Ch20 Electromagnetic Induction

1. Electromagnetic
Induction

2. Define and use
magnetic flux,
20.1 Magnetic Flux
 cosBAAB

2

3. • is defined as the scalar
product between the
magnetic flux density, B and
the vector of the surface
area, A
where,
: magnetic flux
: angle between B and A
B: magnetic flux density
A: area of the coil
• Scalar quantity
• Unit: T m2 or Wb
Magnetic Flux, : a measure of the number of magnetic
field lines that cross a given area
A
B  = 90
BA
 = 0
3
Direction of
vector A always
perpendicular to
the surface area,
A.
 cosBAAB


4. A flat surface with area 3.0 cm2
is placed in a uniform magnetic
field. If the magnetic field
strength is 6.0 T, making an
angle 30° with the surface area,
find the magnetic flux through
this area.
Example 20.1
Solution:

30 
30
coil
B

Normal A

Using: cosAB
23
mT109.0 

θ = 90 – 30 = 60°
4

5. REVISION: Magnetic Flux  cosBAAB

5




6. • 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
• 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.
REVISION: Magnetic Flux
A

area

90
I
I I
SN
 = 2.96×10–5 Tm26

7. The three loops of wire are all in
a region of space with a uniform
magnetic field. Loop 1 swings
back and forth as the bob on a
simple pendulum. Loop 2 rotates
about a vertical axis and loop 3
oscillates vertically on the end of
a spring. Which loop or loops
have a magnetic flux that
changes with time? Explain your
answer.
Example 20.2
7

8. Only loop 2 has a changing
magnetic flux.
Reason:
Loop 1 moves back and forth, and
loop 3 moves up and down, but
since the magnetic field is uniform,
the flux always constant with time.
Loop 2 on the other hand changes
its orientation relative to the field as
it rotates, hence its flux does
change with time.
Solution:
8

9. 9
1A 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.
[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.
[2.96×10–5 T m2]
Exercise 20.1

10. (d) Derive and use induced emf:
- in straight conductor,
- in coil,
- in rotating coil,
(a) Use Faraday’s experiment to
explain induced emf
(b) State Faraday’s law and use
Lenz’s law to determine the
direction of induced current
(c) Use induced emf,
20.2 Induced emf
dt
d

 sinBlv
dt
dB
NA
dt
dA
NB
tNAB  sin
10

11. History – Faraday’s experiment to induced emf
• In this experiment, Faraday
hoped by using a strong
enough battery, a steady
current in X would produce a
current in a second coil Y but
failed
• Current carrying conductor
 magnetic field
• Magnetic field  electric
current????
• The diagram below shows the
apparatus used by Faraday in
his attempt to produce an
electric current from a
magnetic field
11

12. • This is Faraday’s apparatus for
demonstrating that a magnetic
field can produce a current
• A change in the field produced by the
top coil induces an EMF and, hence,
a current in the bottom coil
• When the switch is opened and
closed, the galvanometer registers
currents in opposite directions
• No current flows through the
galvanometer when the switch
remains closed or open
REVISION: Faraday’s experiment to induced emf
OBSERVATIO
N
GALVANOMETER
DEFLECTION
Switch on YES
Switch of YES
Steady current NO
Faraday’s experiment used a magnetic field that was changing
because the current producing it was changing
12

13. • Faraday concluded that although
a steady magnetic field
produces no current, a
changing magnetic field can
produce an electric current
• Such a current is called an
induced current
• We therefore say that an
induced current is produced by
a changing magnetic field
• The corresponding emf required
to cause this current is called an
induced emf
Faraday’s experiment to induced emf
OBSERVATIO
N
GALVANOMETER
DEFLECTION
Switch on YES
Switch of YES
Steady current NO
13

14. • a magnetic
field that is
changing
because
the magnet
is moving
A changing magnetic field induces an emf
14

15. Electromagnetic Induction is the
production of induced e.m.f.s or induced
currents whenever the magnetic flux
through a loop, coil or circuit is changed
meaning of changing in magnetic flux
a relative motion of
loop & magnet
field lines are ‘cut’
number of magnetic
field lines passing
through a coil are
increased or decreased
To increase
induced e.m.f
Use a
stronger
magnet
can
increase
magnetic
flux
Push the
magnet
faster
into the
coil to
increase
speed
The
area of
the coil
is
greater
The
number
of turns
increas
ed
Electromagnetic Induction
15

16. • states that an induced electric
current always flows in such a
direction that it opposes the
change producing it
Lenz’s Law
• states that the magnitude of the
induced emf is proportional to the
rate of change of the magnetic flux
Faraday’s Law
• These two laws are summed up in the relationship
• For a coil of N turns
• Since d = final - initial
• The negative sign indicates that the direction of induced emf always oppose
the change of magnetic flux producing it (Lenz’s law)
dt
dΦ

dt
d
N
Φ

 
dt
N if ΦΦ 

16

17. 17
Solution:A coil of wire 8 cm in diameter
has 50 turns and is placed in a
B field of 1.8 T. If the B field is
reduced to 0.6 T in 0.002 s ,
calculate the induced emf.
Example 20.3





 

t
N
dt
dΦ
N initialfinalB
 





 

t
BB
NA initialfinal
 





 













t
BBd
N initialfinal
2
2
V151
dt
d
N
Φ

Note:
 To calculate the magnitude of induced
emf, the negative sign can be ignored

18. 18
Solution:
By applying the Faraday’s law
equation for a coil of N turns ,
thus
The magnetic flux passing
through a single turn of a coil is
increased quickly but steadily at
a rate of 5.0102 Wb s1. If the
coil have 500 turns, calculate the
magnitude of the induced emf in
the coil.
Example 20.4
12
sWb1005
turns;500




.
dt
d
N
dt
d
N
Φ

  2
100.5500 

V25

19. • When an emf is generated by a change in
magnetic flux according to Faraday's Law, the
polarity of the induced emf is such that it
produces a current whose magnetic field
opposes the change which produces it
• The induced magnetic field inside any loop of
wire always acts to keep the magnetic flux
in the loop constant
• In the examples below, if the B field is
increasing, the induced field acts in
opposition to it
• If it is decreasing, the induced field acts in the
direction of the applied field to try to keep it
constant
Lenz’s Law
19

20. 20
Lenz’s Law

21. 21
Lenz’s Law
N
I
Induced current
flows in
counterclockwise
N
S
Induced current
flows in clockwise

22. 22
dt
dΦ


23. • Faraday's Law : change in
magnetic flux  INDUCED
CURRENT/ EMF, the
polarity of the induced
emf is such that it produces
a current whose magnetic
field opposes the change
which produces it
• The induced magnetic field
inside any loop of wire
always acts to keep the
magnetic flux in the loop
constant
Lenz’s Law
23
• In the examples below, if the B field is increasing,
the induced field acts in opposition to it
• If it is decreasing, the induced field acts in the
direction (SAME DIRECTION) of the applied field
to try to keep it constant

24. 24
Lenz’s Law
N
I
Induced current
flows in
counterclockwise

25. 25
Lenz’s Law
Induced current
flows in clockwise
N
S

26. 26
Another great example of Lenz's law is to take a copper tube
(it's conductive but non-magnetic) and drop a piece of steel
down through the tube. The piece of steel will fall through, as
you might expect. It accelerates very close to the acceleration
due to gravity.
Now take the same copper tube and drop a magnet through it. You
will notice that the magnet falls very slowly. This is because the
copper tube "sees" a changing magnetic field from the falling
magnet. This changing magnetic field induces a current in the
copper tube. The induced current in the copper tube creates its
own magnetic field that opposes the magnetic field that created it.

27. Faraday’s Law dt
dΦ

27

28. 28
• From
Induced emf in a plane coil by changing
area in B
dt
d
N
Φ
  cosBAΦand
 
dt
cosBAd
N

 
  






dt
dA
cosNB 
Stretching the coil reduces the area of the coil  magnetic flux through coil is decreased
and, current is induced in the coil
Flux
through coil
is
decreased

29. 29
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.
Example 20.5

30. 30
Solution:
dt
d
 
dA
NB
dt
 
final initial( )A A
NB
t

 

2
(0 )r
NB
t

 

2
(0 (0.12) )
(1)(0.15)
0.20

  V104.3 2


31. 31
• From
Induced emf in a plane coil by changing
magnetic field, B strength
dt
d
N
Φ
  cosBAΦand
 
dt
cosBAd
N

 
As the magnetic field strength, B is increasing or decreasing with time, the magnetic
flux through the area changes, therefore induces an emf in the coil
  






dt
dB
cosNA 

32. 32
A circular coil has 200 turns and
diameter 36 cm. the resistance of
the coil is 2.0 Ω. A uniform magnetic
field is applied perpendicularly to
the plane of the coil. If the field
changes uniformly from 0.5 T to 0 T
in 0.8s.
(a) Find the induced e.m.f. &
current in the coil while the field
is changed.
(b) Determine the direction of the
current induced.
Example 20.6
Solution:
1. Calculate area, A = r2
2. Determine emf,
3. Determine I from
dt
dB
NA
IR
A36.6

33. 33
Solution:A narrow coil of 10 turns and diameter
of 4.0 cm is placed perpendicular to a
uniform magnetic field of 1.20 T. After
0.25 s, the diameter of the coil is
increased to 5.3 cm.
(a) Calculate the change in the area of
the coil.
(b) If the coil has a resistance of 2.4 ,
determine the induced current in
the coil.
Example 20.7

0
A

B

B

A

Initial Final
  






dt
dA
cosNB 

34. 34
Solution: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
Example 20.8
B


0A

Wb108.0 3
initial


V106.1 2

  






dt
dB
cosNA 

35. 35
Calculate the current through a
37 Ω resistor connected to a
single turn circular loop 10 cm in
diameter, assuming that the
magnetic field through the loop
is increasing at a rate of 0.050
T/s. State the direction of the
current.
Example 20.9

36. 36
• Calculate, 
Solution:
dt
dΦ

2
2













d
A,
dt
dB
A 
R
I


I induced
I induced
V. 4
10933 

A10x1.06 5-

R
I


37. 37
• Induced emf in a plane coil:
– by changing area in
magnetic field
– by changing magnetic field
strength
  






dt
dA
cosNB 
  






dt
dB
cosNA 
dt
dΦ


38. 38
Induced emf in a straight conductor moving
through a magnetic field
• As a conductor is moved
through a magnetic field,
current is induced and the
bulb is lightened up

39. 39
• Consider a straight conductor of
length l is moved at a speed v to the
right on a U-shaped conductor in a
uniform magnetic field B that points
out the paper
• This conductor travels a distance dx
=vdt in a time dt
• The area of the loop increases by
an amount:
Induced emf in a straight conductor moving through
a magnetic field
ldxdA 

40. 40
• This induced emf is called
motional induced emf
• The direction of the induced
current or induced emf in the
straight conductor can be
determined by using the
Lenz’s Law
• If B field is increasing, the induced field
acts in opposition to it
• If B is decreasing, the induced field acts in
the direction (SAME DIRECTION) of the
applied field to try to keep it constant
• According to Faraday’s law, the
e.m.f. is induced in the conductor and
its magnitude is given by
 angle between v and B
Induced emf in a straight conductor moving through
a magnetic field
dt
d

dt
dA
B
dt
dx
Bl v
dt
dx
and
 sinBlvBlv

41. 41

42. 42
Consider the arrangement
shown below. Assume that R = 6
Ω, L = 1.2 m & a uniform 2.50 T
magnetic field is directed into the
page.
(a) At what speed should the bar
be moved to produced a
current of 0.5A in the resistor
(b) what is the direction of the
induced current?
Example 20.10

43. 43
(a)  = Blv sin ,  = IR
(b) From Lenz’s Law
Solution:
1
sm1 
v

44. 44
(a) Calculate the motional
induced emf in the rod.
(b) If the rod is connected in
series to the resistor of
resistance 15 , determine
(i) the induced current and its
direction
(ii) the total charge passing
through the resistor in two
minute
A 20 cm long metal rod CD is
moved at speed of 25 m s1
across a uniform magnetic field
of flux density 250 mT. The
motion of the rod is
perpendicular to the magnetic
field as shown.
Example 20.11
C
D
B

1
sm52 

45. 45
Induced emf in a rotating coil
An ac generator / dynamo: transforms mechanical energy into electric energy

46. 46
• By applying the equation of
Faraday’s law for a coil of N
turns, thus the induced emf is
given by
• Consider a coil of N turns each
of area A and is being rotated
about a horizontal axis in its
own plane at right angle to a
uniform magnetic field of flux
density B.
• As the coil rotates with the
angular speed ω, the
orientation of the loop changes
with time.
Induced emf in a rotating coil
cosBA t and
tBA cos
dt
d
N


 tBA
dt
d
N cos
 t
dt
d
NBA cos
tNBA  sin

47. 47
• The emf induced in the loop
varies sinusoidally in time
• The induced emf is maximum
when hence
where
Induced emf in a rotating coil
 NBAmax
T
f


2
2 
Note:
This phenomenon was
the important part in the
development of the
electric generator or
dynamo.

48. 48
iii) The emf induced in a coil
varies sinusoidally with time.
iv) Maximum voltage (ξ max =
NBA ω) is produced when
the coil is parallel to the
magnetic field (sin ωt = 1).
v) No voltage exists when the
coil is perpendicular to the
magnetic field
• The graph show that :
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.
Induced emf in a rotating coil

49. (a) Define self-inductance
(b) Apply self-inductance,
for coil and solenoid
20.3 Self Inductance







dt
dI
L


50. 50
• Running a changing
current (by changing R),
creates a changing
magnetic field, which
creates an induced emf
that fights the change
• Unit: Henry (V s A-1)
Self Induction – the production of e.m.f. in a circuit due to
the change of current in the circuit itself

51. 51
 When the switch S is closed, a current I
begins to flow in the solenoid
 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 does
magnetic flux linkage
Self Induction
S R
II
NS

52. 52
• According to the Faraday’s law, an emf
has to be induced in the solenoid itself
since the flux linkage changes
• In accordance with Lenz’s law, the
induced emf opposes the changes that
has induced it and it is known as a back
emf
• For an increasing current, the direction of
the induced field and emf are opposite to
that of the current, to try to decrease the
current
• For the current I increases:
Self Induction
indε
- +
NS
I
I
SN
Direction of the induced emf is in the
opposite direction of the current I.

53. 53
• If the current is decreasing, the
direction of the induced field
and emf are in the same
direction as the current, to try
to increase the current
• This coil is said to have self-
inductance (inductance)
• A coil that has inductance is
called an inductor use to store
energy in the form of magnetic
field
• For the current I decreases:
Self Induction
+ -
indε
NS
I IindI
indI
NS
Direction of the induced emf is in the
same direction of the current I.

54. 54
(a) A current in the coil
produces a magnetic
field directed to the left
Self Induction
(b) If the current increases,
the increasing magnetic
flux creates an induced
emf having the polarity
shown by the dashed
battery
(a)The polarity of the
induced emf reverses if
the current decreases

55. 55
• From the Faraday’s law, thus• From the self-induction
phenomenon, we get
where
L: self inductance of the coil,
unit: Henry (H) or Wb A-1
I: current
Self Inductance, L: the ratio of the self induced e.m.f. to
the rate of change of current in the coil
ILΦ
LILΦ
dt
d L

 LI
dt
d

dt
dI
L
dtdI
L
/


• The value of the self-inductance depends on
(a) the size and shape of the coil
(b) the number of turn (N)
(c) the permeability of the medium in the coil ()

56. 56
• Therefore the self-inductance
of the solenoid is given by
• The magnetic flux density at the
centre of the air-core solenoid is
given by
• The magnetic flux passing through
each turn of the solenoid always
maximum and is given by
Self Inductance of a solenoid
l
NI
B 0


0cosBA
A
l
NI






 0
l
NIA0

I
N
L

 






l
NIA
I
N
L 0
l
AN
L
2
0


57. 57
Suppose you wish to make a
solenoid whose self-inductance
is 1.4 mH. The inductor is to
have a cross-sectional area of
1.2 x 10 -3 m2 and a length of
0.052 m. How many turns of wire
needed?
Example 20.14
Induced emf of 5.0 V is
developed across a coil when
the current flowing through it
changes at 25 A s-1. Determine
the self inductance of the coil.
Example 20.12
H.20
dt
dI
L
Example 20.13
If the current in a 230 mH coil
changes steadily from 20.0 mA
to 28.0 mA in 140 ms, what is
the induced emf?
220 turns
l
AN
L
2
0


58. 58
Solution
a. The change in the current is
Therefore the inductance of the
solenoid is given by
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.
Example 20.15
;A5.2;0m;100.8turns;500 fi
2
 
IIlN
V012.0s;35.0  dt
if IIdI  05.2 dI
A5.2dI
dt
dI
L 






35.0
5.2
012.0 L
H1068.1 3
L

59. 59
c. The final magnetic flux linkage
is given by
b. By using the equation of self-
inductance for the solenoid,
thus
l
AN
L
2
0

  
2
27
3
100.8
500104
1068.1 





A
24
m1028.4 
A
  ffL LI
  5.21068.1 3

  Wb102.4 3
fL



60. • An inductor is a circuit
component (coil or solenoid)
which produced an self
induced emf
• Function of an inductor:
(1) to control current
(2) store energy in form of
magnetic field
• Back emf produce in an
inductor is given by:
Derive and use the energy
stored in an inductor,
20.4 Energy Stored In
Inductor
2
2
1
LIU 
Inductor
dt
dI
L

61. 61
• The total work done while the current is
changed from zero to its final value is
given by
and analogous to
in capacitor
• For a long air-core solenoid, the self-
inductance is
• Therefore the energy stored in the
solenoid is given by
• The electrical power P in
overcoming the back emf in
the circuit is given by
Energy Stored In
Inductor
IP 
dt
dI
LIP 
dILIPdt 
dILIdU 
 
IU
IdILdU
00
2
2
1
LIU 
2
2
1
CVU 
l
AN
L
2
0

2
2
1
LIU  








l
AIN
U
22
0
2
1 

62. 62
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.
How much energy is stored in a
0.085-H inductor that carries a
current of 2.5 A?
Example 20.16
2
2
1
LIU 
Example 20.17
A steady current of 2.5 A in a coil of
500 turns causes a flux of 1.4 x 10-4
Wb to link (pass through) the loops of
the coil. Calculate
(a) the average back emf induced in
the coil if the current is stopped in
0.08 s
(b) the inductance of the coil and the
energy stored in the coil
(inductor).
Example 20.18
HL 3
1082.1 

J2
1005.1 

V1125.0

63. 63
Solution:
(a)
1. Calculate A
2. Use
(b) Use
A solenoid of length 25 cm with
an air-core consists of 100 turns
and diameter of 2.7 cm.
Calculate
(a) the self-inductance of the
solenoid, and
(b) the energy stored in the
solenoid, if the current flows
in it is 1.6 A.
(Given 0 = 4  107 H m1)
Example 20.19
l
AN
L
2
0

2
2
1
LIU 
H1088.2 5
L
J1069.3 5
U

64. • Mutual Induction: emf
induced in a circuit by a
changing current in another
nearby circuit
(a) Define mutual inductance
(b) Use mutual inductance
between two coaxial
solenoids or a coaxial coil
and a solenoid
20.5 Mutual Inductance
l
ANN
M 210



65. 65
The induced
current in this loop
that caused by the
change of current
in neighbouring
loop  Mutual
Induction
NS N S
Increasing current in loop 1
produces a change in
magnetic flux. This changes
is experienced by loop 2
that placed nearby
According to
Faraday’s law,
emf is induced
in loop 2 to
oppose the
changes.
12

66. 66
• In mutual induction, the e.m.f. induced in one coil is
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
• The induced e.m.f. in coil 2 is
• If vice versa, the induced emf in coil 1, 1 is given by
where
Mutual Inductance, M: the ratio of induced emf in a
coil to the rate of change of current in another coil
dt
dI1
2 
dt
dI
M 1
122 
dt
dI
M 2
211  MMM  2112
Keep In mind
emf induced in coil
2 is due to the
current change in
coil 1
emf induced in
coil 1 by
changing
current in coil 2

67. 67
• From Faraday’s Law
and
• Since M12 = M21 = M
• Rearrange,
Mutual Inductance, M
dt
dI
M 1
122 
dt
dI
M 1
2

dt
dI
M 2
211 
dt
dI
M 2
2

dt
dI
dt
dI
M
2
1
1
2 

dt
d
N 2
22


dt
d
N
dt
dI
M 2
2
1
12


  22112 dNdIM
1
22
12
I
N
M


22112  NIM
2
11
21
I
N
M


2
11
1
22
I
N
I
N
M





68. 68
Solution:
(a) Using
(b) From
Two coils, X & Y are magnetically
coupled. The emf induced in coil Y
is 2.5 V when the current flowing
through coil X changes at the rate
of 5 A s-1. Determine:
(a) the mutual inductance of the coils
(b) the emf induced in coil X if there
is a current flowing through coil Y
which changes at the rate of 1.5
A s-1.
Example 20.20
1
552 
 As
dt
dI
,V. x
Y








dt
dI
M
X
Y
H5.0M
dt
dI
M Y
X

V75.0

69. 69
• 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
• When a current I1 flows in the
primary coil (N1), it produces a
magnetic field B1,
• Magnetic flux,
Mutual inductance, M between two coaxial
solenoids
l
IN
B 110
1


l
I1 I1
N1N2
A
N1: primary coil
N2: secondary coil

0cos11 AB
l
AIN 110
1



70. 70
• If no magnetic flux leakage,
thus
• If the current I1 changes, an
emf is induced in the
secondary coils, therefore the
mutual inductance occurs and
is given by
Mutual inductance, M between two coaxial
solenoids
21 
l
AIN
I
N
M 110
1
2
12








1
22
12
I
N
M


l
ANN
MM 210
12



71. 71
Solution:
N1= 1000; l = 50×10-2 m;
d1 = 3×10-2 m, N2 = 50;
(a) Using
(b) In secondary coil,
Primary coil of a cylindrical former
with the length of 50 cm and
diameter 3 cm has 1000 turns. If
the secondary coil has 50 turns,
calculate:
(a) its mutual inductance
(b) the induced emf in the secondary
coil if the current flowing in the
primary coil is changing at the
rate of 4.8 A s-1.
Example 20.21
11
84 
 As.
dt
dI
l
ANN
M 210


HM 5
1088.8 

dt
dI
M 1
2

V4
1025.4 


72. 72
Solution:
(a) Using
(b) Using
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.
Example 20.22
P
PS
l
ANN
M 0


dt
dI
M P
S


73. 73
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 self-inductance of coil C and the energy stored in C
before D is moved near to it
(b) Calculate the mutual inductance of the coils
(c) If the current in C decreasing uniformly from 3.0 A to zero in
0.25 s, calculate the induced emf in coil D.
Example 20.23

74. 74
(b) The mutual inductance of the
coils is given by
(a) The self-inductance of coil C
is given by
and the energy stored in C is
Solution
Wb;25.0Wb;75.0A;0.3 DCC I
turns5000turns;1000 DC  NN
C
CC
C
I
N
L

   
0.3
75.01000
C L
H250C L
2
CCC
2
1
ILU 
  2
0.3250
2
1

J1125C U
C
DD
I
N
M


  
0.3
25.05000

H417M

75. 75
Given
The induced emf in coil D is given by
Solution
Wb;25.0Wb;75.0A;0.3 DCC I
turns5000turns;1000 DC  NN
  A0.30.30s;25.0 C  dIdt
dt
dI
M C
D 
  
25.0
0.3
417


V5004D 

76. 76
Next Chapter…
CHAPTER 21 :
Alternating current