Electromagnetic Induction

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 17:
Electromagnetic induction

1

LEARNING OUTCOME:

17.1 Magnetic flux (1/2 hour)

At the end of this chapter, students should be able to:
 Define and use magnetic flux,



  B  A  BAcos

2

17.1.1 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,

 (7.1)

Φ  B  A  BAcos

where  : 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 7.2a 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

 BAcos 0 4
  BA maximum

 
A B

  90

area

Figure 7.2b

 From the Figure 7.2a, the angle  is 90 thus the magnetic flux

is given by Φ  BAcos

 BAcos90

0

Note:

 Direction of vector A always perpendicular (normal) to 5
the surface area, A.

 The magnetic flux is proportional to the number of
field lines passing through the area.

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 SR Q
II

SP

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,   From the figure,  = thus the initial
A B
magnetic flux through the coil is

Finally,  From the figure,  = thus the final
B

 magnetic flux through the coil is
A

Therefore the change in magnetic flux through the coil is 7

Φ  Φf  Φi

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:

17.2 Induced emf

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.

 Apply formulae,    d

dt

 Derive and use induced emf:
I) in straight conductor,   lvB sin

ii) in coil,    A dB OR   B dA

dt dt

iii) in rotating coil,   NAB sin t

10

17.2.1 MAGNETIC FLUX

17.1.1(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

S v

I N
Move towards the coil

I

Figure 7.1b

v0

No movement

12

Figure 7.1c

N v

I S
Move away from the coil
N
I
I
Figure 7.1d

v

S

Move towards the coil

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

v increases induced emf increases

v decreases induced emf decreases

Therefore v is proportional to the induced emf.

15

Example 3 :

Figure 7.4 16
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.

Solution :

17

17.2.2 INDUCED EMF

17.2.2(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 (7.2)

   dΦ OR dt

dt

where dΦ : change of the magnetic flux

dt : change of time
 :induced emf

 The negative sign indicates that the direction of induced emf 18

always oppose the change of magnetic flux producing it
(Lenz’s law).

(7.3)

dΦ  Φf  Φi , then eq. (7.3) can be written as

(7.4)

where Φf : final magnetic flux
Φi :initial magnetic flux

  N dΦ and Φ  BAcos 19

dt

  N d BAcos  (7.5)

dt

  NAcos  dB 

 dt 

 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   N dΦ and Φ  BAcos

dt

  N d BAcos 

dt

  NBcos  dA  (7.6) 20

 dt 

 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
  N dΦ and   IR
dt
I IR  N dΦ
IR (7.7)

dt

Note: Figure 7.5

 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 :


BB


AA

  0 Final 25

Initial

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

17.2.2 (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
solenoid increases as the rule.

bar magnet is moved N I
towards it.
North pole

 An emf is induced in the I 28
solenoid and the
galvanometer indicates that Figure 7.6
a current is flowing.

 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: X X X QX X X X X

 Consider a straight conductor PQ X XX X X X X X
is placed perpendicular to the X X FX X X X
magnetic field and move the 
conductor to the left with constant Xv X

velocity v as shown in Figure 7.7. X XXXX XXX

 When the conductor move to the I
left thus the induced current
needs to flow in such a way to X XXXX XXX
oppose the change which has
induced it based on lenz’s law. X X X PX X X X X 29
Hence galvanometer shows a Figure 7.7
deflection.

 To determine the direction of the induced current (induced
emf) flows in the conductor PQ, the Fleming’s right hand

(Dynamo) rul(emisotuisoend) as shown in Figure 7.8.


B Note:
Thumb – direction of Motion

First finger – direction of Field

induced I OR Second finger – direction of induced

induced emf 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 30
placed in the magnetic field then this conductor will
experience magnetic force.

 Its direction is in the opposite direction of the motion.

3rd experiment:

 Consider two solenoids P and Q arranged coaxially closed to

each other as shown in Figure 7.9a.  ind

S N N Q S
+ -
P I
I ind I ind
I Switch,S

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 NS N
-Q +
P
I Iind I ind
I Switch,S
Figure 7.9b 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 34
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.

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

17.2.3 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 XXXX XXX

Area, A

X X lX X X X XX
X X vX X
XX XI indX

X X X X X X X X ind

X X X X X x X QX X

Figure 7.11

 When the conductor moves through a distance x in time t, the 37

area swept out by the conductor is given by

A  lx

 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  BAcos and   0

Φ  Blx cos 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

  Bl dx and dx  v

dt dt

  Blv (7.8) 38

 In general, the magnitude of the induced emf in the straight
conductor is given by

  lvB sin between  (7.9)
v
where θ : angle 
and B

 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
Note: 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, 39

  NlvB sin (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. 40

ii. the total charge passing through the resistor in two minute.

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

17.2.4 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 an axis as shown
in Figure 7.13. B

N Sω 42

A

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   BAcos and   t

  BAcost

 By applying the equation of Faraday’s law for a coil of N turns,

thus the induced emf is given by

  N d

dt

 N d BAcost
dt
  NBA d cos t 

dt

  NBA sint (7.11)

where t : time

 The induced emf is maximum when sin t  1 hence

 max  NBA (7.12)
where   2f  2
43
T

 Eq. (7.11) also can be written as

  NBA sin (7.13)



where  : angle between A and B

 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  εmax sin ωt

 max

Note: 0 0.5T T 1.5T t
This phenomenon 2T
Figure 7.14
was the important   max 
B
part in the
44
development of

the electric

generator or

dynamo.

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 From the figure, the angle  is

 Therefore the induced emf is given by
A

46

Exercise 17.1 :

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
ANS. : 1.77105 V

LEARNING OUTCOME:

17.3 Self-inductance

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

 Define self-inductance.

 Apply formulae

L     0N 2 A
dI dt l

for a loop and solenoid.

49

17.3 SELF-INDUCTANCE

17.3.1 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

I solenoid and generate the magnetic

S I flux linkage.

R

 If the resistance of the variable

resistor changes, thus the current

Figure 7.16a: initial flows in the solenoid also changed,

then so too does magnetic flux

linkage. 50