Chapter 5_ Electromagnetic Induction

EXAMPLE 5.9 SOLUTION

(a) Calculate the self Given :
inductance of a solenoid N = 300 ; l = 25×10-2 m ;
containing 300 turns if the A = 4×10-4 m2 ; dI/dt = 50 A s-1
length of the solenoid is
25.0 cm & its cross (b) Using:
sectional area is 4 cm2.
|  |= L dI
(b) Calculate the self induced
emf in the solenoid if the dt
current through it is = 1.8110−4 (50)
decreasing at the rate of
50.0 A s-1. = 9.0510−3 V

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5.4 Energy Stored In Inductor

(a) Use the energy stored in an inductor,

U = 1 LI 2
2

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ENERGY STORED IN INDUCTOR

An inductor: An inductor is a coil that is used specifically for
its property of self inductance.

Symbol of inductor:

Function of an (1) to control current
inductor: (2) store energy in the form of magnetic

field.

❑ Power received by the inductor at time t is:

P = I where  = L dI

P = I L dI dt
dt

Pdt = I L dI where dU = Pdt  dU = I L dI (1)

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ENERGY STORED IN INDUCTOR

❑ The total energy, U received by the inductor when the current
increases from 0 to I can be found integrating the equ.(1):

UI Analogous to energy stored
in capacitor
0 dU = L0 IdI
U = 1 CV 2
U = 1L I2 2
2
L = analogous to C
where I = analogous to V
U : energy stored in an inductor
L : self inductance
I : electric current

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FOLLOW UP EXERCISE

1. A 400 turns solenoid has a cross sectional area 1.81×10-3 m2
and length 20 cm carrying a current of 3.4 A.
(i) Calculate the inductance of the solenoid.
(ii) Calculate the energy stored in the solenoid.
(iii) Calculate the induced emf in the solenoid if the current
drops uniformly to zero in 55 ms.

ANS. : 1.82 10−3 H ; ;1.0510−2 J ;0.1125 V

2. Energy stored in an inductor is 30 mJ when current flowing through
the inductor is 50 mA.
(a) Calculate the self inductance of the inductor.
(b) Calculate the new current if the energy stored increases to 4
times the original energy.

ANS: 24 H ; 100 mA

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5.5 Mutual Inductance
(a) Define mutual inductance.
(b) Use mutual inductance

M = N1N2 A

l

between two coaxial solenoids,

56

MUTUAL INDUCTANCE

Definition

Production of an induced emf in
one coil due to the change of
current in another nearby coil
circuit which both circuit are
magnetically coupled

Unit S NN S
Henry (H)

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MUTUAL INDUCTANCE

Consider two circular close- When current, I1 starts to flow in coil 1,
packed coils near each other and
sharing a common central axis. magnetic field lines will begin to build up

inside the coil and this field lines also pass

through coil 2.

If current, I1 changes with time, magnetic

flux through coils 1 and 2 will change with

time simultaneously.

According to Faraday’s law of induction,
due to the change of magnetic flux through
coil 2, an emf is induced in coil 2.

The production of induced e.m.f. in a coil
(coil 2) caused by a change in the current in
a neighbouring coil (coil 1) is known as
G Mutual induction.

58

MUTUAL INDUCTANCE

At the same time, the self-induction
occurs in coil 1, since magnetic flux
through it changes.

In mutual induction, the e.m.f. induced
in one coil is always proportional to
the rate at which the current in the
other coil is changing.

If we assume that the current in coil 1
changes at a rate of dI1/dt, the
magnetic flux will change by dΦ1/dt
and this changes is experienced by
coil 2.

G

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MUTUAL INDUCTANCE

❑ If I1 in coil 1 changes, the magnetic ❑ According to Faraday’s Law, emf

flux through coil 2 will change and an induced in 2nd coil is:

induced emf will occur in coil 2. 2 = −N 2 d 2
❑ Thus, induced emf in coil 2 is
dt

2  dI 1  2 = −M 21 dI 1 Combining these two equations, we
dt dt obtain :

Where: Lenz’s law − M 21 dI 1 = −N 2 d 2

dt dt

• M21 is mutual inductance of coil 2 with  M 21 dI 1 = N 2 d 2
respect to coil 1

• Negative sign is an expression of Lenz’ M 21 = N 2  2
law, indicating that the induced current
I1
will oppose the change in I1.
magnetic flux linkage

through coil 2

Keep In mind 60
emf induced in coil 2 is due to the current change in coil 1

MUTUAL INDUCTANCE

❑ The induced e.m.f. in coil 1 is  M 12 dI 2 = N 1 d1

1  dI 2  2 = −M 12 dI 2 M 12 = N1  1
dt dt
I2
Lenz’s law
Where: magnetic flux linkage
• M12 is mutual inductance of coil 1 with through coil 1

respect to coil 2. DEFINITION :
MUTUAL INDUCTANCE
❑ According to Faraday’s Law, emf
is defined as the ratio of induced emf
induced in 2nd coil is: in a coil to the rate of change of
current in another coil.
1 = −N1 d 1

dt

Combining these two equations, we

obtain :

− M 12 dI 2 = −N 1 d1 Keep In mind

dt dt emf induced in coil 1 is due to the

current change in coil 2 61

MUTUAL INDUCTANCE

The magnetic flux linkage through The Mutual Inductance of 2nd coil
coil 2 caused by the magnetic is
field of the coil 1 is
M 21 = N 2  2
 2 = B1 A...(1)
I1

Where uniform magnetic field M 21 = N 2  o N 1I 1  A
generated at center of the coil is I1 
l

B1 = o N1 I 1 .....(2) M 21 = o N1 N2 A
l
l

Subs. (2) into (1) Where;
μo : permeability of the free space
2 = o N1 I1 ( A) N1 : number of turns in coil 1
l N2 : number of turns in coil 2
l : length of coil 1

A : cross sectional area of coil 1

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MUTUAL INDUCTANCE

Hence, M 21 = N 2  2 and Generalizing, we can omit the
subscript, thus we get :
I1
1 = −M dI 2 and  2 = −M dI 1
M 12 = N1  1 dt dt

I2

It is found that in both expressions Where;
for 1 and 2 , the value of M is the M = Mutual Inductance
same.
M = o N 1 N 2 A
M 12 = M 21
l
For a given pair of coils, the value
of mutual inductance is the same
and does not depend on which
coil carries the current and which
coil experiences induction.

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EXAMPLE 5.10 SOLUTION

Two coils, X & Y are magnetically (a) Using:
coupled. The emf induced in coil
Y is 2.5 V when the current Y = −M dIX
flowing through coil X changes at dt

the rate of 5 A s-1. Determine: M = | Y | = 2.5  M = 0.5H
(a) the mutual inductance of the
dIX 5

coils. dt

(b) the emf induced in coil X if (b) magnitude of the emf induced in
there is a current flowing coil X:
through coil Y which changes
at the rate of 1.5 A s-1. Using:

|X |= M dIY
dt

= 0.5(1.5)

 x = 0.75V

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EXAMPLE 5.11 SOLUTION

Primary coil of a cylindrical Given :
former with the length of 50 cm
and diameter 3 cm has 1000 N1 = 1000 ; l = 50×10-2 m ;
turns. If the secondary coil has
50 turns, calculate : d1 = 3×10-2 m ; N2 = 50 ;

(a) its mutual inductance dI 1 = 4.8 A s−1
(b) the induced emf in the dt

secondary coil if the current (a) Using : o N 1 N 2 (d 2 )
flowing in the primary coil is
changing at the rate of 4.8 M = o N 1 N 2 A = 4
As-1.
ll

4 10−7 (1000) (50) ( (310−2 )2 )
4
=
50 10−2

M = 8.8810−5 H

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EXAMPLE 5.11 SOLUTION

Primary coil of a cylindrical Given :
former with the length of 50 cm
and diameter 3 cm has 1000 N1 = 1000 ; l = 50×10-2 m ;
turns. If the secondary coil has
50 turns, calculate : d1 = 3×10-2 m ; N2 = 50 ;

(a) its mutual inductance dI 1 = 4.8 A s−1
(b) the induced emf in the dt

secondary coil if the current (b) Using :
flowing in the primary coil is
changing at the rate of 4.8 2 = −M dI 1
As-1. dt

= 8.8810−5 (4.8)

 2 = 4.26 10−4 V

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FOLLOW UP EXERCISE

1. 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 mutual inductance of the coils.
(b) if the current in C decreasing uniformly from 3.0 A to zero in
0.25 s, calculate the induced emf in coil D.
( ANS : 417 H ; 5004 V )

2. Suppose that two coils of a common cylindrical former of length, l =
0.1 m and cross-sectional area of 0.05 m2. The number of turns in
the 1st coil is N1 = 100, and the number of turns in the 2nd coil is N2
= 300.
(a) What is the mutual inductance of the two coils?
(b) If the current flowing in the 1st coil increases uniformly from 0
to A in s, what emf is generated in the 2nd coil?
( ANS : 0.0188 H; -1.88 V )

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5.6 Back emf In DC Motor
(a) Explain back emf and its effect on DC motor.

68

BACK EMF IN DC MOTOR

❑ A simple motor has a coil FB
placed inside a magnetic field.
I induced
❑ An external DC voltage ( ex:
battery ) is applied across the + I induced
coil to supply a current to the
coil.

-

FB

Consider a motor rotates when a
current, I flow in it.

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BACK EMF IN DC MOTOR FB

When a current, I flows through the coil I induced

in the magnetic field, magnetic forces will - + I induced
exist
FB
However, as a coil of motor rotate, it will
‘cutting’ the magnetic field lines

Thus, the changes in the magnetic flux
occur

Induced e.m.f called back e.m.f, b is
produced through the coil

[opposes the applied voltage & tends to
reduce the current in the coil]

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BACK EMF IN DC MOTOR

❑ b is directly proportional to Where;
angular velocity () of the coil. b : back emf
V : voltage across the coil
❑ If V is the voltage applied I : current through the coil
across the coil, then the net R : resistance of the coil
voltage driving the motor is
less than V : (1)  I :

Vnet = V −  b  Ib = IV − I 2R

❑ For a motor with a coil of Where;
internal resistance R, the I b : mechanical power
current the motor draws while I 2R : rate of heat loss in coil
in operation is: IV : power supplied by the
source
I = Vnet = V − b

RR

 b = V − IR (1)

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BACK EMF IN DC MOTOR

❑ When a motor is just turned on, there is initially no back emf ; the
current is very large.

❑ As the coil begin to rotate, the induced back emf opposes the
applied voltage & the current in the coil is reduced.

❑ Under normal load conditions, the back emf is less than the applied
voltage ( b <V ).

❑ The larger the load, the slower the motor rotates & the smaller the
back emf is (b  ).

❑ If a motor is overloaded & turns very slowly, the back emf may be
reduced so much that the current becomes very large, this may
burn out the coils.

❑ Thus the back emf plays a vital role in the regulation of a motor’s
operation by limiting the current in it.

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EXAMPLE 5.12 SOLUTION

The coil of an ac motor has a (a) Concept: motor just start up,
resistance of R = 4.1 Ω. The the coil is not
motor is plugged into an outlet rotating. No back emf
where V = 120 V and the coil induced in coil.
develops a back emf of b = 118
V when rotating at normal speed. b = V − IR
The motor is turning a wheel. I = V − b = 120 = 29.27A
Find
(a)the current when the motor R 4.1
first start up.
(b)the current when the motor is (b) A normal speed, the motor
operating at normal speed. develops a back emf 118.0 V.

b = V − IR 73

I = 120 −118
4.1

= 0.49 A

FOLLOW UP EXERCISE

1. A rectangular coil of motor has 10 turns and area 20.0 cm2 rotates
in a uniform magnetic field of magnitude 2.0 T at rate of 5
revolutions per second. Calculate
(a) the maximum induced e.m.f.
(b) the current if resistance of motor is 5  and it supplied by
external voltage of 5 V.
(ANS : 1.26 V ; 0.748 A)

2. A motor has coils with resistance of 20  and is supplied by battery
of 15 V. When motor is running at its maximum speed, the back
emf is 5 V. Calculate the current in the coils when
(a) the motor is first turned on.
(b) the motor has reached maximum speed.
(ANS : 0.75 A ; 0.5A)

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