Download Exercise 5 Solution

Exercise 5 (Electromagnetic induction) Suggested Answers
1. (a) When the magnet moves towards the solenoid, a current passes from X through G to Y. When the magnet is
inside the solenoid, there is no current. When the magnet moves away from the solenoid, a current passes from Y
through G to X.
(b) (i) Q and S.
(ii) Peak voltage = 1 x 50 = 50 mV.
Frequency = 1/T
= 1 / (4×20×10-3) = 12.5 Hz
(iii) (1) Peak voltage increases, frequency increases.
(2) Peak voltage increases, frequency unchanged.
(iv) Wind/running water/sea wave /tidal power.
2. (a) max B = o nImax
 4  10 7 

BA
I
300 1.5
 Np
 N P A 0 n
 200  3  (10 2 ) 2  4  10 7 

 1.36  10 3V
t
t
t
0.1 0.25
EP  N P
(b)
300
 1.5  5.65  10 3 T
0.1
3. (a) Flux linkage = N = NBA cos 
= 200 x 3 x 10-2 x 2 x 10-4 = 0.0012 Wb (1.2 x 10-3 Wb)
(b) (i)  = NBA cos 0o – NBA cos 90o = 0.0012 Wb
t = 1/4 period = (1/4) (1/2) = 1/8 s
N
 0.0012

 0.0096V
t
1/ 8
(9.6mV)
(ii) The induced emf depends on the rate of change of flux linkage or the rate at which the coil cuts the
magnetic field lines.
When  = 0o or 180o or 360o, the plane of coil is parallel is perpendicular to the field lines, its motion does
not cut any magnetic field lines, so the induced emf is zero.
When  = 900, or 270o, the plane of coil is parallel to the field lines, its motion cuts the field lines at the
fastest rate (or the rate of change of flux linkage is the greatest), thus the induced emf is a maximum.
4.
(a)
Induced emf = change in flux linkage / time
E
= 600 x 10-6 x 4.0 x 10-3 / 0.8 =
3 x 10-6 V
Induced current I = E/R = 3 x 106 / 2.6 = 1.15 x 10-6 A
(b)
Direction: clockwise
5. (a)
(b) The resultant magnetic field is perpendicular to and out of the paper. When the current in MN
increases, there is an increase of magnetic field out of paper. By the Lenz’s law, a current flows in clockwise
direction in the frame so as to produce a magnetic field into paper, i.e. to oppose the increase of ‘out of
paper’ magnetic field.
MC ANSWERS
1-5 D D D B A
6-10
CEEDC
11-15 A B C A A
16-20 D B D D C
Explanations to selected mc
5.
At O, the velocity is zero, so there is no cutting of magnetic field lines, hence no emf is induced.
When the bar swings down, PE is changed into KE, so the speed is increasing, hence the cutting of magnetic
field lines is increasing, so induced emf is increased.
When the bar swings down after reaching the highest point, the velocity changes direction, so the emf is in
opposite direction, hence the emf induced is a.c.
8. Since the period of a.c. is doubled, so the rate of rotation of the coil must be halved. This will also half the
amplitude of the emf, but since the amplitude is unchanged, so the no. of turns must be doubled.
11. For a search coil, the maximum emf induced E  NAB0
Now, N  2N, A  A/2, B0 2B0, f  f/2, so E is unchanged
12. E  N
d
50
 3
 1.88V
dt
8
13. When the wire frame is moving inside the magnetic field, the flux does not change as it is totally inside the
magnetic field. However, when it is moving out of the magnetic field, the flux into the paper is decreasing, so
induced current flows in clockwise direction so as to increase the flux into paper. Once it comes completely out of
the field, the induced current becomes zero again.
14. Induced emf = Blv = 1 x 10-5 x 60 x 150 = 0.09V
BA
d

17. Induced electromotive force  =
=
= Bv = B
v.
sin 60
t
t

Bdv
Then the current in the resistor = I  
R R sin 60
18. According to Fleming’s right hand rule, current flows from X to Y through the rod. So Y is at a higher voltage
(just like the positive terminal of battery) and current flows from Y to X through the rails.