Download Electromagnetic Induction 1

Electromagnetic Induction 1
1
A copper bar rests on two parallel conducting rails as shown. A uniform magnetic field is pointing into the
paper.
rail
copper bar

G
F

rail
magnetic field
A student applied a horizontal force F on the bar and pulls the bar to the right.
(a) Draw a free body diagram to show all horizontal forces acting on the bar.
(3 marks)
(b) Find the direction of the current flowing along the bar.
(1 mark)
Answer:
magnetic force
copper bar
F
friction
(a)
(3A)
(b) The current flows upwards along the bar.
(1A)
2
A bar magnet and an iron bar of the same weight are dropped through two aluminium tubes from the same
height as shown. The air resistance acting on the bar magnet and the iron bar is negligible.
(a) Which of the them will fall faster? Explain your answer briefly.
(2 marks)
(b) If the aluminium tubes are replaced by plastic tubes, what is your answer in (a)?
(2 marks)
Answer:
(a) When the bar magnet falls through the aluminium tube, current is induced in the tube so as to oppose
the motion of the magnet (1A). Hence the iron bar will fall faster than the bar magnet (1A).
(b) The changing magnetic field produced by a falling magnet would not induce a current in a plastic tube
(1A). Therefore, both bars would fall at the same rate (1A).
3
A light bulb is connected to a simple d.c. generator as shown.
B
C
A
device X
D
bulb
carbon
brushes
(a) Name device X.
(b) What is the direction of the current?
(c) State two ways to increase the brightness of the bulb.
(1 mark)
(1 mark)
(2 marks)
Answer:
(a) Device X is a commutator /half-ring.
(b) The current flows in the direction DCBA.
(c) The brightness of the bulb can be increased by
1. rotating the coil faster.
2. using stronger magnets.
3. winding the coil on a soft iron core.
4. using a coil of larger surface area.
5. using a coil of more number of turns.
(any two of the above)
(1A)
(1A)
(2A)
4.
A circular metal loop of resistance 0.75 Ω and radius 3 cm is placed in a uniform magnetic field
perpendicularly as shown in the left figure. The magnitude of the field changes as shown in the graph on the
right.
3 cm
(a) What is the direction of the current in the loop from
(i) 0 to 0.1 s;
(ii) 0.1 to 0.3 s, and
(iii) 0.3 to 0.6 s?
(1 mark)
(b) Draw the variation of the current I in the loop with time t in the following graph paper.
(5 marks)
Answer:
(a) (i)
clockwise
(ii) no induced current
(iii) anticlockwise
(b) Applying Faraday’s law of induction,
from 0 to 0.1 s,  
Applying R 
I

t

A B
t
(1A)
π0.03 0.75  0.25
 0.014 137 V
0.1  0
2

V
, the size of the induced current is
I
(1M)
(1M)
0.014 137
 0.01884 A  18.8 mA
0.75
From 0.1 to 0.3 s, there is no change in magnetic flux. There is no induced e.m.f and hence no current
is induced in the loop.
Similarly, from 0.3 to 0.6 s,  

t

A B
t
π0.03 1.5  0.25
 0.011 78 V
0.6  0.3
2

The size of the induced current is
I
0.011 78
 0.01570 A  15.7 mA
0.75
Take the clockwise direction to be positive. The graph is
(1A for suitable scale of the graph + 1A for axes and labels + 1A for correct shape of the graph)
5.
A 100 turn coil of radius 4 cm is coaxial with a long solenoid as shown.
100 turn coil
solenoid
The solenoid has 200 turns per centimetre and the current in the solenoid drops 2.5 A in size in a time
interval of 2 s.
(a) Find the e.m.f. induced in the coil.
(3 marks)
(b) Assume the coil is made of thin enough wire and is closely packed. Estimate the length of the wire. (2
marks)
(c) The resistance per metre of the wire is 0.03 Ω. Find the current induced in the coil.
(2 marks)
Answer:
(a) Applying B  μ 0 nI , the change in magnitude of the magnetic field inside the solenoid is


 200 
B  μ 0 nI  4π  10 7  
2.5  0.06283 T
 0.01 
Applying Faraday’s law of induction, the induced e.m.f. in the coil is

(1M)

NAB 100  π  0.04 2  0.06283
(1M + 1A)


 0.015 79 V  15.8 mV
t
20
(b) Assume the coil is made of thin enough wire and is closely packed. The length of the wire used is
estimated to be
 2πr  N
 2π  0.04  100
(1M + 1A)
 25.13
 25.1 m
(c) The resistance of the wire used is R  0.03  25.13  0.7539 
Applying I 
V 0.01579

 0.02094 A  20.9 mA .
R 0.7539
(1M + 1A)
Thus, the current induced in the coil is 20.9 mA.
6
A long rectangular conducting loop of width W, resistance R and mass m is released from rest under the
influence of Earth’s gravity in a uniform magnetic field B inside a long magnetic tunnel as shown in the left
figure.
uniform magnetic field
W
magnetic
tunnel
v
ground
As the loop emerges from the uniform field, the loop falls at a uniform speed v as shown in the right figure.
(a) Find the decrease in the magnetic flux through the loop in a time interval of Δt as the loop emerges
from the field.
(2 marks)
(b) Find the current induced in the loop.
(3 marks)
(c) Hence, prove that the uniform speed v 
mgR
where g is the acceleration due to gravity.
B 2W 2
(2
marks)
Answer:
(a) In a time interval of Δt, the distance travelled by the loop is vΔt.
Hence, the decrease in the magnetic flux through the loop is
  BA  Bvt W
(1M + 1A)
(b) By Faraday’s law of induction, the e.m.f. induced in the loop is


t

Bvt W
 BvW .
t
Applying I 
I
BvW
R
(1M)

, the current induced in the loop is
R
(1M + 1A)
(c) As the loop emerges from the uniform field, the loop falls at a uniform speed v. The magnetic force on
the upper horizontal arm balances the gravitational force on the loop.
(1M)
We have
mg  BIW
 BvW
mg  B
 R
mgR
v 2 2
BW
2

W

(1A)
A conducting wire with an insulating handle is bent into a shape as shown.
5 cm
2 cm
insulating
handle
The wire is placed in a uniform magnetic field of magnitude 0.15 T and is connected to an external circuit as
shown. It is then rotated through 180° anticlockwise in a time interval of 0.02 s as seen by the observer.
uniform magnetic field
hinge
observer
2.5 Ω
(a) (i) Find the change in magnetic flux in the process.
(2 marks)
(ii) The resistance of the wire between the hinges is 0.05 Ω. Find the average current induced in the
circuit.
(3 marks)
Now, the wire is rotated at a constant angular speed of 50 revolutions per second anticlockwise as seen by
the observer.
(b) (i) Find the period of the induced e.m.f.
(1 mark)
(ii) Hence, sketch the variation of the induced e.m.f. ε in the loop with time t.
(3 marks)
Answer:
(a) (i)
The change in magnetic flux is
   BA  0.15  0.02  0.05  2  3 10 4 Wb
(1M + 1A)
(ii) Applying Faraday’s law of induction, the average induced e.m.f. in the circuit is


t

3  10 4
 1.5  10  2 V  15 mV
0.02
Applying I 

, the average current I induced in the circuit is
R
(1M)
1.5  10 2
(1M + 1A)
 5.882  10 3 A  5.88 mA
0.05  2.5
The frequency of the induced e.m.f. should be the same as that of the rotation of the wire, i.e. 50
Hz. Therefore, the period of the induced e.m.f. is
1
(1A)
T   0.02 s
f
I
(b) (i)
(ii)
(1A for correct shape of the graph + 1A for correct period + 1A for correct axes and labels)