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Definition – is defined as the production
of an induced e.m.f.
e.m.f. in a conductor/coil
whenever the magnetic flux through the
conductor/coil changes.
UNIT 7: ELECTROMAGNETIC
INDUCTION
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1
7.1 The Phenomenon of Electromagnetic
Induction
{
Consider some experiments were conducted by Michael Faraday that
led to the discovery of the Faraday’s law of induction as shown in
figures 7.1a, 7.1b, 7.1c, 7.1d and 7.1e.
v =0
No movement
Fig. 7.1a
v
S
N
I
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Move towards the coil
I
Fig. 7.1b
2
SF017
v =0
No movement
Fig. 7.1c
v
N
S
I
Move away from the coil
I
Fig. 7.1d
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3
v
S
N
I
Move towards the coil
I
Fig. 7.1e
{
From the experiments :
When the bar magnet is stationary, the galvanometer not show any
deflection (no current flows in the coil).
z
When the bar magnet is moved relatively towards the coil, the
galvanometer shows a momentary deflection to the right. When the
bar magnet is moved relatively away from the coil, the
galvanometer is seen to deflect in the opposite direction (Fig.7.1d).
Therefore when there is any relative motion between the coil and
the bar magnet , the current known as induced current will flow
momentarily through the galvanometer. This current due to an
induced e.m.f across the coil.
z
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4
SF017
{
Conclusion :
z
When the magnetic flux through a coil changes (magnetic field lines
been cut) thus the induced e.m.f. will exist across the coil.
z
The magnitude of the induced e.m.f. depends on the speed of the
relative motion where when
v increase
induced e.m.f.
e.m.f. also increase
v decrease
induced e.m.f.
e.m.f. also decrease
v is proportional to the induced e.m.f.
e.m.f.
7.2 Faraday’s law and Lenz’s law
7.2.1 Faraday’s law of induction
{
States “ the magnitude of the induced e.m.f.
e.m.f. is proportional to the
rate of change of the magnetic flux.”
flux.
Mathematically,
dΦB
dΦB
or ε = −
dt
dt
dΦB : change of magnetic flux
dt : change of time
ε : induced e.m.f.
ε ∝−
where
(7.2a)
The negative sign indicates that the direction of induced e.m.f. always
5
oppose the change of magnetic flux producing it (Lenz’s law).
SF027
{
For a coil of N turns, eq. (7.2a) can be written as
dΦB
(7.2b)
dt
dΦB = Φ f − Φi , then eq. (7.2b) can be written as
ε = −N
{
Since
ε = −N
{
{
(Φ f − Φi )
dt
where
Φ f : final magnetic flux
Φi : initial magnetic flux
From the definition of magnetic flux,
ΦB = BA cos θ then eq. (7.2a) also can be written as
d ( BA cos θ )
ε =−
dt
Note :
if the coil is connected in series to a resistor of resistance R and the
induced e.m.f ε exist in the coil as shown in figure 7.2a.
z
Therefore the induced current I is
given by
dΦ
ε = − B and ε = IR
Fig. 7.2a
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I
R
I
dt
dΦB
IR = −
dt
6
SF017
z
z
To calculate the magnitude of induced e.m.f., the negative sign can
be ignored.
If the coil has N turns, then each of turns will have a magnetic flux,
ΦB of BAcos θ through it, therefore the magnetic flux linkage
(refer to the combined amount of flux through all the turns) is given
by
magnetic flux linkage = NΦB
{
Example 1 :
A rectangular coil of sides 10 cm x 5.0 cm is placed between N and S
poles with the plane of the coil parallel to the magnetic field as shown
in figure below.
R
Q
N
S
I
I
S
P
If the coil is turned by 90° about its rotation axis and the magnitude of
magnetic flux density is 1.0 T, find the change in the magnetic flux
through the coil.
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7
Solution: A=(10x10-2)(5.0x10-2)=50x10-4
Initially,
r
A
r
B
m2,B=1.0 T
From the figure, θ =90° thus the
initial magnetic flux through the coil is
Φi = BA cos θ
Φi = 0
Finally,
r
B
r
A
From the figure, θ =0° thus the final
magnetic flux through the coil is
Φ f = BA cos θ
Φ f = 50 x10 −4 Wb
Therefore the change in magnetic flux through the coil is
∆ΦB = Φ f − Φi
∆ΦB = 50 x10 −4 Wb
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8
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{
Example 2 :
The magnetic flux passing through a coil of 1000 turns is increased
quickly but steadily at rate of 2.0 x 10-2 Wb s-1. Calculate the induced
e.m.f. in the coil.
dΦB
= 2.0 x10 − 2 Wb s -1
dt
By applying the Faraday’s law equation for a coil of N turns , thus the
Solution:
N=1000 turns,
induced e.m.f. is
{
dΦB
dt
ε = −20 V
ε = −N
Example 3 :
A circular shaped coil 3.0 cm in radius, containing 20 turns and have a
resistance of 5.0 Ω is placed perpendicular to a magnetic field of flux
density of 5.0 x 10-3 T. If the magnetic flux density is reduced steadily to
zero in time of 2.0 ms, calculate the induced current flows in the coil.
Solution:
N=20 turns, r=3.0x10-2 m, R=5.0 Ω, Bi=5.0x10-3 T
, Bf=0, dt=2.0x10-3 s
The area of the circular shaped coil is
A = πr 2
A = 2.8 x10 −3 m 2
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initially,
r
B
r
A
9
From the figure, θ =0° thus the
change in magnetic flux through the
coil is
dΦB = Φ f − Φi
dΦB = B f A cos θ − Bi A cos θ
dΦB = − Bi A
By applying the Faraday’s law equation for a coil of N turns , thus
dΦB
and ε = IR
dt
(− Bi A)
IR = − N
dt
I = 3.0 x10 −2 A
ε = −N
{
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Example 4 : (exercise)
A flat coil having an area of 8.0 cm2 and 50 turns lies perpendicular to a
magnetic field of 0.20 T. If the flux density is steadily reduced to zero,
taking 0.50 s, find
a. the initial flux through the coil.
b. the initial flux linkage.
c. the induced e.m.f. (Lowe&Rounce,pg.206,no.1)
Ans. : 1.6 x 10-4 Wb, 80 x 10-4 Wb, 16 mV
10
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Direction of
induced current –
Right hand grip
rule.
North pole
N
7.2.2 Lenz’s law
{
States “ an induced electric current always flows in such a
direction that it opposes the change producing it.”
it.
{
This law is essentially a form of the law of conservation of energy.
energy
{
An illustration of lenz’s law can be shown by using the experiments
below.
First experiment : (figure 7.2b)
I
I
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X
X
X
X
X
X
z
In figure 7.2b the magnitude of the
magnetic field at the solenoid increases as
the bar magnet is moved towards it.
z
An e.m.f is induced in the solenoid and
galvanometer indicates that a current is
flowing.
z
To determine the direction of the current
through the galvanometer which
corresponds to a deflection in a particular
sense, then the current through the
solenoid seen is in the direction that make
the solenoid upper end becomes a north
pole. This opposes the motion of the bar
magnet and obey the lenz’s law.
11
Fig. 7.2b
Second experiment : Consider a straight conductor PQ is placed
perpendicular to the magnetic field and move
X X QX X X X X
the conductor to the left with constant velocity v
X X X X X X X as shown in figure 7.2c.
z
When the conductor move to the left thus the
r
r
induced current needs to flow in such a way
X v X XFBX X X X
to oppose the change which has induced it
X X X X X X X
based on lenz’s law. Hence galvanometer
I
shows a deflection.
X X X X X X X
z
To determine the direction of the induced
current (e.m.f.) flows in the conductor PQ,
X X PX X X X X
the Fleming’s right hand (Dynamo) rule is
Fig. 7.2c
used as shown in figure 7.2d.
r
B
r
ν (motion)
z
Therefore the induced current flows from Q to P
as shown in fig. 7.2c.
z
Since the current flows in the conductor PQ and
is placed in the magnetic field then this conductor
will experience magnetic force.
Fig. 7.2d
z
Its direction is in opposite direction of the motion.
Important
induced I or e.m.f.
Only for the
straight
conductor.
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Thumb – direction of Motion
First finger – direction of Field
12
Second finger – direction of Induced current or Induced e.m.f.
.
e.m.f
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Third experiment : Consider two solenoids P and Q arranged coaxially
closed to each other as shown in figure 7.2e.
ε ind
S
N
P
I Switch, S
I
N
+
I ind
Q
S
-I
ind
Fig. 7.2e
z
z
z
SF027
z
z
At the moment when the switch S is closed,
closed current I begins to
flow in the solenoid P and producing a magnetic field inside the
solenoid P. Suppose that the field points towards the solenoid Q.
The magnetic flux through the solenoid Q increases with time.
time
According to Faraday’s law ,an induced current due to induced
e.m.f. will exist in solenoid Q.
The induced current flows in solenoid Q must produce a magnetic
field that oppose the change producing it (increase in flux). Hence
based on Lenz’s law, the induced current flows in circuit consists of
solenoid Q is anticlockwise (fig. 7.2e) and galvanometer shows a
deflection.
13
At the moment when the switch S is opened,
opened the current I starts
to decrease in the solenoid P and magnetic flux through the
solenoid Q decreases with time.
time According to Faraday’s law ,an
induced current due to induced e.m.f. will exist in solenoid Q.
The induced current flows in solenoid Q must produce a magnetic
field that oppose the change producing it (decrease in flux). Hence
based on Lenz’s law, the induced current flows in circuit consists of
solenoid Q is clockwise (fig. 7.2f) and galvanometer seen to
deflect in the opposite direction of fig.7.2e.
ε ind
S
N
P
I Switch, S
I
S
-
I ind
Q
N
+
I ind
Fig. 7.2f
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14
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{
Example 5 :
A single turn circular shaped coil has resistance of 10 ohm and area of
its plane is 5.0 cm2. It moves towards the north pole of a bar magnet as
shown in figure below.
If the average rate of change of magnetic flux density through the plane
of the coil is 0.50 T s-1, determine the induced current in the coil and
state the direction of the induced current observed by the observer
shown in figure above.
Solution:
N=1 turn, R=10 Ω, A=5.0x10-4 m2,
dB
= 0.50 T s -1
dt
By applying the Faraday’s law equation for a coil of N turns , thus
dΦB
o
where Φ B = BA cos 180 and ε = IR
dt
d (− BA)
NA  dB 
IR = − N
I=
I = 2.5 x1015−5 A
 
dt
R  dt 
ε = −N
SF027
Based on the lenz’s law, hence the direction of induced current is
clockwise as shown in figure below.
SI
{
ind
N
Example 6 : (exercise)
A bar magnet is held above a loop of wire in a horizontal plane, as
shown in figure below.
The south end of the magnet is toward the loop of the
wire. The magnet is dropped toward the loop. Find the
direction of the current through the resistor
a. while the magnet falling toward the loop and
b. after the magnet has passed through the loop and
moves away from it.
(Serway&Jewett, pg.991, no.15)
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16
SF017
7.3 Induced E.m.f. in a linear conductor.
{
Consider a linear (straight) conductor PQ of length L is moved
perpendicular with velocity v across a uniform magnetic field B as
shown in figure 7.3a.
r
P
X X X X X X X XB
X
X
X X
rX
v
X X
X
X X
X
X X
X
X
X X
X X
X
X
X
X
LX
X
X
X
X
X
X X
X
X
X X X x X QX X
Fig. 7.3a
X
{
When the conductor moved through a distance x in time t, the area
swept out by the conductor is given by
{
Since the motion of the conductor is perpendicular to the magnetic field
B hence the magnetic flux cut by the conductor is given by
A = Lx
Φ B = BA cos 0 o
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{
Φ B = BLx
17
According to Faraday’s law, the e.m.f. is induced in the conductor and
its magnitude is given by dΦ
ε=
B
dt
d (BLx )
ε=
dt
dx
dx
ε = BL
=v
and
dt
dt
ε = BLv
{
In general, the magnitude of the induced e.m.f. in a linear conductor is
given by
(7.3a)
ε = BLv sin θ
where θ : angle between
In vector form,
form
(
r r
ε = L• v×B
{
{
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)
r
r
v and B
(7.3b)
The induced e.m.f. exist in the linear conductor when cutting the
magnetic flux is also known as motional induced e.m.f.
e.m.f.
The direction of the induced current due to induced e.m.f. flows in
the linear conductor can be determine by using Fleming’
Fleming’s right hand
rule (based on lenz’s law).In case of figure 7.3a, the induced current
18
flows from P to Q.
SF017
{
{
{
Note that the eq. (7.3a) also can be used for the rectangular coil of one
turn moved across the uniform magnetic field.
For a rectangular coil of N turns, ε = NBLv sin θ
(7.3c)
Example 7 :
A 20.0 cm long metal rod PQ is moved at speed of 100 m s-1 across a
uniform magnetic field of flux density 100 mT. The motion of the rod is
perpendicular to the magnetic field as shown in figure below.
r
P
a. Calculate the motional induced e.m.f
B
in the rod.
100 m s −1b. If the rod is connected in series to the
resistor of resistance 10.0 Ω, determine
i. the induced current and its direction.
ii. the total charge passing through the
Q
resistor in one minute.
iii. the electrical energy dissipated through the resistor in one minute.
Solution:
L=20.0x10-2 m, v=100 m s-1,B=100x10-3 T, θ=90°
a. By applying the equation of motional induced e.m.f in the linear
conductor, thus the induced e.m.f. is ε = BLv sin θ
ε = 2.00 V
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b. Given R=10.0
19
Ω
i. From the Ohm’s law , thus
ε = IR
I = 0.200 A
Direction : using Fleming’
Fleming’s right hand rule
ii. Given t=60.0
From P to Q
s
The total charge flows through the resistor is
Q = It
Q = 12.0 C
iii. Given t=60.0 s
By using the equation of electrical energy, thus
E = εIt or E = I 2 Rt
E = 24.0 J
{
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Example 8 : (exercise)
A linear conductor of length 20 cm moves in a uniform magnetic field of
flux density 20 mT at a constant speed of 10 m s-1. The velocity makes
an angle 30° to the field but the conductor is perpendicular to the field.
Determine the induced e.m.f. across the two ends of the conductor.
Ans. : 2.0 x 10-2 V
20
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7.4 Induced E.m.f. in a Rotating Coil
{
Consider a rectangular coil of N turns, each of area A, being rotated
mechanically with constant angular velocity ω in a uniform magnetic
field of flux density B about an axis as shown in figure 7.4a.
N
{
θ
r
A
S
r
B
Fig. 7.4a : side view
When the vector of area, A is at an angle θ to the magnetic field the
flux ΦB through each turn of the coil is given by
ΦB = BA cos θ and θ = ωt
ΦB = BA cos ωt
{
where t : time that has elapsed since θ = 0
By applying the equation of Faraday’s law for a coil of N turns, thus the
induced e.m.f. is given by
ε = −N
SF027
o
dΦB
dt
21
d (BA cos ωt )
dt
d (cos ωt )
ε = − NBA
dt
ε = −N
ε = NBAω sin ωt
{
From the eq. (7.4a), the induced e.m.f. varies with time, t where
Induced e.m.f maximum
therefore
ε max = NBAω where
{
sin ωt = 1 or ωt = θ = 90 o
2π
ω = 2 πf or ω =
T
Conclusion : A coil rotating with constant angular velocity in a uniform
magnetic field produces a sinusoidally alternating e.m.f.
e.m.f. is shown by
the graph in figure 7.4b.
ε (V)
ε = εmax sin ωt
ε max
0
SF027
(7.4a)
− ε max
0.5T
T
1.5T
2T
t
Fig. 7.4b : induced e.m.f.
e.m.f. ε against time t graph
22
SF017
{
Note :
This phenomenon was the important part in the development of
the electric generator or dynamo.
z
Eq. (7.4a) also can be written as
z
ε = NBAω sin θ
(7.4b)
r
r
where θ : angle between A and B
{
Example 9 :
A rectangular coil of 200 turns has size 10 cm x 15 cm. It rotates at a
constant angular velocity of 600 r.p.m. in a uniform magnetic field of
flux density 20 mT. Calculate
a. the maximum e.m.f. produced by the coil.
b. the induced e.m.f. at the instant when the plane of the coil makes an
angle of 60° with the magnetic field.
Solution: N=200
turns, A=(10x10-2)(15x10-2)=150 x 10-4 m2
, B=20x10-3 T , ω = 600 x 2π = 20π rad s -1
60
a. By applying the equation of maximum induced e.m.f. for rotating coil,
thus
ε max = NBAω
εmax = 3.77 V
SF027
b.
r
B
From the figure, θ=90°-60°=30°
60 o
θ
{
23
Hence the induced e.m.f. is
r
A
ε = NBAω sin θ and NBAω = εmax
ε = εmax sin θ
ε = 1.88 V
Example 10 : (exercise)
A coil of area 0.100 m2 is rotating at 60.0 rev s-1 with the axis of rotation
perpendicular to a 0.200 T magnetic field.
a. If the coil has 1000 turns, find the maximum e.m.f. generated in it.
b. What is the orientation of the coil with respect to the magnetic field
when the maximum induced e.m.f. occurs?
(Serway&Jewett, pg.991, no.15)
{
SF027
Ans. : 7.54 kV
Example 11 : (exercise)
A circular coil has 50 turns and diameter 1.0 cm. It rotates at a constant
angular velocity of 25 rev s-1 in a uniform magnetic field of flux density
50 µT. Determine the induced e.m.f. when the plane of the coil makes
an angle 55° to the magnetic field.
Ans. : 1.77 x 10-5 V
24
SF017
7.5 Self-induction and Self-inductance
7.5.1 Self-induction
{
Consider a solenoid which is connected to a battery , a switch S and
variable resistor R, forming an open circuit as shown in figure 7.5a.
{
When the switch S is closed, a current
I begins to flow in the solenoid.
S
N
I
{
{
{
I
The current produces a magnetic field
whose field lines through the solenoid
and generate the magnetic flux linkage.
If the resistance of the variable resistor
changes, thus the current flows in the
solenoid also changed, then so too
Fig. 7.5a
does magnetic flux linkage .
According to the Faraday’s law, an e.m.f. has to be induced in the
solenoid itself since the flux linkage changes.
In accordance with Lenz’s law, the induced e.m.f. opposes the change
that has induced it and it is therefore known as a back e.m.f.
e.m.f
εind
For the current I increases :
S
R
S
{
SN
N
I
SF027
{
Fig. 7.5b : initial
I ind
I
+
SN
I
I ind
I
Fig. 7.5c : I increases
Direction of the induced e.m.f.
e.m.f. is in
the opposite direction of the current I.
{
+
N SS
For the current I decreases :
S
I
Fig. 7.5d : initial
I
I ind
I
25
εind
NN
I
I ind
Fig. 7.5e : I decreases
Direction of the induced e.m.f.
e.m.f. is in
the same direction of the current I.
{
{
{
SF027
This process is known as self-induction.
SelfSelf-induction is defined as the process of producing an induced
e.m.f.
e.m.f. in the coil due to a change of current flowing through the
same coil.
This effect can be shown by the current I against time t graph for
resistor and solenoid in figure 7.5d.
26
SF017
I
Resistor
SelfSelf-induction
effect
Solenoid
0
t
Fig. 7.5d
7.5.2 Self-inductance, L
{
From the process of the self-induction, we get
ΦB ∝ I
ΦB = LI
where
{
(7.5a)
L : self - inductance of the coil
I : current
From the Faraday’s law,
dΦB
dt
d (LI )
ε=−
dt
ε=−
ε = −L
SF027
{
dI
dt
(7.5b)
27
From the eq. 7.5b,
SelfSelf-inductance is defined as the ratio of the self induced e.m.f.
e.m.f.
to the rate of change of current in the coil.
If the coil has N turns, hence
z
{
dΦB
dt
dI
dΦB
L =N
dt
dt
L ∫ dI = N ∫ dΦB
ε = −N
LI = NΦB
NΦB
L=
I
{
{
{
and
ε = −L
dI
dt
(7.5c)
Self-inductance is a scalar quantity and its unit is henry (H).
(H)
Unit conversion :
1 H = 1 Wb A-1 = 1 T m 2 A-1
The value of the self-inductance depends on
the size and shape of the coil
the number of turn (N)
z
the permeability of the medium in the coil (µ).
z
z
SF027
28
SF017
{
{
A circuit element which possesses mainly self-inductance is known as
an inductor.
inductor It is used to store energy in form of magnetic field.
field
The symbol of inductor in the electrical circuit is shown in figure 7.5e.
Fig. 7.5e
7.5.3 Self-inductance of a Solenoid
{
The magnetic flux density at the centre of the air-core solenoid is given
N
by
B = µ0 nI and n =
B=
{
{
µ0 NI
l
l
where
N : number of turns
l : length of the solenoid
The magnetic flux passing through the solenoid is given by
ΦB = BA cos 0 o
 µ NI 
ΦB =  0  A
 l 
ΦB =
µ0 NIA
l
Therefore the self-inductance of the solenoid is given by
NΦB
I
N  µ0 NIA 
L= 

I  l 
L=
SF027
{
L=
µ0 N 2 A
l
or
L = µ0 n 2 29
Al
For the medium-core solenoid :
L=
µN 2 A
l
and
µ = µr µ0
or
where
µ µ N 2 A µ : relative permeability
r
L= r 0
µ0 : permeability of free space
l
µ : permeability of medium
A : area of the solenoid
{
Example 12 :
At an instant, the current in an inductor increases at the rate of
0.06 A s-1 and back e.m.f. of 0.018 V was produced in the inductor.
a. Calculate the self-inductance of the inductor.
b. If the inductor is a solenoid with 300 turns, find the magnetic flux
through each turn when the current of 0.80 A flows in it.
Solution:
ε=0.018 V, dI = 0.06 A s -1
dt
a. By applying the equation below, thus
ε=L
SF027
dI
dt
L = 0.30 H
30
SF017
b. Given N=300
turns, I=0.80 A
By using the equation below, thus
L=
{
NΦB
I
ΦB = 8.0 x10 −4 Wb
Example 13 : (exercise)
An e.m.f. 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. Find the
magnetic flux through each turn of the coil.
{
(Serway&Jewett, pg.1025, no.6)
Ans. : 19.2 µWb
Example 14 : (exercise)
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.
(Given µ0 = 4π x 10-7 H m-1)
Ans. :188 µT, 33.3 nWb, 0.375 mH
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31
7.6 Energy Stored in an Inductor
{
Consider a coil of self-inductance L. Suppose that at time t the current
in the coil is in the process of building up to its stable value I at a rate
dI/dt. The magnitude of the back e.m.f. ε is given by
ε=L
{
dI
dt
The power P in overcoming this back e.m.f. is given by
P = Iε
dI
dt
Pdt = LIdI and Pdt = dU
(7.6a)
dU = LIdI
P = LI
{
The total energy stored in the inductor, U, as the current increases
from 0 to I can be found by integrating the eq. (7.6a). Thus
I
∫ dU = L ∫ IdI
0
U=
SF027
1 2
LI
2
(7.6b) analogous to
U=
1
CV 2
2
32
SF017
{
For a long air-core solenoid, the self-inductance is
µ0 N 2 A
L=
l
Therefore the energy stored in the solenoid is given by
1
U = LI 2
2
{
Note :
a.
U=
1  µ0 N 2 AI 2 


2 
l

I constant
A
B
VAB = 0
B
-
VAB > 0
B
+
VAB < 0
ε =0
b.
I increases
A
+
ε
c.
I decreases
A
-
SF027
{
(7.6c)
ε
33
Example 15 :
An 8.0 cm long solenoid with an air-core consists of 100 turns of
diameter 1.2 cm. Find
a. the self-inductance of the coil, and
b. the energy stored in the coil,
if the current flows in it is 0.77 A.
(Given µ0 = 4π x 10-7 H m-1)
Solution:
N=100 turn, l=8.0x10-2 m, d=1.2x10-2 m, I=0.77 A
a. By using the equation of self-inductance for solenoid, thus
πd 2
µ0 N 2 A
and A =
4
l
2
2
µ N πd
L= 0
4l
L = 1.8 x10 −5 H
L=
b. By applying the equation of energy stored in the inductor, thus
1 2
LI
2
U = 5.3 x10 −6 J
U=
SF027
34
SF017
{
{
{
SF027
Example 16 : (exercise)
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.
(Given µ0 = 4π x 10-7 H m-1)
Ans : 0.039 H m-1, 4.4 x 10-2 J m-1
Example 17 : (exercise)
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 e.m.f. is 0.016 V.
a. Calculate the self inductance of the inductor.
b. If the inductor is a solenoid with 400 turns and the current flows in it is
0.720 A, calculate
i. the magnetic flux through each turn.
ii. the energy stored in the solenoid.
Ans. : 0.250 H, 0.450 mWb, 64.8 mJ
Example 18 : (exercise)
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.
35
Ans. : 19 A s-1
7.7 Mutual induction and Mutual inductance
7.7.1 Mutual induction
{
Consider two circular close-packed coils near each other and sharing a
common central axis as shown in figure 7.7a.
{
A current I1 flows in coil 1, produced
by the battery in the external circuit.
{
{
SF027
If the current I1 changes with time, the
magnetic flux through coils 1 and 2 will
change with time simultaneously.
{
Due to the change of magnetic flux
through coil 2, an e.m.f. is induced in
coil 2. This is in accordance to the
Faraday’s law of induction.
{
In other words, a change of current in
one coil leads to the production of an
induced e.m.f. in a second coil which is
magnetically linked to the first coil.
I1
Fig. 7.7a
The current I1 produces a magnetic
field lines inside it and this field lines
also pass through coil 2 as shown in
figure 7.7.a.
36
SF017
{
{
{
This process is known as mutual induction.
Mutual induction is defined as the process of producing an
induced e.m.f.
e.m.f. in one coil due to the change of current in another
coil.
At the same time, the selfself-induction occurs in coil 1 since the
magnetic flux through it changes.
7.7.2 Mutual inductance, M
{
From the figure 7.7a, consider coils 1 and 2 have N1 and N2 turns
respectively.
{
If the current I1 in coil 1 is changes, the magnetic flux through coil 2 will
change with time and an induced e.m.f will occur in coil 2, ε2 where
ε2 ∝ −
{
dI 1
dt
ε 2 = − M 12
(7.7a)
If vice versa, the induced e.m.f. in coil 1, ε1 is given by
dI 2
dt
M 12 = M 21 = M
ε1 = − M 21
{
dI 1
dt
(7.7b)
Mutual inductance
where
Mutual inductance is a scalar quantity and its unit is henry (H).
(H)
SF027
37
{
{
Mutual inductance is defined as the ratio of induced e.m.f.
e.m.f. in a coil
to the rate of change of current in another coil.
From the Faraday’s law for a coil of N turns,
ε2 = − N 2
dΦ2
dt
dI 1
dΦ2
= N2
dt
dt
M 12 ∫ dI 1 = N 2 ∫ dΦ2
M 12
M 12 I 1 = N 2Φ2
NΦ
M 12 = 2 2
I1
{
M 21 =
N 1Φ1
I2
(7.7c)
Since M12=M21=M , eq. (7.7c) can be written as
M=
SF027
and
N 2Φ2 N 1Φ1
=
I1
I2
(7.7d)
38
SF017
7.7.3 Mutual Inductance for two coaxial solenoids
{
Consider a long solenoid with length l and cross sectional area A is
closely wound with N1 turns of wire. A coil with N2 turns surrounds it at
its centre as shown in figure 7.7b.
A
N1: primary coil
N1
N2
I1
N2: secondary coil
I1
l
Fig. 7.7b
{
When current I1 flows in the primary coil (N1), there exist a magnetic
field B1 and thus the magnetic flux, Ф1.
{
For solenoid, Boutside=0, hence
Φ1 = Φ2
SF027
Therefore
Φ1 = B1 A
and
B1 =
(if no flux leakage)
39
µ0 N 1 I 1
l
µ N I 
Φ1 =  o 1 1  A = Φ2
 l 
{
If the current I1 changes, an e.m.f is induced in secondary coils,
therefore mutual inductance occurs and is given by
N 2Φ2
I1
N  µ N I A
M = 2 o 1 1 
I1 
l

M=
M=
SF027
µo N1 N 2 A
l
(7.7e)
40
SF017
{
Example 19:
A current of 2.0 A flows in coil P and produced a magnetic flux of
0.6 Wb in it. When a coil S is moved near to coil P coaxially, a flux of
0.2 Wb is produced in coil S.
Given that, coil P has 100 turns and coil S has 200 turns.
a. Calculate self-inductance of coil P and the energy stored in P
before S is moved near to it.
b. Calculate the mutual inductance of the coils.
c. If the current in P decreasing uniformly from 2.0 A to zero in 0.4 s,
calculate the induced e.m.f. in coil S.
Solution: NP =100
turns, NS =200 turns, IP=2.0 A,
ΦP=0.6 Wb ,ΦS = 0.2 Wb
a.
Self Inductance of coil P is
N PΦ P
IP
LP = 30 H
LP =
The energy stored in coil P is
U P = 21 LP I P2
U P = 60 J
SF027
b.
41
Mutual Inductance is
N S ΦS
IP
M = 20 H
Given dI P = (0 − 2.0 ) = −2.0 A
M=
c.
and
dt = 0.4 s
The induced e.m.f. in coil S is
dI P
dt
ε S = 100 V
ε S = −M
{
SF027
Example 20 :(exercise)
The primary coil of a solenoid of radius 2.0 cm has 500 turns and length
of 24 cm. If the secondary coil with 80 turns surrounds the primary coil
at its centre, calculate
a. the mutual inductance of the coils
b. the magnitude of induced e.m.f. in secondary coil if the current in
primary coil changes at the rate 4.8 A s-1.
Ans. : 26.32 mH ;126 mV
42
SF017
7.8 Transformer
{
{
An electrical instrument is used to increase or decrease the e.m.f.
e.m.f. or
voltages of an alternating current.
Consider a structure of the transformer as shown in figure 7.8a.
laminated
iron core
{
{
NP
turns
primary coil
NS
{
turns
If NP > NS the transformer is
a stepstep-down transformer.
transformer
If NP < NS the transformer is
a stepstep-up transformer.
transformer
The symbol of transformer in
circuit is shown in figure 7.8b.
secondary coil
Fig. 7.8a
{
{
The working principle of transformer.
(refer to mutual inductance)
The characteristics of an ideal transformer:
a. Zero resistance of primary coil.
b. No magnetic flux leakage from the iron core.
c. No dissipation of energy and power.
power
Fig. 7.8b
SF027
43
{
By referring to mutual inductance, the induced e.m.f. in the primary and
secondary coil is given by
dΦP
dt
εP = − N P
and
{
εS = − N S
(7.8a)
dΦS
dt
(7.8b)
For an ideal transformer, there is no flux leakage so that
dΦP dΦS
=
dt
dt
SF027
same for both primary and
secondary coils.
{
By dividing eq. (7.8a) with (7.8b),
{
For an ideal transformer, the electrical power is given by
εP N P
=
εS N S
PP = PS
I P εP = I S εS
εP I S
=
εS I P
where
PP : power of primary
PS : power of secondary
44
SF017
{
In general:
εP VP NP I S
= =
=
εS VS NS I P
(7.8c)
7.8.1 Energy losses in transformers
{
Although transformers are very efficient devices, small energy losses do
occur in them owing to four main causes:
z
Resistance of coils
The wire used for the primary and secondary coils has resistance and
so ordinary (I2R) heat losses occur.
Overcome : The transformer coils are made of thick copper wire.
wire
z
Eddy current
The alternating magnetic flux induces eddy currents in the iron core.
This current causes heating and dissipation of power in the core.
Overcome : The effect is reduced by using laminated core as shown
in figure 7.8c and 7.8d.
Fig. 7.8c
Fig. 7.8d
SF027
45
Hysteresis
The magnetization of the core is repeatedly reversed by the
alternating magnetic field. The resulting expenditure of energy in
the core appears as heat.
Overcome : By using a magnetic material (such as Mumetal) which
has low hysteresis loss.
loss
z
Flux leakage
The flux due to the primary may not all link the secondary. Some
of the flux loss in the air.
Overcome : By designing the iron core suitably.
Example 21:
The primary coil of a transformer has 1200 turns and the secondary coil
has 60 turns. The primary coil is connected to an a.c. supply of 240 V. A
resistor of resistance 3.0 Ω is connected to the secondary coil. Assume
that there are no loss of power and magnetic flux, calculate the current
flows in the secondary circuit.
z
{
Solution: NP =1200 turns,
For ideal transformer, N P
NS
NS =60 turns, εP = 240 V, R = 3.0 Ω
=
εP
εS
N 
SF027
ε S =  S ε P
 NP 
ε S = 12 V
46
SF017
Thus the current flows in the secondary coil is
ε S = IS R
I S = 4.0 A
Example 22: (exercise)
A transformer, assumed to be 100% efficient, is used with a supply
voltage of 120 V. The primary winding has 50 turns. The required output
voltage is 3000 V. The output power is 200 W.
a. Name this type of transformer.
b. Calculate the number of turns in the secondary winding.
c. Calculate the current supplied to the primary winding
Ans. : 1250 turns, 1.67 A
Example 23: (exercise)
A transformer with a 100 turns primary coil and a 500 turns secondary
coil is connected to a supply voltage of 2.0 V. Calculate the output
voltage and the maximum current in secondary coil if the current in
primary coil is to be limited to 0.10 A.
Ans. : 10 V, 0.020 A
{
{
SF027
47
7.9 Back e.m.f. in d.c. motor
{
Fig 7.9a shows a simple d.c. motor.
{
r
F
r
F
{
{
When current, I flows in the coil
of the armature which is in the
magnetic field, magnetic force is
produced and will cause the coil
to rotate as shown in figure 7.9a.
As the coil rotates, its magnetic
flux changes and so an e.m.f. is
induced across the coil.
(Faraday’s law)
By Lenz’s law this induced e.m.f.
opposes the current which is
making the coil turns. Therefore
it is called back e.m.f. (εB)
ε B = NBAω = (NBA2π ) f
Fig 7.9a
{
As the motor speeds up, the back e.m.f.,εB increases because it is
proportional to the frequency, f
SF027
εB ∝ f
εinitial
f
= initial
ε final
f final
(7.9a)
48
SF017
{
{
{
{
When the motor is first switched on, the back e.m.f. is zero: it rises as the
motor speeds up.
When the motor is running freely, the back e.m.f is nearly equal to the
supply voltage and so there will not be much current drawn.
When a load is applied to the motor, the motor slows down, the back
e.m.f. falls, and so the current in the coil increases.
Figure 7.9a also can be simplified into circuit shown in figure 7.9b.
Applying Kirchhoff’s 2nd law:
Motor
εB R
V − ε B = IR
Eq. (7.9b) x I :
Loop L:
(7.9b)
VI − ε B I = I 2 R
L
V
I
VI = ε B I + I 2 R
(7.9c)
where
VI : power supplied
εB I : mechanical power
Fig 7.9b
I2R : power lost as heat in coil
SF027
49
{
Example 24:
A motor rotates at a rate of 5000 rotations per minute. The supply voltage
is 240 V and the resistance of the armature is 4.5 Ω.
a. Calculate the back e.m.f. if the current in the armature is 12 A.
A load is applied to the motor and the speed of the rotation is found to
decrease to 4000 rotations per minute. Calculate
b. the back e.m.f. now.
c. the new current in the armature.
d. the mechanical power produced by the motor.
Solution:
V =240 V, R =4.5 Ω, ωinitial = 5000 rpm
a. By using the equation below:
V = ε B + IR
ε B = V − IR
ε B = 186 V
Given ωfinal
b.
= 4000 rpm
ε initial ωinitial
=
ε final
ω final
SF027
ε final = 148.8 V
50
SF017
c. The new current in the armature is
V = ε B + IR
240 = 148.8 + (4.5) I
I = 20.3 A
d. The mechanical power is Mechanical power = ε I
B
Mechanical power = (148.8 )(20.3)
Mechanical power = 3020 W
{
SF027
Example 25: (exercise)
The resistance of the armature of a d.c. motor is 0.75 Ω. A supply of 240
V is connected to this motor. When the motor rotates freely without load,
the current in the armature is 4.0 A and the rate of rotation is 400 rpm.
Calculate
a. the back e.m.f. produced.
b. the mechanical power generated.
If a load is applied, the current increases to 60 A. Calculate
c. the back e.m.f. now.
d. the mechanical power.
e. the rotation speed of the armature.
Ans. : 237 V, 948 W, 195 V, 11.7 kW, 329 rpm
51