Electromagnetic Induction Practice Questions - DPP for JEE

 Page 1


1. (b)
2. (b) The self-inductance of a coil L = µ
0
n
2
Al where
µ
0
 = permeability of air,
n = number of turns per unit length,
A = Area of cross-section and
l = length of the solenoid
This depends on the geometry of the inductor such as cross-sectional
area, length and number of turns and not on the material, even if it
is made of a super conducting material. If the superconductor is
below the critical temperature, the current will continuously flow
and the inductance may not have the property of inductance any
more. ?   L
1
 = L
2
; L
3
= 0.
3. (a) The induced emf will remain constant only in the case of
rectangular and square loops. In case of the circular and the
elliptical loops, the rate of change of area of the loops during their
passage out of the field is not constant, hence induced emf will not
remain constant for them.
4. (b) Induced e.m.f. in the ring opposes the motion of the magnet.
5. (a) When a north pole of a bar magnet moves towards the coil, the
induced current in the coil flows in a direction such that the coil
presents its north pole to the bar magnet as shown in figure (a).
Therefore, the induced current flows in the coil in the
anticlockwise direction. When a north pole of a bar magnet moves
away from the coil, the induced current in the coil flows in a
direction such that the coil presents its south pole to the bar magnet
as shown in figure (b).
Therefore induced current flows in the coil in the clockwise direction.
6. (d) Here, induced e.m.f.
Page 2


1. (b)
2. (b) The self-inductance of a coil L = µ
0
n
2
Al where
µ
0
 = permeability of air,
n = number of turns per unit length,
A = Area of cross-section and
l = length of the solenoid
This depends on the geometry of the inductor such as cross-sectional
area, length and number of turns and not on the material, even if it
is made of a super conducting material. If the superconductor is
below the critical temperature, the current will continuously flow
and the inductance may not have the property of inductance any
more. ?   L
1
 = L
2
; L
3
= 0.
3. (a) The induced emf will remain constant only in the case of
rectangular and square loops. In case of the circular and the
elliptical loops, the rate of change of area of the loops during their
passage out of the field is not constant, hence induced emf will not
remain constant for them.
4. (b) Induced e.m.f. in the ring opposes the motion of the magnet.
5. (a) When a north pole of a bar magnet moves towards the coil, the
induced current in the coil flows in a direction such that the coil
presents its north pole to the bar magnet as shown in figure (a).
Therefore, the induced current flows in the coil in the
anticlockwise direction. When a north pole of a bar magnet moves
away from the coil, the induced current in the coil flows in a
direction such that the coil presents its south pole to the bar magnet
as shown in figure (b).
Therefore induced current flows in the coil in the clockwise direction.
6. (d) Here, induced e.m.f.
 
7. (b) At t = 2s, B = 4T,  
A = 20 × 30cm
2
= 600 × 10
–4
 m
2
,   
= – 100 × 10
–4
 m
2
/s
= – [4 × (– 100 × 10
–4
) + 600 × 10
–4
 × 2]
= – [– 0.04 + 0.120] = – 0.08 V
8. (d) f = n BA cos ? = 10 B a
2
 cos ?t
9. (a) As the magnetic field increases, its flux also increases into the
page and so induced current in bigger loop will be anticlockwise.
i.e., from D to C in bigger loop and then from B to A in smaller
loop.
10. (c)
[Since effective resistance R of bridge is  
so total resistance = 1 + 3 = 4?]
? v = 2 cm s
–1
Page 3


1. (b)
2. (b) The self-inductance of a coil L = µ
0
n
2
Al where
µ
0
 = permeability of air,
n = number of turns per unit length,
A = Area of cross-section and
l = length of the solenoid
This depends on the geometry of the inductor such as cross-sectional
area, length and number of turns and not on the material, even if it
is made of a super conducting material. If the superconductor is
below the critical temperature, the current will continuously flow
and the inductance may not have the property of inductance any
more. ?   L
1
 = L
2
; L
3
= 0.
3. (a) The induced emf will remain constant only in the case of
rectangular and square loops. In case of the circular and the
elliptical loops, the rate of change of area of the loops during their
passage out of the field is not constant, hence induced emf will not
remain constant for them.
4. (b) Induced e.m.f. in the ring opposes the motion of the magnet.
5. (a) When a north pole of a bar magnet moves towards the coil, the
induced current in the coil flows in a direction such that the coil
presents its north pole to the bar magnet as shown in figure (a).
Therefore, the induced current flows in the coil in the
anticlockwise direction. When a north pole of a bar magnet moves
away from the coil, the induced current in the coil flows in a
direction such that the coil presents its south pole to the bar magnet
as shown in figure (b).
Therefore induced current flows in the coil in the clockwise direction.
6. (d) Here, induced e.m.f.
 
7. (b) At t = 2s, B = 4T,  
A = 20 × 30cm
2
= 600 × 10
–4
 m
2
,   
= – 100 × 10
–4
 m
2
/s
= – [4 × (– 100 × 10
–4
) + 600 × 10
–4
 × 2]
= – [– 0.04 + 0.120] = – 0.08 V
8. (d) f = n BA cos ? = 10 B a
2
 cos ?t
9. (a) As the magnetic field increases, its flux also increases into the
page and so induced current in bigger loop will be anticlockwise.
i.e., from D to C in bigger loop and then from B to A in smaller
loop.
10. (c)
[Since effective resistance R of bridge is  
so total resistance = 1 + 3 = 4?]
? v = 2 cm s
–1
11. (d) The magnetic field is increasing in the downward direction.
Therefore, according to Lenz’s law the current I
1
 will flow in the
direction ab and I
2
 in the direction dc. 
12. (a)
emf induced per unit turn = 
13. (d) Rate of decrease of area of the semicircular ring
Acoording to Faraday’s law of induction induced emf
The induced current in the ring must generate magneticfield in the
upward direction. Thus Q is at higher potential.
14. (c) The area swept by radius OC in one half circle is pr
2
/2.
 
The flux
change in time T/2 is thus  (pr
2 
B/2). The induced emf is then e =
pr
2
B/T = B?r
2
/2 
The induced current is then I = e/R = B?r
2
 /2R
15. (c) = QR
16. (a) If we consider the cylindrical surface to be a ring of radius R, there
will be an induced emf due to changing field.
Page 4


1. (b)
2. (b) The self-inductance of a coil L = µ
0
n
2
Al where
µ
0
 = permeability of air,
n = number of turns per unit length,
A = Area of cross-section and
l = length of the solenoid
This depends on the geometry of the inductor such as cross-sectional
area, length and number of turns and not on the material, even if it
is made of a super conducting material. If the superconductor is
below the critical temperature, the current will continuously flow
and the inductance may not have the property of inductance any
more. ?   L
1
 = L
2
; L
3
= 0.
3. (a) The induced emf will remain constant only in the case of
rectangular and square loops. In case of the circular and the
elliptical loops, the rate of change of area of the loops during their
passage out of the field is not constant, hence induced emf will not
remain constant for them.
4. (b) Induced e.m.f. in the ring opposes the motion of the magnet.
5. (a) When a north pole of a bar magnet moves towards the coil, the
induced current in the coil flows in a direction such that the coil
presents its north pole to the bar magnet as shown in figure (a).
Therefore, the induced current flows in the coil in the
anticlockwise direction. When a north pole of a bar magnet moves
away from the coil, the induced current in the coil flows in a
direction such that the coil presents its south pole to the bar magnet
as shown in figure (b).
Therefore induced current flows in the coil in the clockwise direction.
6. (d) Here, induced e.m.f.
 
7. (b) At t = 2s, B = 4T,  
A = 20 × 30cm
2
= 600 × 10
–4
 m
2
,   
= – 100 × 10
–4
 m
2
/s
= – [4 × (– 100 × 10
–4
) + 600 × 10
–4
 × 2]
= – [– 0.04 + 0.120] = – 0.08 V
8. (d) f = n BA cos ? = 10 B a
2
 cos ?t
9. (a) As the magnetic field increases, its flux also increases into the
page and so induced current in bigger loop will be anticlockwise.
i.e., from D to C in bigger loop and then from B to A in smaller
loop.
10. (c)
[Since effective resistance R of bridge is  
so total resistance = 1 + 3 = 4?]
? v = 2 cm s
–1
11. (d) The magnetic field is increasing in the downward direction.
Therefore, according to Lenz’s law the current I
1
 will flow in the
direction ab and I
2
 in the direction dc. 
12. (a)
emf induced per unit turn = 
13. (d) Rate of decrease of area of the semicircular ring
Acoording to Faraday’s law of induction induced emf
The induced current in the ring must generate magneticfield in the
upward direction. Thus Q is at higher potential.
14. (c) The area swept by radius OC in one half circle is pr
2
/2.
 
The flux
change in time T/2 is thus  (pr
2 
B/2). The induced emf is then e =
pr
2
B/T = B?r
2
/2 
The induced current is then I = e/R = B?r
2
 /2R
15. (c) = QR
16. (a) If we consider the cylindrical surface to be a ring of radius R, there
will be an induced emf due to changing field.
? 
? Force on the electron 
? acceleration = 
As the field is increasing being directed inside the paper, hence there
will be anticlockwise induced current (in order to oppose the
cause) in the ring (assumed). Hence there will be a force towards
left on the electrons.
17. (a)
18. (b)
[ ]
19. (b) The individual emf produced in the coil 
? The current induced will be 
But   
20. (b) 
Page 5


1. (b)
2. (b) The self-inductance of a coil L = µ
0
n
2
Al where
µ
0
 = permeability of air,
n = number of turns per unit length,
A = Area of cross-section and
l = length of the solenoid
This depends on the geometry of the inductor such as cross-sectional
area, length and number of turns and not on the material, even if it
is made of a super conducting material. If the superconductor is
below the critical temperature, the current will continuously flow
and the inductance may not have the property of inductance any
more. ?   L
1
 = L
2
; L
3
= 0.
3. (a) The induced emf will remain constant only in the case of
rectangular and square loops. In case of the circular and the
elliptical loops, the rate of change of area of the loops during their
passage out of the field is not constant, hence induced emf will not
remain constant for them.
4. (b) Induced e.m.f. in the ring opposes the motion of the magnet.
5. (a) When a north pole of a bar magnet moves towards the coil, the
induced current in the coil flows in a direction such that the coil
presents its north pole to the bar magnet as shown in figure (a).
Therefore, the induced current flows in the coil in the
anticlockwise direction. When a north pole of a bar magnet moves
away from the coil, the induced current in the coil flows in a
direction such that the coil presents its south pole to the bar magnet
as shown in figure (b).
Therefore induced current flows in the coil in the clockwise direction.
6. (d) Here, induced e.m.f.
 
7. (b) At t = 2s, B = 4T,  
A = 20 × 30cm
2
= 600 × 10
–4
 m
2
,   
= – 100 × 10
–4
 m
2
/s
= – [4 × (– 100 × 10
–4
) + 600 × 10
–4
 × 2]
= – [– 0.04 + 0.120] = – 0.08 V
8. (d) f = n BA cos ? = 10 B a
2
 cos ?t
9. (a) As the magnetic field increases, its flux also increases into the
page and so induced current in bigger loop will be anticlockwise.
i.e., from D to C in bigger loop and then from B to A in smaller
loop.
10. (c)
[Since effective resistance R of bridge is  
so total resistance = 1 + 3 = 4?]
? v = 2 cm s
–1
11. (d) The magnetic field is increasing in the downward direction.
Therefore, according to Lenz’s law the current I
1
 will flow in the
direction ab and I
2
 in the direction dc. 
12. (a)
emf induced per unit turn = 
13. (d) Rate of decrease of area of the semicircular ring
Acoording to Faraday’s law of induction induced emf
The induced current in the ring must generate magneticfield in the
upward direction. Thus Q is at higher potential.
14. (c) The area swept by radius OC in one half circle is pr
2
/2.
 
The flux
change in time T/2 is thus  (pr
2 
B/2). The induced emf is then e =
pr
2
B/T = B?r
2
/2 
The induced current is then I = e/R = B?r
2
 /2R
15. (c) = QR
16. (a) If we consider the cylindrical surface to be a ring of radius R, there
will be an induced emf due to changing field.
? 
? Force on the electron 
? acceleration = 
As the field is increasing being directed inside the paper, hence there
will be anticlockwise induced current (in order to oppose the
cause) in the ring (assumed). Hence there will be a force towards
left on the electrons.
17. (a)
18. (b)
[ ]
19. (b) The individual emf produced in the coil 
? The current induced will be 
But   
20. (b) 
.
( Q W
2
 & W
1
 are magnetic flux)
21. (1) Given : 
?
Induced current, 
At t = 1 s,
22. (190) The back e.m.f. in a motor is induced e.m.f., which is maximum,
when speed of rotation of the coil is maximum
23. (0.15) Induced emf = vB
H 
l = 1.5 × 5 × 10
–5
 × 2
= 15 × 10
–5
= 0.15 mV
24. (2) L = 2mH, i = t
2
e
–t
E = 
when E = 0,
–e
–t
 t
2
 + 2te
–t
 = 0
or,  2t e
–t
 = e
–t 
t
2
?  t = 2 sec.
25. (2.4p × 10
–4
)