Download Exercise 4 (Electromagnetism)

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
It is used to reverse the direction of current flowing through the coil (reverse the direction of forces acting
on the wires). Hence the coil will continue to rotate in the same direction.
(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.
4  10 7  2
3. (a) Force on RHS wire = BIl 
 1  0.4  8  10 7 N (toRHS )
2 (0.2)
Force on LHS wire = BIl 
4  10 7  2
 1 0.4  3.2  10 7 N (toLHS )
2 (0.5)
Net force = (8 – 3.2) x 10-7 = 4.8 x 10-7 N (to RHS)
(b) Net torque = 0 since the moment arm is zero.
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
6.
 0 NI
4  10 7  100  5
= 0.00314 T
0.2

(b) To measure the magnetic field, place the hall probe in a known magnetic filed B0 first. Record the
hall voltage V0.
Then place the hall probe inside the solenoid. Rotate it slowly to obtain the maximum reading.
V
The unknown magnetic field can be calculated by B =
B0.
V0
OR To measure the magnetic field, place the hall probe in the magnetic field. Rotate it slowly to
obtain the maximum reading. The current and other parameters such as n, Q and t can be obtained from
the manufacturers. The magnetic field can be calculated using the formula B = VnQt/I.
(a)
B=
(c)
V=
=
0.00 314 1
BI
=
= 1.57 V
25
nQt
110 1.6 10 19 1.25 10 3
MC
1-5 C A A D B
6-10 E B B C C
11-15 B B D D A
16-17 C D
Explanations to mc
11. VH 
BI
 I 2 since B   0 nI (the same current I passes through t he slice and the solenoid)
nqt
Now I  I/2, so VH  VH /4
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’
T’ = 2m / qB
T = 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. After the switch is closed, downward force on magnet = 1112 – 1000 = 112 g (don’t concern the unit)
So upward force on wire = 112 g.
When the current is reversed and doubled, the upward force on magnet = 112 x 2 = 224 g
So new reading = 1000 – 224 = 776 g
17. Use the equation V = BI / nQt, t is 5 mm.