Exercise 4 (Electromagnetism) Suggested Answers
1. (a) (i) Sprinkle iron filings on the board and tap the board gently. The magnetic field pattern is shown by
the pattern of the filings.
(ii)
(b) (i) When S is pressed, current flows through the coil and the soft iron is magnetised. It attracts the
spring towards the left. The hammer strikes the left metal plate to produce the first note. When S is released, the
iron core is demagnetised, the hammer springs back to strike the right metal plate to produce the second note.
(ii) Replace one metal plate with another made of a different metal / Vary the length of one of the two
metal plates / Stick a lump of plasticine to one plate.
(iii) Statement 1 is correct because copper is not a magnetic material. Statement 2 is incorrect. If the
polarities of the battery are reversed, the soft iron core will still be magnetised.
2. (a) (i) End Q is a south pole.
(ii) The coil rotates in a clockwise direction.
(b) Commutator / split ring
The commutator is used to reverse the direction of current for every half cycle so that the force acting on
the coil reverses every half cycle. Thus the direction of rotation of the coil remains unchanged.
(c)Increase the emf of the battery / Increase the no. of turns of the solenoid / Insert a soft iron core inside
the paper cone/ Increase the area of the coil.
(d) If the terminals of the battery are reversed, the polarities of the solenoids and the current flowing
through the coil are reversed at the same time. So the direction of forces acting on the coil will remain
unchanged. Hence the motor will still function properly.
3. (a) The student is incorrect. By right hand grip rule, the magnetic field at points on RS due to each wire
always point downwards. They will not cancel out each other and there should be no neutral point on RS.
(b) The neutral point should be on the left of wire R.
Let the distance between the neutral point and wire R be d.
BR
2.5 A
3A
BS
R
S
d
2m
Consider the neutral point. Applying B 
μ0I
, we have
2πr
μ 0  2 .5
μ0  3

2 πd
2 πd  2 
2.5d  2  3d
d  10 m
The neutral point should be 10 m on the left of wire R.
4. (a)
(b) Assume the magnetic field is perpendicular to the current in the wire.
By Fleming’s left hand rule, the magnetic force on the coil points to left.
(c) Period T = 2 ms
=> f = 1/T = 1/2 x 10-3 = 500 Hz
(d) Yes, a loudspeaker has similar structure as microphone.
When sound waves hit the diaphragm, the diaphragm together with the coil vibrates in and out in the
magnetic field. By electromagnetic induction, an a.c. of the same frequency as the incident sound waves is set
up. The loudspeaker therefore functions as a microphone.
5.
(a) -z direction
(b) The direction of velocity is always changing, so the velocity is not a constant. Acceleration is defined as
the rate of change of velocity, so the electron is accelerating.
(c) Magnetic force
qvB = mv2/r
Bqr 0.43  10 3  1.6  10 19  74  10 3
v

 5.59  106 ms 1
31
m
9.11  10
6.
(a)
(b)
From the graph in (a), the slope is

1.875  10 3  0
 7.50  10 4 N A 1
2.50  0
Assume that the wire is rigid so the magnetic field is perpendicular to the current in the wire.
Applying F = BIL, the slope of the graph in (a) is equal to BL. Therefore, the magnitude B of the magnetic
field between the magnets is
slope 7.50  10 4
B

 0.015 T
L
0.05
7. (a)
−10 kV
0V
uniform
magnetic field
X
hot filament
electric field
lines
radius of
curvature
cathode anode
fluorescent
screen
(b) By the law of conservation of energy, the gain in kinetic energy of an electron equals the loss in electric
potential energy. We have
1 2
mv  qV
2
v
(c)


2qV
2  1.60 10 19  10 000 

 5.93  10 7 m s 1
31
m
9.11 10
(i)
The electron beam is deflected downwards in the uniform magnetic field. By Fleming’s left hand
rule, the magnetic field points into the plane of the paper.
(ii) The electrons move in a circular path in the field. The magnetic force provides the centripetal force
of the motion. We have
mv 2
 qvB
r
mv
r
Bq
(d)
where m, q and v are the mass, charge and speed of the electron respectively.
Therefore, the radius of curvature r of the path is inversely proportional to magnitude B of the field.
(i) The uniform magnetic field can be created between two coils carrying the same size of current in the
same sense of direction.
(ii) By controlling the magnitude and direction of the uniform magnetic field, we can control the radius
of curvature of the path in the field and hence the degree of deflection of the beam. Thus, the vertical
movement of the spot produced can be controlled.
MC
1-5 C A A D B
6-10 D B B C C
11-15 D B D D B
16-18 A C C
Explanations to mc
1. Magnetic field lines start from north pole and magnetic field is the strongest at the poles.
2. Two current carrying straight currents in the same direction attract each other.
10. Efficiency = power output / power input = (mgh/t) / IV = (3 x 10 x 1/5) / (2 x 12) = 25%
11. Current is the same in both wires because of conservation of charge. Thus, nAvq is the same in both wires,
in other words, A v is the same in both wires (n and q are the same in both wires). Thus (1) is correct.
Now the cross-sectional area of LHS is 4 times that of RHS (since A is proportional to radius2 or diameter2),
so the velocity of RHS is 4 times that of LHS.
12. When the charge has come out, it must have travelled for half a period.
qvB = mv2 /r
=> qB = mv/r = m = m (2) / T’ where T’ is the period
T’ = 2m / qB
So generally, required time T = half period = T’/2 = m/qB
Now q  2q, so T  T/2
13. Net B =
0  4
0


 0
2
2  0.5 
Force per unit length =
direction: into paper
F
 2
 BI  0
 8  10 7 N m-1 (direction: to the left)
l

14. Drift velocity v is given by
v = I / nAq =
8
8  10  1  106  e
28
Force on electron = evB = e x
8
x 0.2 = 2 x 10-23 N
8  10  1  106  e
28
15. When the current 2.5A flows from Q to P, downward force corresponds to 13.5-12.5 = 1 g of weight.
When the current is reversed and becomes 1.5 A, the force is changed to upwards and the magnitude
corresponds to (1.5/2.5) x 1 g = 0.6 g of weight.
So the new reading = 12.5 g – 0.6 g = 11.9 g
16. In uniform circular motion, only the direction of velocity changes, the magnitude of velocity (i.e. speed)
does not change. So the KE does not change. Force is perpendicular to velocity, hence no work is done on the
particle. This is consistence with the fact that KE remains unchanged.
The period of circular motion is independent of the speed of particle. Refer to notes.
18. Net force = 0 since the electrons travel in a straight line.
Thus electric force = magnetic force
qE = qvB
q(V/d) = qvB