decreases. In both cases the opposite is true for a negative charge; even though the potential

difference remains the same, the work and potential energy reverse because of the negative

sign on the charge.

9) At the back end of TV picture tube the field, between Aand B points which is 5 cm, is uniform with of 600 kN/C. Find the potential difference between Aand B points .

SOLUTION:Potential differenceV A→ B = V=Eℓ=(600x103

N/C)(0.05m)=30 kVİf( +) B is potantially higher than A, if (-) B is potantially lower than A

The Volt and the Electron Volt CE

The definition of potential difference shows that its units are joules/coulomb. Potential

difference is important enough that 1 J/C has a special name—the Voltt (V). To say that a car

has a 12-V battery, for example, means that the battery does 12 J of work on every coulomb

that moves between its two terminals.

We often use the term voltage to speak of potential difference, especially in describing

electric circuits. Strictly speaking the two terms are not synonymous, since voltage is used

22

even in nonconservative situations that arise when fields change with time. But in common

usage this subtle distinction is usually not bothersome.

Potential Difference Depends on Two Points

Specifically, it is the energy per unit charge involved in moving between those points.

Always think of potential difference in terms of two points..

In molecular, atomic, and nuclear systems it's often convenient to measure energy in

electronvolts (eV), defined as follows: One electron volt is the energy gained by a particle

carrying one elementary charge when it moves through a potential difference of one

volt.

Since one elementary charge is 1.6x10-19 C, 1 eV is 1.6x 10-19 J. Energy in eV is particularly easy to calculate when charge is given in elementary charges.

CAPACITORS

A capacitor is a device for storing electricity. The earliest capacitor was invented almost accidentally by van Musschenbroek of Leyden, in about 1746, and became known as a Leyden jar. A present day, all capacitors consist of two metal platesinsulator in different geometrical forms separated by an The insulator is called the dielectric; in some capacitors it is oil or air

Fig 15

23

CARGING A CAPACITOR

To study the action of a capacitor we need two plates which have the same shape .Two plates are spaced apart by a fixed distance We connect the batteries in series, a two way key(K as shown in fig 16) and a voltmeter to measure their total voltage. If we close the key at A (switch on), the capacitor is connected via the resistor to the batterry, and the potential difference across the capacitor, V, which is measured

Fig 16

by the voltmeter, The potential difference becomes steady when it is equal to the battery voltage V. If we now open the key(switch of), the voltmeter reading stays unchanged (unless the capacitor is leaky) the capcitor is said to be charged, to the battery voltage: its condition does not depend at all on the resistor, whose only purpose was to slow down the charging process, so that we could follow it on the voltmeter.

DISCHARGING A CAPACITOR

We can show that the charged capacitor is storing electricity by discharging it. , if we put a piece of wire across its terminals, a fat spark passes just as the wire makes contact, and the voltmeter reading falls to zero.

If we now recharge the capacitor and then close the key at B, in fig 16, we allow the capacitor to discharge through the resistor R. The potential difference across it now falls to zero as slowly as it rose during charging.

CHARGING AND DISCHARGING PROCESSES

When we connect a capacitor to a battery, electrons flow from the negative terminal of the battery on to the plate A of the capacitor connectedto it (fig17) and , at the same rate, electrons flow from the other plate B of the capacitor towards the positive terminal of the battery. Positive and negative charges thus appear on the plates, and oppose the flow of electrons which causes them. As the charges accumulate, the potential difference between the plates increases, and the charging current falls to zero when the potential difference becomes equal to the battery voltage V.

24

Fig 17

When the battery is disconnected and the plates are joined together by a wire, electrons flow back from plate A to plate B until the positive charge on B is completely neutralized. A current thus flows for a time in the wire, and at the end of the time the charges on the plates become zero.

CAPACITANCE DEFINITION, AND UNITS

Experiments with a ballistic galvanometer, which measures quantity of electricity, show that,

when a capacitor is charged to a potential difference V, the charges stored on its plates±Q , are proportional to V. The ratio of the charge on either plate to the potential difference between the plates is called the capacitance, C, of the capacitor.

Thus C=

QV ………………………………………………….(1)

Q=CV…………………………………………….........(2)

and V=

QC ………………………………………………......(3)

Where Q is coulombs, V is in volts (V) then capacitance C is in farads (F) .One farad (1F) is the capacitance of an extremely large capacitor. In practical circuits,i such as in radio receivers, the capacitance of capacitors used are therefore expressed in microfarads. One

microfarad is one millionth part of a farad, that is , 1μF=10 -6F. It is also quite usual to Express small capacitors, such as those used, in picofarads (pF). A picofarad is one millionth part of a microfarad, that is, 1pF= 10-6μ F = 10-12F.

FACTORS DETERMINING CAPACITANCE, VARIABLE CAPACITOR

We shall now find out by experiment what factors influence capacitance. To interpret our observations we shall require the formula for potential difference:

V=

QC

25

This shows that, when a capacitor is given a fixed charge, the potential difference between its plates is inversely proportional to its capacitance.

. The capacitance of a capacitor decreases when the seperation of its plates is increased, the capacitance is inversely proportional to the separation.

Dielectric. Let us now put a dielectric- a sheet of glass or ebonite between the plates. When we increasethe dielectric by using several sheets together we see that(1) the capacitance increasewith the dielectric thickness.. In practical capacitors the dielectric completely fills the space between the plates. (2)The capacitance is also directly proportional to the area of the plates.

Fig 18A capacitor in which the effective area of the plates can be adjusted is called a variable capacitor. In the type shown in fig. 18, the plates are semicircular and one set can be swung into or out of the other. The capacitance is proportional to the area of overlap of the plates. The plates are made of brass or aluminium, and the dielectric may be air, oil, or mica.

We shall see shortly that a capacitor with paralel plates, having a vacuum ( or air, if we assume the permittivity of air is the same as a vacuum) between them, has a capacitance given by

C=

ε0 A

d

Where C = capacitance in farads (F), A= area of overlap of plates in m2 , d= distance between

plates in m and ε 0 = 8.854×10-12 farad metre -1.

If a material of permittivity ε completely fills the space between the plates, then the capacitance becomes:

C=

εAd

CAPACITANCE VALUES, ISOLATED SPHERE

Suppose a sphere of radius r metre situated in air is given a charge of Q coulombs. We assume that the charge on a sphere gives rise to potentials on and outside the sphere as if all the charge were concentrated at the centre. The surface of the sphere has a potential given by:

26

V=

Q4 πε0 r

QV = 4 πε 0 r C=

QV =4ε 0 r

∴Capacitance, C= 4ε 0 r

CONCENTRIC SPHERES

Faraday used two concentric spheres to investigate the dielectric constant of liquids. Suppose a,b are the respective radii of the inner and outer spheres. Let +Q be the charge given to the iner sphere and let the outer sphere be earthed, with air between them( Fig 19)

Fig 19 The induced charge on the outer sphere is -Q. The potential Va of the iner sphere= potential

due to --Q. =+

Q4 πε0 a

− Q4 πε0 b since the potential due to the charge -Q is -

Q4 πε0 b everywhere

inside the larger sphere.

But Vb = 0 , as the outer sphere is earthed.

∴potential difference, V, = Va – Vb =

14 πεο ( Q

a−Q

b )V=

Q4 πε0

( b−aab )

QV =

4 πε0 ab

b−a C=

4 πε0 ab

b−a

PARALLEL PLATE CAPACITOR

Fig 20

27

Suppose two paralel plates of a capacitor each have a charge numerically equal to Q fig.20.

the surface density σ is then

QV where A is the area of either plate, and the intensity between

the plates, E, is given, by

E=

σε =

QεA C=

QV =

εAd

Now the work done in taking a unit charge from one plate to the other Work= force×distance=E×d where d is the distance between plates. But the work done per unit

charge=V, the p.d. between the plates. V=

σε

d=

QεA

d

Problem:10The plates of a paralel-plate capacitor are 5 mm apart and 2 m2 in area. The plates are in vacuum. A potential difference of 10,000 volts is applied across the capacitor. Compute (a) the capacitance (b) the charge on each plate, and (c) the electric intensity in the space between them.

SOLUTION:

C =

=

= 3.54 x 10-9 C2 N-1 m-1.

1 C2 N-1 m-1 = 1 C2 J-1 = 1 C (J/C)-1

= 1 C V-1 = 1 F,

C = 3.54 x 10-9 F = 0.00354 F.

Q = CVab = (3.54 x 10-9 C V-1) (104 V) = 3.54 x 10-5 C.

E = =

=

= 20 x 105 N C-1;

E =

28

=

= 20 x 105 V m-1.

ε o and its measurement:

We can now see how the units of ε o may be stated in a more convinient manner and how its magnitude may be measured.

Units. From C=

ε0 A

d , we have ε o =

CdA

Thus the unit of ε o=

faradxmeter

(meter )2 = farad metre-1

Now from previous , C=ε OA/d. Thus on charging, the charge stored, Q, is given by

Q=CV=ε 0 VA/d

ARRANGEMENTS OF CAPACITORS

In radio circuts capacitors often appear in arrangements whose resultant capacitances must be known. To derive expressions for these, we need the equation defining capacitance in its three possible forms.

C=

QV V=

QC Q=CV

In paralel. Fig.21 shows three capacitors, having all their left hand plates connected together, and all their right hand plates likewise. They are said to be connected in paralel. If a battery is now connected across them they all have the same potential difference V. The charges on the individual capacitors are respectively.

29

Q1 =C1 V Fig.21

Q2 =C2 V

Q3 =C3 V

The total charge on the system of capacitors is

Q=Q1 +Q2 +Q3 =(C1 +C2 +C3 )V

And the system is therefore equivalent to a single capacitor,of capacitance

C=

QV =C1 +C2 +C3

Thus when capacitors are connected in parallel,their resultant capacitance is the sum of their individual capacitances.It is greater than the greatest individual one.

In series..

Fig 22 Fig 22 shows three capacitors having the right-hand plate of one connected to the left-hand of the next,and so on –connected in series.When a battery is connected across the ends of the system,a charge Q is transferred from the plate H to the plate A,a charge,-Q being left on H.This charge induces a charge +Q on plate G ;similarly,charges appear on all the other capacitor plates,as shown in the figure.(The induced and inducing charges are equal because the capacitor plates are very large and very close together,in effect,either may be said to

30

enclose the other.)The potential differences across the individual capacitors are, therefore,given by

VAB =

QC1 VDF =

QC2 VGH =

QC3

The sum of these is equal to the applied potential difference V because the work done in taking a unit charge from H to A is the sum of the work done in taking it from H to G,from F to D,and from B to A.Therefore

V=VAB +VDF +VGH = Q(

1C1 +

1C2 +

1C3 )

The resultant capacitance of the system is the ratio of the charge stored to the applied potential difference,V. The resultant capacitance is therefore given by

1C =

1C1 +

1C2 +

1C3

Thus,to find the resultant capacitance of capacitors in series,we must add the reciprocals of their individual capacitances.The resultant is less than the smallest individual.

PROBLEM:11

Find the charges on the capacitors in figure shown belove and the potential differences across them.

SOLUTION:

C´ = C2 + C3 = 3µF.

C = = = 1.2µF.

= Q1 + Q2 + Q3 = CV = 1.2 x 10-6 x 120

= 144 x 10-6 coulomb

31

V1 = = 72 volt,V2 = V - V1 = 120 – 72 = 48 volt,Q2 = C2V2 = 2 x 10-6 x 48 = 96 x10-6 coulomb,Q3 = C3V2 = 10-6 x 48 = 48 x10-6 coulomb.

Energy of a Charged Capacitor

In general, the energy stored by a capacitor of capacitance C,carrying a charge Q,at a potential

difference V,is W=

12 (Q2/C) As we know C=

QV V=

QC Q=CV therefore

=

12 QV

=

12 CV2

If C is measured in farad, Q in coulomb and V in volt,then the expressions derived in (1) will give the energy W in joules.

PROBLEM:12

A 12-V battery is connected to a 20 F capacitor.How much electric energy can be stored in the capacitor?

SOLUTION: Energy W=

12 CV2 =

12 (20x10−6F)(12V)2=1.4x10−3J

---------------------------------------

PROBLEM:13) A circular parallel-plate capacitor with a spacing d=3 mm is charged to

produce an electric field strength of 3x106V/m. Find the potantial difference between the

plates and the capacitance value.SOLUTIONThe potantial differece between the plates is

V=E.d=(3x106V/m)(3x10−3

m)=9x103V

The capacitance value is

C=

QV =

1 x 10−6 C9x 103 V =1.11x10−10

F

PROBLEM:14)The plates of a paralel plate capacitor are separated by a distance

d=1mm.What must be the plate area if the capacitance is to be 1F? ε 0 =8.85x10−12F/m

A=

Cdε0 =

(1 F )(1 x10−3 m)8 .85 x 10−12 F /m =1.1x108 m2

32

-------------------------------------------------

PROBLEM:15) A storage capacitor on a random access memory(RAM) chip has a

capacitance of 55x10−15F. If the capacitor is charged to 5.3V, how many excess electrons are

on its negative plate? e=1.6x10−19 C

n=

qe ==

CVe =

(55 x10−15 F )(5 .3 )1. 6 x10−19

=1.8x106electrons

Concentric capacitorAn example, suppose b= 10cm and a=9cm

∴C=

4 πε0 ab

b−a

=

4 π×8 .85×10(−12)×0. 1×0. 09(0 . 1−0 . 09)

= 100pF (approx)-------------------------------------------------------

PROBLEM:16) Determine the equivalent capacitance of combination between points a and b

as shown below. Take C1=6.0 μF ,C2=4.0μF ,C3=8.0μF . If the capacitors are charged by a 12 V battery Determine the charged on each capacitor.

SOLUTIONC2and C3connected parallel C23=C2+C3=4μF +8μF .=12.μFC23 is in series with C1 , So the equivalent capacitance C of the entire cmbination is given by

1Ceq =

1C1 +

1C23 =

16 +

112 =

312 Hence Ceq = 4μF

The total charge that flows from the battery is

Q=CV=(4x10−6)(12)=4.8x10−5

C Both C1and C23 carry the same charge of this Q

33

The voltage across C1then V 1 =

QC1 =

4 .8 x10−5

6 x10−6=8 V

The voltage across the combination C23 is

V 23 =

QC23 =

4 .8 x10−5

12 x10−6=4 V

The actual capacitors C2 and C3 are in parallel,this represent the voltage across each of

them V 2 =V 3 =4V

The charges on C2 and C3

Q2 =C2V 2 =(4x10−6)(4)=1.6x10−5

CQ3 =C3V 3 =3.2x10−5

C---------------------------

PROBLEM:17) The fig. below shows a network of capacitors between the terminals A and B .

a) Reduce this network to a single equivalent capacitor b) Find the total charge supplied by the 12V battery to the capacitor network

SOLUTIONa) First, we can reduce the two paralel capacitors between B and Y to asingle equivalent

capacitor C1:C1 =4μF +8μF =12μF

Next, we can reduce the two paralel capacitors between Y and X to a single equivalent

capacitor C2 :

C2=6μF +2μF =8μFC1 , C2 and the last 24μF capacitors are connected in series and finally,the three capacitors can be reduced to a single equivalent capacitor C:

1C =

112 +

18 +

124 =

14 C=4μF

b) Q=CV=(4μF )(12V)=48 μ C

--------------------------

PROBLEM18) Find the equivalent capacitance of the combination in fig as shown below

34

First, C3 and C4 are in parallel,their equivalent capacitance C34 =C3+C4 =1μF +3μF =4μF

Then C34 in series with C2 C234 =

C2 C34

C2+C34 =

12 x412+4 =3μF

Now C1 and the combination C234 are in paralel connected to A and B. The equivalent capacitance of entire circuit is

C=C1+C234 =4+3=7μF

-----------------------------------------.

PROBLEM:19) Two capacitors, of capacitance 4 μF and 2μF respectively, are joined in series with a battery of e.m.f. 100 volts. The connections are broken and the like terminals of the capacitors are then joined. Find the final charge on each capacitor.

The combined capacitance, C, of the capacitors is given by1C =

1C1 +

1C2

∴ 1C =

14 +

12 =

34 or C=

43 μF

charge on each capacitor = charge on equivalent capacitor

=CV =

43 ¿ 100 =

4003 μ C

When like terminals are joined together, the p.d across each capacitor, which is different at first, becomes equalized. Suppose it reaches a p.d. V. Then, as the total charge remains constant,

İnitial total charge =

4003 +

4003 = final total charge = 4V + 2V.

∴6 V =8003

∴V =4009 V,

35

∴charge on larger capacitor= 4¿400

9 =

16009 μ C

And charge on smaller capacitor = 2¿400

9 =

8009 μ C

------------------------------------------------------------

PROBLEM:20) A 100μF capacitor can tolerate a maximum potantial difference of 20 V,

while a 1μF capacitor can tolerate 300 V. Which capacitor can store the most energy? The most charge?

SOLUTION:

W100 =

12 CV2 =

12 (100μF )(20V)2

=20x103 J=20mJ

W1 =

12 CV2 =

12 (1μF )(300V)2

=45x103 J=45mJ

The smaller capacitor can store more energy. On the other hand, the larger capacitor store more charge

Q100 =CV== (100μF )(20V)=2000C=2C

Q1 =CV=(1μF )(300V)=300C =0.30mC

--------------------------------------------------CE

PROBLEM:21) A wafer of titanium dioxide has an area of 1cm2 and 0.10mm

thickness.Aluminum is evaporated on the paralel faces to form paralel plate capacitor.a) Calculate the capacitanceb)When the capacitor is charged with a 12 V battery,what is the magnitude of charge delivered to each plate?c)What is the free surface charge density?

Dielectric constant of titanium dioxide at room temperature k=173 ; ε 0 =8.85x10−12

SOLUTION:

a)

C=

kε0 A

d =

(173 )(8.85 x10−12 F /m)(10−4 m2)0 .10 x 10−3 m =1.53x10−9

F

b) The battery delivers the free charge Q:

Q=CV=(1.53x10−9F)(12V)=18.4x10−9

C.c) The surface density of free charge

σ =

QA =

18 . 4x 10-9 C10−4 m2

=1.84x10−4C/m

2

36

----------------------------------------------------- I.ENG.

PROBLEM:22) Two insulated spherical conductors of radii 5.00 cm and 10.00cm are charged to potentials of 600 volts and 300 volts respectively. Calculate the total energy of the system. Also calculate the energy after the spheres have been connected by a fine wire. Comment on

the difference between the two results. (N) k=

14 πε0 =9.0x109

N.m2

/C2;

Capacitance C of a sphere of radius r= 4πε o r

For 5 cm radius, or 5¿ 10-2m, C1=

QV =4ε 0 r

C1 = 4πε o¿5×10 -2=

5 x 10−2

9x 109=

59 x10−11

F

Using 4πε 0=1 / (9¿10 9).

For 10 cm radius, C2=

109 ¿ 10-11 F.

Since energy, W=

12 CV2

∴ total energy =

12 ¿

59 ¿ 10-11 x (600)2 +

12

¿159 ¿ 10-11 ¿ 3002

= 15¿10 -7 J.

CURRENT AND RESISTANCE

Electric Current

So far we have been studying static electricity,( electric charges at rest- assumption of electrostatic equilibrium). Now consider situations in which charges are moving. We call a flow of charge an electric currentand an electric currentis defined as charges crossing the given area in time t. Accordingly,its units are coulombs per second (C/s).This units is given the special name ampere(A) after André Marie AmpèreWhen a light bulb is connected between the terminals of a battery by theconducting wire.When such a circuit is formed, charge can flow through the wires of the circuit, from one terminal of the battery to the other Fig 23 .Conventionally from the positive terminal to negative terminal but electron flow from the negative terminal to positive terminal Fig 24

37

Fig 23 Fig 24

The electric current

I=

ΔQΔt

Where Q is the amount of charge that passes through the conductor at any location during the time interval t.

Fig 25Connsider a conductor containing n charges per unit volume,each carrying charge q and

moving with drift speed d . If A is the conductor’s cross-sectional area,then a length ℓ has volume Aℓ and therefore contains nAℓ charges(Fig 25. Since each carries charge q. The total charge is

Q = nAℓ q and t= ℓ / d

I=

ΔQΔt =

nA ℓqℓ /υd =nAqd

The instantaneous current

Where we consider the ratio of charge to small time intervalHere dq is the positive charge that passes in time dt through a surface of conductor

PROBLEM: 23) How many electrons are contained in 3.20x10-6 Coulombs of charge?

SOLUTION: q=ne n=q/n n= =2.0x1013 elektron

RESISTANCE AND OHM’S LAW

Georg Simon Ohm(1787-1854) who established experimentally that the current in a metal wire is proportional to te potential difference V applied to its two ends

I V

38

And the ratio of potential difference V to current I is the Resisatance R of a wire

R=

VI Ohm’s Law

, Ohm’s Law often written in the equivalent form

V=IR I=

VR

Ohm found experimentally that in metal conductor (for ohmic materials), R is costant independent of V. But R is not a constant for many subtances nor for devices which are called nonohmic materials.such as diods,vacuum tubes,transistors and so on. Thus “Ohm’s Law” is not a fundamental law.

24) The voltage in typical houshold wiring is 120 V. How much current does a 100W light bulb draw? What is the bulb’s resistance under these conditions?

I=

PV =

100 W120 V =0.833 A From Ohm’s law R=

VI =

120V0 .833 A =144

We have seen that P=I 2R and P=

V 2

R

R=

V 2

P =

(120 V )2

100 W =144

Resistivity

It is found experimentally that the resıstance Rof a metal wire is directly proportional to its length and inversly proportional to its cross-sectional area A.

That is R=ρ

ℓA

39

Where ρ ,the constant of proportionality is called resistivityand depends on materials used. Typical values of ρ ,whose units are .m are given for various materials in the middle column of table 2

Table 2

The resistivity of a material depends somewhat on temperature.In general, the resıstance of a metals increases with temperature.

ρT =ρ0 [1+α (T−T 0 )]

Where α is the temperature coefficient of resistivity which is shown in the last column of table 2.

Reciprocal of the resistivity, called the conductivity σ (sigma),is σ =

1ρ and has unit of

(.m)−1

PROBLEM: 25) Calculate the resistance at 00

C of a 2m length of copper wire with a

diameter of 1mm. At 00

C is ρ =1.56x10−8.m

SOLUTION:

R=ρ

ℓA =

(1 .56 x10−8 )(2)π (0 .5 x 10−3 ) =

PROBLEM:26) A copper wire 0.50 cm in diameter and 70cm long is used to connect a car battery to the starter motor.What is the wire’s resistance? The resistivity of copper at roomm

temperature 1.68x10−8.m

40

SOLUTION:

R=ρ

ℓA =

(1 .68 x10−8 )(0 .70m)π (0 .25 x 10−2 m) =6x10−4

Electric Power

Electrical enrgy is useful to us beccause it can be easily transformed into other forms of energy.

P=IVThis relation gives us the power transformed by any device( such as motors,electric heaters,toaster,hair dryers….and so on),where I is the current passing through it and V is the potential difference acroross it. The SI unit of electric power is watt

The rate ofenrgy tranformation in a resisatance R can be written,using V=IR in two other ways

P=IV

=I(IR)=I2

R

=(

VR )V=

V 2

R

ELECTRİC CIRCUITS

Electric circuits are basic parts of all electronic device from radio and TV sets to computers and automobiles.To have current in an electric circuits first we need a device such as battery or an electric generator that transforms one type of energy( chemical,mechanical,light…so on) into electric enrgy. such a device is called source of electromotive force or of emf.

The symbol εis usually used for emf .(do notconfuse it withE for electric field) A battery has itself some resistance,which is called its internal resistance;it is usually desinated r.

When no current is drown from the battery the terminal voltage(Vab = Va -Vb )equals the emf. When a current I flows from the battery there is an internal drop in voltage equal to Ir.

Fig 26. Thus the terminal voltage Vab (the actual voltage delivered) is Vab =ε - Ir.

Fgig 26

41

Resistor in Series

Fig 27

When two or more resistors are connected end to end as shown in fig 27 ,they are said to be

connected in series. We let V represent the voltage across all 3 resistors.and V 1 ,V 2 and V 3 be

the potential difference across each of the resistors R1 ,R2 and R3 As shown in fig.

From V=IR V 1 =IR1 V 2 =IR2 V 3 =IR3

V=V 1 +V 2 +V 3 =IR1 +IR2 +IR3 V= I(R1 +R2 +R3 ) R=

VI =R1 +R2 +R3

Req =R1 +R2 +R3

Resistor in Parallel

Another simple way to connect resistors isa in parallel so that the total current from the

source splits into separate branches as shown in Fig.28.Let I 1 , I 2 and I 3 be the currents

through each of the resistors R1 ,R2 ,andR3 respectively.

Fig 28

I =I 1 + I 2+ I 3 I 1=

VR1 I 2=

VR2 I 3 =

VR3

42

VReq =

VR1 +

VR2 +

VR3

When devide out the V from each term

1Req =

1R1 +

1R2 +

1R3

Kirchhoff’s Rules

Some circuits cannot be simplified using series and paralel combinations.When there is more than one source of emf or when circuit element are connected in complicated ways. There are two Kirchhoff’s Rules.

Kirchhoff’s first or junction rulestates that”at any junction point,the sum of all currents entering the junction must equal the sum of all currents leaving the junction”

Fig 29

For example,at the junction point ain fig. 29 I 1and I 2are entering at a point A whereas I 3 is

leaving.Thus Kirchhoff’s junction rule states that I 1+I 2=I 3 or I 1+I 2 -I 3 =0

Kirchhoff’ssecond or closed loop rule is based on on the conservation of energy.It states that “the sum of the changes in potential around any closed path of a circuit must be zero” In other words, round a loop” The algebraic sum of the voltage drop is equal to the algebraic sum of the emf

∑ IR =∑εor ∑∆V=0

27) Two batteries in oposite direction and two external resistors are connected with a wire of negligible resistance as shown below. a) Determine the current Ib)Calculate the power dissipated or generated in each of the circuit elements

43

SOLUTION:

As you see batteriesare connected to drive current in oposite direction aroun the circuit. As

ε1ε2 , the current I flows counterc lockwise which is indicated in the diagram. The battery

of ε1 provides power to the circuit ,whereas 2 absorbspower.

a) In the circuit. ε1- ε2 =I(r1 +r2 +R3 +R4 )

I=

10 V−6 V0 .6+0 .2+15+8 =0.1681 A

The first battery is converting chemical energy into electric energy at a rate

P1 = ε1 x I=10Vx0.168A=1.681 WBattery 2 converting electric energy into chemical energy at a rate

P2 =ε2 x I=6Vx0.1681A=1.008 W

In other words, the first battery is discharging,whereas battery 2 is being charged

The power dissipated in the various resistive elements is

Pr 1=I 2r1 =(0 .1681 A )2x0.60=0.017W

Pr 2 =I 2r2 =(0 .1681 A )2x0.20=0.006W

PR 3=I 2 R3 =(0 .1681 A )2x15=0.424W

PR 4 =I 2 R4 =(0 .1681 A )2x8=0.226W

CEN+IENG+EEE+MTB

MAGNETISM

The word of magnetism came from the region of Magnesia which is called Manisa

44

now in Anatolia. People in this region observed that certain stones (magnetite) attracted bits

of ordinary iron. They also found that the iron itself became magnetized by touching the

magnetic mineral.

Fig 30 shows a type of a magnet. One of the most striking features of magnet is that if

a bar magnet is cut in half, you do not obtain isolated north and south poles. Instead, two new

magnets are produced. If the cutting operation is repeated, more magnets are produced, each

with a north and a south pole. We can continue subdividing each half into two, right down to

the subatomic scale, elementary particles (electron, proton, neutron) themselves act like tiny

magnets. In the eleventh century A.D. the Chinese invented the magnetic compass. This

consisted of a magnet, floating on a buoyant support in a dish of water. The magnet has two

ends which are called poles. The pole of a freely suspended magnet that points toward geo-

graphic north is called the north pole of the magnet. The other pole points toward the south

pole and is called the south pole

Fig 30

It is a familiar fact that when two magnets are brought near one another, each exerts a

force on the other. The force can be either attractive or repulsive and can be felt even when

the magnets don't touch. If the north pole of one magnet is brought near the north pole of a

second magnet, the force is repulsive. Similarly, if two south poles are brought close, the

force is repulsive. But when a north pole is brought near a south pole, the force is attractive.

These results are shown in Fig. 31

45

Fig 31 Fig 32

Magnetic Fields

The region around a magnet where a magnetic force experienced, is called a

magnetic field.As in gravitation or electric field surrounding an electriccharge. The

appearance of a magnetic field is quickly obtained by iron filings and accurately plotted with

a small compass. The grain of iron line up along field lines. Of course, we do not get the

direction of the magnetic field, only the pattern of the field. Figure 32 shows how thin iron

filings reveal the magnetic field lines.

The direction of the magnetic field at a given point can be defined as the direction that

the north pole of a compass needle would point. Figure 33 (a) shows how one magnetic field

line around a bar magnet is found using compass needles. The magnetic field determined in

this way for the field outside a bar magnet is shown in Fig 33 (b). Notice that the lines always

point out from the north pole toward the south pole of a magnet and magnetic field lines

continue inside a magnet. The field round bar is “non-uniform” strength and direction vary

from place to place. The earth’s field locally, however, is uniform Fig 33 (c)

46

Fig 33(a) Fig 33(b)

--- the number of lines per unit area is proportional to the strength of the magnetic

field

Fig 33 c Fig 34

The Earth's magnetic field is shown in Fig. 34.The pattern of field lines is as if there

were an (imaginary) bar magnet inside the Earth.. The Earth's magnetic poles do not coincide

with the geographic poles, which are on the Earth's axis of rotation. The north magnetic pole,

for example, is in northern Canada, about 1300 km from the geographic north pole, or "true

north." This must be taken into account when using a compass. The angular difference

between magnetic north, as indicated by a compass, and true (geographical) north, is called

the magnetic declination. In the U.S. it varies from 0° to perhaps 25°, depending on location.

47

Electric Currents Produce Magnetism

During the eighteenth century, many scientists sought to find a connection between

electricity and magnetism. A stationary electric charge and a magnet wereshown not to have

any influence on each other. But in 1820, Oersted (1777-1851) found that when a compass

needle is placed near an electric wire, the needle deflects as soon as the wire is connected to a

battery and a current flows. As we have seen, a compass needle can be deflected by a

magnetic field.What Oersted found was that an electric current produces a magnetic field.

He had found a connection between electricity and magnetism.

Fig 35 Fig 36

The Magnetic field lines produced by a current in a straight wire are in the form of

circles with the wire at their center. Fig. 35(a). There is a simple way to remember the

direction of the magnetic field lines in this case. It is called a . "right-handrule: you grasp the

wire with your right hand so that your thumb points in the direction of the conventional

current; then your other fingers will encircle the wire in the direction of the magnetic field.

Fig. 35(b). The magnetic field lines due to a circular loop of current-carrying wire can be

determined in a similar way using a compass. The result is shown in Fig. 36.

Force on an Electric Current in a Magnetic Field;

Definition of B

In previous section we saw that an electric current exerts a force on a magnet, such as

48

a compass needle. By Newton’s third law, we might expect the reverse to be true as well; we

should expect that a magnet exerts a force on a current-carrying wire. Experiments indeed

confirm this effect, and it was also first observed by Oersted.

Fig 37 Fig 37( c)

Let us look at the force exerted on a current-carrying wire in detail. Suppose straight wire is placed between the poles of a horseshoe magnet as shown in Fig. 37. When a current flows in the wire, a force is exerted on the wire. But this force is not toward one or the other pole of the magnet. Instead, the force is directed at right angles to the magnetic field direction, downward in Fig.37(a). If the current is reversed in direction, the force is in the opposite direction, Fig. 37(b) It is found that the direction of the force is always perpendicular to the direction of the current and also perpendicular to the direction of the magnetic field, B. This statement does not completely describe the direction, however: the force could be either up or down in Fig. 37(b) and still be perpendicular to both the current and to B. Experimentally, the direction of the force is given by another right-hand rule, as illustrated in Fig.37( c). Orient your right hand so that outstretched fingers can point in the direction of the current, and when you bend your fingers they point in the direction of the magnetic field lines. Then your thumb points in the direction of the force on the wire.

This describes the direction of the force. What about its magnitude? It is found

experimentally that the magnitude of the force is directly proportional to the current I in

the wire, and to the length ℓof wire in the magnetic field (assumed uniform). Furthermore, if

the magnetic field is made stronger, the force is proportionally greater. The force also depends

on the angle θbetween the current direction and the magnetic field (Fig.38). When the current

is perpendicular to the field lines the force is strongest. When the wire is parallel to the

49

Fig 38

magnetic field lines, there in no force at all. At other angles, the force is proportional to sinθ

(Fig.38). Thus for a current I in a wire, with length ℓin a uniform magnetic field B, we have

F IℓBsin

Up to now we have not defined the magnetic field strength precisely. In fact, the

magnetic field B can be conveniently defined in terms of the above proportion so that the

proportionality constant is precisely 1.Thus we have

F=IℓBsin

If the direction of the current is perpendicular to the field (θ = 90°), then the force is

F=IℓB IB

If the current is parallel to the field (θ = 0°), the force is zero. The magnitude of B

can then be defined as B =

F maxIℓ where Fmax is the magnitude of the force on a straight

length ℓ of wire carrying a current I when the wire is perpendicular to B.

The relation between the force F on a wire carrying current I, and the magnetic field B

that causes the force, is vector equation.

F=IℓB

here, ℓ is a vector whose magnitude is the length of the wire and its direction is along the wire

50

(assumed straight) in the direction of the conventional current.

The above discussion applies if the magnetic field is uniform and the wire is straight.

If B is not uniform, or if the wire does not everywhere make the same angle with B. then

above equation can be written

dF=IdℓB . , |

where dF is the infinitesimal force acting on a differential length dℓof the wire. The total force

on the wire is then found by integrating.

F=∫ dF =∫ Id ℓ xB

Above equations can serve as a practical definition of B. An equivalent way to define

B. in terms of the force on a moving electric charge, is discussed in the next Section.

The SI unit for magnetic field B is the tesla (T). From above equations,it is clear that

1T = 1 N/A. m. An older name for the tesla is the "weber per meter square" (1Wb/m2 = 1 T).

Another unit commonly used to specify magnetic field is a cgs unit, the gauss (G):1G=10-4T.

Torque on a Current Loop; Magnetic Dipole Moment

When an electric current flows in a closed loop of wire placed in a magnetic field, as

shown in Fig. 39. the magnetic force on the current can produce a torque. This is the basic

principle behind a number of important practical devices, including voltmeters, ammeters, and

motors. The interaction between a current and a magnetic field is important in other areasas

well, including atomic physics.

Fig 39

51

When current flows through the loop in Fig. 39(a), whose face we assume is parallel to

B and is rectangular, the magnetic field exerts a force on both vertical actions of wire as

shown, F1 and F2 (see also lop view. Fig. 39(b) Notice that, by the right-hand rule, the

direction of the force F1 on the upward current on the left is in the opposite

direction from the equal magnitude (F1=F2=F) force F2 on the descending, current on the

right. These forces give rise to a net torque that tends to rotate the coil about its vertical

axis.

Let us calculate the magnitude of this torque. From Eq. F=IℓB, the force has

magnitude F = IaB. where a is the length of the vertical arm of the coil. The lever arm for

each force is b/2, where b is the width of the coil and the "axis" is at the midpoint. The total

torque is the sum of the torques due to each of the forces, so

τ = IaB.

b2 + IaB.

b2 =IabB= IAB

where A = ab is the area of the coil. If the coil consists of Nloops of wire, the torque on ,N

wires becomes

τ =NIAB

If the coil makes an angle θ with the magnetic field, as shown in Fig. 39( c). the forces

are unchanged, but each lever arm is reduced from ½ b to ½ sin θ. Note that the angle θ is

chosen to be the angle between B and the perpendicular to the face of the co il. Fig.

39( c ). So the torque becomes

τ =NIABsin

The formula, derived here for a rectangular coil is valid for any shape of flat coil.

The quantity NIA is called the magnetic dipole moment of the coil and is considered

a vector: =NIA

where the direction of A (and therefore of μ.) is perpendicular to the plane coil I the green

arrow in Fig. 39( c) consistent with the right-hand rule (cup . right hand so your fingers wrap

around the loop in the direction ot current flow then your thumb points in the direction of

52

(μand A). With this definition of μ, we can rewrite τ =NIABin vector form:

or τ =Bwhich gives the correct magnitude and direction for τ .

Electric Charge and the Magnetic Field

From what we already know we can predict the force on a single moving electric

charge.If N such particles of charge q pass by a given point in time t, they constitute a current

I = Nq/t. We let t be the time for a charge q to travel a distance ℓ in a magnetic field B; then ℓ

= v t where v is the velocity of the particle. Thus, the force on these N particles is. by

Equation F=IℓB= (Nq/t)(vt)B=Nqv B The force on one of the N particles is then

F=qv B

This gives the magnitude of the force on a particle of charge q moving with velocity v

at a point where the magnetic field has magnitude B. The angle between v and B is θ. The

force is greatest when the particle moves perpendicular to B (θ = 900):

F=IℓB IB

The force is zero if the particle moves parallel to the field lines (θ = 00). The

direction of the force is perpendicular to the magnetic field B and to the velocity v of the

particle. In other words, the magnetic force is always at right angels both to the velocity

v of the charge and to the magnetic field B.It is given again by a right-hand rule. Orient

your right hand so that your outstretched fingers point along the direction of motion of the

particle (v ) and when you bend your fingers they point along the direction of B( Fig 39).Then

your thumb will point in the direction of the force.

53

Fig 40

Force on an Electric Charge Moving in a Magnetic Field

We have seen that a current-carrying wire experiences a force when placed in a

magnetic field. Since a current in a wire consists of moving electric charges, we might expect

that freely moving charged particles (not in a wire) would also experience a force when

passing through a magnetic field. The magnetic force changes only direction of the

particle’s velocity not its magnitude.Indeed, this is the case under the influence of the

magnetic field a particle of charge q with velocityv follows circular path or circular motion.

The force exerted on the particle has a magnitude F=qv B

The same force is keeping the charge on circular path in other words this is centripetal

force .Newton’s second law F=ma we have a centripetal acceleration a =

v2

r

Of a particle=

mv2

r =qv B

Radius of circular path r=

mvqB

em of the electron is very important . For electron F=ev B=

mv2

r

54

em =

νB . r

The time T required for a particle of charge q moving with constant speed v to make

one circular revolution in a uniform magnetic field B( v ) is T=

2 π . rν where 2r is the

circumference of its circular path.

CEN

Radius of circular pathr=

mvqB so T=

2 π . rν =

2 π .mqB

Since T is the period of rotation, the frequency of rotation is f=

1T =

qB2 π .m This is

often called the cyclotron frequencyof a particlein the field because this is the rotation

frequency of particles in a cyclotron

MTB-EEE

Sources of Magnetic Field

We will now see how magnetic field strengths are determined for some simple

situations, and discuss some general relations between magnetic fields and their sources. We

begin with the simplest case, the magnetic field created by a long straight wire carrying a

steady electric current. We then look at how such a field, created by one wire, exerts a force

on a second current-carrying wire. Interestingly enough, this interaction is used for the precise

definitions of both the units of electric current and electric charge, the ampere and the

coulomb.

Then we develop an elegant general approach to finding the connection between

current and magnetic field known as Ampere's law, one of the fundamental equations of

physics. We also examine a second technique for determining the magnetic field due to a

current, known as the Biot-Savart law; which does allow us to solve problems more readily in

many cases than Ampere's law.

Finally, we try to understand about iron and other magnetic materials and how they

55

produce magnetic fields.

Magnetic Field Due to a Straight

You might expect that the field strength at a given point would be greater if the current

flowing in the wire were greater; and that the field would be less at points farther from the

wire. This is indeed the case. Careful experiments show that the magnetic field B at a point

near a long straight wire is directly proportional to the current I in the wire and inversely

proportional to the distance r from the wire:

B

Ir

This relation is valid as long as r, the perpendicular distance to the wire, is much less

than the distance to the ends of the wire (i.e., the wire is long). The proportionality constant is

written as μ0/2π; thus.B=

μ0 . I

2 π . r [outside a long straight wire]

The value of the constant μ0, which is called the permeability of free space, is

μ0 = 4π xl0-7 T.m/A

Force Between Two Parallel Wires

Consider two long parallel wires separated by a distance d as in Fig 41 They carry

currents I 1 and I 2 respectively. Each current produces a magnetic field so that each must

exert a force on the other, as Ampère first pointed out. The magnitude of the the magnetic

fieldB1 produced by I 1at the location of the second wire.

B1=

μ0 . I 1

2π .d

56

Fig 41

Operational Definitions of the Ampère

The ampere, the unit of current, is now defined in terms of the magnetic field B it

produces using the defined value of μ0.

In particular, we use the force between two parallel current-carrying wires, to define

the ampère precisely. If I1=I2= lA exactly, and the two wires are exactly 1 m apart, then

Fℓ =

μ0

2 π

I 1 I 2

d =

(4 πx 10−7T .m / A )(1 A )(1 A )2 π (1m ) =2x10−7 N /m

Thus, one ampère is defined as that current flowing in each of two long parallel

conductors 1 m apart, which results in a force of exactly 2 X 10 -7 N/m of length of each

conductor.

The amount of current in a wire can be varied accurately and continuously (by putting

a variable resistor in a circuit).Thus the force between two current-carrying conductors is far

easier to measure precisely. The magnetic field in terms of the force per unit length on a

current-carrying wire, via equation F=IℓB.

Ampère's Law

Equation B=

μ0 . I

2π . r gives the relation between the current in a long straight wire and

the magnetic field it produces. This equation is valid only for a long straight wire. The French

scientist Andre Marie Ampère(1775-1836)proposed general relation between a current in a

wire of any shape and the magnetic field around it. Heconsidered an arbitrary closed path

around a current by using compasses in sequence as shown in Fig. 42, and imagined this path

as being made up of short segments(needle of compasses) each of length Δℓ . First, we take

57

the product of the length of each segment times the component of B parallel to that segment

(call this component Bll). If we now sum all these terms, according to Ampère, the result will

be equal to μ0 times the net current Iencl that passes through the surface enclosed by the path:

Fig 42

∑ B11 Δℓ =μ0 I encl

The lengths Δℓ are chosen so that Bll is essentially constant along each length.

Thesum must be made over a closed path; and Iencl, is the net current passing through the

surface bounded by this closed path. In the limit Δℓ → 0, this relation becomes

∮Bdℓ =μ0 I encl Ampère Law

where dℓ is an infinitesimal length vector and the vector dot product assures that ' the parallel

component of B is taken. Above equation is known as Ampere's law. The integral in above

equation is taken around a closed path.

To understand Ampere's law better, let us apply it to the simple case of a long straight

wire carrying a current I which we've already examined, and which served as an inspiration

for Ampere himself. Suppose we want to find the magnitude of B at some point A which is a

distance r from the wire (Fig. 43). Fig. 43

We know the magnetic field lines are circles with the wire at their center. So to apply

equation ∮Bdℓ =μ0 I encl . We choose as our path of integration a circle of radius r. The choice

58

of path is ours, so we choose one that will be convenient: at any point on this circular path,

Bwill be tangent to the circle. Furthermore, since all points on the path are the same distance

from the wire, by symmetry we expect B to have the same magnitude at each point. Thus for

any short segment of the circle (Fig. 43), B will beparallel to that segment, and hence

μ0 I =∮Bdℓ =B∮ dℓ =B(2r) Ampère Law

where ∮ dℓ = 2πr, the circumference of the circle, and Iencl=I.. We solve for B and obtain

.B=

μ0 . I

2π . r

This is exactly the same equation for the field near a long straight wire as discussed

earlier. Its importance is that it relates the magnetic field to the current in a direct and

mathematically elegant way. Ampere's law is thus considered one of the basic laws of

electricity and magnetism. It is valid for any situation where the currents and fields are steady

and. not changing in time, and no magnetic materials are present.

CEN

Magnetic Field of a Solenoid and a Toroid

A long coil of wire consisting of many loops is called a solenoid. Each loop produces

a magnetic field as was shown in Fig. 44. A toroid, whichis essentially a long solenoid bent

into the shape of a circle (see Fig. 45). In Fig. 44a, we see the field due to a solenoid when the

coils are far apart. Toward the center of the solenoid, the fields add up to give a field that can

be fairly large and fairly uniform. For a long solenoid, with closely packed coils, the field is

nearly uniform and parallel to the solenoid axes within the entire cross section, as shown in

Fig. 44b. The field outside the solenoid is very small compared to the field inside, except near

the ends. Note that the same number of magnetic field lines that are concentrated inside the

solenoid, spread out into the vast open space outside.

Fig44a Fig44b Fig45

59

We now use Ampere's law to determine the magnetic field inside a very long (ideally,

infinitely long) closely packed solenoid. We choose the path a,b,c,d shown in Fig.46.Far from

either end, for applying Ampere's law. We will consider this path as made up of four

segments, the sides of the rectangle: ab, bc, cd, da. Then the Ampere's law, becomes

∮Bdℓ =∫a

b

Bdℓ +∫b

c

Bdℓ +∫c

d

Bdℓ∫d

a

Bdℓ

The field outside the solenoid is so small as to be negligible compared to the field

inside. Thus the first term in this sum will be zero. Furthermore, B is perpendicular to the

segments bc and da inside the solenoid, and is nearly zero between and outside the coils, so

these terms too are zero. Therefore we have reduced the integral to the segment cd where B is

the nearly uniform field inside the solenoid, and is parallel todℓ , so

∮Bdℓ =∫c

d

Bdℓ =Bℓ

where is the length cd. Now we determine the current enclosed by this loop for [the right side

of Ampere's law, equation ∮Bdℓ =μ0 I encl . If a current I flows in the wire of the solenoid,

the total current enclosed by our path abcd is NI where N is the number of loops our path

Fig 46

encircles (five in Fig.45””). Thus Ampere's law gives us

Bℓ =μ0 NI

If we let n = N/ℓbe the number of loops per unit length, then

B =μ0 nI

60

This is the magnitude of the magnetic field within a solenoid. Note that B depends

only on the number of loops per unit length, n, and the current I. The field does t not

depend on position within the solenoid, so B is uniform. It is a good approximation for real

ones for points not close to the ends.

Biot-Savart Law

The usefulness of Ampere's law for determining the magnetic field B due to particular

electric currents is restricted to situations where the symmetry of the given currents allows us

to evaluate ∮

B.dℓ readily. This does not, of course, invalidate Ampere's law nor does it

reduce its fundamental importance. Recall the electric case- where Gauss's law is considered

fundamental but is limited in its use for actually calculating E. We must often determine the

electric field E=k

q

r2 by another method summing over contributions due to infinitesimal

charge elements dq via Coulomb's law: dE =

14 πε0

dq

r2. A magnetic equivalent to this

infinitesimal form of Coulomb's law would be helpful for currents that do not have great

symmetry. Such a law was developed by Jean Baptiste Biot (1774-1862) and Felix Savart

(1791-1841) shortly after Oersted's discovery in 1820 that a current produces a magnetic field.

According to Biot and Savart, a current I flowing in any path can be considered as

many tiny (infinitesimal) current elements, such as in the wire of Fig.47. If dℓ represents any

infinitesimal length along which the current is flowing, then the magnetic field, dB, at any

point P in space, due to this element of current, is given by

dB =

μ0 I

4 π

dℓx r̂

r2Biot-Savart Law

where ris the displacement vector from the element dℓ to the point P, and r̂ is the unit vector

61

in the direction of P (see Fig. 47). Fig. 47

Equation belove is known as the Biot-Savart law. The magnitude of dB is

dB=

μ0 I

4 π

dℓ sin θ

r2 Biot-Savart Law

where θ is the angle between dℓ and r (Fig. 47). The total magnetic field at point P is then

found by summing (integrating) over all current elements:

B=∫ dB

Note that this is a vector sum. The Biot-Savart law is the magnetic equivalent of

Coulomb's law in its infinitesimal form. It is even an inverse square law, likeCoulomb's law.

An important difference between- the--Biot-Savart law and Ampere's law is that in Ampere's

law [∮

B.dℓ =μ0Iencl]. B is not necessarily dueonly to the current enclosed by the path of

integration. But the Biot-Savart law the field dB in equation dB=

μ0 I

4 π

dℓ sin θ

r2

is due only, and entirely, to the current element Idℓ . To find the total B at any point in space,

it is necessary to include all currents.

Magnetic Materials—Ferromagnetism

Magnetic fields can be produced (a) by magnetic materials (magnets) and (b) by,

electric currents. We have studied the latter extensively here so far. Now we take a brief look

at magnetic materials, which are so present in everyday life: Ordinary magnets, iron cores in

62

motors and electromagnets, magnetic recording tape and computer data storage disks, even

the magnetic stripe on credit cards. Only iron and a few other materials such as cobalt, nickel,

gadolinium and certain alloys can be made into strong magnets and show strong magnetic

effects. They are said to be ferromagnetic (from the Latin word ferrum for iron. We now look

more deeply into the sources of ferromagnetism.

Microscopic examination reveals that a magnet is actually made up of tiny regions

known as domains, which are at most about 1 mm in length or width. Each domain behaves

like a tiny magnet with a north and a south pole. In an unmagnetized piece of iron, these

domains are arranged randomly, The magnetic effects of the domains cancel each other out,

so this piece of iron is not a magnet. In a magnet, the domains are preferentially aligned in

one direction (downward in this case). A magnet can be made from an unmagnetized piece of

iron by placing it in a strong magnetic field.

An iron magnet can remain magnetized for a long time, and thus it is referred to as a

"permanent magnet." However, if you drop a magnet on the floor or strike it with a

hammer, you may cause the domains into randomness. The magnet can thus lose some

or all of its magnetism. Heating a magnet too can cause a loss of magnetism, for raising

the temperature increases the random thermal motion of the atoms which tends to randomize

the domains. Above a certain temperature known as the Curietemperature (1043 K for iron),

ferromagnetic materials, are no longer ferromagnetic), they generally are paramagnetic

Paramagnetism and Diamagnetism

All materials are magnetic to some extent. Nonferromagnetic materials fall into two

principal classes: paramagnetic, in which the magnetic permeability μ is slightly greater than

μ0; and diamagnetic, in which, μ. is slightly less than μ0. The ratio of

μμ0 for any material is

called the relative permeability Km:Km=

μμ0

Another useful parameter is the magnetic susceptibility Xm defined as

Xm=Km-1

Paramagnetic substances have Km> 1 and Xm> 0, whereas diamagnetic substances

63

have Km< 1 and Xm< 0.

The difference between paramagnetic and diamagnetic materials can be understood

theoretically at the molecular level on the basis of whether or not the molecules have a

permanent magnetic dipole moment. Paramagnetism occurs in materials whose molecules

(or ions) have a permanent magnetic dipole moment.' In the absence of an' external field, the

molecules are randomly oriented and no magnetic effects are observed. However, when an

external magnetic field is applied, say, by putting the material in a solenoid, the applied field

exerts a torque on the magnetic dipoles , tending to align them parallel to the field. The total

magnetic field (external plus that due to aligned magnetic dipoles) will be slightly greater than

Bo. The thermal motion of the molecules reduces the alignment, however. A useful quantity is

the magnetization vector, M, defined as the. magnetic dipole moment per unit volume,

M=

μV

Where μ is the magnetic dipole moment of the sample and V its volume. It is found

experimentally that M is directly proportional to the external magnetic, field (tending to

align the dipoles) and inversely proportional to the Kelvin temperature T(tending to

randomize dipole directions). This is called Curie's law, after Pierre Curie (1859-1906),

who first noted it: M=C

BT

where C is a constant. If the ratio B/T is very large (B very large or T very-small) Curie's law

is no longer accurate; as B is increased (or T decreased), the magnetization approaches some

maximum value, Mmax. This makes sense of 2 course, since Mmax corresponds to complete

alignment of ail the permanent magnetic dipoles. However, even for very large magnetic

fields, ¿ 2.0T, deviations from Curie's law are normally noted only at very low temperatures,

on the order of a few kelvins.

.Diamagnetic materials (for which μ is slightly less than μ0.) are made up of

molecules that have no permanent magnetic dipole moment. When an external magnetic field

is applied, magnetic dipoles are induced, but the induced magnetic dipole moment is in the

direction opposite to that of the field. Hence the total field will be slightly less than the

external field. The effect of the. external field—in the crude model of electrons orbiting

nucleiis to increase the "orbital" speed of electrons revolving in one direction, and to decrease

64

the speed of electrons revolving m the other direction; the net result is a net dipole moment

opposing the external field, Diamagnetism is present in all materials, but is weaker even than

paramagnetism and so is overwhelmed by paramagnetic and ferromagnetic effects in

materials that display these other forms of magnetism.

Electromagnetic Induction and Faraday’s Law

So far we discussed two ways in which electricity and magnetism are related: (1) an

electric current produces a magnetic field; and (2) a magnetic field exerts a force on an

electric current or moving electric charge. These discoveries were made in 1820-1821.

Scientists then began to wonder: if a magnetic field can produce an electric current? Ten years

later Joseph Henry (1797-1878) and Michael Faraday (1791-1867) independently found that

it was possible. We now discuss this phenomenon and some of its very important applications

such as the electric generator

. Fig.48

Induced EMF

In his attempt to produce an electric current from a magnetic field, Faraday used an

apparatus like- that shown in Fig. 48. It consists of two coils of insulated wire. A coil of wire,

X, was connected to a battery and the other Y to a galvanometer. No currentflowed through

the galvanometer as in all Faraday previous attempts. But Faraday observed the galvanometer

in circuit Y deflect strongly at the moment he closed the switch in circuit X. And the

galvanometer deflected strongly in the opposite direction when he opened the switch.

Faraday realized that an induced currentwas produced only by a change in the magnetic field .

He concluded thata changing magnetic field can produce an electric current! Such a

current is called an induced current. When the magnetic field through coil Y changes, a

current flows as if there were a source of emf in the circuit. We therefore say that an induced

emf is produced by a changing magnetic field.

65

Fig 49

Faraday did further experiments on electromagnetic induction. For example, Fig. 49(ii)

shows that if a magnet is moved quickly into a coil of wire, a current is induced in the wire. If

the magnet quickly removed, a current is induced in the opposite direction. Furthermore, if the

magnet is held steady and the coil of wire is moved toward or away from the magnet, or the

coil is rotated, again induced current flows. When an induced current flows.in a circuit,an

emf is similarly present . Motion or changeis required to induce an emf. It doesn't matter

whether the magnet or the coil moves. It is the relative motion that counts.

Faraday's Law of induction, Direction of emf; Lenz's Law

Faraday investigated quantitatively what factors influence the magnitude of theemf

induced. He found first of all that the more rapidly the magnetic field changes, the greater the

induced emf. But the emf is not simply proportional to the rate of change of the magnetic

field B. Rather the emf is proportional to the rate of change of the magnetic flux. ΦB .

passing through (he circuit or loop of area A. Magnetic flux for a uniform magnetic field is

defined as

ΦB =B¿ A=BAcosθ =BA ( B uniform)

Here B¿ A is the component of the magnetic field B perpendicular to the face of the

loop, and θ is the angle between B and the vector A (representing the area) whose direction is

perpendicular to the face of the loop.

Fig 50 (b)

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These quantities are shown in ) Fig. 50 for a square loop of side ℓ whose area is A =ℓ 2. If

the area is of some other shape, or B is not uniform, the magnetic flux can be written

ΦB = ∫B .dA

As we saw earlier, the lines of B (like lines of E) can be drawn such that the number of

lines per unit area is proportional to the field strength. Then the flux ΦB can be thought of as

being proportional to the total number of lines passing through the loop.

Fig 51

This is illustrated in Fig.50(b) and51 where the loop is viewed from the side (on edge).

For θ = 90", no lines pass through the loop and B = 0, whereas B is a maximum when θ =

00. The unit of magnetic flux is the tesla-meter2; this is calledaweber: lWb = lT.m2.

With this definition of the flux, we can now write down the results of Faraday's

investigations namely, that theemf induced in a circuit is equal to the rateof change of

magnetic flux through the circuit:

ε=-dΦB

dt

This fundamental result is known as Faraday's law of induction, and is one of the

basic laws of electromagnetism.

If the circuit contains N closely wrapped loops, the emfs induced in each add to set

her, so ε=-N

dΦB

dt

The minus sign is placed there to remind us in which direction the induced emf acts.

Experiments show that:( Fig 52 )

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Fig 52

An induced emf gives rise to a current whose magnetic field opposes the original

change in flux. (Lenz's law)

This is known as Lenz's law. Said another way. valid even if no current can flow (as

when a circuit is not complete), is:

An induced emf is always in a direction that opposes the original change in flux

that caused it.

Let us investigate its direction. We can find this out with an arrangement as shown in

Fig.49 If we plunge each pole of a magnet into and out of the coil the needle of the

galvanometer show the results. These result were generalized most elegantly into a rule by

Lenz in 1835. He said that the induced current flows away in such a directionas to opposes

the change which is giving rise to it. What Lenz meant: when the magnet is approaching the

coil, the coil repels it. when the magnet is pulled back from the coil attracts it.

EMF Induced in a Moving Conductor

Another way to induce an emf is shown in Fig. 53. and this situation helps illuminate

the nature of the induced emf. Assume that a uniform magnetic field B is perpendicular to the

area bounded by the U-shaped conductor and the movable rod resting on it. If the rod is made

to move at a speed v, it travels a distance dx = v dt in a timedt. Therefore, the area of the loop

increases by an amount dA=ℓ dx =ℓ vdt in a time dt. By Faraday's law, there is an induced

emf εwhose magnitude is given by

ε=dΦB

dt =BdAdt =

Bℓ vdtdt =Bℓ v

This equation is valid as long as B,ℓ , and v are mutually perpendicular. (If they are

not. we use only the components of each that are mutually perpendicular.) An emf induced on

a conductor moving in a magnetic field is sometimes called a motional emf.

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Fig. 53

We can also obtain Equation ε=dΦB

dt = Bℓ v without using Faraday's law. We saw

in previous lecture that a charged particle moving perpendicular to a magnetic field B with

speed v experiences a force F = qv B. When the rod of Fig. 53( a) moves to the right with

speed v, the electrons in the rod move with this same speed. Therefore, since v¿ B, each

electron feels a force F = qvB. which acts upward as shown in Fig. 53( b). If the rod were not

in contact with the U-shaped conductor, free electrons would collect at the upper end of the

rod. leaving the lower end positive. There must be an induced emf. If the rod does slide on the

U-shaped conductor, the electrons will flow into it. There will then be a clockwise

(conventional) current following in the loop. To calculate the emf, we determine the work W

needed to move a charge q from one end of the rod to the other against this potential differ-

ence W =force x distance = (qvB}(ℓ ). The emf equals the work done per unit charge, so ε=

W/q =qvBℓ q = Bℓ v,just as above

This emf produces an electric field E in the rod which moves the electrons along.

Assuming a uniform E in the rod, then E= ε/ℓ =Bv

Electric Generators

Probably the most important practical result of Faraday's great discovery was the

development of the electric generator or dynamo. A generator transforms mechanical

energy into electric energy. This is just the opposite of what a motor does. Indeed, a generator

is basically the inverse of a motor. A simplified diagram of an ac generator is shown in Fig.

54 A generator consists of many coils of wire (only one is shown) wound on an armature that

can rotate in a magnetic field. The axle is turned by some mechanical means (falling water,

car motor belt), and an emf is induced in the rotating coil. An electric current is thus the

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output of a generator.

Fig 54 Fig 55

In Fig. 54 the equationF=qv Btells us that, with the armature rotating

counterclockwise, the (conventional) current in the wire labeled A on the armature h outward;

therefore it is outward at brush A, as shown. (Each brush presses against a continuous slip

ring.) After one-half revolution, wire A will be where wire C is now in the drawing, and the

current then at brush A will be inward. Thus the current produced is alternating. Let us look at

this in more detail.

Let us assume the loop is being made to rotate in a uniform magnetic field Bwith

constant angular velocity ω. From Faraday's law (Eq. ε=-dΦB

dt ), the induced emf is

ε=-dΦB

dt =-

ddt∫B . dA

=-

ddt [ BA cosθ ]

where A is the area of the loop and θ is the angle between B and A. Since w = dθ/dt, then θ =

θ0 +wt. We arbitrarily take θ0 = 0, so

ε=-BA

ddt (cos wt)=BA wsin wt

If the rotating coil contains N loops,

ε =NBA wsin wt=ε0 wsin wt The maximum emf ε0 =NBA w

Thus the output emf is sinusoidal (Fig55} with amplitude ε = NBAω. Such a rotating

coil in a magnetic field is the basic operating principle of an ac generator.

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The Transformer

A transformer is a device for stepping up or down an alternating voltage. It has

primary and secondary windings, as in induction coils, but no make-and-break (Fig. 56). It

has an iron core, which is made from E-shaped laminations, interleaved so that the magnetic

flux does not pass through air at all; in this way the greatest flux is obtained with a given

current. When an alternating emf. Ep is impressed on the primary winding, it sends an

alternating current through it, which sets up an alternating flux in the core of magnitude BA,

where B is the induction and A is the cross- sectional area. This induces an alternatingemf. in

the secondary Es. If Np, Ns are the number of turns in the primary and secondary coils, their

linkages with the flux Φ p are:

Φ p =NpAB ΦS =NS AB

The magnitude of the emf.. induced in the secondary is Es.=

dΦs

dt =NsA

dBdt

The changing flux also induces a back –emf.. in the primary, whose magnitude is

Ep=

dΦ p

dt =NpA

dBdt

Because the primary winding has inevitably some resistance, the current flowing

through it sets up a voltage drop across the resistance. But in practice this is negligible

compared with the back –emf. due to the changing flux. Consequently we may say that the

voltage applied to the back –emf.. Ep due to the changing flux. Consequently we may say that

the voltage applied to the rimary, form the source of current, is used simply in magnitude to Ep

(This is analogous to saying, in mechanics, that action and reaction are equal and opposite.)

Consequently we have

( emf induced in secondary)/(voltage applied to primary)=

Es

Ep =

N s

N p

Thus the transformer steps voltage up or down according to its “turns-ratio”:

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V S

V P =

N s

N p or

V 2

V 1 =

N 2

N1

(secondary voltage)/(primary voltage)=(secondary turns)/(primary turns)

When a load is connected to the secondary winding, a current flows in it

. This current flows in such a direction as to reduce the flux in the core. At the instant

that the load is connected, therefore, the back-emf.. in the primary falls. The primary current

then increases . the continues until the flux is restored to its original value. The back –emf.. in

the primary is then again equal to the applied voltage, and equilibrium is restore. But now a

greater primary current is flowing than before the secondary was loaded. Thus the power

drawn from the secondary is drawn, in turn, from the supply to which the primary is

connected.

Transformers are used to step up the voltage generated at a power station, from 11000

to 132000 volts for high-tension transmission. After transmission they are used to step it down

again to 33000, to 11000 and to a value safer for distribution (240 volts in houses). Inside a

house a transformer may be used to step the voltage down from 240 to 110. Transformers will

several secondaries are used in, for example, radio-receivers, where several different voltages

are required.

Also it can be written

V p I p =V s I s or

I s

I p =

N p

N s

ELECTERMAGNETIC WAVES

72

The energy E of a photon is equal to a constant h times its frequency v has the value h=6.6260693x10-34 J.s=4.14x10-15 eV.sE = hv From v = c / λ for electromagnetic waves. Where c is the velocity of light and λ is the wave length we have;

E = hv = hcλ

(energy of photon)

where h is a universal constant called Planck’s constant. The numerical value of this constant, to the accuracy known at present, is

h = 6.6260693 (11) X 10-34 J . sA photon wit energy E has momentum with magnitude p given by E = pc. Thus the wavelength λ of a photon and the magnitude of its momentum p are related simply by

P = Ec

= hvc

= hλ

The direction of the photon’s momentum is simply the direction in which the electromagnetic wave is moving.

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