CHAPTER 9 ELECTROMAGNETIC INDUCTION
EXERCISE 43, Page 109
1. A conductor of length 15 cm is moved at 750 mm/s at right-angles to a uniform flux density of 1.2 T.
Determine the e.m.f. induced in the conductor.
Length, l = 15 cm = 0.15 m and velocity, v = 750 mm/s = 0.75 m/s
Induced e.m.f., E = B l v = 1.2  0.15  0.75 = 0.135 V
2. Find the speed that a conductor of length 120 mm must be moved at right angles to a magnetic field of
flux density 0.6 T to induce in it an e.m.f. of 1.8 V.
Induced e.m.f., E = B l v from which, speed, v =
E
1.8

= 25 m/s
B l 0.6  0.12
3. A 25 cm long conductor moves at a uniform speed of 8 m/s through a uniform magnetic field of flux
density 1.2 T. Determine the current flowing in the conductor when (a) its ends are open-circuited, (b) its
ends are connected to a load of 15 ohms resistance.
Induced e.m.f., E = B l v = 1.2  0.25  8 = 2.4 V
(a) If the conductor is open circuited, then no current will flow.
(b) Current, I =
E 2.4

= 0.16 A
R 15
4. A straight conductor 500 mm long is moved with constant velocity at right angles both to its length and to a
uniform magnetic field. Given that the e.m.f. induced in the conductor is 2.5 V and the velocity is 5 m/s,
calculate the flux density of the magnetic field. If the conductor forms part of a closed circuit of total
resistance 5 ohms, calculate the force on the conductor.
Induced e.m.f., E = 2.5 V, velocity, v = 5 m/s and length, l = 500 mm = 0.5 m,
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then if E = B l v then flux density, B =
Current, I =
E
2.5
=1T

l v (0.5)(5)
E 2.5

= 0.5 A
R
5
Force on conductor, F = BIl = (1)(0.5)(0.5) = 0.25 N
5. A car is travelling at 80 km/h. Assuming the back axle of the car is 1.76 m in length and the vertical
component of the earth’s magnetic field is 40 T, find the e.m.f. generated in the axle due to motion.
Generated e.m.f, E = B l v = 40 106 1.76 
80 103
= 1.56 mV
60  60
6. A conductor moves with a velocity of 20 m/s at an angle of (a) 90 (b) 45 (c) 30, to a magnetic field
produced between two square-faced poles of side length 2.5 cm. If the flux on the pole face is 60 mWb,
find the magnitude of the induced e.m.f. in each case.
Induced e.m.f., E = B l v sin 
(a) When  = 90, E = B l v sin 90 =

60 103
 l  v  sin 90 
 2.5 10 2  20  sin 90  = 48 V
2
4
A
 2.5 10
(b) When  = 45, E = B l v sin 45 = 48 sin 45 = 33.9 V
(c) When  = 30, E = B l v sin 30 = 48 sin 30 = 24 V
7. A conductor 400 mm long is moved at 70 to a 0.85 T magnetic field. If it has a velocity of 115 km/h,
calculate (a) the induced voltage, and (b) force acting on the conductor if connected to a 8  resistor.
 115 1000 
(a) Induced voltage, E = B l v sin  = (0.85)(0.4) 
 (sin 70) = 10.206 V or 10.21 V
 60  60 
E
 10.206 
(b) Force on conductor, F = B I l sin  = B     l  sin  = (0.85) 
 (0.4)(sin 70)
R
 8 
= 0.408 N
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EXERCISE 44, Page 110
1. A rectangular coil of sides 8 cm by 6 cm is rotating in a magnetic field such that the longer sides cut the
magnetic field. Calculate the maximum generated e.m.f. if there are 60 turns on the coil, the flux density is
1.6 T and the coil rotates at 1500 rev/min.
Generated e.m.f. E = 2 N B L v sin 
where number of turns, N = 60, flux density, B = 1.6 T,
length of conductor in magnetic field, l = 8 cm = 0.08 m,
 1500
 0.06 
 2 rad / s 
m   1.5 m / s ,
velocity, v = r = 
 60
 2

and for maximum e.m.f. induced,  = 90, from which, sin  = 1
Hence, maximum e.m.f. induced, E = 2 N B l v sin 
= 2  60 1.6  0.08 1.51 = 72.38 volts
2. A generating coil on a former 100 mm long has 120 turns and rotates in a 1.4 T magnetic field. Calculate
the maximum e.m.f. generated if the coil, having a diameter of 60 mm, rotates at 450 rev/min.
Generated e.m.f. E = 2 N B L v sin 
where number of turns, N = 120, flux density, B = 1.4 T,
length of conductor in magnetic field, l = 100 mm = 0.10 m,
 450
  0.06 
 2 rad / s  
m   0.45 m / s ,
velocity, v = r = 
 60
 2

and for maximum e.m.f. induced,  = 90, from which, sin  = 1
Hence, maximum e.m.f. induced, E = 2 N B l v sin 
= 2 120 1.4  0.10  0.451 = 47.50 volts
3. If the coils in problems 1 and 2 generates 60 V, calculate (a) the new speed for each coil, and(b) the flux
density required if the speed is unchanged.
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(a) For Problem 1:
Since E = 2 N B l v sin , then with N, B, l and  being constant, E  v
It was shown that 72.38 V is produced by a speed of 1500 rev/min,
Thus,
Hence,
1 V would be produced by a speed of
1500
= 20.72 rev/min
72.38
60 V would be produced by a speed of 60  20.72 = 1243 rev/min
For Problem 2:
It was shown that 47.50 V is produced by a speed of 450 rev/min,
Thus,
Hence,
1 V would be produced by a speed of
450
= 9.474 rev/min
47.50
60 V would be produced by a speed of 60  9.474 = 568 rev/min
(b) For Problem 1:
Since E = 2 N B l v sin , then with N, l , v and  being constant, E  B
It was shown that 72.38 V is produced at a flux density of 1.6 T,
Thus,
Hence,
1 V would be produced by a flux density of
1.6
= 0.0221 T
72.38
60 V would be produced by a flux density of 60  0.0221 = 1.33 T
For Problem 2:
It was shown that 47.50 V is produced at a flux density of 1.4 T,
Thus,
Hence,
1 V would be produced by a flux density of
1.4
= 0.0295 T
47.50
60 V would be produced by a flux density of 60  0.0295 = 1.77 T
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EXERCISE 45, Page 112
1. Find the e.m.f. induced in a coil of 200 turns when there is a change of flux of 30 mWb linking with it in
40 ms.
Induced e.m.f., E =  N
 30 10 3 
d
= - 150 V
 200 
3 
dt
 40 10 
2. An e.m.f. of 25 V is induced in a coil of 300 turns when the flux linking with it changes by 12 mWb. Find
the time, in milliseconds, in which the flux makes the change.
E N
d
N d 300 12 10 3

from which, time for change, dt =
= 0.144 s or 144 ms
dt
E
25
3. An ignition coil having 10000 turns has an e.m.f. of 8 kV induced in it. What rate of change of flux is
required for this to happen?
E N
E 8 103
d
d

from which, rate of change of flux,
=
= 0.8 Wb/s
dt
N 10 000
dt
4. A flux of 35 mWb passing through a 125-turn coil is reversed in 25 ms. Find the magnitude of the average
e.m.f. induced.
 0.35 103   0.35 103 
d
Magnitude of induced e.m.f., E = N
 125 
 = 3.5 V
dt
25 103


(Note that since the flux is reversed, it changes from 35 mWb to - 35 mWb, which is a change of
35 - - 35, i.e. 70 mWb).
5. Calculate the e.m.f. induced in a coil of inductance 6 H by a current changing at a rate of 15 A/s
Induced e.m.f., E =  L
dI
 6 15  = - 90 V
dt
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EXERCISE 46, Page 113
1. An inductor of 20 H has a current of 2.5 A flowing in it. Find the energy stored in the magnetic
field of the inductor.
Energy stored, W =
1 2 1
2
LI   20   2.5  = 62.5 J
2
2
2. Calculate the value of the energy stored when a current of 30 mA is flowing in a coil of
inductance 400 mH.
Energy stored, W =
2
1 2 1
LI   400 103   30 103  = 0.18 mJ
2
2
3. The energy stored in the magnetic field of an inductor is 80 J when the current flowing in the
inductor is 2 A. Calculate the inductance of the coil.
Energy, W =
1 2
LI
2
from which, inductance, L =
2W 2  80

= 40 H
2
I2
 2
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EXERCISE 47, Page 114
1. A flux of 30 mWb links with a 1200 turn coil when a current of 5 A is passing through the coil. Calculate
(a) the inductance of the coil, (b) the energy stored in the magnetic field, and (c) the average e.m.f.
induced if the current is reduced to zero in 0.20 s.
N  1200  30 10 3

(a) Inductance of coil, L =
= 7.2 H
I
5
(b) Energy stored, W =
1 2 1
2
LI   7.2   5  = 90 J
2
2
(c) Induced e.m.f., E = L
dI
50 
 7.2 
 = 180 V
dt
 0.20 
2. An e.m.f. of 2 kV is induced in a coil when a current of 5 A collapses uniformly to zero in 10 ms.
Determine the inductance of the coil.
Induced e.m.f., E = L
dI
dt
from which, inductance, L =
E
2000
2000 10 10 3


=4H
dI
50
5
dt 10 10 3
3. An average e.m.f. of 60 V is induced in a coil of inductance 160 mH when a current of 7.5 A is reversed.
Calculate the time taken for the current to reverse.
Induced e.m.f., E = L
dI
dt
 7.5  7.5 
i.e. 60 = 160  103  

dt


from which, the time taken for the current to reverse, dt =
160 10 3 15
 40 10 3s = 40 ms
60
4. A coil of 2500 turns has a flux of 10 mWb linking with it when carrying a current of 2 A. Calculate the
coil inductance and the e.m.f. induced in the coil when the current collapses to zero in 20 ms.
Inductance, L =
N 2500 10 10 3

= 12.5 H
I
2
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Induced e.m.f., E = L
dI
 20 
= 1.25 kV
 12.5 
3 
dt
 20 10 
5. Calculate the coil inductance when a current of 5 A in a coil of 1000 turns produces a flux of 8 mWb
linking with the coil.
N 1000  8 10 3

Inductance, L =
= 1.6 H
I
5
6. A coil is wound with 600 turns and has a self inductance of 2.5 H. What current must flow to set up a flux of
20 mWb ?
N
Inductance, L =
I
3
N   600   20 10 

from which, current, I =
= 4.8 A
L
2.5
7. When a current of 2 A flows in a coil, the flux linking with the coil is 80 Wb. If the coil inductance is
0.5 H, calculate the number of turns of the coil.
If L =
N
LI
0.5  2

then number of turns, N =
= 12500
 80 10 6
I
8. A coil of 1200 turns has a flux of 15 mWb linking with it when carrying a current of 4 A. Calculate the
coil inductance and the e.m.f. induced when the current collapses to zero in 25 ms.
Inductance, L =
N 1200 15 10 3

= 4.5 H
I
4
Induced e.m.f., E = L
dI
 40 
 4.5 
= 720 V
3 
dt
 25 10 
9. A coil has 300 turns and an inductance of 4.5 mH. How many turns would be needed to produce a 7.2 mH
coil assuming the same core is used?
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Since inductance, L =
N
then for constant Φ and I, L α N
I
i.e. 4.5 10 3 α 300
i.e. 4.5 10 3 = k(300)
from which, k =
4.5 103
 15 106
300
Hence, when L = 0.72 10 3 H, then
from which, number of turns, N =
L = kN
i.e. 0.72 10 3 = (15 106 )(N)
0.72 10 3
= 48 turns
15 106
10. A steady current of 5 A when flowing in a coil of 1000 turns produces a magnetic flux of 500 Wb.
Calculate the inductance of the coil. The current of 5 A is then reversed in 12.5 ms. Calculate the e.m.f.
induced in the coil.
N 1000  500 10 6

Inductance, L =
= 0.1 H
I
5
Induced e.m.f., E = L
dI
 5  5 
  0.1 
= 80 V
3 
dt
 12.5 10 
11. An iron ring has a cross-sectional area of 500 mm 2 and a mean length of 300 mm. It is wound with 100
turns and its relative permeability is 1600. Calculate (a) the current required to set up a flux of 500 Wb
in the coil, and (b) the inductance of the system, and (c) the induced e.m.f. if the field collapses in 1 ms.
(a) Reluctance of ring, S =
l
300 103

= 298415.52 A/Wb
0  r A 4107 1600  500 106
 S 500 10 6  298415.52

NI = S from which, current, I =
= 1.492 A
N
100
100  = 33.51 mH
N2

(b) Inductance, L =
S 298415.52
2
(c) Induced e.m.f., E =  N
or
E = L
 500 10 6 
d
  100  
= - 50 V
3 
dt
 110 
dI
 1.492 
   33.51 10 3  
= - 50 V
3 
dt
 110 
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EXERCISE 48, Page 116
1. The mutual inductance between two coils is 150 mH. Find the magnitude of the e.m.f. induced in one coil
when the current in the other is increasing at a rate of 30 A/s.
Magnitude of the e.m.f. induced, E 2  M
dI1
 150 103   30  = 4.5 V
dt
2. Determine the mutual inductance between two coils when a current changing at 50 A/s in one
coil induces an e.m.f. of 80 mV in the other.
E2  M
E2
dI1
80 10 3

hence, mutual inductance, M =
= 1.6 mH
dI1
dt
50
dt
3. Two coils have a mutual inductance of 0.75 H. Calculate the magnitude of the e.m.f. induced in
one coil when a current of 2.5 A in the other coil is reversed in 15 ms.
Induced e.m.f., E 2  M
dI1
 2.5  2.5 
  0.75  
= 250 V
3 
dt
 15 10 
4. The mutual inductance between two coils is 240 mH. If the current in one coil changes from
15 A to 6 A in 12 ms, calculate (a) the average e.m.f. induced in the other, (b) the change of flux
linked with the other if it is wound with 400 turns.
(a) Induced e.m.f., E 2  M
(b) E = N
d
dt
dI1
 15  6 
   240  103  
= - 180 V
3 
dt
 12 10 
from which, change of flux, d 
Edt 180 12 103

= 5.4 mWb
N
400
5. A mutual inductance of 0.06 H exists between two coils. If a current of 6 A in one coil is reversed in 0.8 s
calculate (a) the average e.m.f. induced in the other coil, (b) the number of turns on the other coil if the flux
change linking with the other coil is 5 mWb
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(a) Induced e.m.f., E 2  M
(b) E = N
d
dt
dI1
 6  6 
 12 
   0.06  
    0.06  
 = - 0.9 V
dt
 0.8 
 0.8 
from which, number of turns, N =
E
E dt 0.9  0.8


= 144
d d
5 103
dt
6. When the current in the primary coil of 400 turns of a magnetic circuit increases linearly from
10 mA to 35 mA in 100 ms, an e.m.f. of 75 mV is induced into the secondary coil of 240 turns,
which is left open circuited. Determine (a) the mutual inductance of the two coils, (b) the
reluctance of the former, and (c) the self inductance of the secondary coil
(a) E S = M
d Ip
dt
from which,
mutual inductance, M =
(b) M 
ES
75 103
= 0.30 H

dIP  (35  10) 103 
dt  100 103 
N P NS
from which,
S
N N
(400)(240)
reluctance, S = P S 
= 320 000 A/Wb or 320 kA/Wb
M
0.30
NS2 (240)2

(c) Secondary self inductance, LS 
= 0.18 H
S
320 000
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