Electromagnetic Induction- 1 Practice Questions - DPP for NEET

 Page 1


(1) (a) B.A f=
r r
    = (
ˆ
0.02i ) . (
ˆ ˆˆ
30i 16j 23k ++ ) × 10
–4
     = 0.6 × 10
-4
 Wb = 60m Wb
(2) (c) The induced emf
E = – df/dt = – 
d
dt
 (3t
2
 + 2t + 3) × 10
-3
(because given flux is in mWb).
Thus E = ( – 6t – 2) × 10
–3
At t = 2 sec,
E = ( – 6 × 2 – 2) × 10
–3
 = –14 mV .
(3) (a) The direction of current in the solenoid is clockwise.
On displacing it towards the loop a current in the loop
will be induced in opposite sense so as to oppose its
approach. Therefore the direction of induced current
as observed by the observer will be anticlockwise.
(4) (b) When north pole of the magnet is moved away, then
south pole is induced on the face of the loop in front of
the magnet i.e. as seen from the magnet side, a clockwise
induced current flows in the loop. This makes free
electrons to move in opposite direction, to plate a. Thus
excess positive charge appear on plate b.
(5) (d) If electron is moving from left to right, the flux linked
with the loop (which is into the page) will first increase
and then decrease as the eletron passes by. So the
induced current in the loop will be first anticlockwise
and will change direction as the electron passes by .
(6) (c) E = –
dt
df
or  df = – Edt = (0 - f)
or     f = 4 × 10
–3
 × 0.1
            = 4 × 10
–4
 weber
(7) (d) e = 
d
dt
f
 = – 
d
dt
 [10t
2
 + 5t + 1] × 10
–3
   = – [10 × 10
–3
 (2t) + 5 × 10
–3
]
at t = 5 second
e = –[10 × 10
–2
 + 5 × 10
–3
] = [0.1 + 0.005]
|e| = 0.105V
(8) (a) For each spoke, the induced emf between the centre O
and the rim will be the same
e = 
1
2
 BwL
2
 
 
=   BpL
2
 f (Q  w = 2pf)
Further for all spokes, centre O will be positive while
rim will be negative. Thus all emf's are in parallel giving
total emf   e =  BpL
2
 f
independent of the number of the spokes.
Substituting the values
e = 4 × 10
–5
 × 3.14 × (.5)
2
 × 2 = 6.28 × 10
–5
 volt
(9) ( d ) Rate of decrease of area of the semicircular ring
A
–
d
dt
 = (2R) V
According to Faraday’s law of induction, induced emf
e = 
d
dt
f
-
= 
A
–B
d
dt
 = – B(2R V) V)
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
N
M
Q
vt
2R
The induced current in the ring must generate magnetic
field in the upward direction. Thus Q is at higher
potential.
(10) (b) E = B lv
    = 4 × 10
–4
 × 50 × 
360 1000
60 60
´
´
 = 2 V
(11) (c) The flux through the area is
f = BA cos 57º = 42 × 10
-6
 × 2.5 × 0.545
    =  57 × 10
-6
 Wb.
(12) (a) The magnetic flux linked with the loop at any instant of
time t is given by
     f = BAN cos wt
or  f = 10Ba
2
 cos wt
Here N = 10, A = a
2
(13) (b) According to Lenz’s Law
(14) (c) e = Bv l = 0.4 × 10
–4
 × 
3
300 10
3
60 60
´
´
´
  = 10
–2
 V
(15) (b) Magnetic flux passing through the disc is f = BA
= 0.01 
2
weber
meter
 × 3.14 × (15 × 10
–2
 meter)
2
= 7.065 × 10
–4
 weber.
The line joining the centre and the circumference of the
disc cuts 7.065 x 10
-4
 weber flux in one round. So, the
rate of cutting flux (i.e. induced emf)
= flux × number of revolutions per second
= 7.065 × 10
–4
 × 
100
603 ´
 = 3.9 × 10
–4
 volt.
(1 6) (c) The magnetic f lux passing through eac h turn of a coil of area
A, perpendicular to a magnetic field B is given by
f
1
 = BA.
The magnetic flux through it on rotating it through 180º
will be
f
2
 = – BA.(- sign is put because now the flux lines
enters the coils through the outer face)
Page 2


(1) (a) B.A f=
r r
    = (
ˆ
0.02i ) . (
ˆ ˆˆ
30i 16j 23k ++ ) × 10
–4
     = 0.6 × 10
-4
 Wb = 60m Wb
(2) (c) The induced emf
E = – df/dt = – 
d
dt
 (3t
2
 + 2t + 3) × 10
-3
(because given flux is in mWb).
Thus E = ( – 6t – 2) × 10
–3
At t = 2 sec,
E = ( – 6 × 2 – 2) × 10
–3
 = –14 mV .
(3) (a) The direction of current in the solenoid is clockwise.
On displacing it towards the loop a current in the loop
will be induced in opposite sense so as to oppose its
approach. Therefore the direction of induced current
as observed by the observer will be anticlockwise.
(4) (b) When north pole of the magnet is moved away, then
south pole is induced on the face of the loop in front of
the magnet i.e. as seen from the magnet side, a clockwise
induced current flows in the loop. This makes free
electrons to move in opposite direction, to plate a. Thus
excess positive charge appear on plate b.
(5) (d) If electron is moving from left to right, the flux linked
with the loop (which is into the page) will first increase
and then decrease as the eletron passes by. So the
induced current in the loop will be first anticlockwise
and will change direction as the electron passes by .
(6) (c) E = –
dt
df
or  df = – Edt = (0 - f)
or     f = 4 × 10
–3
 × 0.1
            = 4 × 10
–4
 weber
(7) (d) e = 
d
dt
f
 = – 
d
dt
 [10t
2
 + 5t + 1] × 10
–3
   = – [10 × 10
–3
 (2t) + 5 × 10
–3
]
at t = 5 second
e = –[10 × 10
–2
 + 5 × 10
–3
] = [0.1 + 0.005]
|e| = 0.105V
(8) (a) For each spoke, the induced emf between the centre O
and the rim will be the same
e = 
1
2
 BwL
2
 
 
=   BpL
2
 f (Q  w = 2pf)
Further for all spokes, centre O will be positive while
rim will be negative. Thus all emf's are in parallel giving
total emf   e =  BpL
2
 f
independent of the number of the spokes.
Substituting the values
e = 4 × 10
–5
 × 3.14 × (.5)
2
 × 2 = 6.28 × 10
–5
 volt
(9) ( d ) Rate of decrease of area of the semicircular ring
A
–
d
dt
 = (2R) V
According to Faraday’s law of induction, induced emf
e = 
d
dt
f
-
= 
A
–B
d
dt
 = – B(2R V) V)
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
N
M
Q
vt
2R
The induced current in the ring must generate magnetic
field in the upward direction. Thus Q is at higher
potential.
(10) (b) E = B lv
    = 4 × 10
–4
 × 50 × 
360 1000
60 60
´
´
 = 2 V
(11) (c) The flux through the area is
f = BA cos 57º = 42 × 10
-6
 × 2.5 × 0.545
    =  57 × 10
-6
 Wb.
(12) (a) The magnetic flux linked with the loop at any instant of
time t is given by
     f = BAN cos wt
or  f = 10Ba
2
 cos wt
Here N = 10, A = a
2
(13) (b) According to Lenz’s Law
(14) (c) e = Bv l = 0.4 × 10
–4
 × 
3
300 10
3
60 60
´
´
´
  = 10
–2
 V
(15) (b) Magnetic flux passing through the disc is f = BA
= 0.01 
2
weber
meter
 × 3.14 × (15 × 10
–2
 meter)
2
= 7.065 × 10
–4
 weber.
The line joining the centre and the circumference of the
disc cuts 7.065 x 10
-4
 weber flux in one round. So, the
rate of cutting flux (i.e. induced emf)
= flux × number of revolutions per second
= 7.065 × 10
–4
 × 
100
603 ´
 = 3.9 × 10
–4
 volt.
(1 6) (c) The magnetic f lux passing through eac h turn of a coil of area
A, perpendicular to a magnetic field B is given by
f
1
 = BA.
The magnetic flux through it on rotating it through 180º
will be
f
2
 = – BA.(- sign is put because now the flux lines
enters the coils through the outer face)
DPP/ P 44
123
\ change in magnetic flux
D f = f
1
 – f
2
 = – BA – (BA) = – 2BA.
Suppose this change takes in time Dt. According to
Faraday's law , the emf induced in the coil is given by
e = – N
t
Df
D
 = 
2NBA
t D
,
where N is number of turns in the coil. The current in
the coil will be
i = 
e1
RR
=
2NBA
t D
where R is the resistance of the circuit. The current
persists only during the change of flux i.e. for the time
interval Dt second. So, the charge passed through the
circuit is
q = i × Dt = 
2NBA
R
.
Here N = 500, B = 0.2 weber/meter
2
,
A = 4.0 cm
2
 = 4.0 × 10
–4
 meter
2
 and R = 50 ohm.
\  
4
2 500 0.2 4.0 10
q
50
-
´ ´ ´´
=
            = 1.6 × 10
–3
 coulomb.
(17) (a)T
(18) (c) The induced emf between centre and rim of the rotating
disc is
E = 
1
2
BwR
2
 = 
1
2
× 0.1 × 2p × 10 × (0.1)
2
        
 = 10p × 10
–3
 volt,
(19) (a) The induced emf
E = Blv = 0.2 × 10
–4
 × 1 × 180 × 1000/3600
     = 0.2 × 18/3600 = 1 × 10
–3
. V = 1mV
(20) (d) The induced emf is obtained by considering a strip on
the disc fig. Then, the linear speed of a small element dr
at a distance r from the centre is = wr. The induced emf
across the ends of the small element is-
de = B(dr)v = B wr dr
Thus the induced emf across the inner and outer sides
of the disc is
e = 
b
a
B r dr w
ò
 = 
1
2
 Bw (b
2
 – a
2
)
(21) (c) The induced emf e = – (v B). ´
r r
r
l
For the part PX, vB ^
r
r
, and the angle between (v B) ´
r
r
direction (the dotted line in figure and 
r
l
 is (90 – q).
Thus
e
P
 –e
x
 = vB l cos (90 – q/2) = vB l sin (q/2)
Similarly e
y
 – e
p
 = vBl sin (q/2)
Therefore induced emf
between X and Y is e
yx
 = 2 B v l sin (q/2)
(22) (a) Given, area = 10 × 20 cm
2
 = 200 × 10
-4
 m
2
B = 0.5 T, N = 60,  w = 2p × 1800/60
Q e = –
dN
dt
() f
       = –N 
d
dt
(BA cos wt)
      = NBAw sin wt
\ e
max
 = NABw
                  = 60 × 2 × 10
–2
 × 0.5 × 2p × 1800/60 = 113 volt.
Hence, induced emf can be 111 V , 112 V and 113.04  V
(23) (b) The change is flux linked with the coil on rotating it
through 180º is
= nAB – (–nAB) = 2nAB
\ induced e.m.f. = – 
d
dt
f
 = 2nAB/dt (numerically) = 
2 1 0.1
0.01
´´
 = 20 V
The coil is closed and has a resistance of 2.0 W.
Therefore i = 20/2 = 10A.
(24) (d) Initial magnetic flux f
1
 = 5.5 × 10
–4
 weber.
Final magnetic flux f
2
 = 5 × 10
–5
 weber.
\ change in flux
Df = f
2
 - f
1
 = (5 × 10
-5
) – (5.5 × 10
-4
)   = – 50 × 10
–5
weber.
Time interval for this change, Dt = 0.1 sec.
\ induced emf in the coil
e = – N 
t
Df
D
 = – 1000 × 
5
( 50 10 )
0.1
-
-´
 = 5 volt.
Resistance of the coil, R = 10 ohm. Hence induced
current in the coil is
i = 
e 5 volt
R 10ohm
= = 0.5 ampere.
(25) (a),     (26)   (c),     (27)  (b)
(a) The current of the battery at any instant, I = E/R.
The magnetic force due to this current
B
EB
F IBL
R
==
l
This magnetic force will accelerate the rod from its
position of rest. The motional e.m.f. developed in the
rod id B/v.
The induced current,
induced
Bv
I
R
=
l
The magnetic force due to the induced current,
22
induced induced
Bv
FI
R
==
l
From Fleming’s left hand rule, foce F
B
 is to the right
and F
induced
 is to the left.
Net force on the rod = F
B 
– F
induced
.
Page 3


(1) (a) B.A f=
r r
    = (
ˆ
0.02i ) . (
ˆ ˆˆ
30i 16j 23k ++ ) × 10
–4
     = 0.6 × 10
-4
 Wb = 60m Wb
(2) (c) The induced emf
E = – df/dt = – 
d
dt
 (3t
2
 + 2t + 3) × 10
-3
(because given flux is in mWb).
Thus E = ( – 6t – 2) × 10
–3
At t = 2 sec,
E = ( – 6 × 2 – 2) × 10
–3
 = –14 mV .
(3) (a) The direction of current in the solenoid is clockwise.
On displacing it towards the loop a current in the loop
will be induced in opposite sense so as to oppose its
approach. Therefore the direction of induced current
as observed by the observer will be anticlockwise.
(4) (b) When north pole of the magnet is moved away, then
south pole is induced on the face of the loop in front of
the magnet i.e. as seen from the magnet side, a clockwise
induced current flows in the loop. This makes free
electrons to move in opposite direction, to plate a. Thus
excess positive charge appear on plate b.
(5) (d) If electron is moving from left to right, the flux linked
with the loop (which is into the page) will first increase
and then decrease as the eletron passes by. So the
induced current in the loop will be first anticlockwise
and will change direction as the electron passes by .
(6) (c) E = –
dt
df
or  df = – Edt = (0 - f)
or     f = 4 × 10
–3
 × 0.1
            = 4 × 10
–4
 weber
(7) (d) e = 
d
dt
f
 = – 
d
dt
 [10t
2
 + 5t + 1] × 10
–3
   = – [10 × 10
–3
 (2t) + 5 × 10
–3
]
at t = 5 second
e = –[10 × 10
–2
 + 5 × 10
–3
] = [0.1 + 0.005]
|e| = 0.105V
(8) (a) For each spoke, the induced emf between the centre O
and the rim will be the same
e = 
1
2
 BwL
2
 
 
=   BpL
2
 f (Q  w = 2pf)
Further for all spokes, centre O will be positive while
rim will be negative. Thus all emf's are in parallel giving
total emf   e =  BpL
2
 f
independent of the number of the spokes.
Substituting the values
e = 4 × 10
–5
 × 3.14 × (.5)
2
 × 2 = 6.28 × 10
–5
 volt
(9) ( d ) Rate of decrease of area of the semicircular ring
A
–
d
dt
 = (2R) V
According to Faraday’s law of induction, induced emf
e = 
d
dt
f
-
= 
A
–B
d
dt
 = – B(2R V) V)
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
N
M
Q
vt
2R
The induced current in the ring must generate magnetic
field in the upward direction. Thus Q is at higher
potential.
(10) (b) E = B lv
    = 4 × 10
–4
 × 50 × 
360 1000
60 60
´
´
 = 2 V
(11) (c) The flux through the area is
f = BA cos 57º = 42 × 10
-6
 × 2.5 × 0.545
    =  57 × 10
-6
 Wb.
(12) (a) The magnetic flux linked with the loop at any instant of
time t is given by
     f = BAN cos wt
or  f = 10Ba
2
 cos wt
Here N = 10, A = a
2
(13) (b) According to Lenz’s Law
(14) (c) e = Bv l = 0.4 × 10
–4
 × 
3
300 10
3
60 60
´
´
´
  = 10
–2
 V
(15) (b) Magnetic flux passing through the disc is f = BA
= 0.01 
2
weber
meter
 × 3.14 × (15 × 10
–2
 meter)
2
= 7.065 × 10
–4
 weber.
The line joining the centre and the circumference of the
disc cuts 7.065 x 10
-4
 weber flux in one round. So, the
rate of cutting flux (i.e. induced emf)
= flux × number of revolutions per second
= 7.065 × 10
–4
 × 
100
603 ´
 = 3.9 × 10
–4
 volt.
(1 6) (c) The magnetic f lux passing through eac h turn of a coil of area
A, perpendicular to a magnetic field B is given by
f
1
 = BA.
The magnetic flux through it on rotating it through 180º
will be
f
2
 = – BA.(- sign is put because now the flux lines
enters the coils through the outer face)
DPP/ P 44
123
\ change in magnetic flux
D f = f
1
 – f
2
 = – BA – (BA) = – 2BA.
Suppose this change takes in time Dt. According to
Faraday's law , the emf induced in the coil is given by
e = – N
t
Df
D
 = 
2NBA
t D
,
where N is number of turns in the coil. The current in
the coil will be
i = 
e1
RR
=
2NBA
t D
where R is the resistance of the circuit. The current
persists only during the change of flux i.e. for the time
interval Dt second. So, the charge passed through the
circuit is
q = i × Dt = 
2NBA
R
.
Here N = 500, B = 0.2 weber/meter
2
,
A = 4.0 cm
2
 = 4.0 × 10
–4
 meter
2
 and R = 50 ohm.
\  
4
2 500 0.2 4.0 10
q
50
-
´ ´ ´´
=
            = 1.6 × 10
–3
 coulomb.
(17) (a)T
(18) (c) The induced emf between centre and rim of the rotating
disc is
E = 
1
2
BwR
2
 = 
1
2
× 0.1 × 2p × 10 × (0.1)
2
        
 = 10p × 10
–3
 volt,
(19) (a) The induced emf
E = Blv = 0.2 × 10
–4
 × 1 × 180 × 1000/3600
     = 0.2 × 18/3600 = 1 × 10
–3
. V = 1mV
(20) (d) The induced emf is obtained by considering a strip on
the disc fig. Then, the linear speed of a small element dr
at a distance r from the centre is = wr. The induced emf
across the ends of the small element is-
de = B(dr)v = B wr dr
Thus the induced emf across the inner and outer sides
of the disc is
e = 
b
a
B r dr w
ò
 = 
1
2
 Bw (b
2
 – a
2
)
(21) (c) The induced emf e = – (v B). ´
r r
r
l
For the part PX, vB ^
r
r
, and the angle between (v B) ´
r
r
direction (the dotted line in figure and 
r
l
 is (90 – q).
Thus
e
P
 –e
x
 = vB l cos (90 – q/2) = vB l sin (q/2)
Similarly e
y
 – e
p
 = vBl sin (q/2)
Therefore induced emf
between X and Y is e
yx
 = 2 B v l sin (q/2)
(22) (a) Given, area = 10 × 20 cm
2
 = 200 × 10
-4
 m
2
B = 0.5 T, N = 60,  w = 2p × 1800/60
Q e = –
dN
dt
() f
       = –N 
d
dt
(BA cos wt)
      = NBAw sin wt
\ e
max
 = NABw
                  = 60 × 2 × 10
–2
 × 0.5 × 2p × 1800/60 = 113 volt.
Hence, induced emf can be 111 V , 112 V and 113.04  V
(23) (b) The change is flux linked with the coil on rotating it
through 180º is
= nAB – (–nAB) = 2nAB
\ induced e.m.f. = – 
d
dt
f
 = 2nAB/dt (numerically) = 
2 1 0.1
0.01
´´
 = 20 V
The coil is closed and has a resistance of 2.0 W.
Therefore i = 20/2 = 10A.
(24) (d) Initial magnetic flux f
1
 = 5.5 × 10
–4
 weber.
Final magnetic flux f
2
 = 5 × 10
–5
 weber.
\ change in flux
Df = f
2
 - f
1
 = (5 × 10
-5
) – (5.5 × 10
-4
)   = – 50 × 10
–5
weber.
Time interval for this change, Dt = 0.1 sec.
\ induced emf in the coil
e = – N 
t
Df
D
 = – 1000 × 
5
( 50 10 )
0.1
-
-´
 = 5 volt.
Resistance of the coil, R = 10 ohm. Hence induced
current in the coil is
i = 
e 5 volt
R 10ohm
= = 0.5 ampere.
(25) (a),     (26)   (c),     (27)  (b)
(a) The current of the battery at any instant, I = E/R.
The magnetic force due to this current
B
EB
F IBL
R
==
l
This magnetic force will accelerate the rod from its
position of rest. The motional e.m.f. developed in the
rod id B/v.
The induced current,
induced
Bv
I
R
=
l
The magnetic force due to the induced current,
22
induced induced
Bv
FI
R
==
l
From Fleming’s left hand rule, foce F
B
 is to the right
and F
induced
 is to the left.
Net force on the rod = F
B 
– F
induced
.
DPP/ P 44
124
From Newton’s law ,
B induced
dv
FFm
dt
-=
22
EB B v dv
m
R R dt
-=
ll
On separating variables and integrating speed from v
0
to v and time from 0 to t, we have
dvB
dt
E Bv mR
=
-
l
l
    
vt
00
dvB
dt
E Bv mR
=
-
òò
l
l
 
22
EBvB
lnt
E mR
- æö
-=
ç÷
èø
ll
         
22
B
t
mR
E Bv
e
E
-
-
=
l
l
             
t/r
E
v (1e)
B
-
=-
l
where           
2
mR
t
(B)
=
l
(b) The rod will attain a terminal velocity at t, ®¥ i.e.,
when e
–t/t
 = 0, the velocity is independent of time.
T
E
v
B
=
l
(d) The induced current I
induced
 = Blv/R. When the rod
has attained terminal speed.
induced
BE
I E/R
RB
æö
= ´=
ç÷
èø
l
l
The current of battery and the induced current are of
same magnitude, hence net current through the circuit
is zero.
(28) (c) Since both the loops are identical (same area and number
of turns) and moving with a same speed in same
magnetic field. Therefore same emf is induced in both
the coils. But the induced current will be more in the
copper loop as its resistance will be lesser as compared
to that of the aluminium loop.
(29) (c) As the aircraft flies, magnetic flux changes through its
wings due to the vertical component of the earth’s
magnetic field. Due to this, induced emf is produced
across the wings of the aircraft. Therefore, the wings
of the aircraft will not be at the same potential.
(30) (c) Lenz’s Law is based on conservation of energy and
induced emf opposes the cause of it i.e., change in
magnetic flux.