CHAPTER 5: ELECTROMAGNETIC INDUCTION

CHAPTER 5

ELECTROMAGNETIC
INDUCTION

KMM

TOPIC 5 : ELECTROMAGNETIC INDUCTION

5.1 Magnetic flux
5.2 Induced emf
5.3 Self-Inductance
5.4 Energy stored in inductor
5.5 Mutual Inductance

1

5.1 Magnetic flux

Learning outcomes
a) Define and use magnetic flux,

b) Use magnetic flux linkage,

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.

2

Magnetic flux
Magnetic flux is defined as the scalar product between
the magnetic flux density, B with the vector of the area, A.

Mathematically,

where

θ

Scalar quantity. plane
Unit : weber (Wb) or tesla meter squared ( T m2). of coil

1 weber = 1 Wb = 1 T m2 3

Consider a uniform magnetic field B passing through a
surface area A of a single turn coil as shown below.

area area

Ø is maximum, when ϴ = 0o Ø is 0, when ϴ = 90o

Note

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 surface area.

4

Magnetic flux linkage, Φ
Magnetic flux linkage is defined as the product of the
magnetic flux and the number of turns.

5

Example 5.1.1

A rectangular coil of wire is placed in a constant magnetic
field whose magnitude is 0.80 T. The coil has an area of 2.0
m2. Determine the magnetic flux for :
i. θ =0ο , ii. 60 ο and iii. 90 ο.

θ =0ο θ =60ο θ =90ο 6

Solution 5.1.1 B = 0.80 T, A = 2.0 m2

i. θ = 0ο

7

ii. θ = 60ο

8

iii. θ = 90ο

9

Example 5.1.2
A circular coil with a radius of 10 cm and 300 turns is facing
perpendicular to a magnetic field with magnetic flux density of 5.1 mT.
Determine the magnetic flux linkage for this coil.
Solution 5.1.2

θ = 0ο , B = 5.1 m T, r = 10 cm, N = 300 turns

10

5.2 Induced emf

Learning outcomes
a) Explain induced emf by using Faraday’s experiment.

b) State and use Farady’s law,

c) State and use Lenz’s law to determine the direction of induced
current.

d) Apply induced emf in :
i) a straight conductor,

ii) a coil,

iii) a rotating coil

11

Faraday’s Experiment

(a) Bar magnet is held stationary (at rest), no current

flows in the coil. 12

(b) Bar magnet moves towards the coil, current flows in

the coil in one direction(clockwise). 13

(c) Bar magnet moves away from the coil, current flows

in the opposite direction (anticlockwise)to that in (b).

14

CONCLUSION
From this experiment, Faraday concluded that
whenever there is relative motion between a
conductor/coil and a magnetic field, the flux linkage
with a coil changes and this change in flux induces a
voltage across a coil (emf induced).

How To Increase Emf Induced in a Coil

1. By increasing the number of turns (N) in the coil.

2. By increasing magnetic field strength (B) surrounding

the coil.

3. By increasing the speed of the relative motion between

the coil and the magnet. 15

Faraday’s law

Faraday’s law states that the magnitude of the induced
emf is proportional to the rate of change of the magnetic
flux.

Mathematically, where

The negative sign indicates that the direction of
emf always oppose the change of magnetic flux that
induced it (Lenz’s law).
For a coil of N turns, the total induced emf is

SI unit for induced emf: volt (V) 16

Lenz’s law
Lenz’s law states that an induced electric current
always flows in such a direction that it opposes the
change that induced it.
This law is essentially a form of the law of conservation
of energy.

17

How to determine the direction of induced current in a
straight conductor
Right Hand Rule or Fleming’s Right Hand Rule

Oppose motion

Right Hand Rule Fleming’s Right Hand Rule

18

How to determine the direction of induced current in a
coil/solenoid

Right Hand Grip Rule

B induced

I induced

19

A. For a straight conductor /rod Fleming’s Right Hand Rule

As the rod moves to the left, current is induced in the
rod.
By using Fleming’s Right Hand Rule the direction of
induced current in the rod is downward.upward

Thumb

or v

rod Iinduce rail First finger
Second finger
d

Direction of v = direction of Fapplied 20

Right Hand Rule

rail

rod FB

Iinduced

Based on the Lenz’s law, the induced current always
flows in such a direction that it opposes the change
producing it.

To oppose the motion of rod to the left, FB acts to the
right.

By using right hand rule, direction of induced current in

the rod is downward upward. 21

B. For a coil/solenoid A coil
Bfinal = 6 T
∆B= +2 T
Bexternal upwards

Bintial = 4 T I induced

Binduced downwards

As the externel magnetic field increasing (from 4T to 6T),

current is induced in the coil.

To oppose the changes (to mantain B) through the surface

area of coil (B = 4T), thus B induced = -2T (in opposite
direction is produced. .

Using Right Hand Grip Rule, direction B induced is downward,
thus I induced in the coil is clockwise.
22

As the bar magnet moves A solenoid
toward the solenoid,
current is induced in the 23
solenoid.

To oppose the motion of
the bar magnet (to repel) ,
the upper solenoid
becomes a north pole.

Using Right Hand Grip Rule,
direction B induced is upward,
thus I induced in the circuit is
anticlockwise.

Example 5.2.1

If the rail is placed in a uniform magnetic field, determine the
direction of induced current in the rod for each case below.

a) v c) v

xxxx x xxxx x xxxx x xxxx x
xxxx x xxxx x
xxxx x xxxx x d) x x x x x xv x x x x
b) v
xxxx x xxxx x

Solution 5.2.5

(Use Fleming’s Right Hand Rule /Right Hand Rule)

a) Downward c) Downward Upward

b) Upward Downward d) Upward

24

Example 5.2.2
Determine the direction of induced current in the coil/solenoid for
each case below.

Solution 5.2.2
(Use Right Hand Grip Rule)

a) a) Anticlockwise in the
circuit

b) From B to A

b) 25

i. Emf Induced In A Straight Conductor

Consider a straight conductor PQ of length, L moving with
constant velocity v through a uniform magnetic field B
directed into the paper as shown in the figure.

As the conductor moves to the right perpendicular to B, a
magnetic force FB = Bqv sin ϴ directed downward acts on
the electrons in the conductor.

Because of this magnetic force, the free electrons move to
the lower end of the conductor and accumulate there,
leaving a net positive charge at the upper end.

As a result of this separation, an electric field is produced.

The charge stops flowing when the downward magnetic
force is balanced by the upward electric force (equilibrium).

26

In equilibrium P
From
FE X X X X X X X X

X X X++X X X X X (= velocity of rod)

- -X X X X X X X X
(= average velocity
X X X X X X X X of electron)
X X XFXBX X X X

FB X X X-- X X X X X
Q

Equation of emf induced
in a straight conductor

Also called motional emf

27

For a rectangular coil of N turns moves across the uniform
magnetic field, the emf induced is

Where

B = uniform magnetic field
N = number of turns of the coil

L = length of the rod/ one side of a rectangular coil

v = magnitude of velocity of the rod

28

ii. Emf Induced In A coil

How to change the magnetic flux in the coil which
causes induced emf?

i. By moving a magnet towards or away from the coil
(B increase/decrease)

ii. By moving the coil into or out of the magnetic field
(B increase/decrease)

iii. By changing the area of a coil placed in the
magnetic field
(A increase/decrease)

iv. By rotating the coil relative to the magnet /change

the angle

(θ increase/decrease) 29

Equations of emf induced in a coil

θ = 0o A constant

B constant

The induced current I in the coil : Resistance
and of the coil

To calculate the magnitude of

induced emf, the negative sign

can be ignored. 30

iii) Emf Induced In A Rotating Coil
Consider a rectangular coil of N turns, each of area A,
being rotated mechanically with a constant angular
velocity ω in a uniform magnetic field of flux density B
about an axis as shown in figure below.

NS

coil

31

When the vector of area, A is at an angle θ to the
magnetic field B, the magnetic flux φ through each turn
of the coil is given by

and

From Faraday’s law,

Emf induced in
a rotating coil

32

Emf induced in a rotating coil

Where

N = number of turn of the coil
B = uniform magnetic field

A = cross sectional area of the coil

ω = angular velocity

t = time

33

Conclusion :

A coil rotating with constant angular velocity in a uniform
magnetic field produces a sinusoidally alternating emf as
shown by the induced emf ε against time t graph in figure
below.

where

This
phenomenon
was the
important
part in the
development
of the electric
generator or
dynamo.

34

Example 5.2.3
A circular coil of 200 turns is immersed in a uniform perpendicular
magnetic field. If the initial flux is 0.00050 Wb and drops to zero
in 200 ms, calculate the average induced emf.
Solution 5.2.3

N = 200 turns , t = 200 ×10-3 s, φi = 0.00050 Wb, φf = 0 Wb
Average induced emf,

35

Example 5.2.4
A coil of 42 turns and area of 75 cm2 is in a uniform magnetic field
with its plane perpendicular to the field. Calculate the induced emf
in the coil if the magnetic field changes from 0.042 T to 0.075 T in
20 ms.
Solution 5.2.4
N = 42 turns , A = 75 × 10-4 m2, Bi = 0.042 T, Bf = 0.075 T,
t = 20 ×10-3 s
Induced emf in the coil,

36

Example 5.2.5

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 Ω, determine the induced

current in the coil.

Solution 5.2.5

N = 10 turns, a)

B= 1.2 T,
di = 4.0 cm = 4 ×10-2 m,
df = 5.3 cm = 5.3 ×10-2 m,

t = 0.25 s

37

(b) Induced emf,
Induced current,

38

Example 5.2.6

A coil of 150 turns with an area of 36 cm2 is rotating in a uniform
magnetic field of 0.24 T. The axis of rotation is perpendicular to
the filed field and the coil is rotating at 20 Hz. Calculate
a) The maximum induced emf
b) The induced emf when the coil makes an angle of 35o with the

filed field lines.

Solution 5.2.6 N = 150 turns , A = 36 × 10-4 m2, Bi = 0.24 T,
f = 20 Hz, θ = 90 – 35= 55o

a) Maximum induced emf

b)
39

Example 5.2.7
A rod of 1.6 m long is moving at a speed of 5.0 m s-1 in a direction
perpendicular to a 0.80 T magnetic field. Both rod and rails have
negligible resistance. The light bulb, however, has a resistance of 96 Ω.
Find (a) the emf produced by the rod,

(b) the current induced in the circuit.

Solution 5.2.7
L = 1.6 m , v = 5.0 m s-1,
B = 0.80 T, R = 96 Ω

(a)

(b)

40

Example 5.2.8
A single turn of circular shaped coil has a resistance of 20 Ω and an
area of 7 cm2. It moves toward the north pole of a bar magnet as
shown in figure below.

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.

41

Solution 5.2.8

a) Induced emf in the coil
Induced current,

42

b) Direction of the induced current

SN

Based on Fleming’s Right hand Rule, direction of
induced current is clockwise.

43

5.3 Self-Inductance

Learning outcomes
a) Define self-inductance
b) Apply self-inductance,

for coil and solenoid, where:

(i) (ii) (iii)

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.

44

From figure below, when the current in the coil changes,
there is an induced emf produced in the same coil. The
phenomenon is called self-induction.

The current in the coil changes: 45
i.At the instant when the switch is turned on.
ii.At the instant when the switch is turned off.
iii.When the resistance of the rheostat is changed.

Polarity of induced εinduced εinduced
emf in the coil Iinduced

Iinduced

(a) A current in the coil produces a magnetic field directed to the left.

(b) If the current increases (B in the coil increases), the increasing
magnetic flux creates an induced emf in the coil. The direction
of induced emf is opposite to the direction of the current.

(c) If the current decreases (B in the coil decreases), the

decreasing magnetic flux creates an induced emf in the coil.

The direction of induced emf is in the same direction as the

current. 46

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

… (1)

From Farady’s law, for a coil of N turns,

… (2)

(1) = (2)

47

Self- inductance for
coil of N turns

Where

I = current

φ= magnetic flux
Φ =magnetic flux linkage

N = number of turns of the coil

48

Self-inductance at the center of a coil of N turns

The magnetic flux passing through each turn of the
coil always maximum and is given by

From

49