electricity & magnetism · 2018-07-26 · 5 chapter one electrostatics and coulomb’s law 1.0...
TRANSCRIPT
KWAME NKRUMAH UNIVERSITY OF SCIENCE AND
TECHNOLOGY, KUMASI
DEPARTMENT OF PHYSICS
ELECTRICITY & MAGNETISM
2014/2015
LECTURER: DR KWASI PREKO
Tel.: 0242026899
Email: [email protected]
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CONTENT
CHAPTER ONE ........................................................................................... 5 ELECTROSTATICS AND COULOMB‟S LAW .......................................................... 5
1.0 THE STRUCTURE OF THE ATOM ....................................................................... 5
1.1 CONDUCTORS AND INSULATORS .................................................................... 6
1.2. THE ELECTRIC CHARGE .................................................................................... 8
1.3 COULOMB‟S LAW ............................................................................................... 11
1.4 SUPERPOSITION .................................................................................................. 13
1.5 CALCULATION OF THE RESULTANT ELECTROSTATIC FORCE OF TWO
OR MORE CHARGES ................................................................................................. 16
CHAPTER TWO ........................................................................................ 29 ELECTRIC FIELD AND GAUSS‟ LAW .................................................................... 29
2.0 CHARGE DISTRIBUTIONS AND THE ELECTRIC FIELD .............................. 29
2.1 THE ELECTRIC FIELD ........................................................................................ 32
2.1.1 Principle of Superposition .............................................................................. 33
2.2 DIRECTION OF THE ELECTRIC FIELD, E ............................................. 34
2.3 CALCULATION OF THE ELECTRIC FIELD, E ................................................ 34
2.4 ELECTRIC FIELD LINES ..................................................................................... 40
2.6 The Neutral Point (N) ............................................................................................. 45
2.7 ELECTRIC DIPOLE IN AN ELECTRIC FIELD ................................................ 46
2.8 ELECTRIC FLUX ................................................................................................ 47
2.9 GAUSS‟S LAW ................................................................................................... 49
2.10 SOME APPLICATIONS OF GAUSS‟S LAW .................................................... 51
CHAPTER THREE .................................................................................... 58 THE ELECTRIC POTENTIAL .................................................................................... 58
3.0 ELECTRIC POTENTIAL ...................................................................................... 58
3.1 POTENTIAL AND THE ELECTRIC FIELD ........................................................ 59
3.2 POTENTIAL DUE TO A POINT CHARGE ......................................................... 61
3.3 A GROUP OF POINT CHARGES ........................................................................ 63
3.4 POTENTIAL DUE TO A DIPOLE ........................................................................ 66
CHAPTER FOUR....................................................................................... 68 4.0 CAPACITANCE .................................................................................................... 68
4.1 CAPACITORS IN SERIES AND IN PARALLEL ................................................ 73
4.2 DIELECTRICS - AN ATOMIC VIEW .................................................................. 79
4.3 GAUSS‟ LAW IN DIELECTRICS ........................................................................ 82
4.4 ENERGY STORAGE IN CAPACITORS .............................................................. 86
CHAPTER FIVE ........................................................................................ 88 STEADY CURRENTS AND DIRECT CURRENT CIRCUITS ................................. 88
5.0 CHARGE FLOW IN CONDUCTORS: CURRENT AND CURRENT DENSITY
...................................................................................................................................... 88
5.1 ELECTROMOTIVE FORCE AND POTENTIAL DIFFERENCE ....................... 90
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5.2 OHM‟S LAW AND THE CONDUCTION OF ELECTRICITY BY FREE
ELECTRONS ............................................................................................................... 91
5.3 ELECTRICAL RESISTANCE: OHM‟S LAW FOR CIRCUITS .......................... 93
5.4 EQUIVALENT RESISTANCE OF NETWORKS ................................................ 95
5.5 KIRCHOFF´S LAWS ............................................................................................. 97
5.6 SIMPLE R-C CIRCUITS ..................................................................................... 107
CHAPTER SIX ......................................................................................... 113 MAGNETISM AND THE MAGNETIC FIELD ....................................................... 113
6.0 MAGNETISM ..................................................................................................... 113
6.1 MAGNETIC FIELD ............................................................................................. 113
6.2 APPLICATIONS OF MOVING CHARGES IN A MAGNETIC FIELD ........... 117
6.2.1 Charged -particle Linear Momentum Analyzer ............................................ 117
6.2.2 Mass Spectrometer ......................................................................................... 118
6.2.3 Cyclotron ....................................................................................................... 119
6.2.4 Synchrotron .................................................................................................... 121
6.3 Crossed Electric and Magnetic Fields .................................................................. 124
6.4 MAGNETIC DIPOLE IN A MAGNETIC FIELD ............................................ 126
6.5 THE MAGNETIC FIELD OF A CURRENT-CARRYING CONDUCTOR:
THE BIOT-SAVART LAW ....................................................................................... 131
6.6 AMPERE‟S LAW ................................................................................................. 134
6.6.1 Solenoids ........................................................................................................ 137
6.6.2 Magnetic Force between Two Current-Carrying Conductors ....................... 139
CHAPTER SEVEN .................................................................................. 142 ELECTROMAGNETIC INDUCTION ...................................................................... 142
7.0 MOTIONAL ELECTROMOTIVE FORCE ......................................................... 142
7.1 FARADAY‟S LAW OF INDUCTION ............................................................... 146
The following are three equations relating to Maxwell: ......................................... 146
Example 7. 5 ........................................................................................................... 151
7.2 APPLICATIONS OF FARADAY‟S LAW .......................................................... 151
7.2.2 A Rectangular Loop Generator ...................................................................... 152
7.2.3 The Spin-Echo Magnetometer ...................................................................... 154
7.3 LENZ‟S LAW AND EDDY CURRENTS ........................................................... 155
7.4 SELF-INDUCTION AND SELF-INDUCTANCE ........................................... 157
7.4.1 Self-Inductance of a Solenoid ........................................................................ 158
7.4.2 Self-inductance of a Toroid of a Rectangular Cross Section ......................... 159
7.5 LR Series Circuit .................................................................................................. 160
7.6 ENERGY STORED IN INDUCTIVE CIRCUITS .............................................. 164
7.7 MUTUAL INDUCTION ...................................................................................... 166
7.8 OSCILLATIONS IN CIRCUITS CONTAINING A CAPACITOR AND AN
INDUCTOR ................................................................................................................ 170
7.8.1 LC Circuit ...................................................................................................... 170
7.8.2 LCR Circuit .................................................................................................... 172
CHAPTER EIGHT ................................................................................... 174 MAGNETIC PROPERTIES OF MATTER ............................................................... 174
8.0 MACROSCOPIC MAGNETIC PROPERTIES OF MATTER............................ 174
8.1 ATOMIC AND NUCLEAR MAGNETIC MOMENTS ...................................... 174
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8.2 CLASSIFICATION OF MAGNETIC MATERIALS .......................................... 176
8.3 DIAMAGNETISM ............................................................................................... 178
8.4 PARAMAGNETISM ............................................................................................ 179
8.5 FERROMAGNETISM ......................................................................................... 181
8.6 MAGNETIC DOMAINS ...................................................................................... 183
CHAPTER NINE ...................................................................................... 185 A.C. THEORY ............................................................................................................ 185
9.1 ROOT MEAN SQUARE VALUES OF ALTERNATING VOLTAGE AND
CURRENT .................................................................................................................. 186
9.2 RELATIONSHIP BETWEEN THE IRMS, VRMS, AND THE PEAK CURRENTS
AND PEAK VOLTAGES. ......................................................................................... 186
9.2.1 AC CIRCUIT WITH PURE RESISTANCE ............................................. 188
9.2.3 GRAPH OF I AND V FOR A CAPACITOR ............................................ 188
9.3 POWER DISSIPATED IN THE RESISTOR OF A PURE RESISTIVE CIRCUIT
.................................................................................................................................... 189
9.3.1 A.C. CIRCUIT CONTAINING A CAPACITANCE ONLY ....................... 189
9.3.2 VOLTAGE TRIANGLE & IMPEDANCE TRIANGLE ........................... 195
9.3.3 R – C SERIES CIRCUIT ....................................................................... 196
Example 9.5 ................................................................................................................ 196
9.3.4 R – L – C SERIES CIRCUIT ................................................................ 198
9.3.5 R – L – C SERIES RESONANCE .............................................................. 199
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CHAPTER ONE
ELECTROSTATICS AND COULOMB’S LAW
1.0 THE STRUCTURE OF THE ATOM.
The concept of the atom since the time of John Dalton, the English Physcist, as
indivisible particle has been refuted as the atom is now known to comprise of subatomic
particles.
Three kinds of sub atomic particles: the electron, the proton, and the neutron. Other types
of subatomic particles have been observed. Their lives are, however, REFER - i.e.
they have very very short life span. They recombined to form the ordinary subatomic
particles (e, p, n).
Atoms make up ordinary matter. Thus one can also say that matter is made up of a
combination of several sub-atomic particles. The protons and neutrons are always in
closely packed arrangement to form the nucleus. If the nucleus is considered a sphere, its
diameter is of the order of 10 –14
m.
Around the nucleus the electrons are at relatively large distances away.
(Ilustration: A fly in a large building e.g. Great Hall , KNUST)
The Proton:
Has a positive charge equal in magnitude to the electronic charge.
The mass (just like any elementary particle) is expressed in atomic mass units
(a.m.u.)
Note: The a.m.u. is scale based on the mass of an atom of the most abundant isotope of
Oxygen which has been put at Ibamu (i.e 0).
On the amu scale, the mass of „H atom is 1.0081 amu. Thus mass of proton (i.e. nucleus
of hydrogen) is
mp = Mass of „H - mass of electron
= 1.0081 amu – 0.00050 amu = 1.0076 amu
But 1 amu = 1.66 x 10 –27
kg
d = 10-14m
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Thus mp = 1.6726 x 10 –27
kg
The Electron:
The electron has negative charge. The magnitude of the charge on it was determined by
Millikan in the Millikan oil-drop experiment (1909). Earlier (1897) the charge -to- mass
ratio had been determined by J.J. Thomson. The change is 1.603 x 10 –19
coulombs. The
mass is 1/1837 that of a hydrogen atom (i.e. 9.107 x 10 –31
kg).
The Neutron:
Electricity neutral. Its mass is about 1.0090 amu. They are more penetrating than the
electron or the proton.
REASON: Since it has no change, it is not affected by the electric fields around electrons
and nuclei.
1.1 CONDUCTORS AND INSULATORS
Matter may be classified into Conductors, Semiconductors and Insulators.
One of the most striking aspects of the behaviour of matter with respect to electric charge
is provided by the very different properties of electrical conductors and insulating
substances. Electrons are free to move about in, or flow through, a conducting substance,
while in an insulator they are bound to atoms and do not normally move about within the
material. In the case of solids, all metals and a number of substances such as carbon are
conductors, and their electrical properties can be explained by assuming that a number of
electrons are free to wander about the whole volume of the solid instead of being rigidly
attached to one atom. In solid substances of the second class, insulators, each electron is
firmly bound to the lattice of positive ions, and cannot move from point to point. Typical
insulators are sulphur, polystyrene, alumina, glass, wood, stone, waxes and oils.
When a substance has no net electrical charge, the total numbers of positive and negative
charges within it must be equal. Charge may be given to or removed from a substance,
and a positively charged substance has an excess of positive ions, while a negatively
charged substance has an excess of electrons. Since the electrons can move so much more
easily in a conductor than the positive ions, a net positive charge is usually produced by
the removal of electrons. In a charged conductor the electrons move to positions of
equilibrium under the influence of the forces of mutual repulsion between them, while in
an insulator they are fixed in position and any initial distribution of charge remains
almost indefinitely. In a good conductor the movement of charge is almost instantaneous,
while in a good insulator it is extremely slow.
There are, in addition, numerous substances that have a few free electrons but in
concentrations much lower those typical of metallic substances. Such substances conduct
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electrically, but only rather poorly. Their electrical behaviour is intermediate between
that of metallic conductors and good insulators. Such substances are called
semiconductors. Examples of semiconductors are graphite, germanium, and silicon.
How a rod is charged to carry either negative or positive charges.
- Rub ebonite with fur to obtain negative charges.
- Rub glass with silk to obtain positive charges.
Experimental setup:
negatively
charged
Cap copper rod (ebonite rod)
wire OR
leaf Glass rod
(positively changed)
wooden
Insulator support
Electroscope
Fig 1.1: A simple experiment to demonstrate transfer of electric charge
In the above experiment when the charged rod touches the end of the wire, the leaves of
the electroscope diverge. This is an indication that there is a transfer of charge along the
wire. Thus the wire is a conductor.
Repeating the experiment with a silk thread in place of the copper wire, no divergence is
observed. Thus the silk thread is an insulation or dielectric.
In electricity, we would study the motion of charges through material substances.
Generally: Conductors permit the passage of charges through them whiles insulators do
not. Metals in general are good conductors whiles non-metals are insulators.
In a metallic conductor (e.g. refer) a few outer electrons become detached from each
atom and may move freely throughout the metal. These are the free electrons. Thus in a
conductor, the positive nuclei and the other electrons (bound to the nuclei) remain fixed
in position while the free electrons roam about the metal. In an insulator, there exists
either no free electrons or very few indeed.
The phenomenon of charging happens when two substances are rubbed together. But for
a good conductor, the charges easily leak away if it is not supported on an insulator.
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1.2. THE ELECTRIC CHARGE
Mass can create a gravitational field g, which in turn can exert a force mg on a body of
mass m. Electric charge, like mass, is an important inherent property of matter which can
be present in both large and small bodies. Electric charge can also create force fields in
space, and these fields in turn transmit forces to other charged bodies and thereby affect
their motion.
At one time or another, we have experienced or seen effects due to electrified or charged
bodies. We may have walked on a new carpet and as a result experienced a minor shock
upon touching a metal object. We may have rubbed a balloon on the rug and then placed
it on a wall or ceiling, where it stays, apparently attracted by some curious force. If one
places a piece of paper against a blackboard and rubs the palm of the hand over it quickly
a number of times, the paper stays there for a while, attracted to the blackboard. These
are a few of many simple examples which demonstrate that in the process of rubbing
objects together, there can be a net transfer of some entity between the bodies. The entity
transferred is called electric charge. In the process of rubbing, bodies become electrified,
or electrically charged.
Why do bodies become electrified as a result of rubbing? The reason for this can be
partially understood on the basis of the atomic theory of matter. Matter is made up of
atoms which are composed of electrons, protons, and neutrons. The protons and neutrons
of an atom share a very small volume of space called the nucleus of the atom. For light
nuclei, those which contain only a few protons and neutrons, the atomic nucleus has a
radius of about 10-15
m from the nucleus. The radius of a heavy nucleus, however, is
somewhat larger. The distance of the electrons in the atom from the nucleus is typically
about 10-10
m. The electrons are attracted to the nucleus by a force called the electrostatic
force or Coulomb force. This force exists because the electrons and nuclei have electric
charges of opposite sign. It is established experimentally that like charges repel, unlike
charges attract. The electron is negatively charged and the nucleus is positively charged.
The positive charge of the nucleus is entirely due to the charges of protons, since the
neutrons do not have any net electric charge.
Matter is made up of enormous numbers of atoms and molecules and is normally
electrically neutral. This means that equal numbers of protons and electrons are present.
The forces which bind the electrons in an atom to the nucleus is quite strong. However,
these forces can be overcome. The electrons can, therefore, be transferred from one body
to another when two substances are brought into intimate contact. Therefore, in the
process of rubbing two substances together, many electrons may be transferred from one
object to another. When this happens, one of the bodies loses electrons while the other
gains. The one which gains has excess of electrons and the one which loses has a
deficiency. The one with excess is negatively charged, while the one which is deficient is
positively charged. The charge of an object can be regarded as the summation of all the
elementary atomic charges which make up the object. The charge of any elementary
particle, such as the electron, is an intrinsic property of the particle, just as the mass.
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As already seen, bodies may be positively or negatively charged, whenever there is a
deficiency or excess of electrons with respect to protons. Originally, negative charge was
defined by Benjamin Franklin as the type of charge residing on hard rubber (Lucite)
which had been rubbed with fur, while positive charge was the type acquired by glass
electrified by rubbing with silk. An equivalent and more modern definition of the sign of
electric charge would be to say that negatively charged objects have a charge of the same
sign as that of the electron while positively charged objects have a charge of the same
sign as that of the proton.
One of the most important facts about electric charge is that it appears as integral number
of electronic charges. The charge on the electron, therefore, is the smallest possible
quantity of negative charge that can be found in nature. Likewise, the charge of the
proton, which is equal in magnitude but opposite in sign to that on the electron, is the
smallest unit of positive charge to be found in the universe. The electronic charge is
denoted by -e while the charge on the proton is written as +e. The neutron is electrically
neutral and therefore has charge zero. The charges on other elementary particles are
either zero or some integral multiple of the electronic charge. Similarly, the charges
exhibited by ions or atomic nuclei are exact integral multiples of the charge on the
electron or proton. This characteristic occurrence of electric charge in units of an
indivisible elementary charge is referred to as charge quantization, and we say that
electric charge is quantized in units of the electron charge.
Another extremely important feature of electric charge is that electric charge is always
conserved. This means that in any interaction or reaction, the initial and final values of
the total electric charge must be the same. Thus, total electric charge is neither created
nor is it destroyed. A number of processes or reactions between particles or nuclei which
occur in nature are shown below.
(a) Beta decay of the neutron
n p + e- + e
0 +e -e 0
(b) Electron and positron annihilation
e+ + e
- +
+e -e 0 0
(c) Production of carbon 14 by collision of neutrons with nitrogen nuclei
14
7N + n 14
6C + p
7e 0 6e e
In each case, the total charge before the interaction occurs is identical to that which
exists afterward. This is the principle of conservation of charge which can be stated as
follows:
In any interaction, the net algebraic amount of charge remains constant.
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Semiconductors:
These form a group whose conductivity lies between that of conductors and insulators.
Note: Conductivity may be reckoned in terms of current. Our interest at moment lies in
the latter.
So if we are to consider conduction of electrical current then we should look at the
conductivity and resistivity of various conductors.
RA
Resistivity, ℓ = what is the unit? : (Ωm).
L
1 L 1
Conductivity, σ = = = omh/m OR
ℓ RA Ω.m.
Some examples: Resistivity Conductivity
Good conductors (Ω.m) at 200c (mho/m)
Silver 1.6 x 10 –8
6.2 x 107
Copper 1.7 x 10– 8
5.8 x 107
Aluminium 2.7 x 10– 8
3.7 x 107
Manganin 42.0 x 10– 8
0.24 x 107
Platiuum 10.6 x 10– 8
0.94 x 107
Constantan 48.0 x 10– 8
0.21 x 107
Mercury 96.0 x 10– 8
0.10 x 107
Carbon (graphite) 350 – 6, 300 x 10– 8
0.16 – 2.9 x 105
Poor Conductors
Saturated Nacl solution 4.4 x 10 –2
22.6
Distilled water 5.0 x 103 2.0 x 10
-4
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Semiconductors
In sb 5.7 x 10 –5
1.76 x 104
Germanium 0.47 2.1
Silicon 2.3 x 103 4.3 x 10
–4
Insulators
Pyrex glass 1012
10 –12
Paraffin wax 1014
10 –14
Polystyrene 1015
10 -15
Force of Attraction / Repulsion
It has been observed that like changes repel each other while unlike changes attract each
other.
The empirical formula expressing this idea was stated for the first time in 1784 by the
French Physicist Charles Augustin de Coulomb (who lived from 1736 – 1806).
1.3 COULOMB‟S LAW
In 1784, the French Physicist Charles Augustin de Coulomb (1736-1806) discovered the
quantitative force law between two point charges by measuring the forces of attraction
and repulsion using a torsional balance. The actual magnitude of the electric force that
one charged particle exerts on another is given by what is known as the Coulomb‟s law,
which states that
The magnitude of the electric force that a charged particle exerts on another
charged particle is directly proportional to the product of their charges and
inversely proportional to the square of the distance between them. The direction
of the force is along the line joining the particles.
Thus, the magnitude of the force (F) that a particle of charge q1 exerts on another particle
of charge q at a distance r is
F = k(q1q/r2), (1.1)
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where k is a constant. Since force is a vector quantity, Eq. 1.1 can be written in a vector
form as
F = k(q1q/r2)r. (1.2)
r is the unit vector directed from q1 to q (Fig. 1.1).
Note
1. Equation (1.2) holds for point charges or spheres of uniform charge density.
2. The force F acts along the line joining the centres of the two charges.
FIG. 1.2: Two charged particles q1 and q. The unit vector r is directed from q1 to q.
The constant k is traditionally written as
k = 1/4o
with o = 8.85 x 10-12
C2/(N.m
2). Thus the constant k has a value 9.0 x 10
9 N.m
2C
-2. The
quantity o is called the permittivity constant. Thus, the Coulomb‟s law becomes
F = [4o]-1
(q1q/r2)r. (1.3)
This equation applies to particles electrons and protons and also to any small charged
particles, provided that the sizes of these bodies are much less than the distance between
them. Such bodies are called point charges.
In the SI system, the unit of electric charge is the Coulomb. Now, assume q1 = q = 1 C
and r = 1 m. Inserting these values in Eq. 1.1 or 1.3 gives F = 9.0 x 109 N. Thus, the
Coulomb is that amount of charge which when separated by 1 m from a similar charge
leads to a force 9.0 x 109 N.
Permittivity
Refers to the reluctance of a medium to transmit an electrostatic force.
= 1/ 4k. Hence unit of is C2N
-1m
-2
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Permittivity of Free Space 0
Refers to the reluctance of a non-dissipative medium or vacuum to transmit electrostatic
forces. o = 8.8542 x 10-12
C2N
-1m
-2
Relative Permittivity R
This refers to the permittivity of a medium as compared to that of vacuum (or free space).
i.e. r = /o.
The Coulomb C
This is the SI unit of charge and it is defined as the quantity of charge which passes in
one second across any section of a conductor in which a steady current of 1A is flowing.
1 Coulomb = 1 C = 1 As
1.4 SUPERPOSITION
A very important concept in physics is superposition. This is essentially the principle of
additivity of interactions which we used, for example, in figuring out the net gravitational
force due to interaction with a non-point object. When superposition holds, forces tend to
add, and the net force between any two objects is independent of the position of any other
object. The Coulomb force obeys superposition - this is an experimentally observed fact
with no fundamental theoretical justification.
Example 1.1
Consider charges distributed along a straight line as follows:
Fig. 1.3 Charges distributed along a straight line
Use Coulomb's law to calculate the two separate forces on the 6 microcoulomb charge.
The force exerted by the 3 microcoulomb charge is
14
and is directed toward the right. That exerted by the 4 microcoulomb charge is
and is also directed toward the right. Superposing (i.e., adding) these two forces yields
the total force:
The validity of superposition has been checked to limits of experimental accuracy,
leading to the principle of superposition:
The net force exerted by two or more charges on a single charge Q is the vector sum of
the individual forces on Q.
Vector Form Of Coulomb's Law
Forces, including the Coulomb force of course, are vector quantities. We have been
discussing the magnitude of the Coulomb force while only specifying whether it is
repulsive or attractive. This is adequate in some circumstance, e.g., when there are only
two charges or when all charges lie on one line. It is not sufficient, however, when we
want to apply the superposition principle to a collection of charges in a more complicated
geometry. In this case, the vector force exerted by charge q1 on charge q2 is simply
where r12 is the distance between the charges and r12-hat is the unit vector directed from
charge 1 to charge 2. F12 is parallel to this unit vector if the charges have the same sign
(repulsive interaction) and is antiparallel to it if they have a different sign (attractive
interaction). An example will illustrate:
Example 1.2
15
Consider three 1 microcoulomb charges at the vertices of an equilateral triangle, 1 m on a
side. Calculate the net force that the bottom two charges exert on the top charge.
Fig 1.4: Three 1microcoulomb charges at the vertices of an equilateral triangle
By symmetry (all charges are equal, the forces are all repulsive, and the triangle is
equilateral) the force on the top charge must be vertically upward. The net force is the
sum of the vertical components exerted by the lower charges on the upper charge. These
two components are equal by symmetry, and the magnitude of the force on q1 is
Now let's do it the long way using the vector form of Coulomb's law. For F21 in the
figure, we have
We can resolve the unit vector r21-hat into its Cartesian components, with x horizontal
and y vertical:
giving
Similarly, the force F31 is given by
unit vector r31(with hat) is resolved as and the
16
giving
The net force F1 is the sum of these two. The x-components clearly cancel, while the y-
components add to give the simple result deduced by symmetry above.
1.5 CALCULATION OF THE RESULTANT ELECTROSTATIC
FORCE OF TWO OR MORE CHARGES
Methods used
1. Parallelogram law of forces method.
2. Resolution of forces method.
A: The use of the Parallelogram Law of Forces Method
Let us consider two forces F1 and F2 inclined to each other at an angle .
Fig. 1.5: Two forces inclined to each other at angle .
Step 1: Complete the parallelogram of forces for which F1 and F2 form part. When this
is done we have the following fig. below:
Fig.1.6: Parallelogram of forces.
Step 2: Use the Cosine Rule to calculate the resultant force FR of F1 and F2.
If is the angle enclosed by F1 and F2, then, the Cosine Rule gives the resultant
force FR as
F1 and F2 are separated by
F1
F2
17
FR2 = F1
2 + F2
2 - 2F1 F2 cos
Step 3: Use the Sine Rule to find the direction of the resultant force FR.
Fig. 1.7: Use of Sine Rule
Applying the Sine Rule gives
1
R1-R1
sin
Fsin
sin
F
sinF
F
Thus, the resultant force FR acts as an angle .
B: The use of the Rectangular Resolution of Forces Method
Procedure:
1. Choose two suitable perpendicular reference axes e.g. x and y axes.
2. Choose a consistent sign convention along the chosen axes.
3. Find the components of all forces under discussion along the chosen reference.
4. The resultant force is then given by
Direction of the Resultant Force is given by
Worked Examples:
Example 1.3
Consider the three particles A, B and C of charges +1C, +1C and -1C
Respectively arranged at the vertices of an equilateral triangle as shown below:
F F 2
y
2
x RF
x
y1-
F
F tan
18
Fig. 1.8: Three charges at vertices of equilateral triangle
Calculate the total force on the particle A due to particles B and C using
i. the rectangular resolution method
ii. the parallelogram law of forces method
Solution
i. Representation of Forces
ii.
ii. The use of the parallelogram law of forces method
60)sin F - (F 60sinF 60sinF F
60 )cosF (F- 60cosF - 60cosF - F
ACABACAB
ACABACAB
y
x
x
y1-
2
ACAB
2
ACABR
F
F tan
then,, be forceresultant theofdirection Let the
60sin F F 60 cos F F F
19
Let be the direction of the resultant force.
Example 1.4
Find the total force on B due to the charges at A and C
Representation of Forces
A: Rectangular Resolution Method
Let the x-axis of the co-ordinate axes be along BC. Then,
120 cosF 2F - F F F ABACAB2
AC2
R2
x
y1-
2
y
2
xR
BAyBCBA
F
F tan
F F F
60sin F - F ,F - cos60 F F
x
R
AB1-
R
AB
RAB
F
60sin Fsin
F
60sin F sin
F
60sin
F
sin Then,
20
B: Parallelogram Law of Forces Method
Example 1.5
Find the total force on the charge C due to the charges B and A
Representation of Forces
A: Rectangular Resolution Method
120 cosF 2F - F F F BCBABC2
BA2
R2
R
BA1-
RBA F
60sin Fsin
F
60sin
F
sin
x
y1-
2
y
2
xR
CAyCBCA
F
F tan
F F F
60sin F F ,F cos60 F F
x
21
B: Parallelogram Law of Forces Method
Direction of Resultant Force
Example 1.6
In the diagram below point charges of 4C and 3C are placed at A and B respectively.
AC = 9cm, BC = 6cm and ACB = 60O
If a free electron is placed at C, calculate
i. The magnitude and direction of the force it experiencesdue to the charges at A and B.
ii. The acceleration it acquires.
R
CA1-
RCA F
120sin Fsin
F
120sin
F
sin
22
Solution
Substituting values of FCB and FCB in eqn (1.4) gives
FR = 6.8 x 10-13
N
Let the direction of FR be , then,
2
o
CA
2
CACA
CBCACB2
CA2
R2
r4
|q|q
r
qkq F
1.4 60 cosF 2F F F F
N10 12.0 m) 10 (6.0
VmsA 10 9 C10 1.6 C10 3
r4
|q|q F
N10 7.11 m) 10 (9.0
VmsA 10 9 C10 1.6 C10 4
13-
22-
1-1-919-6-
2
o
CBCB
13-
22-
-1-19-19-6
R
CB1-
RCB F
120sin Fsin
F
120sin
F
sin
AC) joining line (with the 38.4
(0.622)sin N 10 x 16.7
0.866 x N 10 x 12.0sin 1-
13-
-131-
2-8
31 -
13-
e
Re
eeR
ms 10 x 1.84 kg 10 x 9.1
N 10 x 16.7
m
F a
)(aon Accelerati Electron´s x )(m Mass Electronic F
23
Try working out the same problem using the rectangular resolution method.
Example 1.7
(a) Compare the magnitudes of the gravitational force of attraction and the electric force
of attraction between the electron and the proton in a hydrogen atom.
(b) According to Newtonian mechanics, what is the acceleration of the electron?
Assume that the distance between the two particles is 5.3 x 10-11
m.
[G = 6.67 x 10-11
N.m2.kg
-2, me = 9.11 x 10
-31 kg, mp = 1.67 x 10
-27 kg]
Solution:
(a) The magnitude of the electric force is given by the Coulomb‟s law:
Fe = [4o]-1
(q1q/r2) = 9.0 x 10
9 x (1.6 x 10
-19)2/(5.3 x 10
-11)2
= 8.2 x 10-8
N.
The gravitational force between the electron and the proton is given by Newton‟s
gravitational law:
Fg = Gmemp/r2 = 6.67 x 10
-11 x 9.11 x 10
-31 x 1.67 x 10
-27/(5.3 x 10
-10)2
= 3.6 x 10-47
N.
The ratio of the electric force to the gravitational force is
Fe/Fg = 8.2 x 10-8
/3.6 x 10-47
= 2.28 x 1039
.
The gravitational force is very small compared with the electric force.
(b) Neglecting the gravitational force, the acceleration of the electron is given as,
using Newton‟s law,
a = F/me = 8.2 x 10-8
/9.11 x 10-31
= 9.0 x 1022
ms-2
.
The significance of Coulomb‟s law goes far beyond the description of the forces between
charged balls or rods. This law, when incorporated into the structure of quantum physics,
correctly describes
(a) the electric forces that bind the electrons of an atom to its nucleus,
(b) the forces that bind atoms together to form molecules, and
24
(c) the forces that bind atoms or molecules together to form solids or liquids.
The electrical force between charged particles is a vector quantity. In the case of only two
charged particles, the force each one experiences is directed along the line connecting the
two particles. If, however, there are many charged particles present, the net force
experienced by a given particle is the vector sum of all the Coulomb forces the other
particles exert on the given particle. The principle is illustrated in Fig. 1.2. The force
vectors Fab, Fac, and Fad represent the individual electrostatic forces experienced by
charge qa arising from the nearby charges qb, qc, and qd.
Fig. 1.9: Superposition of electrostatic forces by vector addition
The total force on charge qa is the resultant or vector sum of these individual forces. That
is,
Fa = Fab + Fac + Fad + .......... = [4o]-1
(qaqr/rai2), (1.5)
where the summation index i represents a, b, c.
Example 1.8
Charges 3.0 x 10-6
C, 4.0 x 10-6
C, and 6.0 x 10-6
C are placed along a line (in the diagram
below). Calculate the total force on the 6.0 x 10-6
C.
3C 4C 6C
1 m 2 m
Solution:
25
3C 4C 6C
F64
1m 2m F63
Let us first consider the forces exerted by the 3.0 x 10-6
C charge.
F63 = [4o]-1
(q3q6/r362) = 9.0 x 10
9 x 3 x 10
-6 x 6.0 x 10
-6/3
2
= 1.80 x 10-2
N (directed to the right).
Next, we consider the force exerted by the 4 x 10-6
C charge.
F64 = 9.0 x 109 x 4 x 10
-6 x 6.0 x 10
-6/2
2
= 5.39 x 10-2
N (directed to the right).
F63 and F64 are parallel. Hence the total force on the 6 x 10-6
C charge is
F6 = F63 + F64 = 7.19 x 10-2
N.
Experiment shows that the presence of a third charge does not influence the Coulomb
force between the other two charges.
The figure below shows three charges q1, q2, and q3. If q1 = -1.0 x 10-6
C, q2 = +3.0 x 10-
6 C, q3 = -2.0 x 10
-6 C and the distance between q1 and q2 is 0.15 m and that between q1
and q3 is 0.10 m and = 30o, determine the force which the other charges exert on q1.
q3 y
Ignoring the signs of the charges,
F12 = [4o]-1
(q1q2/r122)
= 9 x 109 x 1.0 x 10
-6 x 3 x 10
-6/(0.15)
2
= 1.2 N
and F13 = 9.0 x 109 x 1.0 x 10
-6 x 2 x 10
-6/(0.1)
2
q1
F13
F12 q2
x x
26
= 1.8 N
The directions of F12 and F13 are shown in the figure. The components of the resultant
force F1 acting on q1 are
F1x = F12x + F13x = F12 + F13 sin
= 1.2 + 1.8 sin 30
= 2.1 N
and F1y = F12y + F13y = 0 - F13 cos
= - 1.8 cos 30
. .
= - 1.6 N.
Hence F1 = ( F1x2 + F1y
2) = [(2.1)
2 + (-1.6)
2]
= 2.64 N.
Also, tan = F1y /F1x
= 1.6/2.1
= 37.3 o.
Example 1.9
Consider three 1-C charges at the vertices of an equilateral triangle, 1 m on a side. What
is the net force that two of the charges exert on the third?
Solution:
Fig. 1.10: Three 1-microcoulomb charges at the vertices of an equilateral triangle
This array of three charges has left-right symmetry relative to a vertical line through q1.
From this symmetry, the net force of q2 and q3 on q1 will be vertical, and in the upward
27
direction. Note that all the charges have the same sign and all forces are repulsive. The
net force on q1 is, therefore, the vertical component of the force exerted by q2 plus the
vertical component of the force exerted by q3.
For F12, we have
F12 = [4o]-1
(q1q2/r122) = 9.0 x 10
9 x (10
-6)2/1
2
= 9.0 x 10-3
N.
Then F13 = F12 = 9.0 x 10-3
N.
Since the two vertical components are equal (by symmetry), then
F = 2F12 cos 30 = 2 x 9 x 10-3
cos 30 = 1.56 x 10-2
N.
Problem Set 1
3. Figure PS1.2 shows two similar balls, each having
a mass m and carrying a charge q suspended
from the same point by means of silk threads
of equal length l. The inclination of the thread
to the vertical is .
Assuming that is small, show that the separation x
is given by
4. Two 5 g masses hung from a common point from threads 1 m each. If these
positive charges are to remain in equilibrium 1 cm apart, what is the magnitude of
the charge on each mass? (Assume g = 9.81 ms-2
, k = 9.0 x 109 Nm
2C
-2).
1. The figure (PS1.1) shows the
orientation of three charges +Q, -
Q and +q.
Deternime the resultant electric
force on the charge +q due to +Q
and –Q.
Fig. PS1.1
2. Two identical small balls having charges of
+1.0 mC and –0.33 mC are brought in
contact and then moved apart to a distance
of 20 cm.
Detemine the force of their interaction.
PS1.2
mg2
lq x
31
o
2
28
5. The charges and coordinates of two charged particles held fixed in the xy plane
are: q1 = +3.0 C, x1 = 3.5 cm, y1 = 0.50 cm, and q2 = - 4.0 C, x2 = - 2.0
cm, y2 = 1.5 cm.
(a). Find the magnitude and direction of the electrostatic force on q2. (b) Where
could you
locate a third charge q3 = + 4.0 C such that the net electrostatic charge on q2 is
zero?
6. Two identical conducting spheres, fixed in place, attract each other with an
electrostatic force of 0.108 N when separated by 50.0 cm. The spheres are then
connected by a thin conducting wire. When the wire is removed, the spheres repel
each other with an electrostatic force of 0.0360 N. What were the initial charges
on the spheres?
1. Two fixed particles, of charges q1 = +1.0 C and q2 = - 3.0 C, are 10 cm apart.
How far from each other should a third charge be located so that no net
electrostatic force acts on it?
2. Two free point charges +q and +4q are a distance L apart. A third charge is
placed so that the entire system is in equilibrium. (a). Find the location, magnitude
and sign of the third charge. (b) Show that the equilibrium of the system is
unstable.
3. A certain charge Q is divided into two parts q and Q – q, which are then separated
by a certain distance. What must q be in terms of Q to maximise the electrostatic
repulsion between the two charges?
4. A charge Q is fixed at each of two opposite corners of a square. A charge q is
placed at each of the other two corners. (a) If the net electrostatic force on each Q
is zero, what is Q in terms of q? (b) Is there any value of q that makes the net
electrostatic force on each of the four charges zero? Explain.
29
CHAPTER TWO
ELECTRIC FIELD AND GAUSS’ LAW
2.0 CHARGE DISTRIBUTIONS AND THE ELECTRIC FIELD
Note that the discovery of the electron came after Coulomb's work. The common notion
about 'charge' prior to the discovery of charge quantization was that it was some sort of
continuous fluid. For many purposes even today, we think of a charge density (e.g., in a
metal) without worrying too much that 'charge' is quantized. Of course, the reason we do
this is because the fundamental charge is very small, so that, on a macroscopic scale, it
looks quasi-continuous. It is thus useful to develop the tools to understand continuous
charge distributions. As was the case in gravity, this requires application of the
superposition principle.
Consider a 'line charge', that is, a collection of charges spaced very closely on the scale of
any measurement we make, that extends along the x-axis from x = -a to x = +a:
Fig. 2.1: Calculation of
force on a test charge
The magnitude of the force dF exerted by an infinitesimal charge element dq on a 'test
charge' Q located along the y-axis is given by
Since the force is a vector, we need to integrate the x- and y-components separately:
30
and the total force is found by integrating over the
entire charge distribution:
These integrals are general. To apply them to the above specific example, we need only
to insert the limits over which the charge is distributed. Let us say the rod has a total
charge q, and this is distributed evenly over it entire length from +a to -a. The one-
dimensional 'charge density' - charge per unit length - is
Now calculate the force on the test charge a distance y above the origin which we place at
the middle of the rod, as shown:
Fig. 2.2: A collection of charges on the x - axis
There is symmetry about the y-axis, so the net force will clearly not have an x-
component. In the figure, we show a differential charge dxdq
This allows us to change the integration variable to calculate Fy. Also, we have that
Finally, we use the definitions of the trigonometric functions:
31
Inserting these results into the integrals for the total force, we get
These can be solved using the simple substitution dydxandyx 2sectan
(recall that y is a constant for the purposes of this integral!). The x-integral can be seen to
be zero by symmetry; the integrand is odd [I(-x) = -I(x)] and the limits of the integral are
even, so the contribution from -a to 0 precisely cancels that from 0 to a. The Fy integral is
Using aq 2 , we finally get that
Now make some consistency checks. Note that Fy has the proper units. Also, when y is
much larger than a, the line of charge 'looks' like a point, and the result reduces to the
simple Coulomb law, as expected.
Finally, let's see what happens when a goes to infinity but we maintain a constant
charge density. We might not think that the force is finite under these circumstances, but
it clearly must be since this limit really corresponds to y being much less than a - we are
just very close to a finite line of charge. In this case, we find that
Note, that in this limit, the force law is inverse first power rather than inverse square. We
will do quite a few problems like this in the next few weeks.
32
2.1 THE ELECTRIC FIELD
From Coulomb‟s law, a charge q´ exerts a direct force on another charge q even though
these charges are separated by a distance and are not touching. It can be assumed that
each charge generates a permanent, static disturbance in the space surrounding it, and that
this disturbance exerts forces on the other charges. Thus, a charge q´ generates a field
which fills the surrounding space and exerts forces on any other charges that it touches.
This disturbance or field is called electric field.
To discover the field at a given position, a positive test charge qo is placed at that
position. The test charge experiences an electric force F. The electric field E, at that
point, is then defined as the electric force experienced per unit positive charge, that is
E = F/qo. (2.1)
In this equation, F is the sum of all Coulomb forces exerted on the charge qo by the other
charges or distributions of charges. The force F is directly proportional to qo. Hence the
division of F by qo yields a quantity which is independent of the magnitude and sign of
qo. The electric field E may be considered as the “environment” of qo. The force F = qoE
is described as a combination of properties of the particle of charge qo and its
environment (E). E is a vector quantity since F is a vector quantity. The force F on the
test charge qo indicates the existence of the electric field. A measurement of the force,
together with a knowledge of qo, yields the magnitude of the electric field. The test
charge qo must be sufficiently weak so that it does not disturb the distribution of electric
charges that are producing the field.
According to Coulomb‟s law, a point charge qo exerts a force ([1/4o](qoq1/r2)r) on a
charge q1. Hence, the electric field generated is
E = F/qo = [1/4o](q1/r2)r. (2.2)
Thus, the electric field surrounding a charge or distribution of charges is a function of
position. It must be noted that Eq. 2.2 is not a definition of an electric field. It merely
describes the electric field for the specific case of a point charge.
The SI units of the electric field are newtons per coulomb (NC-1
) or volts per meter
(Vm-1
).
For any distribution of point charges, the electric field can be calculated by superposition
of the individual fields of the point charges; in much the same way as is done for electric
forces. That is,
E =E1 + E2 + E3 + ............. (2.3)
33
2.1.1 Principle of Superposition
In the last lecture, we learned that the Coulomb force obeys the principle of
superposition. The definition of the electric field then requires that it obey superposition
as well, since it is proportional to the net force on the particle. Thus, if we have a
collection of charges q1, q2, q3, ..., then the electric field at position ro due to the i-th
charge is
and the total electric field is
To illustrate, consider the electric field on the axis produced by two 1 microcoulomb
charges located on the x-axis 2 m apart. Place the left charge at the origin and the right
charge at x = 2m. The field created by the left charge is
While that created by the right charge is
The fields from these two charges, at a point between them, are oppositely directed, and
this expression is written for this situation - the signs would be different for points not
between the charges.
The net field at a point between the charges is
34
Note that at the point x = 1 m, precisely between the two charges, the field is zero and the
net force on a test charge would vanish. The forces exerted by the two charges would be
equal and opposite, irrespective of the magnitudes of the charges, so long as they are the
same.
2.2 DIRECTION OF THE ELECTRIC FIELD, E
To test experimentally whether an electric field exists at a point we place small positive
test charge at that point. If this charge experiences an electric force, then, an electric field
exists at that point.
The direction of E at a point is the direction of the force acting on an infinitesimally small
positive test charge placed at that point.
Examples
2.3 CALCULATION OF THE ELECTRIC FIELD, E
Consider two charges qA and qB placed at the points
A and B and separated by a distance r.
The electric field strength at the point A is given by
EA = F/qA EA = qB qA / 4or2qA = qB / 4or
2
Likewise the electric field strength at the point B is given by
EB = F/qB = qA qB / 4or2qB = qA / 4or
2
1. Determine the directions of the fields due to all the charges under
electrostatic consideration.
2. Once these directions are marked use the magnitudes of all charges in the
computation of the electric field due to the charges under consideration.
3. The resultant electric field at the given point may then be calculated by
invoking either of the following methods:
(a). The Parallelogram law of vectors method or
(b). The Rectangular resolution method.
35
Worked Examples
Example 2.1
ABCD is a rectangle whose diagonals AC
and BD intersect at O. Point charges of
40 C and – 40 C are placed at A and
B respectively.
If AB = 8 cm and BC = 6 cm. Calculate
The magnitude and direction of the electric
Intensity at O.
Solution:
Representation of Electric Force Fields
Applying the Pythagoras theory gives
OA = OB = [42
+ 32] = 5 = r1 , cos = 3/5, sin = 4/5
Rectangular Resolution Method
Ex = EA sin -– EB sin = 0
Ey = EA cos + EB cos = [EA + EB] cos = 2 EA cos
= 2kq/r12 cos = [(2 9.0 10
9 Nm
2C
-2 40 10
–6)/ 0.05m] (3/5)
2
= 5.18 10 6
N C –1
The resultant field ER = [(Ex )2 +( Ey)
2] = 2 EA cos = 5.18 10
6 N C
–1
36
Let the direction of ER be , then,
tan = Ey / Ex = = 90o
Hence, the resultant field ER is 5.18 10 6
N C –1
and is directed perpendicular to
the x-Cartesian axis.
Example 2.2
Two point charges, one of +36C and the other of –36C are separated by a distance of
8.0 cm. Compute the electric field from a point 6.0 cm directly above the positive
charge.
Solution:
E+ = kq/r2 = k (36 10
–6 C
)/( 6 10
–2m)
2 = k(1.00 10
–2 Cm
–2). Likewise,
E – = kq/r2 = k (36 10
–6 C
)/( 10 10
–2m)
2 = k(0.36 10
–2 Cm
–2).
Ex = k(0.36 10–2
Cm–2
) (8 cm/10 cm) = k(0.29 10–2
Cm–2
).
Ey = k[1.00 10–2
) – (0.36 10–2
)(6 cm/10 cm)] = k(0.78 10–2
Cm–2
).
The magnitude of E is given by
E = (Ex2 + Ey
2) = k(0.83 10
–2 Cm
–2).
= (9.0 109
Nm2C
–2)( 0.83 10
–2 Cm
–2) = 7.5 10
7 N/C
Let be the direction of the resultant electric field E relative to the x-axis, then,
= tan – 1
(Ey/Ex) = tan – 1
(0.78/0.29) = 70o.
Problem set 2
1. Three point charges of +Q, –Q and +q are arranged in an equilateral triangle of side
a. The charges +Q and –Q lie on the x-axis with +Q and –Q placed at the left and
right edges of the base of the triangle respectively. Sketch i. a diagram of the arrangement.
ii. The field lines due to +Q and –Q.
Determine the magnitude and direction of the force that acts on +q due to the presence of +Q
and –Q.
37
2. Calculate the magnitude of the point charge that would create an electric field of 1.00
N/C at points 1.00 m away.
3. Two opposite charges of equal magnitude 2.0 10 –7
C are help 15 cm apart. What
are the magnitude and direction of the electric field E at a point midway between the
charges?
4. An atom of plutonium-239 has a nuclear radius of 6.64 fm and the atomic number Z
= 94. Assuming that the positive charge of the nucleus is distributed, what are the
magnitude and direction of the electric field at the surface of the nucleus due to the
positive charge?
5. A particle of charge –q is located at the origin of the x-axis. (a). Determine the
location of a Second particle of charge –4q1 so that the net electric field of the two
particles is zero at the point x = 2.0 mm. (b). If, instead, a particle of charge +4q1 is
placed at that location what is the direction of the net electric field at x = 2.0 mm?
6. Calculate the direction and magnitude of the electric field at the point P in the fig.
below.
7. In the fig. below two point charges q1 = +1.0 10 – 6
C and q2 = +3.0 10 – 6
C are
separated by a distance d = 10 cm. Plot their net electric field E (x) as a function of x
for both positive and negative values of x, taking E to be positive when the vector E
points to the right and negative when E points to the left.
Example 2.3
A total amount of charge Q is distributed uniformly along the circumference of a thin
glass ring of radius R. Determine the electric field on the axis of the ring.
Solution:
Consider the ring as made up of infinitesimal line elements ds, in the figure below. Each
of the elements can be treated as a point charge.
38
Fig. 2.3: A thin ring with charge distributed along its circumference
The field generated by the line element ds has both a horizontal and a vertical component.
For any given line element there is an equal line element on the opposite side of the
ring‟s centre which contributes an electric field of opposite horizontal component. All the
horizontal components, therefore, cancel pair wise. Consequently, the net electric field is
vertical.
Charge per unit length along the circumference = Q/2R.
Hence, the charge in the line element ds, is dQ = (Q/2R)ds.
The electric field contributed by the charge element dQ at a height z above the plane of
the ring has a magnitude
dE = [1/4or2] dQ
= [1/4o](Q/2R)ds[1/(z2 + R
2)].
This field has a vertical component
dEz = [1/4o](Q/2R)ds[cos /(z2 + R
2)].
Consequently, the net electric field is
Ez = [1/4o](Q/2R)[cos /(z2 + R
2)]ds
= [1/4o](Q/2R)[cos /(z2 + R
2)]ds
= [1/4o](Qcos )/(z2 + R
2) since ds = 2R
Therefore,
Ez = [1/4o]Qz/(z2 + R
2)3/2
. (2.4)
39
In vector notation,
E = [1/4o][Qz/(z2 + R
2)3/2
]z. (2.5)
Note that for z = 0, E = 0; and for z » R, E Qz/(4oz2), which is the charge for a point
charge.
Example 2.4
What is the electric field generated by a large flat sheet, such as a sheet of paper, carrying
a uniform charge density of Cm-2
?
Solution:
Fig. 2.4: Ring segment of a large flat sheet
Assume that the sheet is uniformly large. The sheet can be regarded as made up of a
collection of many concentric rings. The figure above shows one of such rings with a
radius R.
Area of this ring = 2RdR
Charge of element, dQ = (2RdR).
The ring produces a vertical electric field (see Eq. 2.4)
dE = [1/4o](2RzdR)/(z2 + R
2)3/2
. (2.6)
The net electric field is therefore
E = (2z/4o)
RdR/(z2 + R
2)3/2
,
where R = 0 is the radius of the smallest ring and R = is the radius of the largest. Using
u = R2, the integral becomes
40
RdR/(z2 + R
2)3/2
= 1/z.
Therefore,
E = (2z/4o)(1/z) = /2o. (2.7)
In vector notation,
E = (/2o)z. (2.8)
This electric field is proportional to the charge density, and is constant. This means that
the electric field is independent of the distance from the sheet. Although the result is
strictly valid only for the case of an infinitely large sheet, it is also a good approximation
for a sheet of finite size, provided we stay within a distance much smaller than the size of
the sheet and away from the vicinity of the edges.
2.4 ELECTRIC FIELD LINES
It is very useful to have a geometric way of representing electric fields. Electric fields can
be visualised by drawing so-called field lines, or lines of force. The electric field lines
have the following properties:
1. Electric field lines originate on positive charges and terminate on negative
charges (or go to infinity). They do not begin or end on a charge-free point in
finite space.
2. At each point in space, the field line through that point is tangential to the electric
field vector E at that point.
3. The density of the lines of force is a measure of the magnitude of the electric
field.
It should be noted that lines of force can never intersect since the force at any point can
have only one direction. Figure 2.1 below shows the field lines of some charges. Figure
2.1(a) shows the electric field lines of a positive charge while Fig 2.1(b) shows the lines
for a negative charge.
41
Fig. 2.5: (a) Electric field lines of a positive point charge. (b) Electric field lines
of a negative point charge.
Since the magnitude of the electric field is directly proportional to the amount of electric
charge, the number of field lines that is drawn emerging from a positive charge must be
proportional to the amount of the charge.
(a)
(b)
(c)
Fig. 2.6: (a) Field lines generated jointly by a positive and a negative charge of equal
magnitude, (b) field lines of unequal positive and negative charges, and (c) those of a
large, uniformly charged sheet with a positive charge density.
In all cases, the electric field lines start on positive charges and end on negative charges.
The positive charges can therefore be considered as sources of field lines and the
negative charges as sinks.
The above pictures of field lines enable us to develop some intuitive feeling for the
spatial dependence of the electric fields surrounding different arrangements of electric
charges. The electric field lines, in addition to providing us with a pictorial representation
of the electric field, are also useful in some computations of the electric fields of given
charge distributions. To help us illustrate this with an example, we have to define density
of field lines.
42
Density of field lines is the number of lines of force crossing a unit area perpendicular to
their direction, that is, the number of lines intercepted by a small area A erected
perpendicularly to the lines (Fig 2.7) divided by the magnitude of this area.
Fig. 2.7: A small area A intercepts some field lines.
[density of lines] = [number of intercepted lines]/A.
The area A must be small compared with the distance over which the electric field varies
appreciably, but large compared with the spacing between the field lines, so that it
intercepts a fairly large number of field lines.
Example 2.5
Using the concept of field lines and symmetry arguments, obtain the electric field of a
large flat sheet carrying a uniform charge density of Cm-2
.
Solution:
Suppose that the sheet is horizontal and its charge is positive, as in Fig. 2.6 (c). The field
lines must start on the charges on the sheet. Some of the field lines must go in the upward
direction and some in the downward direction. The pattern of field must respect the
charge distribution. The upper and lower surfaces of the sheet are physically equivalent.
Therefore, symmetry requires that the pattern of the field lines in the space above the
sheet be the mirror image of the pattern below the sheet. Now, let us consider one of the
field lines starting on the sheet and going in the, say, upward direction. For an infinite
sheet, the portions of the sheet to the right of the field line and the left of the field line are
physically equivalent. Hence, symmetry requires that the field line must be a straight
vertical line. Finally, since the charge is uniformly distributed, the field lines must also be
uniformly distributed. The pattern of the lines must, therefore, consist of uniformly
spaced vertical lines. Correspondingly, the electric field is of uniform magnitude
throughout all the space, and its direction is vertically upward in the space above the
sheet and vertically downward in the space below.
Now, consider an area A of the sheet.
43
The amount of charge within this area = A.
Hence, the number of field lines starting on this amount of charge = A/o.
Here, the convention that q/o lines emerge from each charge q is used.
The number of field lines going in the upward direction is half of this, or A/2o.
Density of field lines = number of lines/area = /2o.
The density of field lines equals the magnitude of the electric field. Consequently,
E = /2o,
in agreement with Eq. 2.7
2.5 The Electric Field of an Electric Dipole
Consider two point charges +q and -q separated by a distance d. Such a configuration of
equal and opposite charges is called an electric dipole. The product of the positive charge
q and the distance d between the charges q and -q is called the electric dipole moment p
(p = qd).
Figure 2.8 shows a positive and a negative charge of equal magnitude q placed a distance
d = 2a apart. Determine the electric field E due to these charges at a point P, a distance r
along the perpendicular bisector of the line joining the charges. Assume r » a.
Fig. 2.8: Electric field of a dipole at a point P.
From Fig. 2.4, the resultant field E is
E = E1 + E2.
The magnitude of the field E1 is
44
E1 = E2 = [1/4o]q/[a2 + r
2].
The vector sum of E1 and E2 points vertically downwards and the magnitude is
E = 2E1cos .
From Fig. 2.4, we find that
cos = a/(a2 + r
2)½.
Hence
E = (2/4)[q/(a2 + r
2)]a/(a
2 + r
2)½
= [1/4o]2aq/(a2 + r
2)3/2
.
If r » a,
E = (1/4o)2aq/r3 = (1/4o)dq/r
3, (2.9)
where 2aq = dq is the electric dipole moment.
Generally, the electric field produced by the dipole at points distant in comparison to the
size of the dipole may be determined by considering only the field in two dimensions.
The centre of the dipole is at the origin, while the point P is now some arbitrary
observation point at which the electric field is to be determined. The electric field in the y
and x directions are
Ey = -(p/4o)(1 - 3cos2),
Ex = (p/4o)(3cossin). (2.10)
One might ask whether there are any real systems in nature that act like electric dipoles
and therefore produce the type of fields calculated above. One such example is provided
by the hydrogen molecule. Although the separation of charge is more complicated than
that of the simple dipole discussed above, at large enough distances the electric fields are
very nearly approximated by those in Eq. 2.10. In addition to naturally occurring dipoles,
there are also electric dipoles which can be created when atomic systems are subjected to
the influence of external electric field. For example, the presence of an electric field can
bring about a separation of positive and negative charges in insulating material. This
manifests itself by establishing within the material many microscopic atomic dipoles
which in turn produce their own electric fields. This phenomenon is particularly
important in the study of devices known as capacitors and is, therefore, of interest in the
theory of electric circuits.
It is important to note that the electric field produced by an electric dipole has 1/r3
dependence while the electric field produced by a single point charge has 1/r2
45
dependence. The strength of the field due to a dipole falls more rapidly than that of a
point charge because of their near-cancellation of the fields produced by the individual
charges that constitute the dipole.
2.6 The Neutral Point (N)
This represents an imaginary point in an electric field where the resultant electric field
due to two like charges is zero and where an electric charge placed at that point
experiences no resultant force.
Fig. 2.9: The neutral point, N
Let us consider two positive point charges q1 and q2 separated by a distance r. Let a neutral point
N be at a distance y from q1. Further, let us assume that a small positive test charge is placed at
the neutral point N.
If we represent the force exerted by the charge q1 on the test charge q by FN1 and the
corresponding force exerted q2 by FN2, then, at the point N, q experiences no force. This implies
that FN1 = FN2 .
From equation (2.11a) we may deduce that
1. If q1 = q2 , then, y = r [1 + 1]- 1
= ½ r.
This means that the neutral point occurs at exactly midway between q1
and q2.
2. If we let q2 = 4q1, say. Then, y = r [(4)1/2
+ 1] –1
= 1/3 r.
This shows that the neutral point N is always closer to the smaller charge.
(2.11a) 1 q
qr y
q
q 1 -
y
ror
q
q
y
y -r
q
q
y
y) -(r
y) -(r
q
y
q
y) -(r
qkq
y
qkq
1
21
1
2
21
1
2
1
2
2
1
2
2
2
2
2
2
1
2
2
2
1
N
46
2.7 ELECTRIC DIPOLE IN AN ELECTRIC FIELD
An electric dipole moment can be regarded as a vector p. The magnitude of the dipole
moment is the product 2aq of the magnitude of either charge q and the distance 2a
between the charges. The direction of the dipole moment for such a dipole is from the
negative to the positive charge.
Let us consider the influence of an external electric field on a dipole. Figure 2.10 shows
an electric dipole in a uniform external electric field E. The dipole moment p of the
electric dipole makes an angle with the external field.
Fig. 2.10: An electric dipole in a uniform external electric field.
The force on the charge q due to the external field is F+ = qE. Similarly, the force on the
charge -q is F = -qE. The net force on the dipole is, therefore,
F = F+ + F = qE - qE = 0.
It is seen from above that a uniform electric field exerts no net force on an electric dipole.
But the forces F+ and F do not have the same line of action. Therefore, they produce a
net torque . The force F+ on the charge q attempts to align p in the direction of the
external field E. The torque due to F on -q also attempts to align p with E. The two
torques do not cancel.. The net torque about an axis though O is given by
= 2F(a sin) = 2aF sin
= 2aqE sin
= pE sin. (2.11b)
Thus an electric dipole placed in a uniform external electric field E experiences a torque
which tends to align it with the field. Eq. 2.11b can be written in the vector form
= p x E. (2.12)
To change the orientation of an electric dipole in an external field, work must be done by
the external agent. This work is stored as potential energy in the system made of the
47
dipole and the arrangement used to set up the external field. Let the initial value of be
o. Work required to turn the dipole axis to an angle is
W = dW = d = Up,
where is the torque exerted by the agent that does the work and Up is the potential
energy stored in the system. Therefore,
Up = pE sin d = pE sin d
= pE-cos .
The initial angle o is chosen to have any convenient value. In this case, o is chosen to
be 90o, since we are only interested in charges in potential energy. Thus
U = -pE cos, (2.13)
or in vector form
U = -p.E. (2.14)
2.8 ELECTRIC FLUX
Consider a rectangular surface of area A which is immersed in a uniform electric field E
(Fig. 2.11)
Fig. 2.11: Flat rectangular surface immersed in a uniform electric field.
The perpendicular to the surface makes an angle with the field lines.
The electric field makes an angle with the surface. The electric-field vector has a
component tangential to the surface and a component normal to the surface. The electric
flux through the surface is defined as the product of the area A and the magnitude of
the normal component of the electric field. That is,
= EnA. (2.15)
The normal component En can be written as Ecos. Hence
48
= EA cos. (2.16)
The quantity Acos is the projection of the area A onto a plane perpendicular to the
electric field; that is, Acos is that part of the area A that effectively faces the electric
field. According to Section 2.1, the magnitude of the electric field is numerically equal to
the number of field lines intercepted by a unit area facing the electric field. Hence,
EAcos must be numerically equal to the number of field lines intercepted by A. The
electric flux through an area is, therefore, equal to the number of lines intercepted by
the area. Flux and the number of intercepted lines are equal if and only if the electric field
and the number of lines per unit area are equal; the latter is true if we adopt the
convention that q/o lines emerge from each charge q.
More generally, let us consider an arbitrary surface in a non-uniform electric field (Fig.
2.12). The electric flux can now be defined by subdividing the surface onto infinitesimal
plane areas dS . Within each area, the field is assumed to be nearly constant. The flux
through one of such infinitesimal areas is
d = E cos dS.
The total flux through the surface is obtained by summing all these infinitesimal
contributions,
= Ecos dS. (2.17)
Fig. 2.12: An arbitrary surface immersed in a non-uniform electric field
As a further illustration of the relation between a field E and the flux , consider the field
produced by a positive point charge at the origin. The electric field lines run radially
outward and are assumed to be continuous (Fig. 2.13). In the absence of other charges,
the electric field lines continue into infinity. Then the number of field lines passing
through a given spherical shell is independent of the radius of the shell. From our picture
of electric flux as proportional to the number of field lines passing through a given
surface, the flux is independent of the radius of the shell.
49
But the area of the shell of radius r is 4r2. So the number of field lines per unit area, and
therefore E, falls off as the inverse square of the distance, in agreement with Eq. 2.2.
Fig. 2.13: Field lines directed radially outward from a positive point charge intersecting
concentric spherical shells
2.9 GAUSS’S LAW
Let S be a closed surface surrounding a charge q, and let q be a distance r from a small
area dS on the surface S at A (see Fig. 2.14).
Fig. 2.14: Illustrating Gauss‟ theorem
The electric field strength E at A has the value
E = q/4or2.
50
The number of lines of force passing through an element of area dS is
E.dS = Ecos dS = qcos dS/4or2,
where the outward normal to the surface element makes an angle with E. Now the solid
angle subtended by dS at O is
d = cos dS/r2.
This makes
Ecos dS = qd/4o.
The total number of lines of force passing through the whole surface, therefore, becomes
Ecos dS = (q/4o)d
A closed surface subtends a total angle of 4 at any point within the volume enclosed by
the surface. Hence
Ecos dS = q/o. (2.18a)
If there are a number of charges q1, q2, q3,...., qn inside S, the resultant electric field
intensity E at any point is the vector sum of the intensities due to each separate charge. In
this way,
Ecos dS = q/o. (2.18b)
Eq. 2.18a is known as the Gauss‟s law. It is seen that the integral of the normal
component of E over the surface is equal to the total charge enclosed, divided by o,
irrespective of the way in which the charge is distributed.
Gauss’s law and Coulomb’s Law
Coulomb‟s law can be deduced from Gauss‟s law. For a point charge q at the origin, and
a spherical surface, we have a spherical symmetry (Fig. 2.15). Therefore, the electric field
E must be radial in direction and its magnitude is the same everywhere on the surface. At
the position of an element of area dA on the surface, E is parallel to dA and E.dA = EdA,
and hence
E.dA = EdA.
51
Fig. 2.15: A spherical Gaussian surface of radius r surrounding a point charge q.
The magnitude of the field is constant over the surface. Therefore,
E.dA = EdA = E.4r2.
From Gauss‟s law
E.dA = q/o.
Hence
E.4r2 = q/o
and
E = (1/4o)q/r2.
From the definition of E, the force on the test charge qo is
F = (1/4o)qqo/r2,
which is Coulomb‟s law.
2.10 SOME APPLICATIONS OF GAUSS’S LAW
Gauss‟s law can be used to calculate the electric field E if the symmetry of the charge
distribution is high. The following guidelines can be considered when applying Gauss‟s
law to calculate the electric field in a given charge distribution. There are two major
steps.
1. Evaluate the electric flux, = dAE
(a) Determine the pattern of the field lines from the symmetry of the charge distribution.
(b) Construct a Gaussian surface to take advantage of the symmetry. This means that the
surface could be arranged so that E is either parallel to dA or perpendicular to dA
everywhere over the surface. Thus, for any given surface element dA, either
E.dA = 0 or E.dA = EdA.
52
(c) Further exploiting the symmetry of the charge distribution, arrange the Gaussian
surface so that E is constant over that portion where A and dA are parallel. Then
dAE = dAE = E dA .
(d) Finally, the electric flux is given by E = EA where A is the area over which E and
dA are parallel. Note that the vector dA is perpendicular to the surface!
2. Evaluate q, the charge enclosed by your choice of Gaussian surface. By Gauss‟s law,
the flux calculated in step 1 is equal to q/o
Example 2.6
Electric Field at a Charged Conducting Plane.
Consider a conducting plane - a metal plate that is in static equilibrium with positive
electric charge distributed uniformly over its surface. Assume that the conducting plane is
infinite in extent, and thus avoid the edge effects. Since the plane is conducting, the
tangential component of the electric field at the surface must vanish. Otherwise, there
would be currents on the surface, contrary to the assumption of static equilibrium. Then if
E (E 0) has no tangential component, it must be normal.
Because the charge distribution is uniform on a uniform conducting plane, it can be
argued that the electric field must be normal to the surface by symmetry. For the
Gaussian surface of integration, a closed circular cylinder of cross-sectional area A and
height h, is taken (Fig. 2.16).
Fig. 2.16: An infinite sheet of charge pierced by a cylindrical Gaussian surface.
The bottom surface of the cylinder is inside the conductor, where E = 0. Since E is
tangential to the curved surfaces of the cylinder, there is no contribution from the curved
sides of the cylinder to the surface integral. The only remaining contribution is from the
top of the cylinder. At the top surface of the cylinder, E is parallel to dA and constant in
magnitude over the surface. The electric flux is given by
E = dAE = E dA = Er2.
53
By Gauss‟s law, the electric flux E = q/o.
Hence,
Er2 = q/o.
Now, if the surface charge is , then q = r2.
Hence
E = /o. (2.19)
The field directly above a charged conducting plane is normal to the plane and the
strength of the electric field is proportional to the surface density.
Example 2.7: A sheet of charge.
Figure 2.17 shows a portion of a thin, non-conducting, infinite sheet of charge, the charge
density being constant. Determine the electric field E at a distance r in front of the
plane.
As in example 2.6, a convenient Gaussian surface is a closed circular cylinder of cross-
sectional area A and height 2r, arranged to pierce the plane as shown.
Fig. 2.17: An infinite sheet of charge.
From symmetry, the electric field E points at right angle to the end caps and away from
the plane. E does not pierce the cylindrical surface. Therefore, there is no contribution to
the flux from this surface. From Gauss‟s law,
dAE = E dA = E(A + A) = 2AE.
Hence
2AE = q/o
and
54
E = q/2Ao = /2o. (2.20)
Comparing Eqs. 2.19 with 2.20 shows that the electric field is twice as great near a
conductor carrying a charge whose surface density is as that near a nonconducting
sheet with the same surface charge density.
Example 2.8:
A Uniform Charged Spherical Shell - Exterior Field
Let us determine the electric field outside a uniformly charged spherical shell of total
charge Q, as shown in Fig. 2.18.
Fig. 2.18: The electric field of a charged conducting sphere.
The electric field is spread out uniformly and there is complete spherical symmetry. By
symmetry, therefore, the field lines must go radially outward. The magnitude of the
electric field E depends only on the distance r from the centre of the sphere.
To apply Gauss‟s law, we choose for the Gaussian surface a spherical shell that is outside
of and concentric with the charged shell. With E and dA parallel over the surface, we
have
dAE = E dA .
The magnitude E is a function only of r and is, therefore, constant over the spherical
surface. The flux integral may then be written
dAE = E.4r2 = Q/o.
So
E = (1/4o)Q/r2.
With unit vector r to indicate direction, we have
E = (1/4o)(Q/r2)r. (2.21)
55
Thus, for points outside a spherically symmetric distribution of charge, the elctric field
has the value that it would have if the charge were concentrated at its centre. This result is
shown in Fig. 2.19 (r > R).
Fig. 2.19: The radial electric field produced by a uniformly charged spherical shell of
radius R.
Just as a uniformly charged spherical shell acts as if all the charge were concentrated at
the centre, the earth attracts any object as if all the mass of the earth were concentrated
at the centre of the earth.
Example 2.9
A Uniformly Charged Spherical Shell - Interior Field.
Now, let us find the electric field inside the charged spherical shell.
Again, by symmetry, if any field exists at all, it must be radial. For an imaginary sphere
that is inside and concentric with the charged sphere, we have
dAE = 0
by Gauss‟s law. No charge is enclosed. Again, writing E = rE (because symmetry
demands this) and integrating over the imaginary sphere (r is constant), we get
E.4r2 = 0
or
E = 0.
Hence E = 0.
It can be concluded that there is no electric field inside a uniformly charged spherical
shell.
Example 2.10
A sphere of radius R has a total charge q which is uniformly distributed over its volume
(Fig. 2.20).
(a) What is the electric field at points inside the sphere?
(b) What is the electric field at points outside the sphere?
56
Fig. 2.20: A sphere with a uniform distribution of charge.
(a) The charge density within the sphere is
= charge/volume = 4/(4R3/3)
= 3q/4R3.
Since the charge distribution is spherically symmetric, the electric field must be radial
and constant in magnitude over any concentric spherical surface of given radius. To
determine the magnitude of the electric field inside the charge distribution, we take a
spherical Gaussian surface of radius r, where r R. On this surface,
dAE = E dA = E.4r2 .
The charge inside this Gaussian surface is
Q = charge x volume = x 4r3/3
= qr3/R
3.
Then Gauss‟s law gives
E.4r2 = qr
3/oR
3,
that is,
E = (1/4o)qr/R3, for r R. (2.22)
(b) The solution is similar to that in Example 6. That is
E = (1/4o)q/r2, for r R.
Example 2.11
Line of Charge
57
Figure 2.21 shows a section of an infinite rod of charge, the linear charge density being
constant for all points on the line. Determine an expression for the electric field at a
distance r from the line.
Fig. 2.21: An infinite rod of charge, showing a cylindrical Gaussian surface.
From symmetry, E due to a uniform linear charge can only be radially directed. As a
Gaussian surface, we choose a circular cylinder over the cylindrical surface. The flux of
the field through this surface is
E = dAE = E.(2rh).
There is no flux through the circular caps because E here lies in the surface at every
point.
The charge enclosed by the Gaussian surface = h.
Now
dAE = q/o.
Hence
E.2rh = h/o
and
E = /2or. (2.23)
The direction of E is radially outward for a line of positive charge.
58
CHAPTER THREE
THE ELECTRIC POTENTIAL
3.0 ELECTRIC POTENTIAL
Electric potential is analogous to gravitational potential, but here one thinks of electric
field. In the electric field round a positive charge, for example, another positive charge
moves from points near the charge to points further away. Points round the charge are
said to have an “electric potential”.
Let us follow the motion of a positive charge q which experiences a total Coulomb force
F. The force F depends on the location of q and is assumed to be due to the presence of
one, several, or perhaps very many charges whose positions are fixed. F could also be
due to a continuous distribution of charge. Assume that q is initially at a location i and
moves to a final position f. Now, work done by the force F is
U = Upf - Upi = - F.dr, (3.1)
where dr is the displacement. The total Coulomb force F may be expressed as the product
of the charge q and the total electric field E produced by all the other charges. Thus, we
can also write
U = -qE.dr. (3.2)
Equation 3.2 can be rewritten as
U/q = -E.dr. (3.3)
The electrostatic potential, V, is then defined as the potential energy per unit positive
charge, while the potential difference, V, is defined as the change in potential energy
per unit positive charge. Thus, we define the potential difference between the points f and
i as the work done in moving a unit positive charge from i to f. That is,
V = Vf - Vi = (Upf - Upi)/q = -E.dr. (3.4)
Normally, the initial point i is chosen to be at a large distance from all charges, and the
potential Vi at this infinite distance is taken as zero. This allows for the definition of the
electric potential at a point. In making Vi = 0, Eq. 3.4 becomes, dropping the subscripts,
V = U/q. (3.5)
Thus, the potential at a point in an electric field is the work done in moving a unit
positive charge from infinity to that point.
59
Potential energy is expressed in joules and charge is measured in coulombs. Hence the
ratio U/q has unit joules/coulomb (J/C). These units of electrostatic potential and
potential difference are more commonly referred to as volts. The volt is defined as the
work done in taking one coulomb of positive charge from one point to another.
From the definition of the volt, if a charge of Q coulomb is moved through a potential
difference of V volt, then the work done W, is given by
W = QV. (3.6)
The electric potential as defined in Eq. 3.5 is a scalar since U and q are scalars. Any
surface, planar, or curved, over which the potential is constant is called an equipotential
surface.
3.1 POTENTIAL AND THE ELECTRIC FIELD
Assume that a positive charge q is moved, by an external agent which exerts a
force F on it, from a point A to a point B along the straight line connecting them in a
uniform electric field E, as shown in Fig. 3.1.
Fig. 3.1: A test charge q is moved from A to B in a uniform electric field E by an
external agent that exerts a force F on it.
The electric force on the charge q is qE. This force points in the direction of the field. To
move the charge from A to B, an external force F of the same magnitude but opposite in
direction must be applied to counteract this electric force. Work which should be done to
move the charge from A to B is
WAB = Fd = qEd
Hence the potential difference between A and B is
VB - VA = WAB/q = Ed. (3.7)
60
Equation 3.7 shows the connection between potential difference and electric field for a
simple case.
What would be the potential if the field is not uniform and in which the charge is moved
along a path that is not straight, as in Fig. 3.2
Fig. 3.2: A charge q is moved from A to B in a nonuniform electric field by an external
agent that exerts a force F on it.
The electric force exerted on the test charge by the electric field is qE. To prevent this
charge from accelerating, an external force F exactly equal to -qE for all positions of the
test charge is applied.
If the charge moves through a displacement dl along the path A to B, the element of the
work done is F.dl. To determine the work WAB done on the charge in moving it from A to
B, the work contributions for all the infinitesimal segments into which the path is divided
are all added up. This gives
WAB = F.dl = -q E.dl. (3.8)
Substituting this expression into Eq. 3.7 gives
VB - VA = WAB/q = - E.dl. (3.9)
If E is known at various points in the field, the two equations above (3.8 and 3.9) could
be used to calculate the potential difference between any two points, or the potential at
any point. The integral in Eq. 3.9 can be replaced by the product
V - El. (3.10)
Equation 3.10 becomes exact for the case for a uniform electric field. From Eq. 3.10, we
have
E - V/l.
In the limit as l0, this approximation becomes
61
E = -dV/dl. (3.11)
The quantity dV/dl is called the potential gradient and is the rate at which the potential
rises with distance. Equation 3.11 shows that the strength of the electric field is equal to
the negative of the potential gradient.
3.2 POTENTIAL DUE TO A POINT CHARGE
Consider two points X and Y near an isolated point charge (Fig. 3.3).
Fig. 3.3: A charge q1 is moved from X to Y in the field set up by a point charge q.
Assume that the charge q1 is moved without acceleration along a radial line from X to Y.
The potential difference between points X and Y can be determined. Since E and dl point
in opposite directions,
E.dl = Ecos 180o dl = -Edl.
q1 is moved in the direction of decreasing r as it is moved towards q. Thus,
dl = -dr.
Hence E.dl = Edr.
Substituting this in Eq. 3.9 gives
VY - VX = -E.dl = -Edr.
The electric field of a point charge is
E = q/4or2
Therefore,
VY - VX = -(q/4o) dr/r2
= (q/4o)(1/rY - 1/rX). (3.12)
62
When we set VX (rX ) = 0 and rY = r, we get the simplest formula
V(r) = q/4or (3.13)
for the potential at a distance r from a point charge q.
Exampe 3.1
Two positive point charges, of 12 C and 8 C respectively, are 0.1m apart. Determine
the work done in bringing them 0.04 m closer.
(Assume 1/4o = 9 x 109 mF
-1.)
Solution
Suppose the 12 C charge is fixed in position. Potential difference between points 0.06
and 0.1 m from it is given by
V = q/4o(1/r6 -1/r10)
Therefore,
V = 12 x 10-6
x 9 x 109 (1/0.06 - 1/0.1)
= 7.2 x 105 V.
Work done in moving the 8-C charge from 0.1 to 0.06 m away from the 12-C charge is
given by
W = QV = 8 x 10-6
x 7.2 x 105
= 5.8 J.
Example 3.2
The electron in a hydrogen atom is at a distance of 5.3 x 10-11
m from the nucleus. What
is the electrostatic potential generated by the nucleus at this distance?
Solution:
The nucleus is (approximately) a point charge with q = +e = 1.6 x 10-19
C. The
electrostatic potential generated by the nucleus is
63
V = q/4or
= 9 x 109 x 1.6 x 10
-19/(5.3 x 10
-11)
= 27 V.
3.3 A GROUP OF POINT CHARGES
There are two general ways to calculate electrostatic potentials.
(i) If the field E is known, the potential difference may be calculated as the negative
of the work done per unit charge by the field in moving a test charge from point X
to Y. This was used in determining the potential of a point charge in Section 3.3.
(ii) Alternatively, the potential due to a given charge distribution may be obtained as
the sum of the potentials due to the individual charges (or the integral over an
assumed continuous charge). In this case, the superposition principle is used.
If E is the resultant of two or more discrete or distributed charges, then, using the
superposition principle, we can write
V = -(E1 + E2 + E3 + ..........). ds
= -E1.ds - E2.ds - E3.ds - .........
or V = V1 + V2 + V3 +.....
The potential V is the algebraic sum of the individual potentials due to the discrete or
distributed charges. Thus, using the superposition principle, we can write the potential
due to a number of discrete charges as the algebraic sum of the potentials due to the
individual charges. Therefore,
V = (1/4o)(q1/r1 + q2/r2 + q3/r3 + ......)
= (1/4o)qi/ri. (3.14)
Here qi is the ith charge and ri is its distance from the point of measurement.
If the charge distribution is a continuous one, rather than a collection of points,
the sum in Eq. 3.14 is replaced by an integral, or
V = dV = (1/4o)dq/r, (3.15)
where dq is a differential element of the charge distribution, r is its distance from the
point at which V is to be calculated, and dV is the potential it establishes at that point.
64
Example 3.3
Three point charges having values of 2 x 10-10
C, 4 x 10-10
C, and -5 x 10-10
C are placed
at the vertices of an equilateral triangle whose sides are each 10 cm. Determine the
electrostatic potential at the centre of the triangle.
Solution:
The charges are equidistant from the centre of the triangle. Using elementary
trigonometry, we find that the distance r is
r = 10/3 cm.
Hence,
VP = V1 + V2 + V3
= (1/4or)q1 + q2 + q3]
= [9 x 109/(0.1/3)][4 x 10
-10 + 2 x 10
-10 - 5 x 10
-10]
= 15.5 V.
Example 3.4
Find the electric field for the points on the axis of a uniformly charged disk whose
surface charge density is .
Fig. 3.4: A charge element dq.
65
Consider a charge element dq consisting of a flat circular strip of radius y and width dy.
The charge element dq is given by
dq = (2y)dy.
Circumference of the strip = 2y
and area of the strip = 2ydy.
Distance of strip from P = r = (r2 + y
2)½.
Now, dV = (1/4o)dq/r = (1/4o)[(2y)dy/(r2 +y
2)½].
Hence the potential V for all points on the axis of the disk is
V = dV = (/2o) (r2 + y
2)
-½ ydy
= (/2o)(a2 + r
2)½ - r.
In the special case of r » a,
(a2 + r
2)½ = r1 + a
2/r
2
½ = r 1 + a
2/2r
2 + .....
r + a2/2r.
Hence,
V = (/2o)r + a2/2r - r) = a
2/4or = q/4or,
where q = a2 is the total charge on the disk.
Example 3.5
A rod of length l has a charge Q uniformly distributed along its length. Determine the
electrostatic potential at a distance x from one end of the rod.
Solution:
Fig. 3.5: A Rod of length, L with uniform charge Q.
66
Consider a small segment dx of the rod. The charge per unit length = Q/l.
Hence, charge in the small segment = (Q/dl)dx.
Distance of the segment from the position A = x - x. (x is a negative quantity).
Potential due to the element at the point A is
dV = (1/4o) [(Q/l)dx /(x - x)].
Hence potential at A is
V(A) = V(x) = (1/4o)[(Q/l)dx /(x - x)]
= (1/4o)Q/l [-ln(x - x)]
= (1/4o)(Q/l) ln[(x + l)/x].
When performing a summation or an integral of the contributions that the charge
elements make to the potential, one should not worry about the direction of the
contributions. The potential is a scalar quantity, with a magnitude but no direction.
However, it should be kept in mind that positive charges make a positive contribution,
and negative charges a negative contribution.
3.4 POTENTIAL DUE TO A DIPOLE
In this section, the electric potential at any point of space due to a dipole is calculated. It
is assumed that the point is not too close to the dipole.
Fig. 3.6: A point P in the field of an electric dipole.
67
A point P is specified by giving the quantities r and . From symmetry, it is clear that the
potential will not change as the point P rotates about the z-axis, r and being fixed. The
net potential generated by the dipole is the sum of the individual potentials,
V = Vn = V1 + V2 = (q/4or1) - (q/4or2) (3.16)
or
V = (q/4o)[(r2 - r1)/r1r2], (3.17)
where r1 and r2 are shown in the figure. Assume that r1 and r2 are much larger than 2a,
that is, the field point P is at a larger distance from the dipole. Under these conditions, r1
and r2 are approximately parallel and approximately equal;
r1 r2 r.
The difference between r1 and r2 is a small quantity,
r2 - r1 2a cos.
With this, Eq. 3.17 yields the following approximation for V:
V (q/4o)[2acos/r2]. (3.18)
The product of q and 2a is the dipole moment, p; hence the approximation for V has the
form
V =(1/4o)[pcos/r2]. (3.19)
It is seen from Eq 3.19 that V vanishes everywhere in the plane = 900. This reflects the
fact that no work is done in bringing a test charge from infinity along the perpendicular
bisector of the dipole. For a given radius, V has its greatest positive value for = 0o and
its greatest negative value for =180o.
68
CHAPTER FOUR
CAPACITORS AND DIELECTRICS
4.0 CAPACITANCE
Consider two conductors A1 and A2 which carry charges of equal magnitude but
opposite sign. The charges -q and +q contained on the respective conductors are
distributed uniformly on the surfaces. The conductor A1 is at a potential V1, while the
conductor A2 is at a potential V2. Any such arrangement of conductors carrying equal and
opposite charges is called a capacitor. A capacitor is any arrangement of conductors that
is used to store electrical charge.
The potential difference between the two conductors is
V = V2 - V1.
The magnitude of the potential difference, V, is proportional to the magnitude of the
charge, q. That is,
V q,
therefore,
V = kq
or k = V/q
and 1/k = C = q/V. (4.1)
C is the capacitance of the two conductors and is defined as the ratio of the charge on
either conductor to the potential difference between them.
The unit of capacitance is the farad (F). 1 farad = 1 F = 1 coulomb/volt. In
practice smaller units, microfarad, picofarad, are used. 1 F equals 10-6
F and 1 pF = 10-
12 F.
The capacitance is large if the conductor is capable of storing a large amount of
charge at low potential. The capacitance of a conducting sphere is
C = q/V = q/[q/4or] = 4or. (4.2)
Thus, the capacitance of the sphere increases with its radius. Eq. 4.2 is derived in detail in
69
The capacitance of any given system of conductors depends on two important factors:
1. The geometric arrangement of the conductors. This includes the size, shape, and
spacing of the conductors as well as their geometric relation to one another.
2. The properties of the medium in which the conductors are placed (air, vacuum,
dielectric, etc).
First, we consider how the capacitance depends on geometry. The simplest example of a
capacitor is that of the parallel-plate capacitor (Fig. 4.1)
Fig. 4.1: Parallel-plate capacitor.
Two conducting plates each having an area A are separated by a distance d. The distance
of separation is small compared to the linear dimensions of the plates. For such a
configuration of conducting plates, the electric field between the plates is quite uniform
except near the edges, where some fringing might occur.
The magnitude of the potential difference between the plates is given by
V = Ed. (4.3)
The electric field between the plates can be expressed as (refer to Eq. 2.19)
E = /o, (4.4)
where is the surface charge density on the plates. Hence
V = d/o = (qd/A)/o,
where = q/A is used. Therefore,
V = qd/oA.
70
But C = q/V.
Hence, C = q/Ed = (q/qd)oA,
or C = oA/d (4.5)
for the capacitance of a parallel-plate capacitor in vacuum.
Example 4.1
The electrical breakdown of air takes place whenever the electric field exceeds 3.0 x 106
Vm-1
. What is the maximum charge a parallel-plate capacitor of capacitance 2 .0 x 10-9
F
can hold if the plates have an area of 1.0 x 10-2
m2?
Solution:
The maximum charge qmax is proportional to the maximum voltage Vmax that can be
applied between the plates:
qmax = CVmax.
But Emax = Vmax/d,
where d is the separation. Since
C = oA/d,
we have
qmax = CEmaxd
= oAEmax
= (8.85 x 10-12
)(1.0 x 10-2
)(3 x 106)
= 2.66 x 10-7
C.
71
Example 4.2
Two long, conducting hollow cylinders of radii ra and rb are coaxial, as shown in the
figure below. Obtain an expression for the capacitance per unit length if the region
between the cylinders contains air.
Fig. 4.2 Two hollow coaxial conducting cylinders
Solution:
Assume that the inner cylinder bears a charge per unit length +. This must reside on the
outer wall of the cylinder shell in order for the field inside the conductor to be zero. The
outer cylinder contains a charge per unit length of -; this charge is located on the inner
boundary. Using the Gaussian surface of radius r,
E.dS = E(2rl) = q/o = l/o.
Hence, E = /2or. (4.6)
The magnitude of the potential difference between the cylinders is
b
a
b
a
r
r
r
r 0 r
dr
2π
λ Edr ΔV
Therefore, V = (/2o) ln(r b/ra). (4.7)
A length l of the cylinder contains a charge q = l. Therefore, this length has a
capacitance
C = l/V = l/[(/2o) ln(rb/ra)].
72
The capacitance per unit length is, therefore,
a
b
o'
r
rn
2
C C
ll
(4.8)
Example 4.3
Two conducting concentric spheres have radii ra and rb, as shown in the figure below.
Determine the capacitance of this device.
Fig. 4.3: Two concentric conducting spheres of radii ra and rb.
Solution:
The electric field can be obtained by using Gauss‟s law in conjunction with the Gaussian
surface shown. In this way, for the region between the spheres, we can write,
q/o =E.dS = E.4r2.
Therefore, E = q/4or2. (4.9)
The potential difference between the spheres is, therefore, found from
V = b
a
r
r Edr =
b
a
r
r (q/4or
2)dr
= (q/4o) b
a
r
r dr/r
2 = (q/4o)[1/rb-1/ra].
Therefore,
C = q/V = q /[(q/4o)(rb-ra)/rarb] = 4orarb /(rb-ra). (4.10)
It is often convenient to describe the capacitance of a single conductor, in which case it is
assumed that the second conductor of opposite charge is located at infinity. The second
73
conductor is then taken as a sphere of very large radius rb (rb ). In this way, we may
obtain, from Eq. 4.10 (using rb » ra)
C = 4ora,
as the capacitance of a sphere of radius ra.
4.1 CAPACITORS IN SERIES AND IN PARALLEL
Capacitors in electrical circuits may be connected to one another and to other current
elements in a variety of ways. In a circuit diagram, the symbol is used to represent a
capacitor with a fixed capacitance, while the symbol is used to represent a capacitor
whose capacitance can be varied. In a variable capacitor, provision is made for adjusting
the spacing between the parallel plates of the capacitor, by turning a dial, for instance.
The variable air capacitor is used in radio receivers for tuning to the different
wavelengths of commercial broadcasting stations. There are other ways of building
capacitors whose capacitance can be adjusted.
Let us now consider the effect of connecting two different, initially uncharged, capacitors
in the manner shown in Fig. 4.4, having the right-hand plate of one connected to the left-
hand plate of the next, and so on.
+q -q +q -q
AB C (a)
1 2 3 4
V
+q -q
A C (b)
1 4
Figure 4.4: (a) Two capacitors in series, (b) a single equivalent capacitance.
If a voltage is applied between the points A and C, charges appear on the four conducting
surfaces. A charge +q appears on the plate 1 and the opposite charge -q appears on plate
4. Also with a charge +q on plate 1, a compensating charge -q appears on plate 2. This
leaves plate 3 with a charge +q, since the sum of the charges on the plates 2 and 3 must
be zero. When capacitors are connected as in Fig. 4.4 (a), they are said to be in series.
This combination of capacitors may be regarded as a single capacitor having some
74
equivalent capacitance, C, as shown in Fig. 4.2(b). Then C is related to VAC by means
of
VAC = q/C.
But it is seen that
VAC = VAB + VBC
or VAC = q/C = q/C1 + q/C2.
Therefore, the equivalent capacitance, C, must be given by
1/C = 1/C1 + 1/C2. (4.11)
Generally, for N capacitors connected in series, the equivalent capacitance is
1/C = 1/C1 + 1/C2 + 1/C3 + .............+ 1/CN. (4.12)
Thus, to find the resultant capacitance of capacitors in series, the reciprocals of their
individual capacitances must be added. The resultant is less than the smallest individual.
A group of capacitors may also be connected in parallel, as shown in Fig. 4.5. In this
case, all the left-hand plates are connected together and the right-hand plates likewise.
Fig. 4.5: Capacitors in parallel.
If charge is fed into this combination via the two terminals, some of the charge is stored
on the first capacitor and some on the second. Since the top plates and the bottom plates
of the capacitors are joined by conductors, the potentials of these plates are equal, and the
potential differences across both capacitors are also equal;
V = q1/C1 and V = q2/C2 .
75
Therefore, the net charge on the system of capacitors is
Q = q1 + q2 = C1V + C2V
= (C1 + C2)V.
And the system is, therefore, equivalent to a single capacitor, of capacitance
C = Q/V = C1 + C2.
Hence, for a group of parallel capacitors, the equivalent capacitance is obtained by
summing the values of the individual capacitances. For N capacitors in parallel, the
equivalent capacitance is
C = C1 + C2 + C3 + ...........+ CN. (4.13)
Example 4.4
Reduce the capacitance network in the figure below to a single equivalent capacitor.
C2
C1 C3
a o
C4
C6 C5
b o
Fig. 4.6: Series and parallel connection of capacitors.
Solution:
Step 1: The equivalent capacitance of the parallel capacitors 2, 3, 4 is found.
C = C2 + C3 + C4
76
Step 2: Find the net capacitance of the series capacitors, C and C5.
1/C = 1/C + 1/C5,
hence
C = CC5 /(C + C5).
Step3: Find the net capacitance of the parallel capacitors, C and C6.
C = C + C6.
a oo b
C1 C
Step 4: Determine the equivalent capacitance of the series capacitors, C and C1.
1/Ceq = 1/C + 1/C1.
Therefore,
Ceq = C1C /(C1 + C).
Ceq
a o o b
77
Example 4.5
Three identical capacitors, each having a capacitance of 1.0 F, are respectively charged
by applying 1.0 V, 2.0 V, and 4.0 V across them. The sources of voltages are then
removed leaving the three capacitors charged. Next, they are connected to one another in
parallel by connecting the negatively charged plates to one another and likewise
connecting the positively charged plates.
(a) Determine the charges and voltages across each of the capacitors after this
connection is made.
(b) What is the capacitance of the combination?
Solution:
(a) Initial charges on individual capacitors are
q1 = C1V1 = 1 x 10-6
x 1 = 1 x 10-6
C,
q2 = C2V2 = 1 x 10-6
x 2 = 2 x 10-6
C,
and q4 = C4V4 = 1 x 10-6
x 4 = 4 x 10-6
C.
After the connection is made, the voltage across each of the capacitors in identical since
they are connected in parallel.
Let V be the unknown voltage across each of the capacitors and q1, q2 and q4 their
respective charges. Then
V = q1/1F = q2/1F = q4/1F.
Therefore, each of the capacitors, being identical, carry the same charge. Thus,
q1 + q2 + q4 = 7 x 10-6
C = 3q1,
and q1 = q2 = q4 = (7 x 10-6
)/3 = 2.33 x 10-6
C.
The voltage V is, therefore,
V = (2.33 x10-6
) /(1 x 10-6
) = 2.33 V.
(b) The capacitance of the combination is
C = C1 + C2 + C4 = 3.0 F.
78
Example 4.6
Two capacitors of capacitance 3.0 F and 6.0 F are connected in series across a 12-V
battery.
(a) What are the charges and the potential differences on each of them?
(b) They are then disconnected from the battery and connected together in parallel.
Determine the resultant potential difference for both polarities of connection.
Solution:
(a) The charges acquired during the series connection are found by using the relation
q = CV.
The equivalent capacitance, C, has the value
C = C1C2 /(C1 + C2) = (3.x 10-6
x 6 x 10-6
)/[(3 + 6) x 10-6
]
= 2.0 x 10-6
F.
Hence, charge on the combination and also on each of the capacitors is
q = CV = 2 x 10-6
x 12 = 2.4 x 10-5
C.
(b) After the battery is removed, the capacitors can be separated. Assume that each
capacitor retains its charge until they are connected together. Two possible parallel
connections are illustrated below.
q1 q1
q2 q2
q1 + q2 = 2q qtot = q1 + q2 = 0
If positive plates are connected, the combined capacitor has a charge of 2q = 4.8 x 10-5
C.
The equivalent capacitance is then C = C1 + C2 = 9 x 10-6
F.
Hence, voltage across the combination is
V = 2q/(C1 + C2) = (4.8 x 10-5
) /(9 x 10-6
) = 5.33 V,
79
and the charges on the respective capacitors are
q1 = CV = 3 x 10-6
x 5.33 = 1.6 x 10-5
C
q2 = CV = 6 x 10-6
x 5.33 = 3.2 x 10-5
C.
When the combination is made with the polarities opposite, the net charge is zero and,
therefore, the voltage is zero. In this case the separate capacitors are restored to an
uncharged condition.
4.2 DIELECTRICS - AN ATOMIC VIEW
So far, in dealing with the problems of electrostatics, it has been assumed that the space
surrounding the electric charge consisted of a vacuum, which has no effect on the electric
field, or of air, which has only an insignificant effect on the electric field. It was
discovered by Michael Faraday that for a fixed geometry, the capacitance of a condenser
can be increased by replacing the vacuum by a dielectric, that is, by an insulating
substance.
Let us first consider how the capacitance of the parallel-plate capacitor is altered if part of
the space between the plates is replaced by a conductor, as shown in Fig. 4.7.
d
a1 a2
Fig. 4.7: Parallel-plate capacitor into which an uncharged conducting plate has been
inserted.
Initially, the capacitance has the value
Co = oA/d.
Let the distance of the inserted conductor from one plate be a1 and a2 from the other. The
electric field within the conductor vanishes. Hence, the conducting material must itself
somehow establish an electric field in its interior, which exactly cancels the field
produced by the capacitor. In a conductor, this is accomplished by a physical separation
80
of free charges, which produces a positive charge density at one surface and a negative
charge density at the other. There are, therefore, in effect, two capacitors in series with
capacitances
C1 = oA/a1 and C2 = oA/a2,
and the equivalent capacitance is
C =(1/C1 + 1/C2)-1
= oA /(a1 + a2). (4.14)
a1 and a2 must necessarily be less than d. This means that C is much larger than Co.
Any conductor contains charges that are free to migrate in response to applied electric
fields. The separated charges produce an electric field in the interior of the conductor.
The same amount of charge is stored on the plates, but by an external potential difference
smaller than was originally. This means that the capacitance, which is the ratio of the
stored charge to the potential difference, has increased.
Now, let us consider the case where we have an insulating substance or dielectric
material such as glass or mica inserted between a parallel-plate capacitor. Suppose that
the slab of the dielectric material completely fills the space between the plates (Fig. 4.8).
Fig. 4.8: Parallel-plate capacitor filled with an insulator.
The dielectric contains a large number of atomic nuclei and electrons. These positive and
negative charges balance each other, so the material is electrically neutral (Fig. 4.8(a)). In
an insulator, all the charges are bound, that is, the electrons are confined within their
atoms or molecules, and they cannot wander about as in a conductor. However, in
response to the force exerted by the uniform electric field, Eo, provided by the plates, the
charges move very slightly without leaving their atoms (Fig. 4.8(b)). The positive and
negative charges are pulled in opposite directions in the electric field. These opposite
81
pulls tend to separate the positive and negative charges and thereby create electric dipoles
within the dielectric. The slab, as a whole, although remaining electrically neutral,
becomes polarized, as Fig. 4.8(b) suggests. The net effect is a pile-up of positive charge
on one face of the slab and of negative charge on the other face. The positive induced
surface charge must be equal in magnitude to the negative induced surface charge since
the slab as whole remains neutral. In this process electrons in the dielectric are displaced
from their equilibrium positions by distances that are considerably less than an atomic
diameter. There is no transfer of charge over macroscopic distances such as occurs when
a current is set up in a conductor.
The induced surface charges appear in such a way that the electric field set up by them,
E, opposes the external electric field, Eo. The resultant electric field in the dielectric, E,
is the vector sum of Eo and E. That is,
E = Eo + E.
It points in the same direction as Eo but is smaller in magnitude.
The weakening of the electric field is evidence in the reduction in potential difference
between the plates of a charged isolated capacitor when a dielectric is introduced between
the plates. If a dielectric slab is introduced into a charged parallel-plate capacitor, then
Eo/E = Vo/Vd = r, (4.15)
where Vo is the potential difference between the plates without a dielectric and Vd is the
potential difference with dielectric slab between capacitor plates. r is the dielectric
constant of the slab. r is a constant which depends on the dielectric. It is independent of
the geometry of the capacitor. It is also practically independent of the potential difference
across the device, so long as the interior field does not become excessively large.
The details of the mechanism of displacement and separation of charge depend on the
dielectric. In some dielectrics, such as glass, polythylene, or other solids, the creation of
dipole moments involves distortion of the molecules or atoms. By tugging on the
electrons and nuclei in opposite directions, the electric field stretches the molecule while
producing a charge separation within it (see Fig. 4.9).
Fig. 4.9: Distortion of molecules produced by electric field
82
In other dielectrics, such as distilled water or carbon dioxide, the creation of dipole
moments results mainly from a realignment of existing dipoles. In such dielectrics, the
molecules have permanent dipole moments, which are randomly oriented in the absence
of an external electric field. The randomness of the dipoles means that, on the average,
there is no charge separation in the dielectric. When the dielectric is placed in an external
electric field, the permanent dipoles experience a torque that tends to align them with the
electric field (Fig. 4.7). Random thermal motions oppose this alignment, and the
molecules achieve an average equilibrium state in which the average amount of
alignment is approximately proportional to the strength of the electric field.
Fig. 4.10: The electric field produces an (partial) alignment of already distorted
molecules.
4.3 GAUSS’ LAW IN DIELECTRICS
The presence of a dielectric increases the capacitance by a factor r,
C = q/Vd = rq/Vo = rCo. (4.16)
In a parallel-plate capacitor filled with a dielectric,
C = rCo = roA/d. (4.17)
Figure 4.11 shows a parallel-plate capacitor both with and without a dielectric. It is
assumed that the charge q on the plates is the same in each case. The dashed lines
represent a Gaussian surface whose top and bottom caps are of the same shape and size as
the capacitor plates.
Without any dielectric in the capacitor, Gauss‟s law gives
or o dAE = oEoA = q
83
Eo = q/oA.
(4.18)
Fig. 4.11: A parallel-plate capacitor (a) without, and (b) with a dielectric.
With a dielectric between the plates, Gauss‟s law gives
o dAE = q - q = oEA
or E = (1/oA)(q - q). (4.19)
q - q is the net charge within the Gaussian surface. q is the free charge and q is the
induced (bound) surface charge.
Now, from Eq. 4.15, we have
E = Eo/r.
Combining this with Eq. 4.18, we have
E = Eo/r = q/roA. (4.20)
Inserting Eq. 4.20 in Eq. 4.19 gives
q/roA = q/oA - q/oA (4.21a)
84
or free /ro = (1/o)(free - bound) (4.21b)
where free = q/A is the free charge density and bound = q/A is the bound charge density.
Eq. 4.21 can be rewritten as
q = q(1 - 1/r) (4.22a)
or bound = (1 - 1/r)free . (4.22b)
This shows that the induced (or bound) surface charge q is always less in magnitude than
the free charge q and is equal to zero if no dielectric is present , that is, if r = 1.
Now Gauss‟s law for the capacitor with a dielectric can be written in the form
o dAE = q - q,
which can be rewritten as
o dAEε r = q, (4.23)
using Eq. 4.22a. Gauss‟s law in dielectrics relates the total electric field E to the free
charge q. The effect of the bound charge is implicitly contained in the factor r appearing
on the left side of the equation. Although the preceding discussion has focused on
arrangements of conductors and dielectrics with high symmetry, the above modified
version of Gauss‟s law (Eq. 4.23) turns out to be valid for conductors and dielectrics of
any shape whatsoever.
Equation 4.21b can be rewritten in the form
E = (1/o)(free - bound)
and hence
free = oE + bound. (4.24)
The induced charged density is called the electric polarization, P. The vector
combination oE + P is called the displacement vector, D. In the more general case,
therefore,
D = oE + P. (4.25)
The vectors D and P can both be expressed in terms of E alone. We can write
q/A = ro(q/roA).
85
Using Eq. 4.20, the above expression gives
D = roE. (4.26)
Also, P = q/A = (q/A)1 - 1/r)
= D(1 - 1/r) = oE(r -1).
Hence P = o(r -1)E. (4.27)
Combining Eqs. 4.23 and 4.26, Gauss‟s law in the presence of dielectrics can be rewritten
as
o D.dA = q. (4.28)
From the above discussion, it is seen that in a dielectric a mechanism exists for reducing
an applied external electric field. The presence of bound charges accounts for this
reduction. This same mechanism, therefore, also reduces the potential difference between
the conducting plates of a capacitor, for a given fixed charge, and thereby leads to an
increase in the capacitance.
Example 4.7
A parallel-plate capacitor is made of two strips of aluminium foil with a plate area of 0.75
m2. The plates are separated by a layer of polythylene 2.0 x 10
-5 m in thickness. The
dielectric constant of polythylene is 2.3. Suppose that the potential difference of 30.0 V is
applied to this capacitor. Determine
(a) the magnitude of the free charge on each plate,
(b) the magnitude of the bound charge on the surface of the dielectric,
(c) the electric field in the dielectric, and
(d) the polarization of the dielectric.
Solution:
(a) With r = 2.3, the capacitance is
C = roA/d = 2.3 x 8.85 x10-12
x 0.75/(2 x 10-5
)
= 7.6 x 10-7
F.
Hence, magnitude of the free charge on each plate is
qfree = CV = 7.6 x 10-7
x 30
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= 2.3 x 10-5
C.
(b) Bound charge is qbound = Abound = (1 - 1/r)Afree
= (1 -1/r)qfree
= (1 - 1/2.3) x 2.3 x 10-5
= 1.3 x 10-5
C.
(c) Electric field in the dielectric is
E = Efree / r = (1/r)(free /o) = qfree /roA
= 2.3 x 10-5
/(2.3 x 8.85 x 10-12
x 0.75)
= 1.5 x 106 Vm
-1.
(d) Polarization of the dielectric is
P = o(r - 1)E
= 8.85 x 10-12
x 1.3 x 1.5 x 106
= 1.76 x 10-5
Vm-1
.
4.4 ENERGY STORAGE IN CAPACITORS
Capacitors store not only electric charge but also electric energy. Imagine that a parallel-
plate capacitor is charged gradually, starting with an initial charge q = 0 and ending with
a final charge q = Q. When the plates carry charges q, the potential difference between
the plates is q/C.
Work that must be done to increase the charge on the plates by dq is
dU = Vc dq = (q/C)dq.
Total work that must be done to charge the capacitor is then
U = (q/C)dq = (1/C) qdq
= Q2/2C. (4.29)
From the relation q = CV, we can also write Eq. 4.29 as
U = CV2/2, (4.30)
87
or U = QV/2. (4.31)
For a parallel-plate capacitor C = Ao/d and V = Ed. Hence
U = (1/2)[Ao/d](Ed)2
= AdoE2/2.
It should be noted that Ad is the volume between the plates of the capacitor and that it
coincides with the volume where the electric field exists. Hence, the energy density,
which is the stored energy per unit volume is
u = U/Ad = oE2/2. (4.32)
For a capacitor filled with a dielectric, Eq. 4.32 becomes
u = roE2/2. (4.33)
Equation 4.33 is a general result and is true for all electric fields, although it was derived
for a parallel-plate capacitor.
88
CHAPTER FIVE
STEADY CURRENTS AND DIRECT CURRENT CIRCUITS
5.0 CHARGE FLOW IN CONDUCTORS: CURRENT AND
CURRENT DENSITY
Metallic substances represent the most important and familiar class of conductors. The
valence electrons of the atoms of a metal are not bound strongly to the individual atoms
but are, instead, free to move within the conductor. These electrons are called free
electrons. This state of affairs arises on account of the interaction between the atoms that
constitute the crystal lattice of the metal. It is important to note that the number of free
electrons is balanced by an equal number of positive charges on the metal ions in the
conductor. The conductor as a whole is, in general, electrically neutral and bears no net
charge. However, the positive charges on the metal ions are fixed within the crystal
lattice of the metal and cannot move like the mobile free electrons.
When an electric field exists within the conductor, the free charge distribution moves,
setting up a flow of electric current. The flow of current in a conductor need not be
constant with respect to time. But when the current is constant with respect to time, a
steady current flow, or a direct current (dc) is said to have been established.
Consider a uniform conductor carrying a current of charge that is uniform and constant at
all points within the conductor. The current I carried by the conductor is then defined as
the total mobile charge passing a fixed plane normal to the conductor per unit time.
Thus, if a quantity of charge q crosses the shaded area shown in Fig. 5.1, in time t,
then
I = q/t. (5.1)
Fig. 5.1: Concept of current as a flow of charge.
For currents that vary with time, the current at time t is defined as the limit of q/t as the
time interval becomes infinitesimally small. Hence
I(t) = lim (q/t) = dq/dt. (5.2)
t0
89
Current is a scalar quantity. The unit of current is the ampere (A) which is defined as the
flow of charge of one coulomb per second.
Another quantity related to current is the current density, which expresses the strength or
concentration of charge flow at any point in a conduction medium. The current density is
a vector quantity that points in the direction of charge flow (Electric field) at the given
point. Its magnitude is determined by taking the limit of the charge flow, or current, per
unit area through a small area a oriented perpendicular to the direction of the current
at the point as the area a approaches zero. Mathematically, this means that the
magnitude of the current density vector j is given by
j = lim (I /a) = dI / da. (5.3)
a0
In the case of a conductor within which the time rate of flow of free electrons is uniform,
the current density j is the same throughout the conductor. In this case,
I = jda = jda = jA.
Hence,
j = I /A. (5.4)
Thus, when the current through the conductor is uniform, the current density is simply the
ratio of the current to the area normal to the direction of the current flow.
When an electric field is established in a conductor, each electron experiences a force F =
-eE, and an instantaneous acceleration a = F/m. The direction of flow of the electrons is
opposite to the electric field (Fig. 5.2).
Fig. 5.2: Electron flow in a conductor.
The kinetic energy of an electron increases during acceleration. The speed of the electron
is, however, limited because it collides with the ions or atoms that make up the
conductor. Some kinetic energy is lost in such collisions. The resulting motion of the
90
electron is zigzag, but the average velocity is in the direction opposite to the electric field.
This resultant average velocity is called the drift velocity and is denoted by vD.
Now, consider a conductor of uniform circular cross-sectional area A and free-electron
density n. During the time interval t all electrons move a distance vDt. This means that
all those electrons within a distance vDt pass through the cross-sectional area. The
number of electrons which pass through the cross-sectional area is
N = nAvD t.
The charge carried by each electron is -e. Therefore, the quantity of charge which passes
through the cross section is
Q = -Ne = - enAvDt
and the current is
I = Q/t = - enAvD (5.5)
The current is proportional to the drift velocity. The current density is then given by
j = I/A = - envD (5.6)
5.1 ELECTROMOTIVE FORCE AND POTENTIAL DIFFERENCE
Consider a battery connected to a closed conducting circuit containing a motor that is
doing work to lift a weight. It is evident that certain forces are acting on the mobile
charges within the conducting circuit and within the wires that comprise the coils inside
the motor. If this were not so, no current would ever be established in the circuit. It is also
evident that the work done by the motor is caused by the flow of current through its
windings and, therefore, ultimately by the forces that cause the current to flow.
Consider a circuit, such as Fig. 5.3, in which a battery of emf E and internal resistance r
is connected to an external resistance R so that a steady current I flows in the circuit.
When a charge q passes right round the circuit including the battery, the energy delivered
by the battery comes from the chemical energy inside the battery. The current, in passing
round the circuit, converts this energy into heat. Some of the heat is dissipated in the
external resistance R and some in the internal resistance r.
The emf of a battery or any other generator can be defined as the total energy per
coulomb of positive charge it delivers round a circuit joined to it.
If a device of emf E passes a steady current I for a time t, then the charge that it circulates
is
q = It.
91
Hence, electrical energy liberated is, W = qE = ItE,
Fig. 5.3: Electromotive force and internal resistance
and the electrical energy, P = W/t = ItE/t = IE. (5.7)
Therefore,
electromotive force = power/current.
Thus, the emf of a device can be defined as the ratio of the electrical power which it
generates to the current it delivers.
Now, assume that charge q flows in time t, through a conductor from one point to
another. Then, since
I = q/ t,
the power supplied by the emf is
P = W/t = q.V/t = IV, (5.8)
where V is the potential difference between the two points. Comparing Eqs. 5.7 and 5.8,
it is observed that potential difference resembles electromotive force in that both can be
defined as the ratio of power to current. The unit of emf is, therefore, 1 watt/ampere or 1
volt. The emf of a source is numerically equal to the power it generates when it delivers a
current of one ampere.
Sources of emf other than batteries (for example, electric generators, fuel cells, solar
cells), while relying on different internal processes to produce and maintain their terminal
potential differences, behave in essentially the same way when incorporated into
conducting circuits.
5.2 OHM’S LAW AND THE CONDUCTION OF ELECTRICITY BY
FREE ELECTRONS
When a source of emf such as a battery is connected to an electrical circuit which
contains a conducting substance, a steady current flows through the conducting
substance. For most conducting substances, the current density j at any point is directly
proportional to the electric field E within the conductor at that point, provided that the
magnitude of the field is not excessively large. Thus,
92
j = E, (5.9)
where is a constant of proportionality which expresses the current density obtained per
unit electric field intensity. The value of depends on the particular conducting
substance in the circuit and also, usually to a lesser extent, on the temperature. The
quantity is referred to as the electrical conductivity of the material. The units of
electrical conductivity are those of current density divided by electric field intensity,
AV1
m-1
. Equation 5.9 is universally referred to as Ohm’s Law, often written in the form
E = (1/)j = j, (5.10)
where = 1/. (5.11)
is the electrical resistivity and has dimensions of volt-meter/ampere (or ohm-metre,
m).
Steady currents flow in conducting materials when an electric field is present by virtue of
the fact that mobile electrons in the conductor are set in motion by the field. We should,
therefore, be able to express the forces on the free electrons in terms of the applied field,
write their equation of motion using Newton‟s second law, and finally find their velocity,
from which the current density may be obtained. In this way, Ohm‟s law can be derived
from first principles using a free electron model as a basis of the calculations.
Let us first consider the motion of a single electron in a uniform conducting substance in
which a constant electric field E is suddenly applied at a time t = 0. The force acting on
the electron is
F = -eE. (5.12)
Using Newton‟s second law of motion, the acceleration is given by
a = dv/dt = -eE/m. (5.13)
Assuming that the electric field E is constant, Eq. 5.13 represents a uniformly accelerated
motion, and the velocity v, is therefore, given by
v = vo + at = vo - eEt/m. (5.14)
The first term on the right is the field-independent part and represents the effect of the
thermal motion of the electrons. The second term expresses the drift velocity acquired by
the electron due to the action of the field. Since the average value of vo for the thermal
motion is zero, and since each electron in the substance acquires the same drift velocity
-eEt/m in the same time interval, the average value of v is
v = -eEt/m. (5.15)
93
Thus, from Eq. 5.6, the current density is
j = -nev = ne2Et/m. (5.16)
In any event, if the electron starts from rest at t = 0, its subsequent drift speed in a
particular direction increases linearly with time, until the electron undergoes a collision at
time t = c, at which time the velocity in that direction drops, on the average to zero.
After the collision, the process is repeated, and the drift speed rises linearly until it is
again abruptly reduced to zero by a second collision at some later time. This cycle of
events goes on indefinitely. During the time interval from t = 0 to t = c, the drift velocity
increases linearly with time from zero to a maximum of vm = -eEc/m at t = c. Hence,
the average electron drift velocity is
v = -eEc/2m. (5.17)
The free drift time between successive collisions may not always be the same since,
because of the random nature of the electron‟s thermal motion, there will be instances
where successive collisions occur only after much longer time intervals. The average free
time is therefore c and the average electron drift speed over such a time is
v = -eEc/2m. (5.18)
The current density may now be written as
j = -nev = ne2cE/2m. (5.19)
If the average time c between collisions is independent of the electric field E, Eq. 5.19
indicates that the average current density is proportional to the electric field, that is,
Ohm‟s law is obeyed. Equation 5.19 can then be written as
j = E
in agreement with Ohm‟s law, where the conduction is now expressed as
= ne2c/2m. (5.20)
5.3 ELECTRICAL RESISTANCE: OHM’S LAW FOR CIRCUITS
Consider the flow of a steady direct current through a conductor of uniform cross-
sectional area and conductivity . A constant electric field E is established in the
conductor. The relation between the current density and the electric field, assuming that
Ohm‟s law is obeyed, is
j = E.
The potential difference dV across any short segment of length dl is
94
dV = -E.dl = -Edl. (5.21)
Thus the difference in potential between the ends of the conductor of length l is
V = dV = V2 - V1 = -E dl,
or V = -El. (5.22)
The minus sign reflects the fact that the potential change from V1 to V2 is a potential drop
rather than a potential rise. Thus, Ohm‟s law can be written as
j = I / A = V/ l
or I = (A/l)V = (A/l)V,
where = 1/ is the electrical resistivity. Hence,
I = (A/l)V = V/(l/A)
or I = V/R, (5.23)
where R = l/A = l/A. (5.24)
The quantity R is called the electrical resistance of the length of conducting material
under consideration. From Eq. 5.24, it is observed that the resistance of any conductor of
uniform cross section and constitution is directly proportional to its length and inversely
proportional to its area. Equation 5.23 expresses Ohm‟s law in terms of the current,
resistance and potential drop.
Equation 5.23 may be written as R = V/I. Hence the electrical resistance of a conductor
has units of volts/ampere or volt-second/coulomb which are generally referred to as ohm
(). The unit of resistivity is the ohm.meter (.m) and that of the conductivity is 1/ohm-
meter (-1m
-1).
Example 5.1
A wire of resistivity 1.7 x 10-8
m commonly used for electrical intallation in homes has
a radius of 1.3 x 10-3
m.
(a) What is the resistance of a piece of this wire 25 m long?
(b) What is the potential drop along this length of wire if it carries a current of 10 A?
95
Solution:
(a) Cross-sectional area of the wire is
A = r2 = (1.3 x 10
-3)2 = 5.3 x 10
-6 m
2.
Resistance is
R = l/A = (1.7 x 10-8
x 25) / (5.3 x 10-6
)
= 0.08 .
(b) By Ohm‟s law,
V = IR = 10 x0.08
= 0.80 V.
5.4 EQUIVALENT RESISTANCE OF NETWORKS
A number of ideal emfs can be arranged in series as shown in Fig. 5.4. A charge, in
passing through the cells in series, experiences a gain in potential energy equal to the sum
of the potential energy gain associated with each individual cell. Therefore, the potential
rise associated with the array, V, is equal to the sum of the potential rises of the
individual cells.
o o
1 2 3
V1 V2 V3
V
Fig. 5.4: Sources of emf in series.
For emf in series,
V = V1 + V2 + V3 +..........+ Vn . (5.25)
Hence,
= 1 + 2 + 3 +.........+ n. (5.26)
96
When sources of identical emfs are connected in parallel, as shown in Fig. 5.5, the
negative terminals are all at a common potential, and the positive terminals are likewise
at a common potential that is higher by the cell potential Vo. Hence, for identical cells in
parallel,
= Vo. (5.27)
Fig. 5.5: Sources of emf in parallel
When unidentical cells are connected in parallel, very large circulation currents flow, and
the observed potential differences depend critically on the resistance of the circuit
conductors and the internal resistances of the cells.
Resistances may be combined in series and parallel arrangements. The arrangement in
Fig. 5.6 is a series one. In this arrangement, the same current I flows through all the
resistors. If the current is caused by an applied potential difference V across the
resistors,
R1 R2 R3
o o
V1 V2 V3
V
Fig. 5.6: Resistances in series
then for n resistors in series,
V = V1 + V2 + V3 +.............+ Vn. (5.28)
97
Thus, using Ohm‟s law, we have
IR = I(R1 + R2 + R3 + ..........+ Rn)
from which
R = R1 + R2 + R3 + ..........+ Rn, (5.29)
where R is the equivalent resistance of a series array of resistances.
Figure 5.7 shows resistors connected in parallel. In this arrangement, the potential
difference across all the resistors has the same value, V.
Fig. 5.7: Resistors in parallel
For an array of n resistors in parallel,
I = I1 + I2 + I3 + ............. + In. (5.30)
For such an arrangement of resistors, the total current is equal to the potential drop V
divided by the equivalent resistance R. That is,
I = V/R = V(1/R1 + 1/R2 + 1/R3 + ....... + 1/Rn),
from which
1/R = 1/R1 + 1/R2 + 1/R3 + ....... + 1/Rn. (5.31)
It is observed from Eqs. 5.29 and 5.31 that for resistances in series, the equivalent
resistance is always greater than that of any of the individual resistors, while for
resistances in parallel, the equivalent resistances is always smaller than any of the
individual resistances.
5.5 KIRCHOFF´S LAWS An electrical circuit is usually a complicated system of electrical conductors. For such
complex networks, a more general set of principles referred to as Kirchoff’s laws for dc
circuits, is used in the analysis.
98
5.5.1 Kirchoff´s First Law
This may be stated as either of the following:
1. The algebraic sum of currents of all wires meeting at a point is zero i.e. I = 0.
Or
2. The algebraic sum of the currents flowing towards a point equals the algebraic sum
of the currents flowing away from it.
Illustration
I5 + I4 – I1 – I2 – I3 = 0 Or
I5 + I4 = I1 + I2 + I3
The First Law and Charge Conservation
The first law is a consequence of the principle of conservation of charge
I1t + I3t = I2t I1 + I3 = I2
5.5.2 Kirchoff´s Second Law
The algebraic sum of the voltage drops (potential differences) in a closed circuit or loop
is equal to the algebraic sum of the emf.
i.e. (IR) = V
The Second Law and Conservation of Electrical Energy
The second law illustrates the principle of conservation of electrical energy. This is
illustrated overleaf:
99
Fig. 5.8: Illustration of Kirchoff‟s second law.
Let us assume that the batteries have negligible internal resistances, then, the principle
of conservation of electrical energy gives
(E2 – E1 )It = I2 (
R1 + R2 + R3 )t
E2 – E1 = IR1 + IR2 + IR3
(5.32)
Equation (5.32) implies that
Algebraic sum of emfs = Algebraic sum of potential differences of IR´s
Sign Convention
1. The emf is negative if the direction of motion is from the positive terminal to the
negative terminal of the battery (or cell) i.e. in moving in the direction + – the emf
is negative. It is considered positive in the reverse direction i.e. – +
2. The current is negative if the direction of motion is opposite to the direction of
current. If the motion is in the same direction as the current direction it is
considered positive
Example 5.2
Determine the current strengths in the circuit below:
Fig. 5.9: Series and parallel combination of circuit elements.
Solution:
For current loop ABCDA,
0.5i1 + 3i2 + 1i1 = 4 + 2
1.5i1 + 3i2 = 6
100
i1 + 2i2 = 4 (1)
For loop FDCEF,
1i3 – 3i2 = 3 – 2
i3 – 3i2 = 1 (2)
Further, at a junction such as C, we can write
i1 = i2 + i3 (3)
Substituting equation (3) in (1) gives
3i2 + i3 = 4 (4)
3 x eqn. (4) gives
9i2 + 3i3 = 12 (5)
3 x eqn (2) gives
3i3 – 9i2 = 3 (6)
Adding eqns (5) and (6) gives
6i3 = 15 i3 = 15/6 = 2.5A
Substituting i3 in eqn (2) gives
2.5 – 3i2 = 1
3i2 = 2.5 –1 = 1.5 i2 = 1.5/3 = 0.5A
Substituting values of i1 and i2 in eqn (3) gives
i1 = (0.5 + 2.5)A = 3.5A
Example 5.3
Two batteries A and B having emfs of 6V and 2V respectively and internal resistances of
2 and 3 respectively are connected in parallel across a 5 resistor. Calculate:
i. the current through each battery.
ii. The terminal voltage
101
Solution:
Fig. 5.10: Internal resistance of batteries connected in parallel
At the junction C,
i3 = i1 + i2 (1)
For loop ABCFA,
2i1 – 3i2 = 6 – 2 = 4 (2)
For loop FCDEF,
3i2 + 5i3 = 2 (3)
Substituting i3 in eqn (3) gives
3i2 + 5(i1 + i2) = 2
8i2 + 5i1 = 2 (4)
5 x eqn (2) gives
10i1 – 15i2 = 20 (5) 2 x eqn (4) gives
16i2 + 10i1 = 4 (6)
(5) – (6) gives
– 3i2 = 16 i2 = – 16/31 = – 0.52A
Substituting i2 in (2) gives
2i1 + 3(0.52) = 4
i1 = (4 – 1.56)/2 = 1.22A
Eqn (1) then gives,
i3 = i1 + i2 = 1.22A – 0.52A = 0.7A
Current i1 in 2 resistor = 1.22A.
Current i3 in 5 resistor = 0.7A.
Current i2 in 3 resistor = 0.52A in a direction opposite to the direction shown on the
circuit diagram.
102
Terminal voltage = 5i3 = 5 x 0.7A = 3.5 V
Example 5.4
Fig. 5.11: (a) Batteries with negligible internal resistance In the above network the internal resistances of the cells are negligible. Calculate
i. the current through the 4 resistor ii. The current through the 2V cell
iii the pd across the 5 resistor.
Solution:
Fig. 5.11: (b) Loop distribution of current in circuit of fig. 5.10 (a)
For loop ADEHA,
2.5i1 – 5i3 – 6i2 = 2 –3 = –1 (1)
For loop ABGHA,
2.5i1 + 7.5(i1 + i3) = 2
10i1 + 7.5i3 = 2 (2)
For loop ACFHA,
2.5i1 – 5i3 + 4(i2 – i3) = 2
2.5i1 + 4i2 – 9i3 = 2 (3)
6 x (3) gives
15i1 + 24i2 – 54i3 = 12 (4)
4 x (1) gives
103
10i1 – 20i3 – 24i2 = 4 (5)
(4) + (5) gives
25i1 – 74i3 = 8 (6)
2.5 x (2) gives
25i1 + 18.75i3 = 5 (7)
(6) – (7) gives
– 92.75i3 = 3 i3 = – 0.0323A
The negative value of i3 implies that the current flows in the direction opposite to that
indicated in the circuit diagram.
Eqn. (2) gives
10i1 + 7.5( – 0. 0323 ) = 2
10i1 – 0.242 = 2 i1 = 2.242/10 = 0.2242 A
For loop CFEDC,
6i2 + 4(i2 – i3) = 3 6i2 + 4i2 – 4i3 = 3
i2 = (3 + 4i3)/10 (8)
Substituting i3 in eqn. (8) gives
i2 = [3 + 4 (–0.0323)]/10 = 0.287 A
i. Current through the 4 resistor = i2 – i3 = 0.287A – ( – 0.0323) = 0.319 A.
Current through the 5 resistor = i3 = 0.0323A.
ii. Current through the 2V cell = i1 = 0.2242 A.
iii. Pd across the 5 cell = i3R = 5 x 0.0323A = 0.1615 V
Example 5.5
A 1.5 -V battery having a 1- internal resistance is connected to a 10- resistor.
Determine the current in the circuit.
Solution:
Fig. 5.12: Internal resistance of a battery
104
Applying the Kirchoff‟s voltage rule (or loop rule),
- Ir -IR = 0.
Solving for I yields
I = /(r + R) = 1.5 /(1 + 10)
= 0.136 A.
Example 5.6
A cell having an emf 18 V and an internal resistance of 10 is connected to
two 100- resistors as shown in the figure below. Currents at points A and B in the 100-
resistors are each 0.15 A. Determine the currents at the points labelled C, D, E, and F.
Fig. 5.13: A battery with appreciable internal resistance
Solution:
Current is constant in any continuous segment of a circuit. Hence, current at C is the
same as that at A, and the current at D is the same as that at B.
That is, IA = IC = 0.15 A
and IB = ID = 0.15 A.
Using Kirchoff‟s junction (current) rule,
IE = IA + IB = 0.15 + 0.15 = 0.30 A.
Also,
105
IC + ID = IF
that is,
IF = 0.15 + 0.15 = 0.30 A.
Example 5.7
In the dc circuit below, calculate the currents flowing in each branch of the circuit and
the potential difference VAB between the points A and B.
Fig. 5.14: A d.c circuit of four loops
Solution:
For loop EBCAE, starting at E and applying the loop law (Law 2),
= p.d.
24 -12 + 8 = 6I1 + 4I2 + 4I3
Hence, 6I1 + 4I2 + 4I3 = 20. (i)
For loop CADC, starting at C,
8 - 6 = 4I3 + 3I5 + 2I6
Therefore, 4I3 + 3I5 + 2I6 = 2. (ii)
For loop BCDB, starting at B,
-12 + 6 = 4I2 - 2I6 - 3I4.
Therefore,
106
-4I2 + 3I4 + 2I6 = 6 (iii)
At points A, C and D, we write, applying the junction (current) rule,
I1 - I2 - I4 = 0 (iv)
I2 + I6 - I3 = 0 (v)
I4 + I5 - I6 = 0 (vi)
Solve (iv) for I1, (v) for I3, and (vi) for I5. Insert the values of I1, I3, and I5 so obtained into
(i), (ii), and (iii) respectively. We then have
-4I2 + 3I4 + 2I6 = 6 (vii)
4I2 - 3I4 + 9I6 = 2 (viii)
7I2 + 3I4 + 4I6 = 20 (ix)
Add (vii) and (viii) to obtain
11I6 = 8.
Hence, I6 = 8/11 = 0.73 A.
Substitute this value into (viii) and (ix) and then add these two equations to eliminate I4.
This gives
11I2 + 13I6 = 22
11I2 = 22 - 13I6 = 22 - 13 x 8/11 = 12.55.
Hence, I2 = 1.14 A.
Substitute the values of I2 and I6 into (vii)
-4 x 1.14 + 3I4 + 2 x 0.73 = 6
3I4 = 6 - 4.56 - 1.46 = 0.02
I4 = - 0.007 A.
Hence I1 - 1.14 + 0.007 = 0
I1 = 1.133 A.
Also, 1,14 + 0.73 - I3 = 0, I3 = 1.87 A
and -0.007 + I5 - 0.73 = 0, I5 = 0.737 A.
107
5.6 SIMPLE R-C CIRCUITS
So far, we have discussed purely dc or steady state situations. We now want to consider
the situation in which the currents and potential differences in the circuit are time-
dependent. We consider Fig. 5.15 which illustrates the charging of a capacitor by a
source through a resistor R.
Fig. 5.15: A simple series R-C circuit; the emf charges the condenser C through the
resistor R.
Suppose that the capacitor C is initially uncharged. At time t = 0, the switch S is thrown
to the left to complete the circuit, as shown in the figure. A current I flows in the
direction shown but decreases with time as the condenser charges, falling off
asymptotically to zero as the potential difference across the capacitor approaches the emf
.
The potential difference across the capacitor at any time t is related to the charge q on the
capacitor at that time by
Vc = q/C. (5.33)
Since the potential difference across the resistance is IR, we may write, using Kirchoff‟s
loop law,
IR + q/C = . (5.34)
The charge q = 0 at time t = 0. Hence the instantaneous value of I at t = 0 is given by
I(0) = Io = /R. (5.35)
The time rate at which the charge q accumulates on the plates is related to the current I by
dq/dt = I. (5.36)
108
Thus, Eq. 5.34 could be written a
Rdq/dt + q/C = , (5.37)
which is a differential equation. Equation 5.34 can be solved by first differentiating both
sides with respect to time and replacing dq/dt with I. Using Eq. 5.36, we get
dI/dt = I/RC. (5.38)
Multiplying through by dt and dividing both sides by I, we get
dI/I = -(1/RC)dt. (5.39)
This equation can be integrated between the limits t = 0 when I = Io = /R, and a later
time t when I = I(t).
dI/I = -(1/RC) dt.
Therefore,
ln (I/Io) = -t/RC. (5.40)
Using Eq. 5.35, Eq. 5.40 may finally be written as
I(t) = Io e-t/RC
= (/R) e-t/RC
. (5.41)
Now, using Eq. 5.37, the charge on the condenser can be determined:
dq = I(t)dt = (/R) e-t/RC
dt.
Integrating this between the limits t = 0 when q = 0 and a later time t when q = q(t), we
have
dq = (/R) e-t/RC dt,
which gives
q(t) = C(1 - e-t/RC
), (5.42)
109
Fig. 5.16: Plots of (a) current versus time and (b) charge on the condenser versus time
for the circuit in Fig. 5.15.
The current and charge on the condenser are plotted as functions of time in Fig. 5.16,
according to Eqs.5.41 and 5.42. At time t = 0, there is no potential difference across the
condenser since it is initially uncharged. The entire potential difference created by the
emf appears across the resistor initially, therefore, giving rise to an initial current /R.
The potential difference across the condenser builds up as charge flows into it. This
results in a correspondingly smaller potential difference across the resistance and hence a
smaller current. As the condenser becomes fully charged, the potential drop across it
approaches the value of the emf and the potential drop across the resistor approaches
zero, as does the current in the circuit. Finally, the condenser is charged to a potential
difference V = volts, and no current flows. It can be shown that after a time t = RC
seconds has elapsed, the current has decreased to 1/e of its initial value and the difference
between the charge and its final value has dropped to 1/e of its initial amount. The
quantity RC is referred to as the time constant.
Now, we consider the case where after the capacitor is fully charged, the switch S is
thrown to the right-hand position, as in Fig. 5.17, so that the capacitor discharges through
a second resistance R.
Fig. 5.17: A simple R-C circuit in which the charged capacitor C discharges through the
resistor R’.
110
Kirchoff‟s loop rule gives
-IR + q/C = 0. (5.43)
In this case the capacitor behaves like a source of emf in that the potential difference
across its plates is what makes the current flow. However, it should not be regarded as a
source of emf because it does not convert nonelectrical energy into electrical energy. It
merely transforms one form of electrical energy into another in converting the potential
energy of the electrostatic field between the plates into drift kinetic energy of electrons in
a conducting circuit. In this case, however, the time rate at which the charge on the plates
changes is negative, because the charge on the plates is continually decreasing with time.
The change in charge dq is related to the current by
-dq/dt = I. (5.44)
Differentiating Eq. 5.43 with respect to time and using Eq. 5.44 to relate I and dq/dt, we
get
dI/dt = -I/RC, (5.45)
which may be rearranged to give
dI/I = -(1/RC)dt. (5.46)
This may now be integrated from t = 0 when q = qo = C to a later time t when I = I(t).
We find
dI/I = -(1/RC) dt,
hence
I(t) = Io e-t/RC
= (/R) e-t/RC. (5.47).
From Eqs. 5.44 and 5.47, the charge q on the capacitor can be expressed as
dq = -I(t)dt = -(/R) e-t/RC dt, (5.48)
which may be integrated from t = 0 when q = qo = C to a later time t when q = q(t).
Therefore, we find that
dq = -(/R) e-t/RC dt,
which, finally gives
q(t) = C e-t/RC
. (5.49)
111
In this case, both the current and the charge on the capacitor die out exponentially with
time constant RC, as illustrated by Fig. 5.18.
Fig. 5.18: Plots of (a) current versus time and (b) charge on the capacitor versus time
Example 5.8
In the circuits in Figs. 5.15 and 5.17, a capacitor of capacitance C = 2 .0 x 10-6
F is
charged by an emf of 100.0 V through a resistance R = 5.0 x 103 .
(a) How long does it take to charge the capacitor to a potential difference equal to 99
percent of that created by the emf?
(b) If the condenser is subsequently discharged through a resistance of 2.0 x 104 ,
(i) what is the initial current, and
(ii) what would the current be after 0.15 seconds?
Solution:
(a) During the charging process, the potential difference is
Vc = (1 - e-t/RC)
).
Since Vc = 0.99, we have
0.99 = 1 - e-t/RC
or e-t/RC
= 0.01
112
and hence
t = -RC ln (0.01) = RC ln(100)
= 5 x 103 x 2 x 10
-6 ln (100)
t = 0.0461 s.
(b)
(i) If the capacitor is charged to the potential difference = 100 V, and then discharged
through a resistance R = 2 x 104 , the current is given as a function of time by
I(t) = (/R) e-t/RC
For t = 0, I(0) = /R = 100/(2 x 104) = 5.0 x 10
-3 A.
(ii) After t = 0.15 s,
I = (/R) e-t/RC = 5 x 10
-3 exp [-0.15/(2 x 10
4 x 2 x 10
-6)]
I = 1.176 x 10-4
A.
113
CHAPTER SIX
MAGNETISM AND THE MAGNETIC FIELD
6.0 MAGNETISM
The phenomenon of magnetism has been known for at least as long as static electricity.
The magnetic forces exerted by permanently magnetized substances, such as magnetite,
upon objects made of iron was known to the ancient Greeks, and the Chinese were using
magnetic compasses by about 1000 A.D. The word magnetism comes from the district of
Magnesia in Asia Minor, which is one of the places where magnetites were found.
Magnets hold a certain fascination for many people, perhaps because it is possible to
actually feel magnetic force. If you hold a magnet in each hand, you sense forces exerted
by one magnet on the other, even though the magnets are not in touch. If you place an
insulating material such as glass between the two magnets, the forces persist. In fact, the
forces exist even if the magnets are in a vacuum.
The sources of the magnetic force in a magnet are concentrated in regions called poles.
The forces between poles can be attractive or repulsive. Two types of poles exist which
are defined as N and S poles. Two N or two S poles repel each other, but an N pole and an
S pole attract each other. Thus, like poles repel while unlike poles attract.
If a magnet is cut into two pieces, a separate N pole and a separate S pole are not
obtained. Instead, two small magnets, each having an S pole and an N pole are obtained.
This happens no matter how many times the magnets are cut. This leads to the conclusion
that the elementary magnetic entity is a magnetic dipole having one S pole and one N
pole.
If iron filings are sprinkled in the vicinity of a bar magnet, the filings become magnetized
and produce a pattern. The needle of a magnetic compass placed in the vicinity of the
magnet aligns itself with the iron filings. The poles of the bar magnet produce this
alignment by exerting forces on the poles of the magnetized iron filings. These forces can
be envisioned as being transmitted to the iron filings and the compass via a magnetic
field created by the magnet.
6.1 MAGNETIC FIELD
All magnetic fields, even those associated with permanent magnets, arise from the flow
of electric currents. The magnetic fields of permanently magnetized substances arise from
currents that circulate within the atoms of the substance and which adds up on the surface
of the magnet to an equivalent surface current, as illustrated by Fig. 6.1. It is this current
that represents the source of the magnetic field.
114
Fig. 6.1: Currents within the atoms of a magnet act as the source of the magnetic field.
The space around a magnet or a current-carrying conductor is defined as the site of a
magnetic field. The basic magnetic vector is B. The magnetic field can be represented by
the lines of B. The magnetic field vector is related to its lines in the following way:
1. The tangent to a line of B at any point gives the direction of B at that point.
2. The lines of B are drawn so that the number of lines per unit cross-sectional area
is proportional to the magnitude of the magnetic field vector B.
The flux B for a constant magnetic field can be defined as the scalar product of the
magnetic field and the area.
B = B.A. (6.1)
The flux can be interpreted as the number of magnetic field lines passing through an area
A. For a non-uniform magnetic field, the flux at any point in the field is
dB = B.dA. (6.2)
For the special case of B and dA parallel, we can write
B = dB /dA. (6.3)
Thus the magnitude of the magnetic field at a given position can be inferred from the
number of magnetic field lines per square meter at that position. The greater the
concentration of the magnetic field lines, the larger the magnetic field.
The net number of magnetic field lines, or net magnetic flux penetrating a closed surface,
is obtained by integrating Eq. 6.2 over the surface.
B = S B. dA. (6.4)
115
The magnetic field lines of a moving charge form closed loops, that is, the magnetic field
lines do not begin or end anywhere. In general, the magnetic field lines of any kind of
magnetic field always form closed loops, since there is no known “magnetic charge” that
could act as source or sink of magnetic field lines.
Mathematically, these features of the magnetic field lines can be expressed in terms of a modified
version of Gauss‟s law. Consider a closed surface of arbitrary shape. The number of magnetic
field lines that enter this surface is exactly equal to the number that leave, that is, the magnetic
flux through the closed surface is zero:
∫ B.dA = 0 (6.5)
Consider a beam of electrons produced by a cathode-ray tube which move in a uniform
magnetic field.
1. There is a unique direction in space along which the moving charges experience
no magnetic force. This direction lies along a line perpendicular to the pole faces.
2. The magnitude of the magnetic force is directly proportional to the product of the
charge, the speed, and the sine of the angle between the velocity and the zero-
force direction
(i.e. F qvsin).
3. The direction of the magnetic force is perpendicular to both the velocity and the
zero-force direction.
4. Negative and positive charges moving in the same direction are deflected in
opposite directions.
Exhaustive experiments disclose the above four facts about the magnetic force on the
moving charges in the magnetic field.
To determine the magnitude and direction of a magnetic field at some point in the
vicinity of moving charges or currents, a test charge q is placed at this point and launched
with a velocity v in some direction. This charge experiences a magnetic force depending
on its velocity. The force on the charge is given by
F = qv x B. (6.6)
By repeating this procedure several times, it is discovered how the force depends on the
direction of the velocity v. In one direction, the force is zero; this is the direction parallel
or antiparallel to B. In the direction perpendicular to this direction, the magnitude of the
field is
B = F/qv. (6.7)
116
Hence, in the special case of the velocity perpendicular to the magnetic field, the
magnitude of the field is the force per unit charge and unit velocity. From Eq. 6.6, the
force on the test charge is always perpendicular to both B and v, in the direction specified
by the right-hand rule.
According to Eq. 6.7. the magnetic field has units of
force/(charge.velocity) = newtons/[coulomb.(meter/second)]
= newtons/(ampere.meter)
= weber/meter2.
This unit is called a tesla and the symbol is T.
Example 6.1
A 5.0-MeV proton moves perpendicularly to a uniform magnetic field of strength 1.5 T.
Determine the
(i) Speed of the proton, and
(ii) force that acts on the proton. (Mass of proton = 1.7 x 10- 27
kg).
Solution:
(i) Kinetic energy of the proton,
K = 5.0 x 106 x 1.602 x 10
-19
= 8.01 x 10-13
J.
Using the relation, K = mv2/2, we get
v = (2K/m)½,
where v is the speed of the proton.
Hence, v = 2 x 8.01 x 10-13
/(1.672 x 10-27
)½
= 3.09 x 107m/s
(ii) Now, force on the proton is
F = qvBsin
117
F = 1.602 x 10-19
x 3.09 x 107 x 1.5 sin90
o
=
= 7.425 x 10-12
N.
6.2 APPLICATIONS OF MOVING CHARGES IN A MAGNETIC
FIELD
In addition to its fundamental importance as a definition of the magnetic vector B, the
relation F = qv x B has many practical applications and is involved in a variety of natural
phenomena.
6.2.1 Charged -particle Linear Momentum Analyzer
The magnetic force on a charged particle is always perpendicular to its instantaneous
velocity. The work done to cause displacement ds is
dW = F.ds = Fcos ds,
where is the angle between F and ds. It is seen that there is no work done by the
magnetic force because the magnetic force and the displacement are perpendicular. If the
magnetic field is the net force on the particle, then there will be no change in the kinetic
energy of the particle. The magnetic force can change the direction of the velocity but not
the speed.
Consider a positively charged particle moving in a uniform magnetic field directed
perpendicular to the velocity of the charge (Fig. 6.2)
Fig. 6.2: A positively charged particle moving in a uniform magnetic field.
The magnetic force does not work on the particle. Therefore, if the magnetic force is the
net force, the particle executes a uniform circular motion with radial acceleration
ar = v2/r.
Using Newton‟s second law to relate the magnetic force, qvB, we have
118
mv2/r = qvB. (6.8)
Therefore, the linear momentum, p, is
p = mv = qrB. (6.9)
If the strength of the magnetic field and the charge of the particle are known, the
momentum of the particle can be determined, after the radius of the circular path has been
measured. This principle is used to measure the linear momentum of charged particles.
6.2.2 Mass Spectrometer
A mass spectrometer is an instrument used to measure the mass of a particle. One type of
mass spectrometer enables us to measure the ratio of mass and charge by using magnetic
principles. In this type of spectrometer, a charged particle of known velocity enters a
uniform magnetic field with the direction perpendicular to the direction of the magnetic
field (Fig. 6.3).
Fig. 6.3: A mass spectrometer.
The magnetic force on the particle causes it to move in a circular trajectory of radius r.
According to Eq. 6.8, the mass-to-charge ratio is given by
m/q = rB/v. (6.10)
By placing an appropriate charged-particle detector in the spectrometer, the radius r can
be determined. The velocity of a particle can be determined by accelerating it through a
known potential difference V. Using the work-energy principle, we can write
qV = mv2/2. (6.11)
119
Squaring the terms of Eq. 6.10 and using Eq. 6.11 to eliminate the velocity, we obtain, for
the mass-to-charge ratio,
m/q = B2r
2/2V. (6.12)
Thus, by measuring B, r, and V, m/q can be determined. If the charge of the particle is
known, then the mass can be computed.
If two atoms with the same charge but different masses are accelerated through the same potential
difference and passed into the same magnetic field, then the ratio of their masses is related to their
radii of curvature by
m1/m2 = (r1/r2)2. (6.13)
6.2.3 Cyclotron
The cyclotron is a device for the acceleration of protons, deuterons, or other charged
particles used in high-energy collision experiments. It is made up of an evacuated
chamber placed between the poles of a large electromagnet. Within the chamber there is a
flat metallic can cut into two D-shaped pieces, or dees (Fig. 6.4). An oscillating high-
voltage generator is connected to the dees. This creates an oscillating electric field in the
gap between the dees. The frequency of the voltage generator is adjusted so that it
coincides with the cyclotron frequency.
An ion source at the centre of the cyclotron releases charged particles. The charged
particles are then accelerated across the gap between the dees. Once inside the dees, the
particles are shielded electrically, and coast at constant speed while being bent in a
circular path by the magnetic force
F = qvB.
Fig. 6.4: Cyclotron
Now, from Eq. 6.8, we have
120
qvB = mv2/r,
r/v = m/qB.
The period of revolution is
T = 2r/v = 2m/qB
Therefore, the frequency is
f = 1/T = qB/2m. (6.14)
It should be noted that the frequency of revolution is independent of the radius and the
speed of the charge. This frequency is called the cyclotron frequency.
When the ions re-enter the gap between the dees the polarity of the potential difference
between has changed and the particles are again accelerated which increases their speed.
Again they coast and bend in a circle. However, the radius of this second circle is larger
than that of the first because the speed is greater. After performing about a hundred
revolutions the particles reach the edge of the system and enters a subsidiary electric field
that deflects them out of the circle to strike a target.
Each acceleration of a charge q through a potential difference V imparts an energy qV
to the charge. Two accelerations are produced for each revolution. Hence an energy
produced per revolution is 2qV. Typically, a potential difference of about 7 x 104 V is
used. Assume a proton of charge q = 1.6 x 10-19
C executes about 200 revolutions. The
energy of the proton upon emerging from the cyclotron is, therefore, about 30 MeV.
The cyclotron fails to operate at high energies because the assumption that the frequency
of rotation of an ion circulating in a magnetic field is independent of its speed, is true
only for speeds much less than that of light. As the speed of the accelerated particles
increases, relativistic effects are encountered, so that the cyclotron frequency can no
longer be considered independent of speed. The relativistic mass
m = mo /[1 - v2/c
2]
½ (6.15)
increases with velocity so that at high enough speeds,
f = (qB/2mo )(1 - v2/c
2)½
(6.16)
decreases with velocity. (c is the velocity of light.) Thus the ions get out of step with the
electric oscillator, and eventually the energy of the circulating ions stops increasing.
Another difficulty associated with the acceleration of the particles to high energies is that
the size of the magnet that would be required to guide such particles in a circular path is
very large. For a 30-GeV proton, for example, in the field of 1.5 T, the radius of
curvature is 65 meters. Incidentally, a 30-GeV proton has a speed equal to 0.99998c.
121
6.2.4 Synchrotron
Fig. 6.5: Synchrotron
Contemporary high-energy particle accelerators called synchrotrons, utilize varying
magnetic fields, varying frequency, and a fixed circular path. In this machine the particles
are accelerated inside a large annular ring (Fig. 6.5 ). The magnetic field is then only
required over a limited region, and a ring-shaped magnet can be used to hold the particles
in the channel. Since the acceleration is to take place at fixed radius, the magnetic field
must be steadily increased as the particles gain energy. At one point of the ring the
particles pass through a hollow cylindrical electrode connected to a high-frequency
generator, and are accelerated both on entering and leaving it. For this to happen they
must enter the electrode at the moment when it is at a large negative potential (with
respect to the earthed frame of the apparatus), and they must leave it after precisely half
an oscillation of the potential. The frequency must, therefore, be steadily increased as the
particles speed up, and this increase must be kept in step with the variations of magnetic
field.
Example 6.2
The energy of the doubly charged -particles of mass 6.64 x 10-27
kg is 6.068 MeV.
(a) What is their velocity?
(b) What magnetic field applied perpendicular to a collimated beam of such particles
would bend the beam into a circle of radius 0.40 m?
Solution:
122
(a) The energy of the -particle = 6.048 x 106 x 1.6 x 10
-19
= 9.677 x 10-13
J.
Thus, the kinetic energy, K = mv2/2 = 9.677 x 10
-13 J
Hence velocity of -particle, v = (2K/m)½
= [2 x 9.667 x 10-13
/(6.64 x 10)]½
= 1.707 x 107 ms
-1.
(b) The centripetal acceleration of a particle with the velocity v about a circle of radius r
is
a = v2/r.
By Newton‟s second law,
F = mv2/r = qvB
or B = mv/qr = 6.64 x 10-27
x 1.707 x 7/(2 x 1.5 x 10
-19 x 0.4)
= 0.886 T.
Example 6.3
In one type of mass spectrometer the charged particles pass through a velocity selector
before entering the magnetic field. In another the particles pass through a strong electric
field before entering the magnetic field. Compare the ratio of the radii of singly-charged
lithium ions of masses 6u and 7u in the two cases.
Solution:
The magnetic force on the ion of charge q with a velocity v in a magnetic field of strength
B is
F = qvB. (i)
This is the required centripetal force. Hence
F = mv2/r, (ii)
where m is the mass of the ion and r is the radius of the circle traversed by the ions. By
equating (i) and (ii),
123
qvB = mv2/r
or v = qrB/m. (iii)
When the ions have passed through a velocity selector, both lithium ions have the same
velocity in the field. Further, they have the same charge and are in the same magnetic
flux density. Thus, using (iii),
qBr6 /m6 = qBr7/m7
r6 /r7 = m6 /m7 = 6/7 = 0.857.
If the ions pass through a strong electric field, they come out with the same kinetic
energy. But from (iii),
mv2/2 = q
2B
2r
2/2m.
Therefore, since q and B are the same for both isotopes,
q2B
2r6
2/2m6 = q
2B
2r7
2/2m7
or r62/r7
2 = m6 /m7
r6 /r7 = [m6 /m7] = (6/7) = 0.926.
Example 6.4
A cyclotron has an oscillator frequency of 1.14 x 107 Hz and a radius of 0.60 m. (a) What
magnetic induction is required to accelerate protons of mass 1.67 x 10-27
kg and charge
1.6 x 10-19
C?
(b) Calculate the final energy that they acquire.
(c) Determine the error made by assuming that the mass of the protons remains constant.
Solution:
(a) The velocity of the protons in the cyclotron is
v = qBr/m,
where q is the charge, B is the magnetic field intensity, r is the radius of the circle and m
is the mass of the proton.
Now, circumference of the circle = 2r,
124
and time for one revolution = T.
Hence v = 2r/T = 2r
Therefore, 2r = qBr/m
or B = 2m/q
= 2 x 1.14 x 107 x 1.67 x 10
-27/(1.6 x 10
-19)
= 0.748 T.
(b) The final energy of the proton is
mv2/2 = q
2B
2r
2/2m
= (1.6 x 10-19
)2 x (0.748)
2x (0.6)
2/[2 x 1.67 x 10
-27]
= 1.54 x 10-12
J.
(c) Since E = mc2, this energy is equivalent to an increase in mass
m = E/c2 = 1.54 x 10
-12/[3 x 10
8]2
= 1.71 x 10-29
kg
Hence, the error in this is
(m/m) x 100 % = [1.71 x 10-29
/(1.67 x 10-27
) x 100 % = 1.02 %.
6.3 Crossed Electric and Magnetic Fields
An electric field can coexist with a magnetic field in a region of space. A moving charge
in such a region experiences an electric force FE = qE as well as a magnetic force
FB = qv x B. The total force is the superposition of the electric and magnetic forces, and
so
F = FE + FB
= q(E + v x B). (6.17)
The combination of electric and magnetic forces in Eq. 6.17 is called the Lorentz force.
Figure 6.6 shows electric field and magnetic field lines at right angles to one another.
Suppose that a positively charged particle enters the field region from left to right
x x x x x x B
x x x x x x
x x x x x x
125
x x x x x x
x x x x x x
E
Fig. 6.6: Electric field and magnetic field at right angles.
The crosses show the tails of the magnetic field vectors. A positively charged particle
moves from left to right.
The magnetic force qv x B is then opposite to the electric force qE. These forces cancel if
the velocity has the right magnitude. That is,
F = qE - qvB = 0.
Therefore,
E = vB,
or v = E/B. (6.18)
This cancellation of the magnetic and electric fields is the basic principle behind the
velocity selector ( or velocity filters) often used in physics laboratories to select particles
of some desired velocity from a beam containing particles with a large variety of
velocities.
According to the free-electron model of conduction (Section 5.2), conduction electrons in
an isolated conductor can move randomly about, like molecules in a gas. The conductor
confines the electrons within its boundaries. If the isolated conductor is placed in a
magnetic field, each conduction electron experiences a force, F = qv x B, because
conduction electrons are moving charged particles. But because the electron motion is
random, the vector sum of the forces on all the electrons is zero, and the conductor as a
whole does not experience any net force. However, if a current is established in the
conductor, then an ordered motion among the conduction electrons is created, and the
vector sum of the forces of all the electrons is not zero, but is instead related to the
electron drift velocity. The conductor now experiences a net force.
Now suppose that a current I is established in a straight wire of uniform cross-section.
This wire is placed in a uniform magnetic field B (Fig. 6.7).
Fig. 6.7: A straight wire of uniform cross-section carrying a current I is placed in a
uniform magnetic field.
126
The amount of charge of flowing through any cross-section of the wire at any time, t is
q = It. This charge experiences a force
F = qv x B
= Itv x B, (6.19)
where v represents the drift velocity of the charges. If t is the time required for the
charges to travel the length l of the wire, then
l = vt.
If l has the same direction as v (and the current), we can write
F = Il x B. (6.20)
Assume that the magnetic field is perpendicular to the length of the wire. The field, then,
is a maximum and has the magnitude
F = IlB.
There is no net force on the wire if the magnetic field is parallel (or antiparallel) to the
length of the wire.
6.4 MAGNETIC DIPOLE IN A MAGNETIC FIELD
A magnetic dipole in a uniform magnetic field experiences a torque. Analogous to the
torque on an electric dipole in an electric field (Section 2.7), the torque on a magnetic
dipole in a magnetic field and the magnetic field itself are related by a property of the
dipole called the magnetic dipole moment () that is defined by the relation
= x B. (6.21)
.
and B can be measured, hence can be deduced .
torque/magnetic field = Newton.meter/[Newton/(ampere.meter)]
= A.m2 or JT
1.
The magnetic dipole moment points from the S pole to the N pole.
A current in a circular loop produces a magnetic field pattern like that produced by a
magnetic dipole having N and S poles. A current-carrying loop placed in a magnet field
127
experiences forces and a torque similar to those experienced by a magnetic dipole such as
a compass.
Consider a rectangular loop of wire of length a and width b placed in a magnetic field of
strength B, with sides C and E normal to the field direction (Fig. 6.8).
FC
E
B
FE
(b)
Fig. 6.8: A rectangular coil carrying a current I placed in a uniform external
magnetic field.
The normal NN to the plane of the loop makes an angle with the direction of B.
Assume the current to be as shown in the figure. The net force on the loop is the resultant
on the four sides of the loop. On the side D the displacement vector b points in the
direction of the current and has the magnitude b. The angle between the displacement
vector and B is 90o - . Thus the magnitude of the force on this side is
FD = IbBsin(90o - ) = IbBcos.
128
The forces FD and FG have the same magnitude but point in opposite directions. Thus
these two forces taken together have no effect on the motion of the loop. The net force
they provide is zero and, since they have the same line of action, the net torque due to
these forces is also zero.
The forces FC and FE, too are oppositely directed so that they do not move the coil bodily.
However, as Fig. 6.8b shows, they do not have the same line of action if the coil is in the position
shown. There is a net torque which tends to rotate the coil clockwise. The torque is represented
by the vector . The magnitude of the forces FC and FE is IaB. The torque caused by the forces
FC and FE about the axis XX is
= 2(IaB)(b/2)sin
= IabBsin
= IABsin,
where A is the area of the coil (= ab). Assume the coil has N turns. Hence, the torque on the entire
coil is
=N = NIABsin. (6.22)
This equation holds for all plane loops of area A, whether they are rectangular or not. A torque on
a current loop is the basic operating principle of the electric motor and of most electric meters
used for measuring current or potential difference.
Work is done when a magnetic field exerts a torque on a magnetic dipole and causes it to rotate
(Fig. 6.9).
Axis of rotation
Fig. 6.9: A magnetic dipole in a uniform magnetic field feels a torque and rotates.
The work done by a magnetic torque is
dW = d
= Bsin d.
From Fig. 6.9, we see that
= /2 - .
Hence d = -d.
129
Therefore, the work done is
dW = -Bsin d.
The work done on the magnetic dipole by the magnetic field is negative and the magnitude
increases as the angle between and B increases.
The dipole is capable of doing external work and hence possesses potential energy when the
magnetic dipole moment and the magnetic field vectors are not parallel. For convenience, the
potential energy is chosen to be zero when the axis of the dipole is perpendicular to the magnetic
field ( = /2). The potential energy relative to this position is
U = - dW = - -d
= Bsin d = B(-cos)
Hence
U = -Bcos
which can be written in the form
U = -.B. (6.23)
The potential energy of the magnetic dipole is a maximum when and B are antiparallel and at a
maximum when and B are parallel.
Example 6.5
A 0.3 m long wire carrying a current of 25 A makes an angle of 60o to a magnetic field of flux
density 8.0 x 10-4
T. What is the magnitude of the force on this wire?
Solution:
Force on the wire , F = IlBsin
where l is the length of the wire, B is the flux density, and is the angle between the
directions of l and B.
Hence, F = 25 x 0.3 x 8.0 x 10-4
x sin60o
= 5.2 x 10-3
N.
Example 6.6
A simple electric motor consists of a rectangular coil of wire that rotates on a longitudinal
axle in a magnetic field of 0.50 T. The coil measures 10 x 20 cm; it has 40 turns of wire,
and the current in the wire is 8.0 A.
130
(a) As a function of the angle between the lines of B and the normal of the coil,
what is the torque that the magnetic field exerts on the coil?
(b) To keep the sign of the torque constant, a slotted sliding contact, or commutator,
mounted on the axle reverses the current in the coil whenever passes through
and radians. Plot this torque as a function of .
Solution:
(a) The torque on the coil is
= NIABsin
= 40 x 8.0 x 0.1 x 0.2 x 0.5 x sin
= 3.2sin Nm
(b) If the torque always has the same sign (say positive), we can write it as
= 3.2 sin.
(Students are to plot against as an exercise).
Example 6.7
A circular coil of 100 turns has an effective radius 0.05 m and carries a current 0.1 A.
How much work is required to turn it in an external magnetic field of intensity 1.5 T from
a position in which = 0o to one in which = 180
o?
Solution:
The work required is the difference in energy between the two positions.
W = U=180 - U=0
= (-Bcos180o) - (-Bcos0)
= 2NIAB = 2NIr2B
= 2 x 100 x 0.1 x x (0.05)2 x 1.5
= 0.236 J
131
6.5 THE MAGNETIC FIELD OF A CURRENT-CARRYING
CONDUCTOR: THE BIOT-SAVART LAW
In our study of electrostatics, it was observed that Coulomb‟s law which describes the
electric field of point charges was simply the way in which experimental observations
regarding electrostatic forces on charged bodies could be summarized. The situation is
the same with regard to the magnetic field produced by steady currents. The magnetic
forces created by actual currents experimentally are observed and then a mathematical
expression is found for the magnetic field that agrees with the results of all the
observations. The Biot-Savart law was observed in this way.
The Biot-Savart law tells us that the element of magnetic induction dB associated with a
steady current I in a segment of conductor described by the vector dl is:
1. in a direction perpendicular both to dl and to the position vector r from the
segment of conductor to the point P at which the field is being measured, as
illustrated in Fig. 6.10;
2. directly proportional to the length dl of the segment and to the current I it carries;
3. inversely proportional in magnitude to the square of the distance r between the
current element and the point P;
4. proportional to the sine of the angle between the vectors dl and r.
In a mathematical form this law can be written as
dB = km(I/r2)dl x r, (6.24)
where r is a unit vector in the direction of the vector r.
132
Fig. 6.10: The current-element dl establishes a magnetic field contribution dB at point P.
The constant of proportionality, km, is given the value
km = o/4 = 10-7
T m A-1
or o = 4 x 10-7
T m A-1
, (6.25)
where o is the magnetic permeability constant. Equation 6.24 gives the magnetic field generated
by a short segment of a wire. The magnetic field generated by a wire of any length and shape can
be calculated by integrating this equation along the wire.
Example 6.8
Determine the value of B at the centre of a narrow circular coil of N turns and radius r due to the
current I through the coil.
Solution:
The radius r is constant for all the elements l and the angle is constant and equal to
90o.
Length of the wire in the coil = 2rN.
Field at the centre is
B = dB = (o/4)I dl sin90 /r2
= (oI/4r2) dl = (oI/4r
2).2rN.
Thus B = oNI/2r.
This equation shows that
(a) B 1/r when I and N are constant,
133
(b) B N when I and r are constant, and
(c) B I when r and N are constant.
Example 6.9
A very long thin straight wire carries a steady current I. What is the magnetic field at
some distance from the wire?
Solution:
Magnetic field from a long straight current-carrying conductor
From the diagram above
α tan R x
Hence
dα α sec R dx 2
Also
α Rsec r
and
α cos φ sin
Making these substitutions in the equation
2
0
r
sinφidx
4π
μ dB
we have
dα α cos R
i
4π
μ
α sec R
α cos dα αsec R i
4π
μ dB 0
2
2
0
To account for the entire infinite length of the conductor, we integrate from = -90o to
90o. Therefore,
R 2π
iμ B 0
134
This equation shows that the magnetic field of a long straight wire, at a point near it, is
inversely proportional to the distance of the point from the wire.
Example 6.10
A hydrogen atom consists of a proton and an electron separated by about 5.0 x 10-11
m. If
the electron moves around the proton in a circular orbit with a frequency of 1013
s-1
,
determine the magnetic field at the position of the proton due to the motion of the
electron.
Solution:
The motion of the electron is equivalent to an electric current. That is, the charge q moves
by a point on the orbit in time, T, where T is the period of revolution. Thus the equivalent
current is
I = q/T = e,
where = 1/T. The magnetic field at the centre of the circular loop of radius R is
dB = (oI/4) dl x R/R3.
All the infinitesimal contributions to B from the infinitesimal circuit elements are in the
direction perpendicular to the plane of the orbit. The magnitude of the magnetic field is
dB = (oI/4) dl Rsin90o/R
3 = (oI/4) dl/R
2.
Hence, the total magnetic field is
B = dB = (oI/4R2) dl = (oI/4R
2).2R
= oI/2R
= 4 x 10-7
x 1.6 x 10-19
x 1013
/( 2 x 5 x 10-11
)
= 2.0 x 10-2
T.
6.6 AMPERE’S LAW
In the calculation of magnetic flux density B so far, only the Biot-Savart law has been
used. Another useful law for calculating B is Ampere’s Law which states that, If a
continuous closed line or loop is drawn round one or more current-carrying
conductors, and B is the flux density in the direction of an element dl of the loop,
then for free space
135
∫ B.dl = oI. (6.26)
Unlike the integration of the Biot-Savart law, the path of integration for the Ampere‟s
law does not follow the current The path may or may not follow a magnetic field line.
Here, symmetry is exploited and a path is chosen such that
(a) the component B parallel to dl is constant,
(b) B.dl = 0 because B = 0,
(c) B.dl = 0 because B is perpendicular to the path, or
(d) a combination of these three options is present.
For the calculation of magnetic fields, Ampere‟s laws plays a role similar to that of
Gauss‟s law for the calculation of electric fields. Provided that the distribution of currents
has sufficient symmetry, Ampere‟s law completely determines the magnetic field.
Example 6.11
Deduce the magnetic field of a current on a very long thin straight wire by means of
Ampere‟s law.
Solution:
By consideration of symmetry, the magnetic field lines of such a current can have the
following forms: concentric circles, radial lines, or parallel lines in the same direction as
the wire. Radial lines would require that the field lines start on the wire; this is
impossible, because the field lines must form closed loops. Parallel lines in the direction
of the wire are likewise inconsistent with closed loops. Thus, the field lines must be
concentric circles. Also, by symmetry, the magnitude of the magnetic field is constant
along each circle.
Taking a path that follows one of these circles, of radius r, we see that
∫ B.dl = ∫ Brd = 2rB.
The Ampere‟s law then gives
2rB = oI.
Hence B = oI/2r.
Equation (6.26) is the statement of Ampere´s law.
136
Note: 1. The integral on the left-hand side of (6.26) implies that we take the component of B in the
direction of the path element dL along a closed loop. This loop is called the Amperian loop.
2. The current i represents the current enclosed by the loop about the conductor.
3. Ampere´s law holds for any closed path about any configuration of conductors if the current
i is taken to be the net current enclosed by the loop.
4. Equation (6.26) implies that B cos dL = oi.
Application of Ampere´s Law
Ampere´s law is very useful in finding the magnetic field situations of high geometrical
symmetry.
1. When the loop encloses the conductor or a current the line integral equals oi i.e.
B . dL = oi where i is the net current enclosed by the Amperian loop.
2. When the loop does not enclose any current or when the loop does not enclose the conductor
the line integral equals zero i.e. B . dL = 0. This means that currents outside the loop
makes no contribution at all to B . dL.
3. In the diagram below the current enclosed by the Amperian loop (i.e. for r < R) is given by
(r2/R
2)i. Hence Ampere´s law becomes
B . dL = o (r2/R
2)i. m
137
B (2r) = o (r2/R
2)i = o (r
2/R
2)i
B = ( oi/2R2)r
Example 6.12
A very long straight wire has a circular cross section of radius R. The wire carries a
current Io uniformly distributed over the cross-sectional area of the wire. What is the
magnetic field (a) inside the wire, and (b) outside the wire?
Solution:
Consider a circular path of radius r inside the wire. Symmetry indicates that the magnetic
field lines are circles. Taking a path of integration along one of these circles, it is found
that
∫ B.dl = 2rB.
The current is uniformly distributed over the volume of the wire. Therefore, the amount
of current I intercepted by the area within the circular path is in direct proportion to this
area,
I = Ior2/ R
2 = Io r
2/R
2.
Ampere‟s law then leads to
2rB = oI
= oIor2/R
2
Hence B = oIor/2oR2. (6.27)
When r = 0, that is at the centre of the wire, the magnetic field is zero. The magnetic field
reaches a maximum value at the surface of the wire where r = R.
Outside the wire, r R. The magnetic fiels is, therefore,
B = oIo / 2r.
6.6.1 Solenoids
A solenoid is a thin conducting
wire wound in a tight helical
coil of many turns (Fig. 6.11)
138
Fig. 6.11: A Solenoid.
A current in this wire produces a strong magnetic field within the coil.
Consider a solenoid as approximately a series of circular current loops. If the wire is
loosely wound (Fig. 6.11), some magnetic field lines encircle the individual turns of the
coil. Other field lines combine to form the axial magnetic field of the solenoid. If the wire
is wound more tightly, the number of field lines encircling each individual turn is reduced
and the desired axial field is strengthened (Fig. 6.12). The detailed calculation of the
magnetic field of a solenoid is rather messy because the individual magnetic fields of the
rings of current must be added vectorially. To attempt this calculation, the magnetic field
of an ideal solenoid is considered. An ideal solenoid is infinitely long and so tightly
wound so that the current distribution on the surface of the solenoid is nearly uniform
(that is, the current is essentially a cylindrical sheet).
To determine this magnetic field, we begin with an appeal to symmetry. The ideal
solenoid has translational symmetry (along the axis of the solenoid) and rotational
symmetry (around the axis). For consistency with these symmetries, the magnetic field
lines inside the solenoid will then have to be, as in Example 6.11, either concentric
circles, or radial lines, or lines parallel to the axis. Concentric circles would require the
presence of a current along the axis, which is unacceptable; radial lines would require
that the field lines start on the axis, which is impossible. Thus the field lines inside the
solenoid must all be parallel to the axis. These field lines emerge from the end of the
solenoid and return to the other end (Fig. 6.11).
To apply Ampere‟s law, we consider a rectangular path, as shown in Fig. 6.12, and
evaluate the integral of B along the path. The line segments CD and EF are parallel to the
z-axis, while DE and FC are perpendicular to the z-axis. Therefore the integral around the
closed path can be expressed as a sum of four integrals:
∫ B.dl = Bzdz + Brdr + Bzdz + Brdr. (6.28)
F E
oooooooooooooooooooooooooooooooooooo
C D
z
139
B
ooooooooooooooooooooooooooooooooooooo
Fig. 6.12: A path of integration for the calculation of the axial magnetic field inside a
solenoid.
The horizontal side external to the solenoid (EF) does not contribute to the integral (B =
0); and the vertical sides (DE and FC) do not contribute to the integral since B is
perpendicular to the path. Thus
Brdr = Bzdz = Brdr = 0.
Hence, only the horizontal side within the solenoid contributes to the magnetic field. The
magnitude of the magnetic field is constant along this side. If the length of this side is l,
∫ B.dl = Bzdz = Bl.
For a current Io, and a winding of N turns in the distance CD = l, the current enclosed by
the path of integration is IoN. Ampere‟s law then gives
Bl = oNIo
or B = oIo(N/l). (6.29)
The ratio N/l is the number of turns of the wire per unit length, commonly represented by
n. Thus,
B = oIon. (6.30)
This result is independent of the length of the vertical side of the path of integration;
hence B has the same magnitude everywhere inside the solenoid. This shows that the
magnetic field within the ideal solenoid is perfectly uniform. This is why solenoids are so
useful in scientific work. They can produce strong, uniform magnetic fields. Also
because of the strength of the axial field, solenoids have important everyday applications.
For example, solenoids are used in the starters of cars.
6.6.2 Magnetic Force between Two Current-Carrying Conductors
Consider two long straight parallel conductors C and D, a distance d apart. One carries a
current I1 and the other carries a current I2. Such wires exert a force on each other. For
example, when the currents in two long neighbouring straight conductors are in the same
direction, there is a force of attraction between them. If the currents flow in opposite
directions, there is a repulsive force between them. Each conductor has a force on it due
to the magnetic field of the other.
140
Figure 6.13 shows the resultant magnetic flux around two long straight conductors with
parallel currents. Since the field lines act as though they are under tension, we see that the
two conductors attract one another. In Fig. 6.13 the currents are in opposite directions.
Here they tend to push the conductors apart.
Flux density B due to current I1 in C = oI1/2d.
This is the field which acts on the wire D. Hence, the force per unit length on the
conductor D due to the field around C is
F = BI2l = oI1I2/2d, (6.31)
for l = 1 m. Now if I1 = I2 = I, the force per unit length on conductor D is
F = oI2/2d. (6.32)
Fig. 6.13: Two long straight wires carrying currents I1 and I2.
The attraction between two long parallel wires is used to define the ampere, the unit of
current. Taking I1 = I2 = 1 A, and d = 1 m, we have
F = oI2/2d = 4 x 10
-7 x 1
2/(2 x 1)
= 2 x 10-7
Nm-1
.
Thus, the ampere is the current, which flowing in each of two infinitely-long parallel
straight wires of negligible cross-sectional area separated by a distance of 1 meter in
vacuo, produces a force of 2 x 10-7
Nm-1
.
141
It may be noted that the electrostatic force of repulsion between the negative charges of
the moving electrons in the two wires is completely neutralised by the attractive force on
them by the positive charges on the stationary metal ions. Thus the force between the two
wires is only the electromagnetic force, due to the moving electrons (current).
Example 6.13
Calculate the magnetic induction B along the axis of a very long 0.25 m solenoid of 1000
turns of wire if the radius of the coil is 0.02 m and the current in the wire is 15 A.
Solution:
From Eq. 6.29,
B = oI(N/l).
If N = 1000, I = 15 A, and l = 0.25,
B = 4 x 10-7
x 1000 x 15/0.25
= 7.54 x 10-2
T.
Example 6.14
Two long, straight wires, each carrying a current of 9 A in the same direction are placed
parallel to each other. Determine the force that each wire exerts on the other when the
separation distance is 0.1m
Solution:
Using Eq. 6.32, the force per unit length is
F = oI2/2d
= 4 x 10-7
x 92 /(2 x 0.1
= 1.62 x 10-4
Nm-1
.
142
CHAPTER SEVEN
ELECTROMAGNETIC INDUCTION
7.0 MOTIONAL ELECTROMOTIVE FORCE
In the preceding chapter, it was observed that the passage of electric current creates a
magnetic field around the conductor through which it flows. In 1831 Faraday discovered
that the reverse is also true. That is, current could be generated magnetically but such an
effect is observed only when the magnetic flux through the circuit changes with time.
This effect is referred to as electromagnetic induction, and the currents and emfs that are
generated this way are called induced currents and induced emfs. The discovery of
electromagnetic induction led to the possibility of constructing machines that would
convert mechanical energy into electrical energy simply by rotating coils of wire in a
strong magnetic field.
Consider a conductor of length l sliding along and making electrical contact with parallel
metal rails. The conductor moves in a direction perpendicular to its length with a velocity
v, which is perpendicular to the magnetic field B, as shown in Fig. 7.1.
Fig. 7.1: (a) A conductor moving in a magnetic field.
The charges in the moving conductor move relative to the field B and therefore
experience a force
F = qv x B.
The electrons, therefore, flow along the rod, accumulating negative charge on the upper
end (A) and in the rail AD and leaving positive charge on the lower end (B). This
separation of charge opposes the continued flow of electrons. The electron concentration
at A increases until equilibrium is achieved. At equilibrium, the magnetic force is
balanced by the electrostatic force set up by the charge separation.
143
Fig. 7.1: (b) A conducting rod with a resistor in circuit moving through a magnetic field
The emf associated with the rod is the work done by the driving force on a unit positive
charge that passes from the negative end of the rod to the positive end. The driving force
on a unit positive charge is vB, and if the length of the rod is l, the work done by this
force is
= lvB. (7.1)
This is called a motional emf because it is generated by the motion of the rod through the
magnetic field. This motional emf is not present in a static system. The emf depends on
the size of the system ( l) and on the strength of the field.
Now assume that the conducting rails are joined at the right edges by a resistance R to
form a complete circuit (Fig. 7.1b). The electrons accumulating on the rail AD are free to
move through the circuit. Positive charge would flow in the direction opposite to the
electron flow. In terms of potential differences, the induced emf creates a potential
difference along the moving rod, VB > VA.
The quantity lv in Eq. 7.1 is the area swept out by the moving rod per unit time. The
product lvB, therefore, is the magnetic flux swept across by the rod. Equation 7.1 can
then be expressed as
= [time rate of sweeping of magnetic flux]. (7.2)
Equation 7.2 holds for rods and wires of arbitrary shape moving through arbitrary
magnetic fields. Furthermore, the relation between the induced emf and the rate of
sweeping flux holds even if the magnetic field is time dependent. Thus, Eq. 7.2 is of
general validity.
If the moving ends of the rod are in sliding contact with a circular tract connected to the
midpoint via an external circuit (Fig. 7.2) then the rod generates a current in the circuit.
The device then acts as a generator of dc voltage; this device is called a homopolar
generator. For practical applications, homopolar generators are constructed with a
rotating disk rather than a rotating rod, as shown in Fig. 7.3, which rotates between the
poles of a magnet. Connections are made to its axle and circumference. It is assumed, for
simplification, that the magnetic field is uniform over the radius CD of the disc. As the
144
disc rotates, CD continuously cuts the field between the poles of the magnet. For this
straight conductor, the velocity at the end of C is zero and that at the other end D is r,
where r is the length of CD and is the angular velocity of the disc.
Hence,
average velocity of CD, v = (0 + r)/2 = r /2.
Therefore, the induced emf in the conductor CD is
E = Blv = Br(r /2) = Br2 / 2.
= Br2, (7.3)
since = 2, where is the number of revolutions per second.
Fig. 7.2: Moving rod in sliding contact with a circular track.
Fig. 7.3: A homopolar generator.
To understand the origin of the emf, consider an electron between C and D. When the
disc rotates, the electron moves to the left, as shown in Fig 7.3b. The equivalent
conventional current I is then to the right. It is found that the force on the electron drives
it to C so that C obtains a negative charge while D a positive charge. The radius is thus a
generator with D as the positive pole.
The emf induced in the moving conductor of a homopolar generator is always in the same
direction. Homopolar generators are used in applications requiring a large current, but
145
only a fairly small voltage, such as in electroplating. The emf of a homopolar generator is
limited to that induced in one radius of the disc.
Example 7.1
A circular metal disc is placed with its plane perpendicular to a uniform magnetic field of
flux density B. The disc has a radius of 0.20 m and is rotated at 5 rev s-1
about an axis
through its centre perpendicular to its plane. The emf between the centre and the rim of
the disc is balanced by the potential difference across a 10- resistor when carrying a
current of 1.0 mA. Calculate B.
Solution:
The induced emf E = Br2 = B x (0.2)
2 x 5
= 0.2B.
pd across the 10- resistor, V = IR = 1 x 10-3
x 10
= 1 x 10-2
V
Therefore, 0.2B = 1 x 10-2
and B = 1 x 10-2
/0.2
= 1.6 x 10-2
T.
Example 7.2
The conductor in Figure 7.2 is 0.1 m long and moves with a speed of 10.0
ms-1
across a magnetic field of 1.5 T. Calculate (i) the emf, and (ii) the force per coulomb
of charge.
Solution:
(i) The induced emf = Blv = 1.5 x 10 x 0.1
= 1.5 V.
(ii) The force per coulomb of charge is
F/q = Bv
= 1.5 x 10
= 15 T ms-1
.
146
7.1 FARADAY’S LAW OF INDUCTION
The following are three equations relating to Maxwell:
The third of these started with Oerstead's observation that an electric current will produce
a magnetic field and ended with Maxwell's discovery of the displacement current and his
(and Einstein's) unification of electricity and magnetism. Soon after Oerstead's discovery,
many figured searched for a relationship between electric field and 'magnetic current'.
None was found, for reasons that are now obvious. There are no magnetic monopoles and
thus no magnetic current.
Given the sequence with which we have developed these equations, it should be apparent
that they were looking in the wrong place. Maxwell (long after these efforts in the 1820's)
invented/discovered the 'displacement current' that was related only to the time derivative
of the electric flux. While there are no magnetic monopoles and thus no magnetic current,
there certainly is a magnetic field and we can obviously define a magnetic flux which
might vary in time. We thus should be happy to find that a magnetic analog of the
displacement current does indeed exist.
In 1831 (only 11 years after Oerstead), Faraday discovered, sort of by accident,
electromagnetic induction which is the essence of this analog. He found that a change in
current, and thus a change in magnetic field induces an emf in a nearby circuit. Thus the
name electromagnetic induction.
In Section 7.1, we saw that the moving conductor cuts magnetic field lines at the time
rate of Blv. This is the time rate of change of magnetic flux. Equation 7.2 is an expression
of Faraday‟s law of induction which states that:
The induced emf along any moving path in a constant or changing magnetic field
equals the time rate at which magnetic flux sweeps across the path.
For a closed path, Faraday‟s law can conveniently be stated in terms of the magnetic flux
through the area within the path. The net rate of sweeping of the magnetic flux is equal to
the rate of change of the flux intercepted by the area within the path.
147
Hence, Faraday‟s law can be stated as
The induced emf around a closed path in a magnetic field is equal to the time rate of
change of the magnetic flux intercepted by the area of the path.
That is
= -dB /dt. (7.4)
The minus sign is an indication of the direction of the induced emf. As the conductor
moves in the above example, there is a change in the magnetic flux passing through the
loop. The magnitude of the flux is
The time rate of change of this quantity is
We see that the induced emf is equal in magnitude to this time derivative:
This is Faraday's law, with a minor but important correction of sign.
We want to express this result in terms of a line integral of the electric field. An observer
on the rod would experience no magnetic force - the electrons are not moving relative to
himself. Instead, he/she will observe an electric force qE. This electric force must
precisely equal the magnetic force observed by someone at rest relative to the rails:
Using the emf deduced above, we find that the emf developed across the bar is just
That is, it is just the line integral of the electric field. There is no induced emf along the
rails or through any external part of the circuit, so can change this into an integral around
148
a closed loop:
We have showed that the magnitude of the emf is given by the magnitude of the time rate
of change of the magnetic field, and also that the emf is give by the integral around a
loop. We still need to establish the sign relating the time derivative to the emf. This
relates to the convention used in the direction of the path integral for the emf and
direction of the area element dA in the flux integral.
The line integral is over a closed loop, and this loop encloses an area over which the flux
integral is to be evaluated. Use a right-hand rule to relate the direction of the path to the
direction of the area element dA. That is, we curl our fingers in the direction of the path,
and our thumb points in the direction of the area element. When we apply this convention
to the above problem in a counterclockwise fashion, the line integral is positive since the
electric field is directed upward (fields starts on positive charge and electrons are
accelerated opposite to the electric field) and the this is in the same sense as the path ds.
The area element is opposite to the field as drawn, however, so the flux integral must be
less than zero. The appropriate sign is thus minus, and Faraday's law becomes
We have shown Faraday's law to be true only in this special case, but it is true in general.
There being no magnetic monopoles and thus no magnetic current, this law is complete.
Faraday's discovery shifted attention from static to dynamic electric and magnetic
phenomena and thus gave birth to electrodynamics.
The integral form of Maxwell's equations are thus
These relationships are sufficient and necessary to determine the free-space electric and
magnetic fields from static and time-varying source charge and current distributions. In a
material medium, we more information like Ohm's law and various material constants.
We unfortunately will not be able to apply them in this broad context - that is left for a
149
future course. You should know however, that they predict a diverse array of stuff, from
all the static fields we have deduced to light waves of all frequencies. For now, we will
simply look into some of the many interesting manifestations of magnetic induction.
If Eq. 7.4 is applied to a coil of N turns, an emf appears in every coil and these emfs are
added up. If the coil is so tightly wound that each turn can be said to occupy the same
region of space, the flux through each turn is then the same. The flux through each turn is
also the same for ideal toroids and solenoids. The induced emf in such devices is given
by
= -N(dB /dt) = -d(NB)/dt, (7.5)
where NB is the flux-linkage in the device.
Example 7.3
(a) A narrow coil of 20 turns and area 6.0 x 10-2
m2 is placed in a uniform magnetic
field of flux density of 2 x 10-2
T so that the flux links the turns normally.
Calculate the average induced emf in the coil if it is removed completely from the
field in 0.5 s.
(b) If the same coil is rotated about an axis through its middle so that it turns through
60o in 0.4 s in the magnetic field, calculate the induced emf.
Solution:
(a) Flux linking coil initially = NAB = 20 x 6 x 10-2
x 2 x 10-2
Hence, average induced emf = flux change/time
= 20 x 6 x 10-2
x 2 x 10-2
/0.5
= 4.8 x 10-2
V.
(b) When the coil is initially perpendicular to the magnetic field B, flux linking coil is
BP = NAB
When the coil is turned through 60o, the flux linking the coil is
60 = NABcos60o
Therefore, flux change through coil = NAB -NABcos60o
= NAB(1 - cos60o)
= 0.5 NAB.
150
So, average induced emf = flux change/time
= 0.5 x 20 x 6 x 10-2
x 2 x 10-2
/0.4
= 6.0 x 10-2
V.
Example 7.4
A long solenoid of diameter 3 x 10-2
m and 2.0 x 104 turns/m carries a current of 1.5 A.
At its centre is placed a 100-turn, closed-packed coil of diameter 2 x 10-2
m. This coil is
arranged so that the magnetic field at the centre of the solenoid is parallel to its axis. The
current in the solenoid is reduced to zero and then raised to 1.5 A in the other direction at
a steady rate over a period of 5 x 10-2
s. Calculate the emf which appears in the coil while
the current is being changed.
Solution:
The magnetic field B at the centre of the solenoid is given by
B = onIo = 4 x 10-7
x 2.0 x 104 x 1.5
= 3.77 x 10-2
T.
Area of the coil = d2/4 = x (2 x 10
-2)2/4 = 3.14 x 10
-4 m
2.
Initial flux through each turn of coil is
B = BA = 3.77 x 10-2
x 3.14 x 10-4
= 1.18 x 10-5
weber
The flux goes from an initial value of 1.18 x 10-5
weber to a final value of -1.18 x 10-5
weber. Hence,
Change in flux, B = (1.18 x 10-5
) - (-1.18 x 10-5
)
= 2.36 x 10-5
weber.
Hence, induced emf is given by
= - NB /t = - 100 x 2.36 x 10-5
/(5 x 10-2
)
= - 4.72 x 10-2
V.
151
Example 7. 5
A rectangular coil of wire of 150 turns measuring 0.2 x 0.1 m forms a closed circuit. The
resistance of the coil is 5.0 . The coil is placed in an electromagnet, face on to the
magnetic field. Suppose that when the electromagnet is suddenly switched off, the
strength of the magnetic field decreases at the rate of 2.0 T per second. Calculate (i) the
induced emf, and (ii) the induced current in the coil.
Solution:
(i) The induced emf is
= - dB /dt = - NAdB/dt (B = NAB)
= - 150 x 0.2 x 0.1 x 2
= - 6 V.
(ii) The induced current is
I = /R = 6/5
= 1.2 A.
7.2 APPLICATIONS OF FARADAY’S LAW
7.2.1 Flip-Coil Measurement of Magnetic Field
A useful way of measuring magnetic field is to rapidly rotate a coil of wire and
measure the electric charge that is driven through a circuit connected to the coil, as
illustrated in Fig. 7.4. Consider a coil with area A whose plane is perpendicular to a
uniform magnetic field B. The flux through the coil is
B = NAB.
152
Fig. 7.4: A flip-coil for measuring magnetic field.
If the coil is flipped over through 180o, the change in flux, B = -2NAB in the time
interval t. The magnitude of the induced current is
I = /R = -(1/R)dB /dt, (7.6)
where R is the total resistance of the circuit, which consists of the coil, leads, and a
galvanometer. The total charge passing through the galvanometer is, from Eq. 7.6,
Q = -(1/R) (d/dt).dt
= -(1/R) B = 2NAB/R
So, B = RQ/ 2NA. (7.7)
If N, A, and R are known or can be measured, a measurement of Q yields a value of B.
7.2.2 A Rectangular Loop Generator
A rectangular loop of wire of length l and height h is rotated about an axis perpendicular
to an external field B, as shown in Fig. 7.5. The angular velocity, is a constant.
153
Fig. 7.5: A rectangular loop generator.
The flux through the loop is
B = Blh cost (7.8)
for the situation with the loop vertical at t = 0. Then
d/dt = - Blh sint = - . (7.9)
Now, if the loop has a total resistance R, the current is
I = /R = (Blh/R) sint. (7.10)
The time rate of dissipation of electrical energy as heat then is
Pelect = I2R = (2
B2l2h
2/R) sin
2 t. (7.11)
The time rate of doing mechanical work is
Pmech = dW/dt = F.v. (7.12)
v is the velocity of the conductor itself, not the charges through the conductor. Now, the
magnetic force on the top conductor, where whose length is l is
Fmag = Il x B
or Fmag = (Blh/R)lB sin t. (7.13)
This magnetic force is vertically up, perpendicular to l and B. Resolving Fmag into radial
and tangential components, a force must be exerted to overcome Ftan to turn the loop
(Fig. 7.6)
Fig. 7.6: Radial and tangential components of Fmag.
Ftan = Fmag sint
154
= (B2l2h/R) sin
2 t. (7.14)
The radial component does no work because it is always perpendicular to the
displacement. It is therefore ignored. Similarly, the force on the front and backsides of
the loop do no work, and are ignored.
The rate at which work must be done to turn the loop = F.v.
The force, F has a magnitude F = Ftan but is directed in the positive v-direction, or
tangential direction.
Pmech = 2Ftan v = (2B2l2h/R)sin
2 t (h/2)
= (2B
2l2h
2/R) sin
2 t (7.15)
The 2 in the numerator is included to take into account the bottom conductor as well as
the top conductor. Comparing Eq. 7.11 with Eq. 7.15, it is observed that Pelect = Pmech.
Mechanical work is converted into electrical energy and into thermal energy. Energy is,
therefore, conserved.
The above discussion describes the basic principles of the electric generator. Almost all
the electricity used today comes from generators operating on the principles of this
rotating loop and Faraday‟s law.
7.2.3 The Spin-Echo Magnetometer
The spin-echo magnetometer is a magnetic field measuring device. One of its
applications is in measuring magnetic anomalies in the ocean floor. This device consists
of a source of protons, such as a bottle of distilled water, surrounded by a coil of heavy
wire able to carry a large current (see Fig. 7.7). The magnetometer is lowered to the
ocean floor, and a strong pulse of current is sent through the coil or solenoid. This large
current creates a strong magnetic field in the solenoid. Each proton has an intrinsic
magnetic moment. Therefore, the protons align themselves in the strong magnetic field of
the solenoid.
Fig. 7.7: The spin-echo magnetometer.
155
The current is then dropped to zero. Acting like tiny gyroscopes, the aligned protons tend
to precess about the weak magnetic field that remains, as shown in Fig 7.8. This magnetic
field is the magnetic field of the earth, plus the fields created by any magnetized areas of
the ocean floor.
Fig. 7.8: Protons precess about weak magnetic field.
The precessing proton magnetic dipoles induce an alternating emf in the solenoid, by
Faraday‟s law. The frequency of the induced emf is equal to the precession frequency of
the protons, which in turn is proportional to the strength of the remaining magnetic field.
A measurement of this frequency thus gives oceanographers and geophysicists a
measurement of the strength of the magnetic field on the ocean floor.
From such magnetic measurements, earth scientists have concluded that portions of the
earth‟s crust are moving relative to one another with speeds of a few centimeters per year.
These measurements also document very clearly that the earth‟s magnetic field has
experienced many polarity reversals over time.
7.3 LENZ’S LAW AND EDDY CURRENTS
Lenz‟s law concerns the direction of the currents that are induced by flux changes and
their own associated magnetic fields. In the preceding sections, we have mainly discussed
how to calculate induced emfs and currents, using Faraday‟s law, without worrying about
their direction. The direction of any current that is induced by a change in the magnetic
flux through a circuit is uniquely determined by Faraday‟s law. But, why is the minus
sign there? What are its implications?
To answer these questions, let us consider two physical situations. We first take another
look at the moving rod of Section 7.1. As long as the rod is isolated so that there is no
current, the motion of the rod across the magnetic field is unrestrained. However, when
the rod slides on conducting rails and there is a current, an opposing force appears. The
interaction of the external magnetic field and the current leads to a magnetic force on the
rod which opposes the motion that is inducing the current.
Also, consider a bar magnet approaching a conducting ring, as shown in Fig. 7.9
156
Fig. 7. 9: A bar magnet approaching a conducting ring.
Assume the direction from z to x to be positive. Hence, the positive normal for the area of
the ring is in the y-direction and the magnetic flux is negative. As the distance between
the conducting ring and the N pole of the bar magnet decreases, more and more field lines
go through the ring, making the flux more and more negative. Thus dB /dt is negative.
The induced current is as shown.
The current induced in the ring creates an induced secondary magnetic field that is
opposite to the original field inside the ring. This induced magnetic field is similar in
form to the field of a second bar magnet, shown dotted in Fig. 7.9b. This induced magnet
repels the incoming magnet. This opposition is a consequence of the minus sign in
Faraday‟s law, and is formalised as Lenz‟s law. What is called Lenz‟s law is simply a
statement of the physical effects that require that the algebraic sign in Faraday‟s law be
minus rather than plus. Lenz‟s law may be stated as:
When a current is induced in a conductor, the direction of the current is such that the
current’s magnetic effects oppose the change that induced it.
It is important in using Lenz‟s law to understand that the magnetic field of the induced
current is always in opposition not to the primary magnetic flux but to the change in
primary magnetic flux that induces the current.
A discussion of Lenz‟s law would not be quite complete without a brief description of
eddy currents. When a conductor moves in a magnetic field, emfs are induced in all parts
of it that cut the magnetic flux. If the conductor is a large lump of metal, significant emfs
can act round closed paths inside the lump of metal itself. The resistance of the currents
paths in the metal are so low that large currents may flow, although the emfs are not
normally very large. These induced currents circulating inside a piece of metal are known
as eddy currents.
157
The effects of eddy currents may be demonstrated by swinging a pendulum with a thick
copper bob between two poles of an electromagnet (Figure 7.10a). When the
Fig. 7.10: A swinging copper bob between poles of an electromagnet.
electromagnet is switched on, the flux through the bob varies rapidly along its path, and
considerable eddy currents are generated in it, which produce a very marked braking
effect. Thus owing to the generation of eddy currents and the retardation forces they
experience as they enter and leave the magnetic field region, the motion of the pendulum
is quickly damped and its kinetic energy appears as Joule heat within the conductor. The
eddy currents can be prevented in this case by using a bob with a series of slots cut in it.
These slots prevent the formation of large-scale eddy currents and limit those that form to
long, narrow loops between the slots. As a result, the eddy currents are reduced in
magnitude and generate much less Joule heat within the conductor. The retarding forces
experienced by the pendulum as the bob enters and leaves the field region are much
smaller, and the motion of the pendulum suffers very little damping. This method of
breaking up eddy current flow and reducing eddy current heating is used also in
transformers, where the iron cores are made of laminations that are electrically insulated
from one another rather than from a single solid piece of metal.
7.4 SELF-INDUCTION AND SELF-INDUCTANCE
Any flux change experienced by a circuit, even a change in the magnetic flux produced
by the current flowing in the circuit itself, induces an emf in the circuit. Electromagnetic
forces generated by a circuit‟s own current in this way are called self-induced emfs, and
their generation is referred to as self-induction.
Consider an ideal coil, a toroid, or the central section of a long solenoid. In all these cases
the flux B set up in each turn by a current I is the same for every turn. According to
Faraday‟s law, the induced emf in such coils is
= - dB /dt.
For a given coil, provided no magnetic materials such as iron are nearby, the flux linkage
NB is proportional to the current I, or
158
NB = LI. (7.16)
L is the proportionality constant, and is called the self-inductance or simply inductance of
the device. From Faraday‟s law, the induced emf can be written as
= - dB /dt = - d(LI)/dt = - LdI/dt. (7.17)
According to Eq. 7.17, the induced emf is proportional to the time rate of change of
current in the loop. It is worth noting that a constant current, no matter how large,
produces no induced emf, whereas a current changing rapidly produces a large induced
emf.
The units of self-inductance are
units of L = ( units of ) /( units of dI/dt)
= volts / (ampere/second)
= ohm second = .s
= henry (H).
Self-inductance, like capacitance, depends on geometric factors.
7.4.1 Self-Inductance of a Solenoid
The axial magnetic induction B within the turns of a solenoid is
B = onI,
where I is the current in the coil. In the ideal, long solenoid all the magnetic flux
associated with this field passes through every turn of the coil. The flux through each turn
can be written as the product of the magnetic induction and cross-sectional area A of the
coil as
m = BA = onIA. (7.18)
If the current changes with time, the time rate of change of the flux within the solenoid is
dm /dt = onAdI/dt. (7.19)
The self-induced emf is now obtained from Faraday‟s law as
= - Ndm /dt = - oNnA(dI/dt), (7.20)
159
where N is the total number of turns equal to nl, where l is the length of the solenoid. The
self-induced emf can then be written as
= - on2lA(dI/dt). (7.21)
Comparing Eqs. 7.16 and 7.21, the self-inductance is found to be
L = on2lA. (7.22)
With A and l fixed, L is proportional to the square of the number of turns per unit length.
7.4.2 Self-inductance of a Toroid of a Rectangular Cross Section
Figure 7.11 shows a cross section of a toroid. The lines of the magnetic field B for the
toroid are concentric circles. Applying Ampere‟s law,
∫ B.dl = oI
to a circular path of radius r yields
B(2r) = oIoN,
where N is the number of turns and Io is the current in the toroid windings.
Hence
B = oIoN/ 2r.
The flux B for the cross section of the toroid is
B = B.dS = b
aB(hdr) =
b
a(oIoN/ 2r)hdr
= (oIoNh / 2) b
adr/r = (oIoNh/ 2) ln(b/a), (7.23)
where hdr is the area of the elementary strip shown in the figure. Thus the self inductance
is
L NB /Io = (oN2h/ 2) ln(b/a). (7.24)
160
Fig. 7.11: A cross-section of a toroid.
7.5 LR Series Circuit
To illustrate how an inductor in an electrical circuit influences the current, let us consider
a coil having self-inductance L and resistance R in series with a battery of emf o and a
switch S (Fig 7.12a). The resistance R, representing the resistance of the wire in the
windings of the coil, is shown as a separate element in the circuit, so that the effects
associated with the resistance and the self-inductance can be discussed separately. It is
assumed that all the circuit resistance is accounted for by the resistance R and all the
circuit self-inductance by L.
Fig. 7.12: LR Series Circuit
When the switch S is closed, the current I, starts to flow through the circuit. The initial
value of this current is zero. As it flows through the coil, a magnetic field builds up. Since
the flux in the turns of the coil and the rest of the circuit is time-varying, a self-induced
emf is generated;
L = - LdI /dt.
161
The current I produces a field B within the coil and continually increases, since the
current increases with time. In a time interval t, there is a flux change caused by a field
change B. The self-induced emf tends to set up a current that opposes the one that
actually flows; current flows the way it does because O is larger than L. Instead of
causing a potential rise in the circuit as the circuit is traversed in the direction of the
current flow, it causes a potential drop, L = LdI/dt = VL. Using Kirchoff‟s loop law,
O - LdI/dt - IR = 0. (7.25)
This is a differential equation that can be solved to give the current I in the circuit as a
function of time and thus exhibit the way in which current grows when the switch is
closed.
To solve the equation, both sides must be differentiated with respect to time:
L d2I/dt
2 + RdI/dt = 0. (7.26)
Now, let dI/dt be represented by k. Then dk/dt = d2I/dt
2,
hence
Ldk/dt = - Rk
or
dk/k = -(R/L)dt. (7.27)
At t = 0, the current I = 0. Hence at t = 0,
dI/dt = o /L. (7.28)
Integrating Eq. 7.27, we have
dk/k = - (R/L) dt
and kL/o = e-Rt/L
or
k = dI/dt = (o /L)e-Rt/L
. (7.29)
Now integrating from t = 0 when I = 0 to time t when I = I, Eq. 7.30 becomes
dI = (o /L) e-Rt/Ldt,
or
I = (o /R)[1 - e-Rt/L
]. (7.30)
162
The potential difference across the inductor is
VL = -LdI/dt = -o e-Rt/L
= - o e-t/
, (7.31)
where =L/R has the dimensions of time and is referred to as the inductive time constant
of the circuit. For circuits in which the inductance is large and the resistance small, this
time constant is appreciable. The minus sign in Eq. 7.32 reflects the fact that the current
experiences a fall in potential in traversing the inductance.
Now, suppose that the current has after a long time attained the asymptotic value o /R.
The circuit is then open, the battery suddenly removed and then the switch is closed, (as
in Fig. 7.12b). There is no external energy source to sustain the current in the circuit or
the magnetic field it produces, so that the current starts to decrease, and the magnetic flux
enclosed by the turns of the inductor (or by the circuit itself) starts to decrease. As a
result, an emf is induced. The induced emf is generated by the field change which
produces a flux change in the inductor. The emf tends to sustain the current flow through
the inductor and through the rest of the circuit, so long as the current continues to vary
with time. The potential difference contributed by the emf is VL = - L dI/dt. Here dI/dt is
negative.
According to Kirchoff‟s loop law, we have
- LdI/dt - IR = 0 (7.32)
which can be rearranged to give
dI/I = - (R/L)dt. (7.33)
Eq. 7.34 can be integrated from t = 0, at which time the current I = o /R to the time t,
when I = I. Thus
dI/I = - (R/L) dt
or ln (IR /o) = -Rt /L. (7.34)
Therefore,
I = (o /R)e-Rt/L
. (7.35)
The current falls exponentially from o /R to zero. In a time = L/R, the current declines
to 1/e of its initial value.
163
Example 7.6
The current in a coil of wire is changed uniformly according to the relation
I = 0.10t
where t is the time. Experimental measurements show that an induced emf of 1.3 x 10-4
V
is produced across the leads to the coil. The induced emf of 1.3 x 10-4
V is in addition to
any potential difference across the leads that is due to electrical resistance of the coil.
Determine the self-inductance of the coil.
Solution:
From the relation
I = 0.10t,
we have dI/dt = 0.10
Now, given that = - 1.3 x 10-4
V, the self-inductance is
L = - /(dI/dt) = - (- 1.3 x 10-4
/0.10)
= 1.3 x 10-3
H.
Example 7.7
An inductor of inductance 3 H and resistance 6 is connected to the terminals of a
battery of emf 12V and of negligible internal resistance. Determine
(a) the initial rate of increase of current in the circuit,
(b) the rate of increase of current at the instant when the current is 1 A, and
(c) the instantaneous current 0.2 s after the circuit is closed.
Solution:
(a) Using Kirchoff‟s voltage law around the circuit,
o = IR + L(dI/dt)
or dI/dt = o /L - (R/L)I.
164
At t = 0, the initial current is zero. Hence the initial rate of increase of current is
dI/dt = o /L = 12/3 = 4 As-1
.
(b) When I = 1 A,
dI/dt = 12/3 - 6 x 1/3 = 2 As-1
(c) From Eq. 7.31, the current at any time is
I = (o /R) [1 - e-Rt/L
)
At t = 0.2, I = (12/6)[1 - e-6x0.2/3
]
= 0.65 A.
7.6 ENERGY STORED IN INDUCTIVE CIRCUITS
Let us return to the situation illustrated in Fig. 7.12b, where the magnetic field initially
associated with the inductor sustains the flow of current in the circuit for a time as the
magnetic flux decays. The current that flows during this time produces a certain amount
of thermal energy in Joule heating as it passes through the circuit resistance R. The
thermal energy was caused by the passage of current through a resistance and the emf
that caused the current to flow was generated by the change in magnetic flux through the
turns of the inductor as the magnetic field in its neighbourhood died away. The thermal
energy created by Joule heating, therefore, must have been magnetically stored in the
inductor‟s magnetic field.
From the law of conservation of energy, the sum of the magnetic and thermal energies
must always be equal. Therefore, if we denote the magnetic field energy by Um and the
thermal energy by Q and equate magnetic plus thermal energy at time t and t + t,
between which changes dUm and dQ take place, we have
Um + Q = Um + dUm + Q + dQ
or - dUm = dQ. (7.36)
Hence, - dUm /dt = dQ /dt = I2R (7.37)
recalling that the rate at which electrical energy is converted to heat is I2R.
From Eq. 7.33, we find that for this system,
LdI/dt + IR = 0.
165
Multiplying through by I and noting that IdI/dt = d(I2/2)/dt, this becomes
LIdI/dt + I2R = 0,
or - d(LI2) /dt = I
2R = dQ/dt = -dUm /dt. (7.38)
Thus, the energy in the magnetic field of an inductor is
Um = LI2/2. (7.39)
This result has a general applicability, although it was derived by considering a particular
process.
The energy density or energy per unit volume is an important parameter and can be
calculated. We refer to the example of the long. straight solenoid wound with n turns per
unit length in the interior. The current in the solenoid is
I = B/on. (7.40)
The inductance of the solenoid is given by Eq. 7.21 as
L = on2lA.
Hence the magnetic energy associated with the solenoid is
Um = LI2/2 =(1/2)(on
2lA)(B/on)
2
= lAB2/ 2o . (7.41)
The energy density or energy per unit volume associated with the magnetic field is
um = Um/Al = B2/ 2o . (7.42)
This relation, although derived for a solenoid, is also valid for all magnetic fields.
Example 7.8
A large research electromagnet has parallel circular pole faces that are 0.3 m in diameter
and are spaced 0.05 m apart. The electromagnet can produce a maximum field of 1.2 T
that is essentially constant within the cylindrical region between the poles.
(a) How much energy is stored in the magnetic field in this region under these
conditions?
(b) How long could a 60-W bulb be illuminated at rated power by this field energy?
166
Solution:
(a) The total energy in the volume V of the cylindrical region between the poles is
Um = B2V/2o = B
2r2h/ 2o,
where r is the radius of the pole face and h the spacing between the faces. Substituting the
given values, we get
Um = (1.2)2 (0.15)
2(0.05)/[2 x 4 x 10
-7]
= 2.025 x 103 J.
(b) A 60-W bulb consumes 60 J of energy per second. Hence time to consume 2025 J is
t = 2.025 x 103/60 = 33.75 s.
7.7 MUTUAL INDUCTION
If a conductor carrying a time-dependent current is near some other conductor, then the
changing magnetic field of the former can induce an emf in the latter.
Consider two coils X and Y which are close to each other and carrying time-dependent
currents IX and IY, respectively. The current IX generates a magnetic field. The changing
magnetic flux BX through the coil Y induces an emf in this coil. The emf in coil Y is
Y = - dBX/dt. (7.43)
The flux BY depends on the strength of the magnetic field BX in the coil Y produced by
the current IX in the coil X. This field strength is directly proportional to IX; hence the
flux BX is also proportional to IX. Thus,
BX = LYXIX. (7.44)
Equation 7.45 then becomes
Y = - LYX dIX /dt. (7.45)
Here LYX is a constant of proportionality which depends on the size of the coils, their
distance, and the number of turns in each. This constant is called the mutual inductance
of the coils.
Equation 7.47 states that the emf induced in coil Y is proportional to the time rate of
change of current in coil X. The converse is also true, that is, if coil Y carries a current IY,
then the emf induced in coil X is proportional to the time rate of change of the current IY,
167
X = - dBY /dt (7.46)
or X = - LXY dIY/dt. (7.47)
It can be shown that the two constants are equal, that is
LXY = LYX = M. (7.48)
Generally, LYX and LXY are defined by
LXY = dXY /dIY, LYX = dYX /dIX . (7.49)
The mutual induction coefficients can also be expressed in terms of the self-inductances
LX and LY :
LXY = LYX = (LXLY)
½. (7.50)
The principle of mutual induction finds application in transformers. It is often desirable
to change the relative magnitudes of currents and emfs in a system. For instance, the
generation of electrical power is most conveniently effected in a few thousand volts with
large currents. On the other hand, in the transmission of power, especially over long
distances, high voltages and small currents are used to avoid excessive ohmic losses, for
the latter are of the form I2R and so diminish rapidly as the current is decreased. Finally,
devices using power, such as motors, are usually designed for a few hundred volts.
The transmission of currents and emf of the sort indicated are readily made by means of
transformers, which essentially consists of two coils wound on a laminated iron core, so
arranged that the magnetic coupling between them is as nearly perfect as possible, as
shown in Fig. 7.13a. The iron increases the strength of the magnetic field in its interior by
a large factor. Since the field is much stronger inside the iron than outside, the field lines
must concentrate inside the iron. Thus the iron tends to keep the field lines together and
acts as a conduit for the field lines from one coil to the other.
Fig. 7.13: A transformer
Each coil is part of a separate electric circuit (Fig. 7.13b). Consider the transformer to be
ideal, (i) in which there are no energy losses and no leakage of the flux, that is, all the
168
flux links every turn of both primary and secondary, and (ii) in which the number of turns
are so large that the self-inductances of both windings as well as the mutual inductances
between them are effectively infinite. Power is delivered to the primary winding and
withdrawn from the secondary, being transferred from one to the other by
electromagnetic induction. As all the flux links all the turns of both windings, an induced
emf in one is always accompanied by an induced emf in the other, and the amplitudes of
the emfs are in the same ratio as the numbers of turns in the windings. Further, at any
finite frequency the induced emfs can be finite only if the total flux in the core is small,
since by hypothesis the numbers of turns in the windings are very large. Hence a current
in the secondary always accompanies a current in the primary such that the resulting
fluxes are substantially equal and opposite. This requires that the amplitudes of the
currents be in the inverse ratio of the numbers in the windings.
Now, in Fig. 7.13b, the emf 1 of the source must be equal to the induced emf 1,ind
across the primary coil. By Faraday‟s law, the induced emf equals the time rate of change
of flux,
1 = 1,ind = - d1/dt. (7.51)
Similarly, the emf 2 delivered to the load must equal the induced emf 2,ind in the
secondary coil, which, in turn, equals the time rate of change of flux in that coil,
2 = 2,ind = - d2 /dt. (7.52)
Since the magnetic field lines pass through both coils, the fluxes and their time rates of
change are in the ratio N2 /N1, where N2 is the number of turns in the secondary and N1
that in the primary,
d2 /dt = (N2 /N1) d1/dt. (7.53)
Therefore,
2 = - (N2 /N1)d1/dt = (N2 /N1)1 . (7.54)
If N2 > N1, we have a step-up transformer, and if N2 < N1, a step-down transformer.
As long as the secondary circuit is open and carries no current (I2 = 0), an ideal
transformer does not consume electric power. Under these conditions, the primary circuit
consists of nothing but the source of emf and an inductor, that is, it is a pure L circuit. In
such a circuit the power delivered by the source of emf to the inductor averages to zero.
If the secondary corcuit is closed, a current will flow (I2 0). This current contributes to
the magnetic flux in the transformer and induces a current in the primary circuit. The
current in the latter is then different from that in a pure L circuit, and the power is not
169
zero over a cycle. In an ideal transformer, the electric power that the primary circuit takes
from the source of emf exactly matches the power that the secondary circuit delivers to
the load, and thus,
I11 = I22 . (7.55)
Good transformers approach this ideal condition fairly closely; about 99 percent of the
power supplied to the input terminals emerges at the output terminals. The difference is
lost as heat in the iron core and in the windings.
Transformers play a large role in the electric technology. As stated earlier, transmission
lines for electric power operate much more efficiently at high voltage since this reduces
the Joule losses. To take advantage of this high efficiency, power lines are made to
operate several kilovolts. The voltage must be stepped up to this value at the power plant,
and, for safety sake, it must be stepped down just before it reaches the consumer. For
these transformations, banks of large transformers are used at both ends of the
transmission line. Transformers are also widely used in TV receivers, computers, X-ray
machines, and so on, where high voltages are required.
Example 7.9
A long solenoid has n turns per unit length. A ring of wire of radius r is placed within the
solenoid, perpendicular to the axis. What is the mutual inductance of ring and solenoid?
Solution:
If the current in the solenoid windings is I1, the magnetic field B1 = onI1.
Hence, flux through the ring is
B1 = r2onI1.
Comparing with Eq. 7.46, we find
L12 = r2on.
Example 7.10
Door bells and buzzers usually are designed for 12-V ac, and they are powered by small
transformers which step down 220-V ac to 12-V. Assuming that such a transformer has a
primary winding with 1500 turns, calculate the number of turns in the secondary
windings.
170
Solution:
Equation 7.56 applies to the instantaneous voltage. It is therefore also valid for the rms
voltages. Thus,
N2 = N12 /1
= 1500 x 12/220
= 82 turns.
7.8 OSCILLATIONS IN CIRCUITS CONTAINING A CAPACITOR
AND AN INDUCTOR
7.8.1 LC Circuit
A charged capacitor connected to an inductor is an interesting and useful configuration.
The inductor provides a conducting path, which allows the capacitor to discharge. By
discharging, the capacitor converts the electrical energy stored in its electric field to
magnetic energy in the inductor. Once this transfer of energy is complete, the magnetic
field begins to diminish, and the magnetic energy is transformed back to electrical energy
in the capacitor. In the absence of a resistor, this back-and-forth transfer of energy
continues indefinitely.
Let us consider a charged capacitor which is connected through a switch to an inductor
having zero electrical resistance, as shown in Fig. 7.14. Charge flows from the capacitor
when the switch is closed. The resulting current produces in the inductor a magnetic field
and an induced emf that opposes the action of the capacitor. Applying Kirchoff‟s voltage
law, we have
q/C = - LdI/dt. (7.56)
The charge on the capacitor decreases so that
I = - dq/dt.
Therefore, Eq. 7.58 becomes
q/C + L(d2q/dt
2) = 0. (7.57)
If q0 is the initial charge on the capacitor, we can write the solution of Eq. 7.59 as
q = qocos t, (7.58)
171
with = 1/(LC). The frequency = /2 is called the natural frequency of the
oscillation. This is the oscillation frequency when there is no electrical resistance in the
circuit.
Fig. 7.14: LC Circuit
From Eq. 7.60, we get
I = -dq/dt = qosin t. (7.59)
The initial energy of the capacitor is Uo = (qo2/2C). At any instant thereafter its electrical
energy is
UE = q2/2C =(qo
2/2C)cos
2t
= Uo cos t.
If the initial magnetic energy is zero, the magnetic energy stored by the inductor at any
instant is
Um = LI2/2 = (L/2)2
q2
o sin2 t.
Since 2 = 1/LC,
we have Um = (qo2/2C)sin
2 t
= Uo sin2 t.
The total energy at any instant then, is
Utotal = UE + Um
= Uo (sin2 t + cos
2 t)
= Uo .
Thus, the total energy is conserved.
172
7.8.2 LCR Circuit
An LCR circuit consists of a charged capacitor connected to a resistor and an
inductor (Fig. 7.15).
Fig. 7.15: An LCR Circuit.
Thermal energy is produced in the resistor, thereby diminishing the electric and magnetic
energy in the circuit. Eventually the back-and-forth transfer of energy between the
capacitor and the inductor vanishes as all the available energy is converted to thermal
energy.
Applying Kirchoff‟s loop law, we have
q/C - L(dI/dt) -IR = 0. (7.60)
Using I = - dq/dt, we obtain
L(d2q/dt
2) + R(dq/dt) + q/C = 0. (7.61)
This equation represents charge on the capacitor as a function of time.
Assuming that q = qo at t = 0, we have as a solution of Eq. 7.63, for the condition
(1/LC) > (R/2L)2,
q = qo e-Rt/2L
cos t, (7.62)
where
= [(1/LC) -(R/2L)2].
The exponential term continually reduces the amplitude of the oscillations until they
eventually die out (Figure 7.16a). Increasing the resistance makes the damping of the
oscillations more severe and reduces the frequency of the oscillations. At a critical value
of R such that (R/2L)2 = 1/LC, the circuit no longer oscillates, and the capacitor simply
discharges converting its electrical energy directly to thermal energy. The circuit is said
to be critically damped. When (R/2L)2 > 1/LC the circuit does not oscillate and is said to
be overdamped (Fig. 16b).
173
The use of combinations of capacitors and inductors is widespread. They are used to
generate controlled oscillations in a variety of electronic circuits. Tuning circuits such as
those in radios and television sets are often LCR circuits.
174
CHAPTER EIGHT
MAGNETIC PROPERTIES OF MATTER
8.0 MACROSCOPIC MAGNETIC PROPERTIES OF MATTER
The macroscopic magnetic properties of matter are due to the following atomic
properties:
1. An atom is made up of a number of charged particles in constant motion.
Electrons orbit round the nucleus continually whilst within the nucleus protons
orbit round each other.
2. These two forms of orbital motion may be considered flowing electric currents.
These electric currents generate corresponding magnetic fields.
3. In addition to the orbital motions the charged particles within the atom also rotate
(or spin) about their axes. Thus, the electrons, protons and neutrons all spin about
their axes. These spin motions may also be regarded as flowing electric currents
which generate magnetic fields.
4. The magnetic fields arising from the currents flowing in loops within the atom,
nuclei and atomic particles can be described in terms of their corresponding
magnetic dipole moments.
5. These small magnetic dipole moments within a material sample can be aligned to
produce a strong magnetic field especially in the presence of an external
magnetic field e.g. from an electromagnet.
6. The strength of the magnetic field produced, however, depends on how readily the
atomic and subatomic dipoles respond to the external magnetic field.
7. Depending on their magnetic response, materials may be put into the following
categories:
i. Diamagnetic Materials
ii. Paramagnetic Material
iii. Ferromagnetic Materials
8.1 ATOMIC AND NUCLEAR MAGNETIC MOMENTS
An electron moving in an orbit around a nucleus produces an average current along its
orbit. If the electron has a circular orbit with radius r and speed v, then the time taken for
one complete circular motion is 2r/v. The charge moved in this time is e
Let us consider the simple case of an electron e in a circular orbit of radius r around the
nucleus with a speed v. The time for one complete cycle (called periodic time T) is given
by T = 2r/v.
175
Let I = current generated along the orbit, then,
I = charge/time = e/(2r/v) = ev/2r (8.1)
The circulating current I (8.1) will give rise to an orbital magnetic dipole moment m.
m = I (area of orbit)
= ev/2r r2 = evr/2 (8.2)
Let me = mass of the orbiting electron. If L is the angular momentum of the orbiting
electron, then, (8.2) is written as
mem = meevr/2 but L = mevr
m = (e/2me).L (8.3)
m is called the orbital magnetic dipole moment and L refers to the orbital angular
momentum of the orbiting electron.
The orbital magnetic dipole moment for an orbiting electric charge is thus proportional to
the orbital angular momentum.
Equation 8.3 is also valid for periodic orbits and can be obtained using quantum
mechanics. The net magnetic moments of the atoms is the sum of the magnetic moments
of all its electrons. Hence Eq. 8.3 can also be regarded as a relation between the net
orbital angular momentum and the net magnetic moment of the entire atom.
The magnitude of the orbital angular momentum is always some integer multiple of the
constant . Thus, the possible values of the orbital angular momentum are
L = 0, , 2, 3, 4, ......... (8.4)
Because angular momentum exists only in discrete packets, it is said to be quantised. The
constant is the fundamental quantum of the angular momentum, just as e is the quantum
of electric charge.
Besides the magnetic moment generated by the orbital motion of the electrons, we must
also take into account that generated by the rotational motion of the electrons. An
electron may be thought of as a small ball of negative charge rotating about an axis at a
fixed rate. This kind of rotational motion again involves circulating charge and gives the
electron a magnetic moment. The net magnetic moment of the atom is obtained by
combining the orbital and spin moments of all the electrons, taking into account the
directions of these moments.
176
The nucleus of the atom also has a magnetic moment. This is due to (i) the orbital motion
of the protons inside the nucleus, and (ii) the rotational motion of individual protons and
neutrons. The magnetic moment of a proton or neutron is small compared with that of an
electron, and in reckoning the total magnetic moment of an atom, the nucleus can usually
be neglected.
8.2 CLASSIFICATION OF MAGNETIC MATERIALS
We have already seen that a current I in a loop of wire of cross-sectional area A produces
a magnetic dipole
= IA.
The direction of is perpendicular to the plane defined by A. A long solenoid is
equivalent to N closely spaced identical loops of wire, each having the same current I.
Each loop in the solenoid produces an identical magnetic dipole moment IA. The dipole
moment is aligned along the axis of the solenoid. Thus, for a solenoid with N turns, the
net magnetic dipole moment is
= NIA. (8.5)
If the solenoid is placed in a magnetic field, then the solenoid experiences a torque given
by
= x B. (8.6)
In a vacuum, the magnetic field inside a long solenoid is
BE = onI, (8.7)
which may be rewritten as
BE = oNIA/Al,
using n = N/l. Thus, using Eq. 8.5, we have
BE = o /V (8.8)
where V = Al is the volume contained within the windings of the solenoid. If N and A are
known, a measurement of the current I determines the magnetic moment . The magnetic
moment changes with changing current. When the inside of the solenoid is filled with a
material and a magnetic dipole moment is then induced, the magnetic dipole moment of
the solenoid changes. That is, when the interior of the solenoid is filled with some
material, the induced magnetic moments produce an additional contribution to the
magnetic moment. This additional contribution may be denoted as i. Thus,
177
B = o( + i)/V. (8.9)
It is found experimentally that the induced magnetic moment depends on the current in
the solenoid. This dependence can be written as
i = m , (8.10)
where m is the magnetic susceptibility. Depending on the type of material inside the
solenoid, m may be constant (at a particular temperature) or may depend on the current I.
In terms of m, Eq. 8.9 can be written as
B = o(1 + m)/V (8.11)
where is determined from the characteristics of the solenoid and the current in the
solenoid. The magnitude of B is determined from measurements of magnetic flux. Thus
the magnetic susceptibility is determined from the equation
m = (BV/o) - 1. (8.12)
The three main classes of magnetic materials can be described in terms of measurements
of the magnetic susceptibility.
(a) Diamagnetic materials interact weakly with an imposed magnetic field, weaken
the existing magnetic field, and have negative values of m. The magnetic
susceptibility is essentially independent of temperature and solenoid current.
(b) Paramagnetic materials interact weakly with the imposed magnetic field,
strengthen the existing magnetic field, and have positive values of m. m depends
on temperature and is essentially independent of solenoid current.
(c) Ferromagnetic materials interact strongly with an imposed magnetic field,
strengthen the existing magnetic field, and have magnetic susceptibilities that
depend sensitively on the solenoid current.
Now, 1 + m is the relative permeability and is denoted by m. That is,
m = 1 + m. (8.13)
Equation 8.11 then becomes
B = om/V.
om is called the permeability.
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8.3 DIAMAGNETISM
A change in magnetic field lines threading a current loop causes a current to be induced
in the loop. The magnetic flux produced by the induced current always acts to oppose the
change.
Whenever a material is subjected to a magnetic field, magnetic field lines thread the path
of electrons, and the currents and magnetic dipole moments created by the circulating
electrons change. These changes oppose the action of the applied magnetic field, and the
induced magnetic moments orient oppositely to the applied magnetic field, in accordance
with Lenz‟s law. Thus the induced magnetic moments reduce the strength of the applied
magnetic field. In terms of Eq. 8.9, the induced magnetic moment (i) and the magnetic
moment due to the current () have opposite directions. From Eq. 8.10, it follows that m
is negative. Diamagnetism is a property of all materials, but it is a very weak property and is
observed in materials made of atoms that have permanent magnetic dipole moments.
When a diamagnetic material is placed in a magnetic field B, the force experienced by the
electron is -ev x B in addition to the usual electric force within the atom. Assume that the
nucleus produces an electric field E. Then the net force on the electron is -eE - ev x B. To
keep the electron in a circular orbit of radius r,
eE + evB = mev2/r. (8.14)
Using = v/r, Eq. 8.14 can be written as
eE + erB = me2r. (8.15)
In the absence of the magnetic field,
eE = meo2r. (8.16)
Subtracting Eq. 8.16 from Eq. 8.15, we have
eB = me(2 - o
2). (8.17)
The increment of frequency is
= - o. (8.18)
For small magnetic field, is small compared with o. Hence
2 - o
2 = (o + )
2 - o
2
= 2o +()2 2o. (8.19)
Eq. 8.17 then becomes
179
eB 2meo . (8.20)
and o are nearly equal. Thus Eq. 8.20 becomes
= eB/2me. (8.21)
This frequency is called the Lamor frequency; it tells how much faster the electron moves
around its orbit as a result of the presence of the magnetic field.
There is a change in the orbital magnetic moment corresponding to the change in the
orbital frequency. From Eq. 8.2,
= evr/2 = er2o/2.
Hence,
= (er2/2).
Thus the fractional change in the magnetic moment is
/ = /o. (8.22)
Typically, the frequency of an electron in an atom is o 1016
s-1
. If the magnetic field is
B = 1.5 T, then
= eB/2me = 1.6 x 10-19
x 1.5/(2 x 9.1 x 10-31
)
= 1.32 x 1011
s-1
.
As a result,
/ = /o = 1.32 x 1011
/1016
= 1.32 x 10-5
10-5
.
That is, the magnetic moment changes by about 1 part in 105. This gives an indication of
the small size of the diamagnetic effect.
8.4 PARAMAGNETISM
A paramagnetic material is composed of a uniform distribution of atomic magnetic
dipoles sufficiently separated so that the magnetic field of any given dipole does not
influence any of its neighbours. In the absence of magnetic field, the dipoles are
randomly oriented as a result of thermal motions. The net magnetic moment of a
paramagnetic material is, therefore, zero. However, when an external magnetic field is
applied, the dipoles align themselves with the field and produce a net magnetic moment
in the material. This alignment is not perfect, because of the disturbance caused by
180
random thermal motions. But even a partial alignment of the dipoles have an effect on the
magnetic field. The material becomes magnetized and contributes an extra magnetic field
that enhances the original magnetic field. Magnetic alignment can be achieved in two
ways: (i) by lowering the temperature of the specimen or (ii) by increasing the applied
magnetic field.
How does such an increase of magnetic field come about? Consider a piece of
paramagnetic material placed between the poles of an electromagnet. Figure 8.1 shows
the alignment of the magnetic dipoles in such a material.
Fig. 8.1: A piece of paramagnetic material in an electromagnet.
For the sake of simplicity, Fig. 8.1b shows a case of perfect alignment. The magnetic
dipoles are due to small current loops within the atoms. Figure 8.2a shows the alignment
of current loops. Now look at any point inside the material where two of these current
loops (almost) touch. The currents at this point are opposite and cancel. Thus, everywhere
inside the material, the current is effectively zero. However, at the surface of the material,
the current does not cancel. The net result of the alignment current loops is therefore a
current running along the surface of the magnetized material (Fig. 8.2b). The material
consequently behaves like a solenoid; it produces an extra magnetic field in its interior.
This extra magnetic field has the same direction as the original, external magnetic field.
Hence, the total magnetic field in a paramagnetic material is larger than the original
magnetic field produced by the currents of the electromagnet.
The alignment of atomic dipole moment in a paramagnetic specimen enhances the
magnetic dipole moment, and the magnetic field increases. It follows that m is positive.
Fig. 8.2: Alignment of current loops.
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8.5 FERROMAGNETISM
Ferromagnetism is exhibited by five elements - iron (Fe), nickel (Ni), cobalt (Co),
dysprosium (Dy), and gadolinium (Gd) - and some alloys, which usually contain one or
more of these five elements.
The intense magnetization in ferromagnetic materials is due to a strong alignment of the
spin magnetic moments of electrons. In these materials, there exists a special force that
couples the spins of the electrons in adjacent atoms in the crystal. This force (known as
exchange coupling) couples magnetic moments of adjacent atoms together in rigid
parallelism.
Since this special spin-spin force is fairly strong, we must ask, why is it that
ferromagnetic materials are ever found in nonmagnetized state? Why is it that not every
piece of iron is a permanent magnet? The answer is that, on a small scale, ferromagnetic
materials are always magnetized. There are regions in every ferromagnetic specimen that
have near perfect alignment of magnetic dipole moments even when there is no applied
magnetic field. These regions are called magnetic domains. The direction of alignment of
the dipoles varies from one domain to the next (Fig. 8.3). Hence on a large scale, there is
no discernible alignment, because the domains are oriented at random.
The formation of domains results from the tendency of the material to settle into a state of
least energy (equilibrium state). The state of least energy for the spins would be a state of
complete alignment. But such a complete alignment would generate a large magnetic
field around the material. Energetically, this is an unstable configuration. The domain
arrangement is a compromise - the magnetic energy is then small because there is little
magnetic field, and the spin-spin energy is then also reasonably small because most
adjacent spins are aligned.
Fig. 8.3: Magnetic domains in a ferromagnetic material.
182
However, if the material is immersed in an external magnetic field, all dipoles tend to
align along this field. The domains then change in two ways:
1. Those domains with magnetic dipole moments parallel to the magnetic field grow
at the expense of the neighbouring domains (Fig. 8.3). This effect is responsible
for producing a net magnetic dipole moment in a weak applied magnetic field.
2. The magnetic dipole moments of the domains rotate toward alignment with the
applied magnetic field. This is the mechanism of magnetic dipole alignment when
the applied magnetic field is strong.
If all the magnetic dipoles in a piece of ferromagnetic material align, their contribution to
the magnetic field will be very large. Let us consider measurements of the magnetic field
(B) in an iron specimen in a solenoid. Let us assume that initially the specimen is
unmagnetized. It is seen in Fig. 8.4a that as the current increases, the magnetic field (B)
also increases. The field, however, tends to saturate. The non-linear relationship between
B and BE( =onI) means that the magnetic susceptibility is not constant. It depends on BE.
In Fig. 8.4b, it is seen that if BE is decreased from the value labelled b, the magnetic field
measurements are consistently higher than when the current was increased from the zero
value. A remnant magnetic field, Br, persists even when the current in the solenoid
vanishes. The remaining magnetic field results from the alignment of magnetic domains.
This is one way to produce a permanent magnet.
Fig. 8.4: Measurement of magnetic field in an iron specimen in a solenoid.
If the direction of the current in the solenoid is reversed, the magnetic field (B) within the
specimen is reduced steadily from the remanent value Br, as shown in Fig. 8.4c. At a
critical value of BE, called the coercive force (Bc), the magnetic field is zero. The larger
the coercive force, the more difficult it is to demagnetize a ferromagnetic specimen.
Ferromagnetic materials having a large coercive force are said to be magnetically
“hard”, those having a small coercive force are said to be magnetically “soft”. Ordinary
183
iron is magnetically soft and has a coercive force of about 10-4
T. A hard magnetic
material used in the speaker of a high-fidelity system may have a coercive force 20-50
times that of ordinary iron.
In Fig. 8.4d it is found that if the same direction of the current is kept and the magnitude
increased, the magnetic field increases. However, the direction is now reversed from the
starting direction. Again, the magnetic field tends to saturate, attaining the magnitude
labelled c. The saturation indicates that the alignment of the magnetic domains
approaches completion.
When the current is reduced to zero again, the magnetic field intensity decreases. When
the direction of the current is then reversed, the magnetic field is brought back to its
initial saturation, labelled b. The lack of retraceability shown in Fig. 8.4 is called
hysteresis. A closed curve representing measurements of B and BE is called a hysteresis
loop. As a ferromagnetic specimen is cycled around a hysteresis loop, irreversible
changes occur in its domain structure. Work is done by the magnetizing field in order to
alter the domains, and the temperature of the specimen increases. The greater the
hysteresis loop, the greater the amount of work required and the larger the temperature
rise in the specimen.
Many practical electromagnetic devices utilize one or more coils of wire wound around a
ferromagnetic medium. The current in the coil often varies cyclically in direction and
magnitude. This means that the ferromagnetic medium is continually cycled through a
hysteresis loop; energy is lost in each cycle. In order to minimize the energy loss,
materials having a small loop area is used. This condition is generally satisfied by soft
materials. Iron is magnetically soft and is widely used in electromagnetic devices such as
transformers. New alloys with superior magnetic properties have been developed.
Magnetically hard materials are used as permanent magnets. For example, magnets in
stereo speakers, magnets in some types of electric motors, and magnets in mechanical
ammeters and voltmeters are made from “hard” materials. Some alloys which consist
mostly of nickel, iron, cobalt and aluminium, also make good permanent magnets.
“Hard” materials are characterized by broad hysteresis loops. They are hard to magnetize
and hard to demagnetize.
8.6 MAGNETIC DOMAINS
On an atomic scale unmagnetised ferromagnetic materials in a way possess some
magnetism. Their ions possess unpaired electron spins which are strongly coupled
together by the quantum mechanical force called exchange coupling. As a result of this a
ferromagnetic material consists of a large number of groups of atoms or magnetic
domains. In each domain all the electron spins are perfectly aligned to form a fairly large
magnetisation vector for the domain. However, the direction of alignment varies from
one domain to another. In each domain the electron spins are all aligned but due to the
random orientation of the domains a net zero magnetisation results.
184
Fig. Magnetic domains
When the material is placed in an external magnetic field all the magnetic dipoles tend to
align themselves in the direction of the external field.
185
CHAPTER NINE
A.C. THEORY
We recall that for a coil of N turns rotating in a magnetic field of magnetic flux density
B, the magnetic flux is given by = BANcos where A = surface area of the coil
and = angle between the field B and the normal to the surface.
Fig. 9.1: A coil rotating in a magnetic field.
Now the induced emf E = – d/dt = –d/dt(BANcos) = –d/dt(BANcost)
where = t.
E = – BAN (– sint ) = BAN (sint ) (9.1)
Let the maximum emf be Emax. This is the peak voltage and occurs when = t =
90o.
Emax = BAN (9.2)
Equations (6.1) and (6.2) give
E = Emax sin t (9.3)
Likewise,
i = imax sin t (9.4)
Equations (6.3) and (6.4) represent the sinusoidal wave for an a.c.
Writing = 2f in (6.3) gives
E = Emax sin 2f
= Emax sin (21/T) t where f = 1/T. T = periodic time
186
9.1 ROOT MEAN SQUARE VALUES OF ALTERNATING
VOLTAGE AND CURRENT
The effective value or root-mean-square (rms) value of an alternating current is the value
of the direct current which produces in the same conductor the same amount of heat in
the same time.
Let us assume that a direct current of magnitude id is passed through a resistance R. The
rate of heat dissipated = id2R. If by passing a certain alternating current i through the
same resistance R the rate iof heat dissipated exactly equals that of the direct current id ,
then, the alternating current i is said to have a root mean square (rms) value equal to Id.
Instantaneous power = i2R = average power
= ( i12
R + i22R
+ i3
2R
+ i4
2R
+ ….. in
2R )/n = id
2R
= R( i12
+ i22 + i3
2 + i4
2 + ….. in
2 )/n = id
2R
= ( i12 + i2
2 + i3
2 + i4
2 + ….. in
2 )/n = id
2 (9.5)
But the quantity in bracket i.e. ( i12
+ i22 + i3
2 + i4
2 + ….. in
2 )/n is the average or mean
of the squares of the alternating current i.e. i2. Thus from (6.5) we write
i2 = id
2 = I
2rms.
Irms = i2 = (Average of the squares of the instantaneous currents) = Irms
Likewise,
Vrms = V2 = (Average of the squares of the instantaneous voltages) =
Vrms
9.2 RELATIONSHIP BETWEEN THE IRMS, VRMS, AND THE PEAK
CURRENTS AND PEAK VOLTAGES.
For an alternating current,
i = imax sin t
Irms2R = (Average of i
2)R = (Average of imax
2 sin
2 t )R
Irms2 = (Average of i
2) = (Average of imax
2 sin
2 t )
187
But average value of imax2 sin
2 t is given by integrating this value w.r.t. time from t = 0
to t = T where T = period and dividing the resultant by T.
i.e. Irms2 = Average of imax
2 sin
2 t = ( imax
2
T sin
2 t)/ T
= ( imax2
2/ sin
2 t)dt / 2/ (9.6)
where T = 2/ = period = time for 1 cycle.
Using the relation sin2 = ½ (1 - cos2), (9.6) becomes
Irms2 = imax
2
2/ ½ (1 – cos2t) / 2/dt
= imax2. /4
2/ (1 – cos2t)dt
= imax2. /4 [
2/dt –
2/ cos2tdt]
= imax2. /4(2/) – [ (sin 2t)/2]
2/
= ½ . imax2
Irms = imax /2
Likewise,
Vrms = Vmax /2
Example 9.1
The equation of an alternating current is
i = 42.42sin(628t).
Determine
i. its maximum value ii. Its frequency iii. Its rms value
Solution:
Comparing the given equation with the standard equation i = imaxsint = imaxsin2ft
gives
i. imax = 42.42A ii. 2f = 628 f = 628/2 = 100 Hz.
iii. Irms = imax /2 = (42.42 A) /2 = 30 A
Example 9.2
188
What is the equation of a 25 cycle current sine wave having rms value of 30A?
Solution:
Generally, equation i = imaxsint
where imax = 2 Irms = 1.414 30A = 42.42 A.
= 2f = 2 25 = 50. Hence the required equation is i = 42.42 sin50t.
9.2.1 AC CIRCUIT WITH PURE RESISTANCE ONLY
Fig. 9.2: A.C circuit with pure resistance.
For the a.c circuit V = Vmaxsint but V = iR . Vmaxsint = iR.
i = (Vmaxsint)/R but Vmax/R = imax
i = imaxsint
Thus, for a circuit with a pure resistance only,
V = Vmaxsint and
i = imaxsint
This means that the current is exactly in phase with the voltage across the resistance.
9.2.3 GRAPH OF I AND V FOR A CAPACITOR
189
Fig. 9.3: A graph of I and V for a capacitor
Using Imax = Vmax/R and dividing through by 2 gives
(imax )/2 = (Vmax/2)/R
Irms = Vrms/R
9.3 POWER DISSIPATED IN THE RESISTOR OF A PURE
RESISTIVE CIRCUIT
Instantaneous power P = iV where V = Vmaxsint and = imaxsint
P = imax Vmaxsin2t
Average power Pave dissipated over one cycle (i.e. with t = T = 2/ ) is given by
Pave = imax Vmax [ 2/
(sin2 t)/ 2/]dt = ½ imax Vmax
Or Pave = ir2R
where ir is the current passing through the resistor.
9.3.1 A.C. CIRCUIT CONTAINING A CAPACITANCE ONLY
Fig. 9.4: A.C circuit containing only a capacitance.
Let the voltage across the capacitor be V, then,
V = Vmaxsint (9.7)
If q = charge on the capacitor, then, q = CV q = C Vmaxsint.
Now writing i = dq/dt gives
i = d/dt[C Vmaxsint] = (CVmax )cost (9.8)
Comparing eqn (6.8) with the general equation
190
i = imax cost
gives
imax = CVmax (9.9)
Let xc = capacitive reactance, then,
xc = Vmax/imax = 1/C = 1/2fC (with = 2f)
Capacitive reactance refers to the opposition to an a.c. circuit due to a capacitor.
a. POWER DISSIPATED IN A CAPACITANCE
We recall that for the capacitor,
V = Vmaxsint and i = imax cost
Instantaneous power dissipated P = iV
P = imax cost .Vmaxsint = imax Vmax sint cost (9.10)
Using the relation sin2 = 2sincos sincos = ½ sin2, eqn (9.10) gives
P = ½ imax Vmax sin2t
Average power dissipated,
Pave = ½ imax Vmax [ 2/
(sin2t)/ 2/]dt
= ½ imax Vmax/2/ 2/
sin2t dt
but 2/
sin2t dt = 0. Pave = 0
This means that there is no power dissipation in a capacitance. This is because in one half
cycle when the current flows in one direction power is stored in the capacitor. In the other
half cycle when the current direction changes the power stored is released. This results in
no power dissipation.
b. PHASE RELATION BETWEEN CURRENT AND VOLTAGE
V = Vmaxsint (9.11)
i = imax cost
i = imax sin(t + ½) (9.12)
191
Comparing the phases of the voltage and current in Fig.(9.5) we see that the current
leads (i.e. it is ahead of) the voltage across the capacitance by a phase difference of ½ or
said otherwise the voltage lags behind the current by a phase difference of ½.
Fig. 9.5: Phase relation between current and voltage.
Example 9.3
An a.c. circuit with a pure capacitance has a reactance of 8 at 60 Hz. Calculate
i. the current ii. the capacitance of the circuit iii. The reactance when the
frequency is 25 Hz. Assume that the terminals of the capacitance are connected to 110V,
60 Hz supply.
Solution:
xc = 8 , f = 60 Hz, Vrms = 110 V.
i. i = Vrms/xc = 110 V/ 8 = 13.75 A.
ii. xc = 1/C = 1/2fC
C = 1/(2fxc) = 1/(2 . 60 Hz . 8 )
= 332F
iii. When f = 25 Hz,
xC = 1/(2 . 25 Hz .332 x 10–6
F) = 19.17
192
Example 9.4
A potential difference of 100 volts rms is applied to a capacitor of capacitance 20 F.
What must be the frequency if the current through it is 0.628 A?
Solution:
Vrms = 100V, C = 20 x 10–6
F, Irms = 0.628 A
But Irms = Vrms/xc = Vrms/(½ 2 fC) f = Irms/(Vrms.2C)
= (0.628 A)/(2 . 100V . 20 x 10–6
F) = 50 Hz.
c. AC CIRCUIT CONTAINING INDUCTANCE ONLY
Fig. 9.6: A.C circuit with only inductance.
V = applied voltage, E = induced emf. The back emf will oppose at every instant the
variation of current through it and it is equal and opposite to the applied voltage V. i.e. V
= – E.
Let i = imax sint. E = – Ldi/dt = – L . d/dt (imax sint ) = – L imaxcost
But V = – E
V = L imaxcost (6.13)
Comparing (6.13) with the standard equation V = Vmaxcost gives
Vmax = L imax
Vmax/imax = L = 2fL = xL
where xL = inductive reactance = the opposition to an a.c. due to an inductance.
d. POWER DISSIPATED IN AN INDUCTANCE
We recall that for the inductor,
193
V = Vmaxcost and i = imax sint
Instantaneous power dissipated P = iV
P = imax sint .Vmaxcost = imax Vmax sint cost = ½ imax Vmax sin2t
Comparing with our analysis done for the power dissipated in a capacitor we see that the
average power dissipated in a pure inductor = 0.
This is explained as follows: At one time when the current increases the induced emf
produced opposes the increase. In another instance when the current decreases energy s
released to restore the decrease. Hence no energy is dissipated.
e. PHASE RELATION BETWEEN CURRENT AND VOLTAGE
For an inductor,
i = imaxsint
V = Vmax cost
V = Vmax sin(t + ½)
Hence, the voltage always leads the current by a phase difference of ½
f. GRAPH OF I AND V FOR AN INDUCTOR
Fig. 9.7: Graph of I and V for an inductor.
g. REACTANCE
This refers to the opposition to electrical current due to the storage of magnetic or
electrical energy in the circuit.
194
h. INDUCTIVE REACTANCE AND CAPACITIVE REACTANCE
1. xL xc 1/
2. For the same current the voltage in a pure inductive circuit leads the current by 90o
whilst the voltage in a pure capacitive circuit lags behind the current by 90o.
3. For a circuit containing a capacitance and an inductor, xL is considered positive
and xC negative.
4. An aid for remembering the relationship between current (I) and voltage (V) in
both capacitors (C) and inductors (L) is CIVIL; that is for a capacitor (C), current
(I) leads voltage (V) and for an inductor (L) ,voltage (V) leads current(I).
i. AC SERIES CIRCUITS
i. R – L Series Circuit
The total reactance z is found by vector analysis. From the vector diagram above,
V2
= VR2 + VL
2 V = [VR
2 + VL
2]
½
Writing VR = iR, VL = iL gives
V = [(iR)2 + (iL)
2]
½
= i [R2 + (L)
2]
½
i = V / [R2 + (L)
2]
½ = V/[R
2 + xL
2]
½
The quantity [R2 + xL
2]
½ has a unit of resistance and is called impedance (z).
z = [R2 + xL
2]
½
The impedance z is defined as the effective (or resultant) opposition to the flow of an a.c.
in a combination of resistance and reactance.
195
9.3.2 VOLTAGE TRIANGLE & IMPEDANCE TRIANGLE
The vector V leads the vector I by an angle of . The cosine of this angle (i.e. cos) is
called the power factor (pf).
Pf = cos = R/Z
ii. Power dissipated in an R – L Series Circuit
We recall that the average power dissipated in L is zero. Hence for a R–L series circuit all
the power is dissipated in R.
Power P = VI cos
= voltage component of current in phase with V
iii. Active and Reactive Components of a Circuit
The figure above shows the vector diagram of a circuit in which the voltage V leads the
current I by an angle . The current may be resolved into two parts:
1. Icos = voltage-in-phase component.
2. Isin = voltage-out-of-phase component
196
Note:
1. Only the in-phase component Icos determines the power in the circuit.
Hence the component Icos is often called the active or in-phase or energy
component.
2. The component Isin is called the reactive, idle, quadrature or wattless
component. The product VIsin is called the wattless power or reactive
power.
9.3.3 R – C SERIES CIRCUIT
V = (VR
2 + VC
2)
where VR = IR, VC = I/C or XC = 1/C.
tan = XC/R = (1/C)/R = 1/CR
XC = capacitive reactance.
Note:
1. If the voltage leads the current then, the angle is in the first quadrant and is
considered positive.
2. If the current leads the voltage, then, the angle is in the 4th
quadrant and is considered negative.
Z2
= R2 + XC
2
Z = [R2 + (1/C)
2]
Example 9.5
In a series a.c. circuit consisting of 8 inductive reactance and 10 resistance. A current
of 6 A flows.
i. What is the voltage across each part of the circuit?
ii. What is the voltage across the combination?
197
iii. What is the phase difference between the current and the voltage in the
combination?
Solution:
i. VL = IXL = 6A x 8 = 48 V. VR = 6A x 10 = 60 V.
ii. Z = [R2 + XL
2] = [(6)
2 + (10 )
2] = 12.8
V = IZ = 6 A x 12.8 = 76.8 V
iii. tan = XL/R = 0.8 = tan–1
(0.8) = 38.6o
Example 9.6
A current consisting of 20 resistance in series with 0.1 H is connected across a 50 Hz supply.
The current through the circuit is i = 2 x 5.4sin(3.14t). Determine the voltage across the
resistance, reactance and entire circuit.
Solution:
R = 20, f = 50 Hz, XL = L = 2fL = 2 x 50 Hz x 0.1 H = 31.4
Comparing the given equation
i = 2 x 5.4sin(3.14t)
with the standard equation i = imax sint gives
imax = 2 x 5.4 Irms = Imax/2 = (2 x 5.4)/2 = 5.4 A.
Voltage across R = IR = 5.4A x 20 = 108 V.
Voltage across L = IXL = 5.4 A x 31.4 = 169.6V
Voltage across the entire circuit = V = IZ
V = I(R2 + XL
2) = 5.4 A(20
2 + 3.14
2) = 201 V
Example 9.7
198
An inductive coil is connected to a 200V . 50 Hz a.c. supply with 10 A of current flowing
through it dissipates 1000 watts. Calculate
i. impedance ii. reactance iii. Inductance iv. Power factor v. angle of lag.
Solution:
Power dissipated, P = I2R 1000 = 10
2 R R = 10
i. Z = V/I = 200V/10A = 20
ii. XL = (Z2 – R
2) = (20
2 + 10
2) = 17.32 .
iii. XL = L = 2fL L = XL/2f
= 17.32/2 x 3.14 x 50 Hz = 0.0552H
iv. Power factor cos = R/Z = 10 /20 = 0.5
v. = cos–1
0.5 = 60o
9.3.4 R – L – C SERIES CIRCUIT
VL leads I by a phase of 90o. VC lags behind I by a phase of 90
o. The resultant VL – VC
leads I by 90o assuming that VL > VC.
199
V2
= VR2 + (VL – VC)
2
(IZ)2 = (IR)
2 + (IXL – IXC)
2
Z = [R2 + (XL – XC)
2] = [R
2 + (L – 1/C)
2]
Where Z = impedance
I = V/Z = V/[R2 + (L – 1/C)
2]
I lags behind V by an angle where cos = R/Z = R/[R2 + (L – 1/C)
2]
Note:
1. When XL > XC, then, the net reactance XL –XC is positive. This means that is
positive and I lags behind V or said otherwise V leads I.
2. When XL = XC, then, the net reactance XL –XC = 0 Z = R. This means
that R/Z = 1, cos = 1 or = 0. This implies that the current and the voltage are
in phase.
3. When XL < XC the net reactance XL – XC < 0 is negative. This further
implies that the current I leads the voltage V.
9.3.5 R – L – C SERIES RESONANCE
We recall that for an R –L – C series circuit,
Z = [R2 + (XL – XC)
2] = [R
2 + (L – 1/C)
2]
For various frequencies,
L 1/C
but at a specific angular frequency o (or frequency fo) called the resonance frequency
L = 1/C which gives Z = R. Thus a circuit is said to be in resonance when the
inductive reactance is equal to the capacitive reactance. At resonance,
oL = 1/oC
o2LC = 1 o = (1/LC)
2fo = (1/LC) fo = 1/(2LC)
Now let us consider the relation
I = V/Z
1. Since Z = R gives the minimum value of Z, the current I becomes maximum.
Therefore when a system is in resonance the amplitude is as large as possible.
200
2. At resonance V = VR and since the current is maximum the power dissipated i.e.
I2R is maximum.
3. Hence, when the power dissipated in an ac circuit is maximum then resonance
conditions apply to the circuit.
Example 9.8
An a.c. generator supplies current to a variable capacitor in series with an 0.150 H
inductor having a resistance of 10.0. To what value should the capacitor be adjusted to
give resonance when the frequency of the supply is 50.0 Hz. If the supply voltage is 20
V rms calculate
i. the current in the circuit.
ii. the power dissipated at resonance
iii. the power factor of the circuit if the capacitance is adjusted to
80F.
Solution:
i. At resonance fo = 1/2(1/LC)
C = 1/42fo
2L = 1/[4 x 3.14
2 x (50Hz)
2 x 0.150 H] = 67.5
ii. Maximum current flows through the circuit at resonance. At resonance Z = R
= 10.
Imax = V/R = 20 V/10 = 2A.
iii. Maximum power dissipated Pmax = Imax2R = (2A)
2 x 10 = 40 W.
iv. Power factor Pf = R/Z where Z = [R2 + (XL – XC)
2].
Now XL = 2fL = 2 x 3.14 x 50 Hz x 0.150 H = 47.1 .
XC = 1/2fC = 1/(2 x 3.14 x 50 Hz x 80 x 10–6
F) = 39.8 .
Z = [(10)2 + (47.1 – 39.8)
2] = 12.4
Pf = R/Z = 10/12.4 = 0.81
Example 9.9
A lamp of negligible inductance connected in series with an a.c. source. If the frequency
of the source is doubled what will happen to the brightness of the lamp. How can another
capacitor be connected in the circuit to increase the brightness of the lamp?
201
Solution:
Impedance Z = [R2 + (1/C)
2]. If doubles then 1/C decreases Z increases.
Therefore the current increases. The higher the amount of current through the lamp the
brighter the lamp becomes. Hence the lamp will become brighter.
2nd
Part: The current should increase for brightness Z ( = V/I) must decrease. If we
consider the relation
Z = [R2 + (1/C)
2]
We see that for Z to decrease C must increase. Hence the effective capacitance of the
two capacitors increases when they are connected in parallel.