SUGGESTED SOLUTIONS FOR TUTORIAL 7
(ELECTROMAGNETIC INDUCTION)
1. A UHF television loop antenna has a diameter of 11 cm. The magnetic field of a TV
signal is normal to the plane of the loop and, at one instant of time, its magnitude is
changing at the rate 0.16 T/s. The magnetic field is uniform. What emf is induced in
the antenna?
Ans : ( 1.52 Mv )
d  11cm  11 10 2 m  r 
Solution:
d
 55  10 3 m
2
dB
 0.16 T s 1
dt
d
dt
d BA 

dt
dB
 r 2
dt


  55  10 3
 0.16
2
 1.52  10 3 V#
 1.52 mV#
2. A solenoid of length 2.8 cm and diameter 0.75 cm is wound with 160 turns per cm.
When the current through the solenoid decreasing at a rate of 35.0 As-1, what is the
induced emf
(a) in one of the windings?
Ans: ( 3.1 x 10-5 V)
(b) in the entire solenoids?
Solutions:
  2.8  10 2 m
d  0.75 10 2 m  r  3.75  10 3 m
n  160 turns per cm 160  10 2 turns per m
dI
 35.0 As 1
dt
Ans : (14 Mv)
N  n
(a)


 160  10 2 2.8  10 2

 448 turns
 ε = 14 mV for 448 turns, therefore for 1 turn,

14  10 3
1  
 3.1  10 5 V
N
448
 emf in one of the windings
(b)   N
#
d
dt
d BA 
dt
d  0 nI 
 nA
dt
dI
 n 2 A 0
dt
N

 160  10 2
 2.8  10  3.75  10  1.26  10 35.0
2
 14.0  10 3 V
2
3 2
6
#
 14 mV#
 emf in the entire solenoid
3. An emf of 3.5 V is obtained by moving a wire 1.1 m long at a rate of 7 ms -1
perpendicular to the wire and to a uniform magnetic field. What is the intensity of the
field? Ans: ( 0.45 T )
Solution:
  3 .5 V
  1.1 m
v  7 ms 1
  Bv

B
v
3.5

1.17 
 0.45 T #
4. A horizontal wire 0.8 m long is falling at a speed of 5 ms-1 perpendicular to a uniform
magnetic field of 1.1 T, which is directed from east to west. Calculate the magnitude
of the induced emf. Is the north or south end of the wire positive?
Ans : (4.4 V)
Solution:
  0 .8 m
v  5 ms 1
B  1.1T
 from east to west
  Bv
 1.10.85
 4 .4 V
#
 north end is positive #
5. An emf of 8 V is induced in a coil when the current in it changes at the rate of 32 As 1
. Compute the inductance of the coil.
Ans: ( 0.25 H )
Solution:
  8V ,
dI
 32 As 1
dt
dI
dt
dt
L
dI
8

32
 0.25 H
 L
#
6. A steady current of 2.5 A
the inductance of the coil?
Ans: (28 mH)
Solution:
I  2.5 A ,   140 Wb  140  10 6 Wb , N  500
N  LI
N
I
500 140  10 6

2 .5
 0.028 H #
L 

 28 mH

#
7. When the current in a certain coil is changing at a rate of 3 As-1, it is found that an
emf of 7 mV is induced in a nearby coil. What is the mutual inductance of the
combination? Ans: (2.33 mH)
Solution:
dI
 3 As 1 ,
dt
dI
dt
dt
M 
dI
7  10 3

3
 2.33 mH
  7  10 3 V
 M


#
8. The perpendicular component of the external magnetic field through a 50-turns coil of
radius 5.0 cmincreases from 0 to 1.8 T in3.6 s.
(a) If the resistance of the coil is 2.8 Ω, what is the magnitude of the induced current
in the coil?
Ans: ( 0.07 A)
(b) What is the direction of the current if the perpendicular component of the field is
away from the viewer?
Solution:
N = 50
r = 5.0 cm = 5.0 × 10-2 m
B = 0  1.8 T
t = 3.6 s
(a) r = 2.8 Ω
d
dt
=0
 akhir   awal 
N
t
NBakhir A

t
N
2

50 1.8 5.0  10  2 

3.6
 0.2 V
#
Therefore,
  Ir
 0.2
I

 0.07 A #
r 2.8
 magnitude of the induced current in the coil
(b) The direction of the current is counter-clockwise #
9. A step-down transformer has 4000 turns on the primary and 200 turns on the
secondary. If the primary voltage is 2.2 kV, what is the secondary voltage?
Ans : (110 V)
Np = 4000
Ns = 200
Vp = 2.2 kV = 2.2 × 103 V
Np
Ns

Vs 
Vp
Vs
Vp Ns
Np
2.2  10 200

3
4000
 110V
#
10. When the emf for the primary of a transformer is 5.00 V, the secondary emf is 10.0
N
V. What is the transformer turns ratio  2
 N1
ε1 = 5.00 V
ε2 = 10.0 V
N2  2

N1  1

N 2 10.0

 2.00 #
N 1 5.00

 ?

Ans : ( 2.00 )