CHAPTER 5: ELECTROMAGNETIC INDUCTION

Therefore the self-inductance
at the center of the coil,

Where

μο = 4π ×10-7 H m-1

r = radius of the coil
A = cross sectional area of the coil
N = number of turns of the coil

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Self-inductance at the center of the air-core solenoid
The magnetic flux passing through each turn of the
solenoid always maximum and is given by

l

X X XX

From

51 51

Therefore the
self-inductance at the
center the air-core
solenoid

Where

μο = 4π ×10-7 H m-1

l = length of the solenoid
A = cross sectional area of the solenoid
N = number of turns of the solenoid

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Scalar quantity.

Unit : henry (H) , [1 H = 1 Wb A-1= 1 T m2 A-1]

The value of the self-inductance depends on
i. the size and shape of the coil/solenoid,
ii. the number of turn (N),
iii. the permeability of the medium in the coil/solenoid(μ).

A circuit element which possesses mainly self-inductance is
known as an inductor.

The inductor is used to store energy in the form of
magnetic field.
The symbol of inductor in the electrical circuit :

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Equations for self-inductance

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Example 5.3.1
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.
Solution 5.3.1

N = 500 turns, l = 8.0 × 10-2 m, Ii = 0 A, If= 2.5 A, t = 0.35 s,
ε = 0.012 V
a) Inductance of the solenoid,

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b) Cross-sectional area of the solenoid,
From

c) Final magnetic flux linkage through the solenoid,
From

Wb
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5.4 Energy stored in inductor

Learning outcome
a) Derive and use the energy stored in an inductor,

Consider an ε
inductor is
connected in
series to a dc
power supply and
a switch as shown
in figure.

When the switch
is closed.

Switch 57

When the switch is closed, a current will flow from zero
to its steady value, I.
Due to self induction an induced emf (back emf) in
inductor is

As the current starts from its zero value and flowing
against the induced emf ε, the energy will increases
gradually from zero value to U.

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The electrical power absorbed/received by the
inductor is

and … (1)
… (2)
(1)=(2)
59

So, the total energy stored in the inductor when the
current has just reach a maximum constant value I is

Energy stored in an inductor
Energy stored in a solenoid

where

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Example 5.4.1
A solenoid of length 25 cm an air core consists of 100 turns and
diameter of 2.7 cm. Calculate
a) the self-inductance of the solenoid,
b) the energy stored in the solenoid if the current flows in it is 1.6 A.
Solution 5.4.1
N = 100 turns, l = 2.5 × 10-2 m, d = 2.7 × 10-2 m, I =1.6 A

Cross sectional area of
the solenoid,

a) Self-inductance of the solenoid,

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b) The energy stored in the solenoid

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5.5 Mutual inductance between two coaxial

Learning outcome

a) Define mutual inductance.

b) Use mutual inductance,
solenoids.

Mutual induction

Mutual induction is a phenomenon by which a
changing electric current in a coil produces an
induced emf across the other coil.

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Mutual inductance

Consider two coils, Coil 1 and Coil 2, placed near to
each other.

Coil 1 Coil 2

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When the current I1 in Coil 1 changes, the magnetic
flux also changes. This changes is experienced by
Coil 2. Therefore an induced emf ε2 is produced in
Coil 2.

In the similar manner, current I2 in Coil 2 changes, the
magnetic flux also changes. This changes is
experienced by Coil 1and an induced emf ε1 is
produced in Coil 1.

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Mutual inductance, M

Mutual inductance is defined as the ratio of induced
emf in a coil to the rate of change of current in
another coil.

Scalar quantity
Unit : henry (H).

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From Faradays Law, Coil 1
Coil 2

Where

M = mutual inductance
φ1=magnetic flux through coil 1
φ2= =magnetic flux through coil 2
I1 = current flows in coil 1
I2 = current flows in coil 2
N1 = number of turns of coil 1
N2 = number of turns of coil 2

Mutual inductance between two coaxial solenoids
When a current I1 flows in Coil 1 (N1), it produces a magnetic
field B1,
The magnetic flux in Coil 1,

If no magnetic flux leakage,

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Mutual inductance between
two coaxial solenoids

Where

M = mutual inductance
l = length of the solenoid
A = cross sectional area of the solenoid
N1 = number of turns of solenoid 1
N2 = number of turns of solenoid 2

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Equations for mutual inductance

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Example 5.5.1

A coil of 20 turns is wrapped round a solenoid having a length of
18 cm and a total of 24 00 turns. The diameter of the solenoid is
6.0 cm.
a)Determine the mutual inductance of the coil and solenoid
b)If the solenoid carries a current of 2 A, find the magnetic flux in
the coil
c)If the current in the solenoid changes at the rate of 15 A s-1, find the
emf induced in the coil.

Solution 5.5.1

Ncoil = 20 1tu0r-n2 sm, ,NIso=len2oiAd =, 2400 turns, l= 18 × 10-2 m,
d = 6.0 × dI/dt= 15 A s-1

Cross sectional area of
the solenoid,

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a) Mutual inductance, M
From

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b) Magnetic flux in the coil, φcoil

From

74

c) Emf induced in the coil, εcoil

From

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Exercise

1. 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. (3.40 T)

2. A coil with 500 turns has an air-core with a radius of 2 cm. It
is placed in a field of 0.6 T such that there is an angle of 30o
between the field and the normal to the cross-sectional area.
Calculate the magnetic flux and the magnetic flux linkage.
(6.5 ×10-4 Wb, 0.33 Wb)

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3. A circular loop of area 5 × 10-2 m2 rotates in a uniform magnetic
field of 0.2 T. If the loop rotates about its diameter which is
perpendicular to the magnetic field. Find the magnetic flux
linked with the loop when its plane is
(i) normal to the field
(ii) inclined 60o to the field and
(iii) parallel to the field.
(1×10-2Wb, 8.66 ×10-3 Wb, 0 Wb)

4. 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.

(8.0 ×10-3 Wb, 1.6×10-2 V)

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5. 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.
(1.92×10−5 Wb)

6. The induced emf in a single loop of wire has a magnitude of
1.48V when the magnetic flux is changed from 0.850 Wb to
0.110 Wb. How much time is required for this change flux?
(0.5 s)

7. The magnetic flux passing through a single turn of a coil is

increased quickly but steadily at a rate of 5.0 x 10-2 Wb s-1. If

the coil have 500 turns, calculate the magnitude of the induced

emf in the coil. (25 V)

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8. The flexible loop has a radius of 12 cm and is in a magnetic
field of 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. (3.4×10-2 V)

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9. A rectangular coil of 100 turns has a dimension of 10 cm x
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.
(157 V, 124 V)

10. A rectangular coil of area 70 cm2 having 600 turns rotates
about an axis perpendicular to a magnetic field of 0.4 Wb m-2.
If the coil completes 500 revolutions in a minute, calculate the
instantaneous emf when the plane of the coil is
(i) perpendicular to the field
(ii) parallel to the field and
(iii) inclined at 60o with the field.
(0 V, 88 V, 44 V)

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11. 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.
(1.88×10−4 T, 3.33×10−8 Wb, 3.75×10−4 H)

12. 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.
(0.039 H m−1, 4.4×10−2 J m−1)

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13. 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.
(0.250 H, 4.5×10−4 Wb, 6.48×10−2 J)

14. 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. (19 A s−1)

15. An emf of 9.7×10−3V is induced in a coil while the current in a

nearby coil is decreasing at a rate of 2.7 A/s. What is the mutual

inductance of the two coils? (3.6 ×10−3 H) 82