Chapter 5_ Electromagnetic Induction

CHAPTER 5:
ELECTROMAGNETIC

INDUCTION

ELECTROMAGNETIC INDUCTION

❑ Electromagnetic or magnetic induction is the production of an
electromotive force across an electrical conductor in a changing
magnetic field. Michael Faraday is generally credited with the
discovery of induction in 1831, and James Clerk Maxwell
mathematically described it as Faraday's law of induction.

OR

❑ The production of an induced e.m.f in a coil due to the relative
motion of a magnet and the coil (the changing in magnetic field
through coil)

INDUCED CURRENT

The current obtained due to relative motion between magnet or the
coil (the changing in magnetic field through coil) that must induce
an electromotive force (e.m.f)

❑ Induce or induction means producing or transferring something
without direct contact with any physical body. In case of induce
current too, a current is produced into a body without being into
the direct contact of the current carrying conductor …
Induced current is generally produced when current in a
conductor changes , so if you take a magnet towards a
conductor and start moving it to and fro from the conductor , you
will be able to produce induced current

Michael Faraday
The 1st person prove that magnet can
create current

By experiment conducted by Faraday, the deflection of needle in
galvanometer indicates that current is induced in the coil

SUB TOPIC

5.1 Magnetic Flux

5.2 Induced emf
5.3 Self Inductance

5.4 Energy Stored In Inductor

5.5 Mutual Inductance

5.6 Back emf In DC Motor

SUB TOPIC

5.1 Magnetic Flux

5.2 Induced emf

5.3 Self Inductance

5.4 Energy Stored In Inductor

Electromagnetic Induction

Magnetic flux Induced emf Self-inductance Mutual Back
inductance emf

 = B•A Formula Energy Formula 7
 = N
= −  stored
Faraday’s
law L dI 1 M12 = N 2 12
2
 = − d dt U = LI 2 I1

dt Lenz’s law Formula  = −NA dB

 = lvB sin dt

 = NAB sin t  = −NB dA
dt

5.1 Magnetic Flux
(a) Define and use magnetic flux:

 = B • Acos

(b) Use magnetic flux linkage,

 = N

8

MAGNETIC FLUX

Items Explanation number of magnetic field line
Definition
is a measure of the number of passing through an area A.

magnetic field lines linking a

surface of cross-sectional area A



 = B • A  = B Acos

 : magnetic flux 
 : angle bet. Band Normal A

Equation B : magnitude of magnetic 
flux density B
(magnetic field strength)

A : area of surface

Unit T m2 or Wb ( weber ) 9
Quantity scalar quantity

MAGNETIC FLUX


= B•A

 = B Acos



 : angle bet. B and normal A

DIAGRAM  

 = 0  = BA  max .
OR
OR
 = −BA
 = 180

 = 90 =0

10

MAGNETIC FLUX LINKAGE

We now know that the amount of flux Coil with N turns
through one loop of wire is:

 = BA

If we have a coil of wire made up of N
loops of wire the total flux is given by:

N = BAN

NThe total amount of flux, Note:

is called the Magnetic Flux Linkage;  Direction of vector A always
this is because we consider each loop
of wire to be linked with a certain perpendicular (normal) to the
amount of magnetic flux.
surface area, A.
where
B : magnetic field strength  The magnetic flux is proportional to
A : area of the coil the number of field lines passing
θ : angle between B and normal through the area.

of area A 11
N : numbers of turns for the coil

EXAMPLE 5.1 SOLUTION

A flat surface with area 3.0 cm2 is Given:
placed in a uniform magnetic field. If B = 6.0 T
the magnetic field strength is 6.0 T, A = 3.0 cm2 = 3.0 × 10–4 m2
making an angle 30° with the surface θ = 60°
area, find the magnetic flux through
this area.

30

 = B Acos

= 6.0(3 10−4 ) cos 60
= 0.9 10−3 T m2

12

FOLLOW UP EXERCISE

1. A long, straight wire carrying a current of 2.0 A is placed along
the axis of a cylinder of radius 0.50 m and a length of 3.0 m.
Determine the total magnetic flux through the cylinder.

Ans: zero

2. A solenoid 4.00 cm in diameter and 20.0 cm long has 250 turns
and carries a current of 15.0 A. Calculate the magnetic flux
through the circular cross sectional area of the solenoid.

Ans: 2.96×10–5 Tm2

13

5.2 Induced Electromotive Force (emf)

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

(b) State and use Faraday’s law,
(c) State and use Lenz’s law to determine the

direction of induced current.
 = − d

dt

(d) Derived and use induced emf in:

(i) a straight conductor,  = Blv sin

(ii) a coil,  = -NA(dB/dt) or  = -NB(dA/dt)

(iii) a rotating coil,  = NAB sin(t)

14

ELECTROMAGNETIC FARADAY’S
INDUCTION EXPERIMENT

DEFINITION:

Electromagnetic Induction is the

production of induced e.m.f. or

induced currents whenever the
magnetic flux through a loop, coil
or circuit is changed.

NOTED:
No change in magnetic
flux, electromagnetic
induction cannot occur.

15

FARADAY’S EXPERIMENT

NO MOVEMENT BETWEEN MOVEMENT TOWARDS THE LOOP
MAGNET & LOOP

From exp.: From exp.:
➢ No. of magnetic field lines increases
➢ G shows no deflection ➢ Iinduced produced by moving needle

➢ Iinduced = 0 of G

(no current flows in loop) 16

FARADAY’S EXPERIMENT

MOVEMENT AWAY FROM THE LOOP MOVEMENT TOWARDS THE LOOP

From exp.: From exp.:

➢ No. of magnetic field lines decreases ➢ No. of magnetic field lines
➢ Iinduced produced but needle of G
decreases
deflect in the opposite direction. ➢ Iinduced produced but needle of G

deflect in the opposite direction.

17

FARADAY’S EXPERIMENT

Greater area of the coil

Speed of magnet is How to The number of
increased increased turns increased
induced emf?

Increased magnetic flux  = N B Acos
(stronger B)
18

FARADAY’S LAW OF ELECTROMAGNETIC INDUCTION

Faraday’s Law

Definition Equation

The magnitude of induced emf NBAcos
in a closed loop is proportional  = − d
to time rate of change of the
dt
magnetic flux.

Notes: Lenz’s Law  : induced e.m.f
To calculate the magnitude Definition d : change of the magnetic

of induced emf, the flux linkage
negative sign can be dt : change of time
N : number of turns
ignored

the direction of induced emf / current is always in such a direction that is

opposes the change in the magnetic flux that causes it.

19

LENZ’S LAW

When bar magnet moves toward solenoid

APPROACHING Magnitude of B at solenoid increases (  )

N Magnetic flux is changing

e.m.f induced in solenoid

North pole I Current is induced  Indicates by
galvanometer

To determine direction of induced current
Right hand grip rule.

I Since B increase, solenoid create its own B that
oppose the change in magnetic flux.

Therefore, upper end of solenoid becomes N pole.

Direction Iinduced. This oppose is obey the Lenz’s Law.

20

LENZ’S LAW

When bar magnet moves away from solenoid

Magnitude of B at solenoid decreases (  )

Magnetic flux is changing

AWAY e.m.f induced in solenoid

Current is induced  Indicates by
galvanometer

To determine direction of induced current

I Right hand grip rule.

Direction Iinduced. Since B decrease, solenoid create its own B that
North pole oppose the change in magnetic flux.

Therefore, lower end of solenoid becomes N pole.

I

N This oppose is obey the Lenz’s Law. 21

LENZ’S LAW

X XXXX XXX To determine direction of induced current
Q

X XX X X X X X FLEMING’S RIGHT HAND
X XFX X X X RULE
 RIGHT (motion )
Xv X HAND

X XXXX XXX RULE
B
I

X XXXX XXX
P

X XXXX XXX

When straight conductor moves induced I OR
to the left induced emf

with perpendicular to B  From figure, Iinduced flows from Q to P.
at constant speed, v  Since conductor placed in magnetic field,
Current is induced
thus conductor will experience magnetic
To oppose the changing of force.
magnetic flux
 Its direction is in opposite direction to
motion. (or using Fleming’s left hand rule)

22

EXAMPLE 5.2 SOLUTION

Figure shows a permanent As the magnet moves to the right, the
magnet approaching a loop of magnetic flux through the loop
wire. The external circuit increases.
attached to the loop consists
of the resistance R. Find the By Lenz’s law, the induced current must
direction of the induced flow in such a direction that producing
current and the polarity of a magnetic field directed to the left of
the induced emf. the coil. This oppose the change of the
original magnetic flux.

The induced current is anticlockwise :

point A is + ; point B is – . 23

INDUCED EMF

STRAIGHT CONDUCTOR

INDUCED EMF COIL

ROTATING COIL

24

EMF INDUCED IN A STRAIGHT CONDUCTOR
MOVING THROUGH A MAGNETIC FIELD

Metal rod (conductor)

Moves perpendicular to B with 
constant speed
(−) P B
Through a distance x in time, t
 Area swept out, A changed XX XXX X XXX

(A=xl) X X XXX X XX X

Magnetic flux is changing by X I Area, A X
cutting magnetic field, B X X X X X X lX X
X X
Current is induced ind X ind
vX X X X X X X

X XXXX XX

(+)X X X X X xX XQ X X

Metal rail

25

EMF INDUCED IN A STRAIGHT CONDUCTOR
MOVING THROUGH A MAGNETIC FIELD

According to Faraday’s Law;  = − B d (l x) and d x = v

 = − d and  = BAcos dt dt

dt  = − Blv

 = − d (BAcos ) indication of the direction of the

dt induced emf ( Lenz’s Law )

Knowing that B is constant and the angle

between normal of plane with B is 0° , In general, the magnitude of the induced

thus we have: emf is given by

 = − B dA (1)  = Blvsin

dt where

As the rod is pulled, the area of the B : magnetic field strength
circuit increases by an amount l : length of the conductor
v : velocity of conductor
A = l x (2) θ : angle between B and v

Substitute (2) into (1): Note: Direction of Iinduced or induced 
Fleming’s right hand rule.
26

EXAMPLE 5.3 SOLUTION

Consider the arrangement shown (a) Using:
in figure.
 = Blvsin
Assume that R = 6 Ω, L = 1.2m
& a uniform 2.50 T magnetic IR = Blvsin
field is directed into the page.
(a)At what speed should the bar v = IR
be moved to produced a current
of 0.5A in the resistor. Bl sin 
(b)what is the direction of the
induced current ? = 0.5(6) 90
2.50(1.2) sin

v = 1 m s−1

(b) Applying Right Hand Rule, the
direction of the induced current is
from :

b→a→d→c→b

( anticlockwise )

27

EMF INDUCED IN A COIL

 induced   = − d (Faraday’s Law)

dt

if
Change of magnetic flux

By changing

Area of a coil The magnetic field strength

 = − NB dA  = − NA dB

dt dt
28

CHANGING AREA & MAGNETIC FIELD STRENGTH

Stretching Change
the coil magnetic field

will reduces strength
the area

Changes 

Changes 

If: induced  & I If: induced  & I
❑B perpendicular to the plane of
❑B perpendicular to the plane of coil,  coil,  between B & normal A is 0°.

between B & normal A is 0°. ❑Magnitude of A remain constant.

❑Magnitude of B remain constant.

According to Faraday’s Law According to Faraday’s Law

 = − d NBAcos  = − d NBAcos
 = − NB dA  = − NA dB
dt dt dt dt 29

EXAMPLE 5.4 SOLUTION

The flexible loop has a radius of 12 Final × ×
cm and is in a magnetic field of × × A× × ×
strength 0.15 T. The loop is × ×
grasped at point A and B and × ×× × ×
stretched until its area is nearly × ×× × ×
zero. If it takes 0.20 s to close the × ×× × ×
loop, find the magnitude of the × ×× × ×
average induced emf in it during × ××
this time.
× × B×

Apply Faraday’s law:

× × ×A × ×  = − d = − B dA
× ×××× dt dt

× × × × × = − B ( A − A )final initial = − B (0 −  r2 )
× × × × ×
× × × × × t t

× × ×× × = − (0.15) (0 −  (0.12)2 )
× × ×× ×
× × × 0.20
×B ×

= 3.4 10−2 V 30

EXAMPLE 5.5 SOLUTION

A circular coil has 200 turns and (a) Given: N = 200, d = 36 ×10-2 m,

diameter 36 cm. the resistance of B change from 0.5 T→0 T in 0.8 s, R =
the coil is 2.0 Ω. A uniform
2.0 Ω,
magnetic field is applied

perpendicularly to the plane of 0.8 s
the coil. If the field changes

uniformly from 0.5 T to 0 T in

0.8s.

A = r 2 = d 2 =  (36  10−2 )2
4
(a) Find the induced emf & 4
current in the coil while the
field is changed. = 0.1018 m2

(b) Determine the direction of the From Faraday’s Law :
current induced.
 = − d = − d (NBAcos )
31
dt dt

( N & A constant, θ = 0° )

EXAMPLE 5.5 SOLUTION

A circular coil has 200 turns and (a) Given: N = 200 , d = 36 × 10-2m

diameter 36 cm. the resistance of the B change from 0.5 T → 0 T in

coil is 2.0 Ω. A uniform magnetic field 0.8 s,
is applied perpendicularly to the plane
of the coil. If the field changes R = 2.0 Ω,
uniformly from 0.5 T to 0 T in 0.8s.  N & A constant, θ = 0°

 = −NA dB

dt

(a) Find the induced emf &  = − NA(B − B )final initial
current in the coil while the
field is changed. t

(b) Determine the direction of the  = −(200)(0.1018) (0 − 0.5)
current induced.
0.8

 = 12.725 V

32

EXAMPLE 5.5 SOLUTION

A circular coil has 200 turns and (a) Given: N = 200 ,d = 36 × 10-2 m

diameter 36 cm. the resistance of the B change from 0.5 T → 0 T in

coil is 2.0 Ω. A uniform magnetic field 0.8 s,
is applied perpendicularly to the plane
of the coil. If the field changes R = 2.0 Ω,
uniformly from 0.5 T to 0 T in 0.8s.  N & A constant, θ = 0°

 = −NA dB

dt

(a) Find the induced emf &  = − NA(B − B )final initial
current in the coil while the
field is changed. t

(b) Determine the direction of the  = −(200)(0.1018) (0 − 0.5)
current induced.
0.8

 = 12.725 V

 = IR  I =  = 12.725
2
R

I = 6.36 A 33

EXAMPLE 5.5 SOLUTION

A circular coil has 200 turns and (b) Apply Lenz’s law, B from
diameter 36 cm. the resistance of induced current will try to
the coil is 2.0 Ω. A uniform prevent the decrease in flux.
magnetic field is applied By using Right hand grip rule,
perpendicularly to the plane of thus induced current is in
the coil. If the field changes clockwise direction.
uniformly from 0.5 T to 0 T in
0.8s.

(a) Find the induced emf &
current in the coil while the
field is changed.

(b) Determine the direction of the
current induced.

34

FOLLOW UP EXERCISE

1. A circular coil of radius 20 cm is placed in an external
magnetic field of strength 0.2 T so that the plane of the coil is
perpendicular to the field. The coil is pulled out of the field in
0.30 s. Find the average induced emf during this interval.

Ans : 84 mV

2. A 25 turn circular coil of wire has a diameter of 1.0 m. It is
placed with its axis along the direction of earth’s magnetic field
of 50.0 μT, and then in 0.20 s it is flipped 180°. An average

emf of what magnitude is generated in the coil?

Ans : 9.82 mV

3. The plane of a rectangular coil, 5.0 cm by 8.0 cm, is
perpendicular to the direction of a magnetic field B. If the coil
has 75 turns and a total resistance of 8.0 , at what rate must
the magnitude of B change to induce a current of 0.10 A in the
winding of the coil?

Ans : 2.7 T/s 35

EMF INDUCED IN ROTATING COIL

A coil rotates

constant angular velocity, 

In uniform magnetic field

the magnetic flux through the area According to Faraday’s Law:
by the coil changes with time
From: Φ = BA cos θ
e.m.f. & current induced As the θ change, flux change

in the coil θ↓ Φ↓
Flux change → Induced emf /
36
current

EMF INDUCED IN ROTATING COIL

For a coil of N turns, the induced
emf is

 = − N d = − N d (BAcos )

dt dt

Where: B, N and A are constant

If coil has N turns and each has  = − NBA d (cos t)
area, A which rotates with
dt
constant  at an angle  to the
|  |= NBA sin  t
magnetic field, B
Where: t = time
 Thus, the magnetic flux 
through each turns is given by The induced emf is maximum

 = BAcos and  = t when sin t = 1 ,hence
 = BAcost
max = NBA 37
Where:  = 2f = 2

T

THE GRAPH OF E.M.F. INDUCED AGAINST TIME

When a coil rotating with constant angular velocity,  in a uniform

magnetic field, B it will produces a sinusoidally alternating emf as shown

by the graph of induced emf,  against time, t below.

|  |= NBA sin  OR |  |= NBA sin  t

 = t = 90  = t = 0
 max = NBA  =0
38

THE GRAPH OF E.M.F. INDUCED AGAINST TIME

i) The magnitude of induced emf is depends on the angle between
the field and the coil.

ii) The induced emf is an alternating voltage because has positive
value as well as negative value.

iii) The emf induced in a coil varies sinusoidally with time.
iv) Maximum voltage (max = NBA) is produced when the coil is

parallel to the magnetic field.
v) No voltage exists when the coil is perpendicular to the magnetic

field.

39

EXAMPLE 5.6 SOLUTION

Given: A = 0.1 m2 ; B = 0.2 T ;
N = 1000, ω = 60 rev/s

A loop of area 0.10 m2 is rotating (a) Maximum voltage induced
at 60 rev/s with its axis of rotation → sin ωt = 1.0
perpendicular to a 0.20 T
magnetic field.  max = NBA
(a) If there are 1000 turns on = 1000(0.2)(0.1)(120 )

the loop, what is the  max = 7.5103 V
maximum voltage induced in
the loop? (b) When the plane of the loop is
(b) When the maximum induced parallel to the magnetic
voltage occurs, what is the field.
orientation of the loop with
respect to the magnetic field?

40

FOLLOW UP EXERCISE

(1) An AC generator consists of 8 turns of wire, each of area A = 0.09
m2 & the total resistance of the wire is 12 . The loop rotates in a
0.5 T magnetic field at a constant frequency of 60 Hz.
(a) Find the maximum induced emf.
(b) What is the maximum induced current ?
(c) Determine how the induced emf & induced current vary
with time.

Ans : 135.72 V ; 11.31 A

(2) An automobile generator produces 12.0 V when turning at 500
rev/min. What potential difference will it produce at 1200 rev/min?

Ans : 28.8 V

41

5.3 Self-Inductance

(a) Define self-inductance.
(b) Apply self-inductance,

L = −  = oN 2 A (iii)

dI dt l

for a loop and solenoid, where
(i) (ii)

L = N Lcoil = 0N 2 A Lsolenoid = 0N 2 A

I 2r l

42

SELF-INDUCTANCE

Consider a solenoid which is connected ➢ then so too does magnetic flux

to a battery , a switch S and variable linkage. FLUX IS CHANGING!!

resistor R, forming an open circuit as

shown in figure. o According to the FARADAY’s

LAW, an emf has to induced in

SInitial N solenoid itself since the flux
linkage changes.

IS RI ➢ The induced emf opposes the
changes that has induced it &

• When switch S is closed, a current I it is known as a back emf .
(Lenz’s law)
begins to flow.
• Currents produces a magnetic field • This process is known as self-
lines through the solenoid & generate induction.
• Self Induction is the process of
the magnetic flux linkage.

• If the resistance of the variable producing an induced emf in the
changes, thus the current flows in the coil due to a change of current
solenoid also changed. flowing through the same coil. 43

SELF-INDUCTANCE

➢ For the current, I increases - εind
SN
• From equ. B = o I +
2 r Iind SN
IS
 B  Iw, hen I  ; B  I
R Iind
• As the B change,  change

 B  thus   , induced.
• To oppose the change in flux, 

are by decreasing the .

• In order to  , it must flows in
opposite direction as the original
.

• Hence  across solenoid will Direction of the induced emf is in the
opposite direction of the current I.
decrease and I induced flows in
opposite direction of I as well as

the direction of induced.

44

SELF-INDUCTANCE

➢ For the current, I decreases

• B  I, when I  ; B . + -εind
• As the B change,  change.

 B  thus   SS NN

• To oppose the change in flux,  Iind S R Iind
are by increasing the . I I

• In order to  , it must flows in
same direction as the original .

• Hence  across solenoid will Direction of the induced emf is in the
same direction of the current I.
increase and I induced flows in same
direction of I as well as the

direction of induced.

45

SELF-INDUCTANCE

❑ From self-induction phenomenon, Definition

L  I Self Inductance, L is the ratio of the
self induced (back) emf to the rate
L = LI of change of current in the
coil/solenoid.
❑ From Faraday’s Law,
Equation
 = − d L = − d (LI )
L=− 
dt dt
dI / dt
 = −L dI Negative sign indicates
that back emf opposes Depends
dt the increase in current
❑ Size and shape of the coil
Where, ❑ The number of turn (N).
 = induced emf/back emf ❑ The permeability of medium in
dI / dt = the rate of change of
the coil ()
source current
L = Self Inductance Unit
L = magnetic flux linkage
Henry ( H ) or V s A-1 or Wb A-1 46

SELF-INDUCTANCE

SELF INDUCTANCE DIAGRAM FORMULAE
OF A:
 = −N d  and  = −L d I
Coil
dt dt
of N turns
L = N  = L
II

 = NIA and L= N
I
solenoid l

L = o N 2 A or

l L = on2 Al

L : self inductance ; N : number of turns ; ΦL : magnetic flux linkage ; Φ : magnetic flux
in the coils ; I : current flowing in the coil ; μo : permeability of free space ( 4π × 10–7 T
m A–1 ) ; A : cross sectional area of the solenoid ; l : length of the solenoid ; n : number
of turns per unit length (N/l) ; μ : permeability of the material (μ = μr μo)

47

EXAMPLE 5.7 SOLUTION

Induced emf of 5.0 V is Using:  = −L dI
developed across a coil when the
current flowing through it dt
changes at 25 A s-1. Determine
the self inductance of the coil. During calculation, we only
consider the magnitude of the emf
induced.

Minus sign that indicates the
direction of emf ( Lenz’s Law )
can be ignored.

 L= | | = 5

(dI / dt) 25

= 0.2 H

48

EXAMPLE 5.8 SOLUTION

Calculate the magnetic flux Given : N = 300 ;
through the area enclosed by a L = 7.20×10-3 H ;
300 turn, 7.20 mH coil when the I = 10×10-3 A
current in the coil is 10 mA.
Using: L = N

I

  = LI

N
= 7.2 10−3 (10 10−3 )

300
 = 2.4 10 −7 Wb

49

EXAMPLE 5.9 SOLUTION

(a) Calculate the self Given :
inductance of a solenoid
containing 300 turns if the N = 300 ; l = 25×10-2 m ;
length of the solenoid is
25.0 cm & its cross A = 4×10-4 m2 ;dI/dt = 50 A s-1
sectional area is 4 cm2.
(a) Using:
(b) Calculate the self induced
emf in the solenoid if the L = o N 2 A
current through it is
decreasing at the rate of l
50.0 A s-1.
= 4 10−7 (300)2 (4 10−4 )
25 10 −2

= 1.8110−4 H

50