Electromagnetic Induction

 According to the Faraday’s law, an emf has to be induced in
the solenoid itself since the flux linkage changes.

 In accordance with Lenz’s law, the induced emf opposes the
changes that has induced it and it is known as a back emf.

 For the current I incεriendases :

S- N+
NIind I IS

I ind

Figure 7.16b: I increases

Direction of the induced emf is in the

opposite direction of the current I.

51

 For the current I decreases : εind

+ -

SS NN

Iind I I
I ind

Figure 7.16c: I decreases

Direction of the induced emf is in the

same direction of the current I.

 This process is known as self-induction. 52

 Self-induction is defined as the process of producing an
induced emf in the coil due to a change of current flowing
through the same coil.

17.3 SELF-INDUCTANCE

Iinduced Iinduced

(a) A current in the coil produces a magnetic field

directed to the left.

((bb)) IIff tthhee ccuurrrreenntt iinnccrreeaasseess,, tthhee iinnccrreeaassiinngg mmaaggnneettiicc
fflluuxx ccrreeaatteess aann iinndduucceedd eemmff hhaavviinngg tthhee ppoollaarriittyy
sshhoowwnn bbyy tthhee ddaasshheedd bbaatttteerryy..

((cc)) TThhee ppoollaarriittyy ooff tthhee iinndduucceedd eemmff rreevveerrsseess iiff 53
tthheeccuurrrreennttddeeccrreeaasseess..

Self-induction experiment

 The effect of the self-induction can be demonstrated by the
circuit shown in Figure 7.17a.

 switch, S

iron-core lamp A1

coil, L lamp A2

R

Figure 7.17a

 Initially variable resistor R is adjusted so that the two lamps

have the same brightness in their respective circuits with steady

current flowing.

 When the switch S is closed, the lamp A2 with variable resistor R

is seen to become bright almost immediately but the lamp A1 with

iron-core coil L increases slowly to full brightness. 54

 Reason:

 The coil L undergoes the self-induction and induced emf
in it. The induced or back emf opposes the growth of
current so the glow in the lamp A1 increases slowly.

 The resistor R, however has no back emf, hence the lamp

A2 glow fully bright as soon as switch S is closed.

 This effect can be shown by the graph of current I against
time t through both lamps in Figure 7.17b.

I lamp A2 with resistor R 55

I0

lamp A1 with coil L

0t

Figure 7.17b

Example 10 :

A circuit contains an iron-cored coil L, a switch S, a resistor R and

a dc source  arranged in series as shown in Figure 7.18.

The switch S is closed for a long switch, S
time and is suddenly opened.
Explain why a spark jump across the  coil, L
switch contacts S .

Figure 7.18

Solution : R

 When the switch S is suddenly opened, the ………………………
……………………………………… and …..................................
.....………………………..…. which tends to maintain the current.

 This back emf is high enough to …………………………………… 56
…………………………….……………….and a …………………
………………………………………………………………………..

17.3.2 SELF-INDUCTANCE, L

 From the self-induction phenomenon, we get

ΦL  I

ΦL  LI (7.14)

where L : self - inductance of the coil

I : current

L : magnetic flux linkage

 From the Faraday’s law, thus

   dL

dt

  d LI 

dt

  L dI (7.15) 57

dt

 Self-inductance is defined as the ratio of the self induced
(back) emf to the rate of change of current in the coil.

OR L   

dI / dt

 For the coil of N turns, thus

  N d and   L dI

dt dt
 L dI  N d

dt dt

L dI  N  d magnetic flux linkage

LI  N

L  N  L (7.16) 58
II

 It is a scalar quantity and its unit is henry (H).

 Unit conversion :

1 H  1 Wb A1  1 T m2 A1

 The value of the self-inductance depends on

 the size and shape of the coil,

 the number of turn (N),

 the permeability of the medium in the coil ().

 A circuit element which possesses mainly self-inductance is
known as an inductor. It is used to store energy in the form of
magnetic field.

 The symbol of inductor in the electrical circuit is shown in Figure
7.19.

Figure 7.19 59

17.3.3 SELF-INDUCTANCE OF A SOLENOID

 The magnetic flux density at the centre of the air-core

solenoid is given by B  0 NI

l

 The magnetic flux passing through each turn of the solenoid

always maximum and is given by

  BAcos 0

   0 NI  A   0 NIA

l l

 Therefore the self-inductance of the solenoid is given by

L  N L  N  0 NIA 
I
I l 

L  0N 2 A (7.17) 60

l

Example 11 :

A 500 turns of solenoid is 8.0 cm long. When the current in the
solenoid is increased from 0 to 2.5 A in 0.35 s, the magnitude of the
induced emf is 0.012 V. Calculate
a. the inductance of the solenoid,
b. the cross-sectional area of the solenoid,
c. the final magnetic flux linkage through the solenoid.

(Given 0 = 4  107 H m1)

Solution :

a. The change in the current is

Therefore the inductance of the solenoid is given by

61

Solution :
b. By using the equation of self-inductance for the solenoid, thus

c. The final magnetic flux linkage is given by

62

Exercise 17.2 :

Given 0 = 4  107 H m1

1. An emf of 24.0 mV is induced in a 500 turns coil at an instant
when the current is 4.00 A and is changing at the rate of
10.0 A s-1. Determine the magnetic flux through each turn of
the coil.

(Physics for scientists and engineers,6th edition,Serway&Jewett,
Q6, p.1025)

ANS. : 1.92105 Wb

2. A 40.0 mA current is carried by a uniformly wound air-core
solenoid with 450 turns, a 15.0 mm diameter and 12.0 cm
length. Calculate

a. the magnetic field inside the solenoid,

b. the magnetic flux through each turn,

c. the inductance of the solenoid.

ANS. : 1.88104 T; 3.33108 Wb; 3.75104 H

63

3. A current of 1.5 A flows in an air-core solenoid of 1 cm radius
and 100 turns per cm. Calculate

a. the self-inductance per unit length of the solenoid.

b. the energy stored per unit length of the solenoid.

ANS. : 0.039 H m1; 4.4102 J m1

4. At the instant when the current in an inductor is increasing at a
rate of 0.0640 A s1, the magnitude of the back emf is 0.016

V.

a. Calculate the inductance of the inductor.

b. If the inductor is a solenoid with 400 turns and the current
flows in it is 0.720 A, determine

i. the magnetic flux through each turn,

ii. the energy stored in the solenoid.

ANS. : 0.250 H; 4.5104 Wb; 6.48102 J

5. At a particular instant the electrical power supplied to a

300 mH inductor is 20 W and the current is 3.5 A. Determine

the rate at which the current is changing at that instant.

ANS. : 19 A s1 64

LEARNING OUTCOME:

17.4 Mutual inductance

At the end of this chapter, students should be able to:

 Define mutual inductance.

 Derive and use formulae for mutual inductance of two
coaxial coils,

M12  N 2 12  0 N1N2 A
I1
l

 Explain the working principle of transformer and the
effect of eddy current in transformer.

65

17.4 MUTUAL INDUCTANCE

17.4.1 Mutual induction  
B1 B1
 Consider two circular close-
packed coils near each other I1 Coil 2
and sharing a common
central axis as shown in Coil 1
Figure 7.20.

 A current I1 flows in coil 1,

produced by the battery in
the external circuit.

 The current I1 produces a

magnetic field lines inside it
and this field lines also pass
through coil 2 as shown in
Figure 7.20.

66

Figure 7.20

 If the current I1 changes with time, the magnetic flux through

coils 1 and 2 will change with time simultaneously.

 Due to the change of magnetic flux through coil 2, an emf is
induced in coil 2. This is in accordance to the Faraday’s law of
induction.

 In other words, a change of current in one coil leads to the
production of an induced emf in a second coil which is
magnetically linked to the first coil.

 This process is known as mutual induction.
 Mutual induction is defined as the process of producing an

induced emf in one coil due to the change of current in
another coil.
 At the same time, the self-induction occurs in coil 1 since the
magnetic flux through it changes.

67

17.4.2 MUTUAL INDUCTANCE, M

 From the Figure 7.20, consider the coils 1 and 2 have N1 and
N2 turns respectively.

 If the current I1 in coil 1 changes, the magnetic flux through coil

2 will change with time and an induced emf will occur in coil 2,

2 where    dI1
dt
2

2  M12 dI1 (7.21)
dt

 If vice versa, the induced emf in coil 1, 1 is given by

1  M 21 dI 2 (7.22)
dt

 It is a scalar quantity and its unit is henry (H). 68

where M12  M 21  M : Mutual inductance

 Mutual inductance is defined as the ratio of induced emf in a
coil to the rate of change of current in another coil.

 From the Faraday’s law for the coil 2, thus

2  N2 d 2
dt

 M12 dI1  N2 d 2
dt dt

 M12 dI1  N2 d2 magnetic flux linkage magnetic flux linkage

M12 I1  N22 through coil 2 through coil 1

M 12  N22
I1
M  N22  N11 (7.23)
and I1 I2
69
N11
M 21  I2

17.4.3 MUTUAL INDUCTANCE FOR TWO
SOLENOIDS
 Consider a long solenoid with length l and cross sectional area

A is closely wound with N1 turns of wire. A coil with N2 turns

surrounds it at its centre as shown in Figure 7.21.

N1: primary coil
N2: secondary coil

Figure 7.21

 When a current I1 flows in the primary coil (N1), it produces a

magnetic field B1, 0 N1I1 70

B1  l

and then the magnetic flux Ф1,  0 N1I1A

 1  B1Acos 0 1 l

If no magnetic flux leakage, thus

1  2

 If the current I1 changes, an emf is induced in the secondary

coils, therefore the mutual inductance occurs and is given by

M 12  N22
I1

M 12   N2  0 N1I1A
I1
l

M 12  M  0 N1N2 A (7.24) 71

l

Mutual inductance, M

M  N2Φ2  N1Φ1 M  o N2 N1A 1  M12 dI 2
I1 I2 dt
l

72

Example 13 :

A current of 3.0 A flows in coil C and is produced a magnetic flux
of 0.75 Wb in it. When a coil D is moved near to coil C coaxially, a
flux of 0.25 Wb is produced in coil D. If coil C has 1000 turns and
coil D has 5000 turns.
a. Calculate self-inductance of coil C and the energy stored in C

before D is moved near to it.
b. Calculate the mutual inductance of the coils.
c. If the current in C decreasing uniformly from 3.0 A to zero in
0.25 s, calculate the induced emf in coil D.
Solution :

a. The self-inductance of coil C is given by

73

Solution :
a. and the energy stored in C is

b. The mutual inductance of the coils is given by

74

Solution :
c. Given

The induced emf in coil D is given by

75

17.4.4 TRANSFORMER

 is an electrical instrument to increase or decrease the emf
(voltage) of an alternating current.

 Consider a structure of the transformer as shown in Figure 7.22.

laminated iron core

alternating NP NS
voltage source
turns turns
primary coil
secondary coil

Figure 7.22 76

 If NP > NS the transformer is a step-down transformer.
 If NP < NS the transformer is a step-up transformer.

 The symbol of transformer in the electrical circuit is shown in
Figure 7.23.

Figure 7.23
Working principle of transformer
 When an alternating voltage source is applied to the primary

coil, the alternating current produces an alternating magnetic
flux concentrated in the iron core.
 Without no magnetic flux leakage from the iron core, the same
changing magnetic flux passes through the secondary coil and
inducing an alternating emf.
 After that the induced current is produced in the secondary coil. 77

 The characteristics of an ideal transformer are:
 Zero resistance of primary coil.
 No magnetic flux leakage from the iron core.
 No dissipation of energy and power.

78

Energy losses in transformer 79

 Although transformers are very efficient devices, small energy
losses do occur in them owing to four main causes:

 Resistance of coils

The wire used for the primary and secondary coils has

resistance and so ordinary (I2R) heat losses occur.

Overcome : The transformer coils are made of thick
copper wire.

 Hysteresis

The magnetization of the core is repeatedly reversed by
the alternating magnetic field. The resulting expenditure
of energy in the core appears as heat.

Overcome : By using a magnetic material (such as
Mumetal) which has a low hysteresis loss.

 Flux leakage

The flux due to the primary may not all link the secondary.
Some of the flux loss in the air.

Overcome : By designing one of the insulated coils is
wound directly on top of the other rather than having two
separate coils.

LEARNING OUTCOME:

17.5 Energy stored in an inductor

At the end of this chapter, students should be able to:
 Derive and use formulae for energy stored in an inductor,

U  1 LI 2
2

80

17.4 ENERGY STORED IN AN
INDUCTOR

 Consider an inductor of inductance L. Suppose that at time t,

the current in the inductor is in the process of building up to its

steady value I at a rate dI/dt.

 The magnitude of the back emf  is given by

  L dI

dt

 The electrical power P in overcoming the back emf in the circuit

is given by P  I

P  LI dI
dt

Pdt  LIdI and Pdt  dU

dU  LIdI (7.18)

81

UI

 dU  L IdI
00

and analogous to (7.19)

U  1 CV 2 in capacitor
2

L  0N 2 A

l

U  1 LI 2 (7.20) 82
2

Example 12 :

A solenoid of length 25 cm with an air-core consists of 100 turns
and diameter of 2.7 cm. Calculate
a. the self-inductance of the solenoid, and
b. the energy stored in the solenoid,
if the current flows in it is 1.6 A.

(Given 0 = 4  107 H m1)

Solution :
a. The cross-sectional area of the solenoid is given by

Hence the self-inductance of the solenoid is

83

Solution :

b. Given I  1.6 A

By applying the equation of energy stored in the inductor, thus

84