Electromagnetic induction
Code: 28L2A001, Total marks: 1
A solenoid is connected to a centre-zero galvanometer as shown. The pointer of the
galvanometer deflects to the left when a bar magnet with the north pole facing
downwards is moved towards the solenoid.
S
motion of magnet
N
solenoid
G
centre-zero galvanometer
hollow paper cylinder
In which of the following cases would the pointer of the galvanometer deflects to the
right?
(1)
(2)
(3) N
S
N
N
S
G
S
G
G
A.
B.
C.
D.
(1) only
(2) only
(1) and (3) only
(2) and (3) only
Answer: C
Code: 28L2A002, Total marks: 1
The magnitude of a simple d.c. generator’s output voltage varies with time as shown
below.
magnitude of the voltage / V
0
0.05
0.1
0.15
time / s
Which of the following statements is/are correct?
(1)
(2)
(3)
A.
B.
C.
D.
The coil is at the vertical position when t = 0.025 s, 0.075 s and 0.125 s.
The output voltage is reversed at t = 0.025 s, 0.075 s and 0.125 s.
The coil is rotating at a frequency of 10 Hz.
(1) only
(3) only
(1) and (2) only
(2) and (3) only
Answer: B
Code: 28L2A003, Total marks: 1
The magnitude of a simple d.c. generator’s output voltage varies with time as shown
below.
magnitude of the voltage / V
0
0.05
0.1
0.15
time / s
When the coil is rotating at 15 Hz,
(1) a larger force is required.
(2) the maximum magnitude of the output voltage would be larger.
(3) the waveform would be narrower.
A. (1) and (2) only
B. (1) and (3) only
C. (2) and (3) only
D. (1), (2) and (3)
Answer: D
Code: 28L2A004, Total marks: 1
Statement 1:
The output voltage of a simple a.c. generator increases with the rate of
rotation of the coil.
Statement 2: The frequency of the a.c. voltage is the same as the frequency of
rotation of the coil.
A. Both statements 1 and 2 are correct and statement 2 is a correct explanation for
statement 1.
B. Both statements 1 and 2 are correct but statement 2 is not a correct explanation
for statement 1.
C. Statement 1 is correct and statement 2 is incorrect.
D. Statement 1 is incorrect and statement 2 is correct.
Answer: B
Code: 28L2A005, Total marks: 1
A bar magnet moves from rest towards a solenoid and then stops as shown. The
solenoid is connected to a centre-zero galvanometer.
solenoid
N
S
bar magnet
+
G
hollow paper
cylinder
_
Which of the following graphs best represents the reading of the galvanometer?
A. galvanometer reading / mA
B. galvanometer reading / mA
0
time / s
C. galvanometer reading / mA
0
0
time / s
D. galvanometer reading / mA
time / s
0
time / s
Answer: A
Code: 28L2A006, Total marks: 1
R1
R2
G
solenoid A
solenoid B
In the figure above, current is induced in solenoid B
(1) when the resistance of rheostat R1 becomes lower.
(2) when the resistance of rheostat R2 becomes lower.
(3) at the instant the cell is disconnected from solenoid A.
A. (1) and (2) only
B. (1) and (3) only
C. (2) and (3) only
D. (1), (2) and (3)
Answer: B
Code: 28L2A007, Total marks: 1
A simple a.c. generator produces an alternating current when its coil rotates. The
graph below shows how the current varies with time.
current / mA
I
0
t
2t
3t
4t
5t
6t
time / s
-I
The coil now rotates at a higher speed. Which of the following graphs best show how
the current varies with time?
A.
current / mA
I
0
t
2t
3t
4t
5t
6t
2t
3t
4t
5t
6t
2t
3t
4t
5t
6t
2t
3t
4t
5t
6t
time / s
-I
B.
current / mA
I
0
t
time / s
-I
C.
current / mA
I
0
t
time / s
-I
D.
current / mA
I
0
t
time / s
-I
Answer: B
Code: 28L2A008, Total marks: 1
The figure below shows the simplified structure of a d.c. generator.
carbon brush
N-pole
coils
output
S-pole
magnet
carbon brush
The coils are now set to rotate. Which of the following graphs best represents how the
output voltage varies with time?
A. output voltage / V
B. output voltage / V
0
time / s
C. output voltage / V
0
D. output voltage / V
0
0
time / s
time / s
time / s
Answer: C
Code: 28L2A009, Total marks: 1
A bar magnet is held above a coil wound on a paper cylinder as shown. Which of the
following figures correctly shows the direction of the induced current in the coil if the
bar magnet or the coil moves in the ways as indicated by the arrows in the figures?
A.
B.
C.
D.
Answer: B
Code: 28L2A010, Total marks: 1
An aluminium rod XY is placed on a C-shaped copper rod in which AB is parallel to
DC. Rod XY is moved with a uniform velocity to the right across a uniform magnetic
field, whose direction is pointing into the paper as shown above. Which of the
following statements about rod XY are correct?
(1) An external force is required to keep rod XY moving at a uniform velocity.
(2) Y is at a higher potential than X.
(3) Current is induced in the loop AXYD and flows in an anticlockwise direction.
A. (1) and (2) only
B. (1) and (3) only
C. (2) and (3) only
D. (1), (2) and (3)
Answer: D
Code: 28L2A011, Total marks: 1
A conducting rectangular coil ABCD with its end connected to a galvanometer is
placed in a uniform magnetic field between two magnets as shown above. Which of
the following would produce a deflection in the galvanometer?
(1) Move the coil along the y-axis with sides AD and BC parallel to the field lines.
(2) Move the coil along the y-axis with the sides AD and BC perpendicular to the
field lines.
(3) Rotate the coil about the x-axis.
A. (1) only
B. (3) only
C. (1) and (2) only
D. (2) and (3) only
Answer: B
*Code: 28L2C002, Total marks: 9
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
(1M + 1A)
   BA  0.15  0.02  0.05  2  3 10 4 Wb
(ii) Applying Faraday’s law of induction, the average induced e.m.f. in the
circuit is
 3  10 4


 1.5  10  2 V  15 mV
(1M)
t
0.02

Applying I  , the average current I induced in the circuit is
R
2
1.5  10
(1M + 1A)
I
 5.882  10 3 A  5.88 mA
0.05  2.5
(b) (i) 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
(ii)
(1A for correct shape of the graph + 1A for correct period + 1A for correct axes and
labels)
*Code: 28L2C003, Total marks: 10
A right angled triangular conducting loop is moved across a uniform magnetic field
from position 1 to position 5 at a uniform speed v as shown. The height and base of
the loop is h and b respectively and the resistance of the loop is R. The magnetic field
points out of the page and has a magnitude B.
uniform magnetic field
(a) Determine the direction of the induced current in the loop at
(i) position 2;
(ii) position 3, and
(iii) position 4.
(b) (i)
(1 mark)
Find the change in area of the loop enclosed by the field from time = 0 (the
instant that the loop just enters the field) to time t as shown below.
uniform magnetic field
time = 0
time = t
(2 marks)
(ii) Based on the above answer, prove that the e.m.f. induced in the loop
increases linearly with time elapsed t when the loop enters into the field.
(2 marks)
(iii) Hence, prove that the maximum e.m.f. induced in the loop is given by
Bvh

.
(2 marks)
2
(c) Sketch the variation of the induced current I in the loop with time t when the
loop moves from positions 1 to 5.
(3 marks)
Answer:
(a) (i) clockwise
(ii) no induced current
(ii) anticlockwise
(1A)
(b) (i)
The area enclosed by the magnetic field at time = 0.
After time t, the loop travels a distance of vt.
By the properties of similar triangle, the height of the area enclosed by the
vth
field is
.
(1M)
b
Thus, the change in area ΔA of the loop enclosed by the field from time = 0
to time t is
1
A   base   height   0
2
1
 vth 
(1A)
  vt   

2
 b 
v 2t 2 h

2b
(ii) By Faraday’s law of induction, the induced e.m.f. is
 v 2t 2 h 

B
 B A
2b  Bv 2 h





t
(1M + 1A)
t
t 0
t
2b
Hence, the e.m.f. ε induced in the loop increases with time elapsed t.
(iii) From the answer of (b)(ii), the induced e.m.f. increases linearly with time
elapsed t. The maximum e.m.f. is obtained when the whole loop is just
b
enclosed by the field, i.e. vt = b. We have t  .
(1M)
v
Bvh
Hence, the maximum e.m.f.  
.
(1A)
2
(c)
(1A for correct shape of the graph + 1A for correct axes and labels + 1A for correct time
and current values)