is defined as the production of an induced
e.m.f. in a conductor/coil whenever the
magnetic flux through the conductor/coil
changes.
CHAPTER 20:
Electromagnetic induction
(6 Hours)
1
LEARNING OUTCOME:
20.1
Magnetic flux (1/2 hour)
At the end of this chapter, students should be able to:

Define and use magnetic flux,
 
  B  A  BA cos 
2
20.1.1 Magnetic flux of a uniform magnetic field

is defined as the scalar product between the magnetic flux
density, B with the vector of the area, A.
Mathematically,
 
Φ  B  A  BAcos
where
(7.1)
 : magnetic flux


 : angle between the direction of B and A
B : magnitude of the magnetic flux density
A : area of the coil
3

It is a scalar quantity and its unit is weber (Wb) OR tesla
meter squared ( T m2).

Consider a uniform magnetic field B passing through a surface
area A of a single turn coil as shown in Figures
 7.2a and 7.2b.
B

A
area

Figure 7.2a
From the Figure 7.2a, the angle  is 0 thus the magnetic flux is
given by
Φ  BAcos
 BA cos 0
maximum
  BA
4

A

B
  90
area
Figure 7.2b

Note:


From the Figure 7.2a, the angle  is 90 thus the magnetic flux
is given by
Φ  BA cos 
 BA cos 90
0
Direction of vector A always perpendicular (normal) to
the surface area, A.
The magnetic flux is proportional to the number of
field lines passing through the area.
5
Example 1 :
A single turn of rectangular coil of sides 10 cm  5.0 cm is placed
between north and south poles of a permanent magnet. Initially, the
plane of the coil is parallel to the magnetic field as shown in Figure
7.3.
N
S
R
I
Q
I
S
P
Figure 7.3
If the coil is turned by 90 about its rotation axis and the magnitude
of magnetic flux density is 1.5 T, Calculate the change in the
magnetic flux through the coil.
6
Solution :
B  1.5 T
The area of the coil is
Initially,
Finally,

A
 From the figure,  = thus the initial
B magnetic flux through the coil is

B

A
From the figure,  = thus the final
magnetic flux through the coil is
Therefore the change in magnetic flux through the coil is
Φ  Φ f  Φ i
7
Example 2 :
A single turn of circular coil with a diameter of 3.0 cm is placed in
the uniform magnetic field. The plane of the coil makes an angle
30 to the direction of the magnetic field. If the magnetic flux
through the area of the coil is 1.20 mWb, calculate the magnitude of
the magnetic field.
Solution :
The area of the coil is
8
Solution :
The angle between the direction of magnetic field, B and vector of
area, A is given by
Therefore the magnitude of the magnetic field is
9
LEARNING OUTCOME:
20.2
Induced emf (2 hours)
At the end of this chapter, students should be able to:

Use Faraday's experiment to explain induced emf.

State Faraday’s law and Lenz’s law to determine the
direction of induced current.
d

Apply formulae,
 
dt

Derive and use induced emf:
I) in straight conductor,   lvB sin 
dB
ii) in coil,
OR
dA
  A
iii) in rotating coil,
dt
  B
dt
  NAB sin t
10
20.2.1 MAGNETIC FLUX
20.1.1(a) Phenomenon of electromagnetic induction

Consider some experiments were conducted by Michael
Faraday that led to the discovery of the Faraday’s law of
induction as shown in Figures 7.1a, 7.1b, 7.1c, 7.1d and 7.1e.
v0
No movement
Figure 7.1a
11
v
S
N
I
Move towards the coil
I
Figure 7.1b
v0
No movement
12
Figure 7.1c
v
N
S
I
Move away from the coil
I
Figure 7.1d
v
S
N
Move towards the coil
I
I
13
Figure 7.1e


From the experiments:
 When the bar magnet is stationary, the galvanometer not
show any deflection (no current flows in the coil).
 When the bar magnet is moved relatively towards the coil,
the galvanometer shows a momentary deflection to the right
(Figure 7.1b). When the bar magnet is moved relatively
away from the coil, the galvanometer is seen to deflect in the
opposite direction (Figure 7.1d).
 Therefore when there is any relative motion between the
coil and the bar magnet , the current known as induced
current will flow momentarily through the galvanometer.
This current due to an induced e.m.f across the coil.
Conclusion :
 When the magnetic field lines through a coil changes
thus the induced emf will exist across the coil.
14

The magnitude of the induced e.m.f. depends on the
speed of the relative motion where if the
induced emf increases
v increases
v decreases
induced emf decreases
Therefore v is proportional to the induced emf.
15
Example 3 :
Figure 7.4
The three loops of wire as shown in Figure 7.4 are all in a region of
space with a uniform magnetic field. Loop 1 swings back and forth
as the bob on a simple pendulum. Loop 2 rotates about a vertical
axis and loop 3 oscillates vertically on the end of a spring. Which
loop or loops have a magnetic flux that changes with time? Explain
your answer.
16
Solution :
17
20.2.2 INDUCED EMF
20.2.2(a) Faraday’s law of electromagnetic induction

states that the magnitude of the induced emf is proportional
to the rate of change of the magnetic flux.
Mathematically,
dΦ
 
dt
where

OR
dΦ
 
dt
(7.2)
dΦ : change of the magnetic flux
dt : change of time
 : induced emf
The negative sign indicates that the direction of induced emf
always oppose the change of magnetic flux producing it
(Lenz’s law).
18
(7.3)
dΦ  Φ f  Φ i , then eq. (7.3) can be written as
(7.4)
where
Φ f : final magnetic flux
Φ i : initial magnetic flux
dΦ
  N
dt
and
Φ  BA cos 
19
d BA cos  
  N
dt
 dB 
   NAcos   
 dt 

(7.5)
For a coil of N turns is placed in a uniform magnetic field B
but changing in the coil’s area A, the induced emf  is given
by
dΦ
  N
and Φ  BA cos 
dt
d BA cos  
  N
dt
 dA 
   NBcos   
 dt 
(7.6)
20

For a coil is connected in series to a resistor of resistance
R and the induced emf  exist in the coil as shown in Figure 7.5,
the induced current I is given by
I
R
I
dΦ
  N
and   IR
dt
dΦ
(7.7)
IR   N
dt
Figure 7.5
Note:
 To calculate the magnitude of induced emf, the negative sign
can be ignored.

For a coil of N turns, each turn will has a magnetic flux  of
BAcos through it, therefore the magnetic flux linkage (refer to
the combined amount of flux through all the turns) is given by
magnetic flux linkage  NΦ
21
Example 4 :
The magnetic flux passing through a single turn of a coil is
increased quickly but steadily at a rate of 5.0102 Wb s1. If the coil
have 500 turns, calculate the magnitude of the induced emf in the
coil.
Solution :
By applying the Faraday’s law equation for a coil of N turns , thus
22
Example 5 :
A coil having an area of 8.0 cm2 and 50 turns lies perpendicular to a
magnetic field of 0.20 T. If the magnetic flux density is steadily
reduced to zero, taking 0.50 s, determine
a. the initial magnetic flux linkage.
b. the induced emf.
Solution :

B

A
a. The initial magnetic flux linkage is given by
23
Solution :
a.
b. The induced emf is given by
24
Example 6 :
A narrow coil of 10 turns and diameter of 4.0 cm is placed
perpendicular to a uniform magnetic field of 1.20 T. After 0.25 s, the
diameter of the coil is increased to 5.3 cm.
a. Calculate the change in the area of the coil.
b. If the coil has a resistance of 2.4 , determine the induced
current in the coil.
Solution :
Initial

B

B

A

A
 0

25
Final
Solution :
a. The change in the area of the coil is given by
26
Solution :
b. Given
The induced emf in the coil is
Therefore the induced current in the coil is given by
27
20.2.2 (B) LENZ’S LAW
states that an induced electric current always flows in such
a direction that it opposes the change producing it.
 This law is essentially a form of the law of conservation of
energy.
 An illustration of lenz’s law can be explained by
the following experiments.
Direction of
1st experiment:
induced current –
 In Figure 7.6 the magnitude
Right hand grip
of the magnetic field at the
rule.
solenoid increases as the
I
bar magnet is moved
North pole
towards it.
N


An emf is induced in the
solenoid and the
galvanometer indicates that
a current is flowing.
I
Figure 7.6
28

To determine the direction of the current through the
galvanometer which corresponds to a deflection in a particular
sense, then the current through the solenoid seen is in the
direction that make the solenoid upper end becomes a
north pole. This opposes the motion of the bar magnet and
obey the lenz’s law.
2nd experiment:
 Consider a straight conductor PQ
is placed perpendicular to the
magnetic field and move the
conductor to the left with constant
velocity v as shown in Figure 7.7.

When the conductor move to the
left thus the induced current
needs to flow in such a way to
oppose the change which has
induced it based on lenz’s law.
Hence galvanometer shows a
deflection.
X
X
X QX
X
X
X
X
X
X
X
X
X
X
X
X

Xv X

X FX
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X PX
X
X
X
X
I
Figure 7.7
29

To determine the direction of the induced current (induced
emf) flows in the conductor PQ, the Fleming’s right hand
(Dynamo) rule is used as shown in Figure 7.8.

B

 (motion )
Note:
Thumb – direction of Motion
First finger – direction of Field
induced I OR
induced emf

Second finger – direction of induced
current OR induced emf
Figure 7.8
Therefore the induced current flows from Q to P as shown in
Figure 7.7.

Since the induced current flows in the conductor PQ and is
placed in the magnetic field then this conductor will
experience magnetic force.

Its direction is in the opposite direction of the motion.
30
3rd experiment:

Consider two solenoids P and Q arranged coaxially closed to
each other as shown in Figure 7.9a.
 ind
S
N
P
I Switch , S
I
N
+
Q
I ind
S
-I
ind
Figure 7.9a

At the moment when the switch S is closed, current I begins
to flow in the solenoid P and producing a magnetic field inside
the solenoid P. Suppose that the field points towards the
solenoid Q.
31


The magnetic flux through the solenoid Q increases with
time. According to Faraday’s law ,an induced current due to
induced emf will exist in solenoid Q.
The induced current flows in solenoid Q must produce a
magnetic field that oppose the change producing it (increase
in flux). Hence based on Lenz’s law, the induced current flows
in circuit consists of solenoid Q is anticlockwise (Figure 7.9a)
and the galvanometer shows a deflection.
 ind
S
N
P
I Switch, S
I
S
-
I ind
Figure 7.9b
Q
N
+
I ind
32


At the moment when the switch S is opened, the current I
starts to decrease in the solenoid P and magnetic flux through
the solenoid Q decreases with time. According to Faraday’s
law ,an induced current due to induced emf will exist in
solenoid Q.
The induced current flows in solenoid Q must produce a
magnetic field that oppose the change producing it (decrease
in flux). Hence based on Lenz’s law, the induced current flows
in circuit consists of solenoid Q is clockwise (Figure 7.9b) and
the galvanometer seen to deflect in the opposite direction of
Figure 7.9a.
33
Example 7 :
A single turn of circular shaped coil has a resistance of 20  and an
area of 7.0 cm2. It moves toward the north pole of a bar magnet as
shown in Figure 7.10.
Figure 7.10
If the average rate of change of magnetic flux density through the
coil is 0.55 T s1,
a. determine the induced current in the coil
b. state the direction of the induced current observed by the
observer shown in Figure 7.10.
34
Solution :
a. By applying the Faraday’s law of induction, thus
Therefore the induced current in the coil is given by
35
Solution :
b. Based on the lenz’s law, hence the direction of induced current is
clockwise as shown in figure below.
36
20.2.3 INDUCED EMF IN A STRAIGHT CONDUCTOR
 Consider a straight conductor PQ of length l is moved
perpendicular with velocity v across a uniform magnetic field B
as shown in Figure 7.11.

X X X X X X PX X B
X
X
X
v
X
X
X
X
X indX
X
X
X
X
X
X
X
x X QX
X
X
X
X
X
X
X
X
lX
X
X
X
X
X
X
X
X
X
X
X
X
X
Area, A
I
 ind
Figure 7.11

When the conductor moves through a distance x in time t, the
area swept out by the conductor is given by
A  lx
37

Since the motion of the conductor is perpendicular to the
magnetic field B hence the magnetic flux cutting by the
conductor is given by
Φ  BA cos 
Φ  Blx cos 0

and
  0
Φ  Blx
According to Faraday’s law, the emf is induced in the conductor
and its magnitude is given by
d

dt
d
  Blx 
dt
dx
dx
v
  Bl
and
dt
dt
  Blv
(7.8)
38

In general, the magnitude of the induced emf in the straight
conductor is given by
  lvB sin 
(7.9)


θ : angle between v and B



Note:
where
This type of induced emf is known as motional induced emf.
The direction of the induced current or induced emf in the
straight conductor can be determined by using the Fleming’s
right hand rule (based on Lenz’s law).
In the case of Figure 7.11, the direction of the induced current or
induced emf is from Q to P. Therefore P is higher potential than
Q.

Eq. (7.9) also can be used for a single turn of rectangular coil
moves across the uniform magnetic field.

For a rectangular coil of N turns,
  NlvB sin 
39
(7.10)
Example 8 :
A 20 cm long metal rod CD is moved at speed of 25 m s1 across a
uniform magnetic field of flux density 250 mT. The motion of the rod
is perpendicular to the magnetic field as shown in Figure 7.12.

C
B
25 m s 1
Figure 7.12
D
a. Calculate the motional induced emf in the rod.
b. If the rod is connected in series to the resistor of resistance
15 , determine
i. the induced current and its direction.
ii. the total charge passing through the resistor in two minute.
40
Solution :
a. By applying the equation for motional induced emf, thus
b. Given
R  15 
i. By applying the Ohm’s law, thus
By using the Fleming’s right hand rule,
ii. Given
The total charge passing through the resistor is given by
41
20.2.4 INDUCED EMF IN A ROTATING COIL
 Consider a rectangular coil of N turns, each of area A, being
rotated mechanically with a constant angular velocity  in a
uniform magnetic field of flux density B about an axis as shown

in Figure 7.13.
B
N
ω


A
S
coil
Figure 7.13: side view

When the vector of area, A is at an angle  to the magnetic
field B, the magnetic flux  through each turn of the coil is given
by
  BA cos  and   t
  BA cos t
42

By applying the equation of Faraday’s law for a coil of N turns,
thus the induced emf is given by
d
  N
dt
d
  N BA cos t 
dt
d
  NBA cos t 
dt
  NBA sin t

where t : time
The induced emf is maximum when
 max  NBA
where
(7.11)
sin t  1 hence
2
  2f 
T
(7.12)
43

Eq. (7.11) also can be written as
  NBA sin 
(7.13)


 : angle between A and B

where
Conclusion : A coil rotating with constant angular velocity in a
uniform magnetic field produces a sinusoidally alternating emf
as shown by the induced emf  against time t graph in Figure
7.14.
ε V 
ε  ε max sin ωt
 max
Note:
0
This phenomenon
was the important   max
part in the
development of
the electric
generator or
dynamo.
t
0.5T
T
1.5T
2T

B
Figure 7.14
44
Example 9 :
A rectangular coil of 100 turns has a dimension of 10 cm  15 cm. It
rotates at a constant angular velocity of 200 rpm in a uniform
magnetic field of flux density 5.0 T. Calculate
a. the maximum emf produced by the coil,
b. the induced emf at the instant when the plane of the coil makes
an angle of 38 to the magnetic field.
Solution :
The area of the coil is
and the constant angular velocity in rad s1 is
45
Solution :
a. The maximum emf produced by the coil is given by
b.

B

A
From the figure, the angle  is
Therefore the induced emf is given by
46
Exercise 20.1 :
1.
A bar magnet is held above a loop of wire in a horizontal
plane, as shown in Figure 7.15.
The south end of the magnet is toward the
loop of the wire. The magnet is dropped
toward the loop. Determine the direction of
the current through the resistor
a. while the magnet falling toward the loop,
b. after the magnet has passed through the
loop and moves away from it.
(Physics for scientists and engineers,6th
edition, Serway&Jewett, Q15, p.991)
ANS. : U think
Figure 7.15
47
2.
A straight conductor of length 20 cm moves in a uniform magnetic
field of flux density 20 mT at a constant speed of 10 m s-1. The
velocity makes an angle 30 to the field but the conductor is
perpendicular to the field. Determine the induced emf.
ANS. : 2.0102 V
3. A coil of area 0.100 m2 is rotating at 60.0 rev s-1 with the axis of
rotation perpendicular to a 0.200 T magnetic field.
a. If the coil has 1000 turns, determine the maximum emf
generated in it.
b. What is the orientation of the coil with respect to the
magnetic field when the maximum induced emf occurs?
(Physics for scientists and engineers,6th edition,Serway&Jewett, Q35,
p.996)
ANS. : 7.54103 V
4. A circular coil has 50 turns and diameter 1.0 cm. It rotates at a
constant angular velocity of 25 rev s1 in a uniform magnetic field of
flux density 50 T. Determine the induced emf when the plane of
the coil makes an angle 55 to the magnetic field.
48
5
ANS. : 1.7710 V
LEARNING OUTCOME:
20.3
Self-inductance (1 hour)
At the end of this chapter, students should be able to:

Define self-inductance.

Apply formulae
L

dI dt

0 N 2 A
l
for a loop and solenoid.
49
20.3 SELF-INDUCTANCE
20.3.1 Self-induction

Consider a solenoid which is connected to a battery , a switch S
and variable resistor R, forming an open circuit as shown in
Figure 7.16a.
 When the switch S is closed, a current
I begins to flow in the solenoid.
S
N

The current produces a magnetic
field whose field lines through the
solenoid and generate the magnetic
flux linkage.

If the resistance of the variable
resistor changes, thus the current
flows in the solenoid also changed,
then so too does magnetic flux
50
linkage.
I
I
S
R
Figure 7.16a: initial



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 increases :
εind
S
N
I
I
ind
+
N
I S
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 :
+
SS
I ind
I
εind
NN
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.
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.
52
20.3 SELF-INDUCTANCE
Iinduced
Iinduced
(a) A current in the coil produces a magnetic field
directed to the left.
(b)
If the
(b) If
the current
current increases,
increases, the
the increasing
increasing magnetic
magnetic
flux
creates an
an induced
induced emf
emf having
having the
the polarity
polarity
flux creates
shown
shown by
by the
the dashed
dashed battery.
battery.
53
(c)
The
polarity
of
the
induced
emf
reverses
if
(c) The polarity of the induced emf reverses if
the
the current
current decreases.
decreases.
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
R


lamp A2
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
I0
lamp A1 with coil L
0
t
Figure 7.17b
55
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
time and is suddenly opened.
Explain why a spark jump across the
switch contacts S .
switch, S

coil, L
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 ……………………………………
…………………………….……………….and a …………………
………………………………………………………………………..
56
20.3.2 SELF-INDUCTANCE, L

From the self-induction phenomenon, we get
ΦL  I
Φ L  LI
where
(7.14)
L : self - inductance of the coil
I : current
 L : magnetic flux linkage

From the Faraday’s law, thus
d L
 
dt
d
  LI 
dt
dI
  L
dt
(7.15)
57

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
d
  N
dt
dI
d
 L  N
dt
dt
L dI  N d


and
LI  N
N  L
L

I
I
dI
  L
dt
magnetic flux linkage
(7.16)
58



It is a scalar quantity and its unit is henry (H).
Unit conversion :
1 H  1 Wb A 1  1 T m 2 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
20.3.3 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
  BA cos 0
  0 NI 

A
 l 


 0 NIA
l
Therefore the self-inductance of the solenoid is given by
N
L
I
N   0 NIA 
L 

I  l 
L
0 N 2 A
l
(7.17)
60
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 20.2 :
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.
64
ANS. : 19 A s1
LEARNING OUTCOME:
20.4
Mutual inductance (2 hours)
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,
N 2 12 0 N1 N 2 A
M 12 

I1
l

Explain the working principle of transformer and the
effect of eddy current in transformer.
65
20.4 MUTUAL INDUCTANCE

B1
20.4.1 Mutual induction

Consider two circular closepacked coils near each other
and sharing a common
central axis as shown in
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.

B1
I1
Coil 1
Coil 2
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
20.4.2 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
dI
2  
1
dt
 2   M 12

dI1
dt
If vice versa, the induced emf in coil 1, 1 is given by
dI 2
 1   M 21
dt

(7.21)
(7.22)
It is a scalar quantity and its unit is henry (H).
where M 12  M 21  M : Mutual inductance
68


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
d 2
 2  N2
dt
dI1
d 2
 M 12
 N2
dt
dt
M 12 dI1  N 2 d 2
M 12 I1  N 2  2

M 12

N 2 2

I1
and
N11
M 21 
I2
magnetic flux linkage magnetic flux linkage
through coil 1
through coil 2
N 2  2 N11
M

I1
I2
(7.23)
69
20.4.3 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,
B1 
 0 N1 I 1
l
70
and then the magnetic flux Ф1,
1  B1 A cos 0

1 

If no magnetic flux leakage, thus
 0 N1 I 1 A
l
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
N 2 2

I1
M 12
 N 2   0 N1 I 1 A

 
l
 I1 
M 12  M 
 0 N1 N 2 A
l
(7.24)
71
Mutual inductance, M
N 2Φ2 N 1Φ1
M

I1
I2
M
o N 2 N1 A
l
dI 2
 1   M 12
dt
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
20.4.4 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
voltage source
NP
turns
primary coil
NS
turns
secondary coil
Figure 7.22

If NP > NS the transformer is a step-down transformer.

If NP < NS the transformer is a step-up transformer.
76

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
 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.
79
LEARNING OUTCOME:
20.5
Energy stored in an inductor (½ hour)
At the end of this chapter, students should be able to:

Derive and use formulae for energy stored in an inductor,
1 2
U  LI
2
80
23.4 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
dI
 L
dt
The electrical power P in overcoming the back emf in the circuit
is given by
P  I
dI
P  LI
dt
Pdt  LIdI and Pdt  dU
dU  LIdI
(7.18)
81
U
I
0
0
 dU  L  IdI
(7.19)
1
U  CV 2
2
and analogous to
L
1 2
U  LI
2
in capacitor
0 N 2 A
l
(7.20)
82
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
Next Chapter…
CHAPTER 24 :
Alternating current
85