Chapter 38 : Electromagnetic Induction - HC Verma Solution, Physics Class 11 Notes | EduRev

Physics Class 11

JEE : Chapter 38 : Electromagnetic Induction - HC Verma Solution, Physics Class 11 Notes | EduRev

 Page 1


38.1
ELECTROMAGNETIC INDUCTION
CHAPTER - 38
1. (a) 
3 1 2 1 3
E.dl MLT I L ML I T
? ? ? ?
? ? ?
?
(b) 
1 1 2 2 1 3
BI LT MI T L ML I T
? ? ? ? ?
? ? ? ? ?
(c) 
1 2 2 2 1 2
s
d / dt MI T L ML I T
? ? ? ?
? ? ? ?
2. ? = at
2
+ bt + c
(a) a = 
2
/ t Volt
t Sec t
? ? ? ? ? ?
? ?
? ? ? ?
? ? ? ?
b = 
t
? ? ?
? ?
? ?
= Volt
c = [ ?] = Weber
(b) E = 
d
dt
?
[a = 0.2, b = 0.4, c = 0.6, t = 2s]
= 2at + b
= 2 ? 0.2 ? 2 + 0.4 = 1.2 volt
3. (a) ?
2
= B.A. = 0.01 ? 2 ? 10
–3
= 2 ? 10
–5
. 
?
1
= 0
e = 
5
3
d 2 10
dt 10 10
?
?
? ? ?
? ?
?
= – 2 mV
?
3
= B.A. = 0.03 ? 2 ? 10
–3
= 6 ? 10
–5
d ? = 4 ? 10
–5
e =
d
dt
?
? = –4 mV
?
4
= B.A. = 0.01 ? 2 ? 10
–3
= 2 ? 10
–5
d ? = –4 ? 10
–5
e = 
d
dt
?
? = 4 mV
?
5
= B.A. = 0
d ? = –2 ? 10
–5
e = 
d
dt
?
? = 2 mV
(b) emf is not constant in case of ? 10 – 20 ms and 20 – 30 ms as –4 mV and 4 mV. ?
4. ?
1
= BA = 0.5 ? ?(5 ? 10
–2
)
2
= 5 ? 25 ? 10
–5
= 125 ? 10
–5
?
2
= 0 
E = 
5
1 2
1
125 10
t 5 10
?
?
? ? ? ? ?
?
?
= 25 ? ? 10
–4
= 7.8 ? 10
–3
. ?
5. A = 1 mm
2
; i = 10 A, d = 20 cm ; dt = 0.1 s
e = 
0
i d BA A
dt dt 2 d dt
? ?
? ? ?
?
= 
7 6
10
1 1
4 10 10 10
1 10 V
2 2 10 1 10
? ?
?
? ?
? ? ?
? ? ?
? ? ? ?
.
6. (a) During removal,
?
?
= B.A. = 1 ? 50 ? 0.5 ? 0.5 – 25 ? 0.5 = 12.5 Tesla-m
2
?
?
????? ??????????
(ms) 
10   20  30  40  50   t ?
0.01 ?
0.03 ?
0.02 ?
10A ?
20cm 
Page 2


38.1
ELECTROMAGNETIC INDUCTION
CHAPTER - 38
1. (a) 
3 1 2 1 3
E.dl MLT I L ML I T
? ? ? ?
? ? ?
?
(b) 
1 1 2 2 1 3
BI LT MI T L ML I T
? ? ? ? ?
? ? ? ? ?
(c) 
1 2 2 2 1 2
s
d / dt MI T L ML I T
? ? ? ?
? ? ? ?
2. ? = at
2
+ bt + c
(a) a = 
2
/ t Volt
t Sec t
? ? ? ? ? ?
? ?
? ? ? ?
? ? ? ?
b = 
t
? ? ?
? ?
? ?
= Volt
c = [ ?] = Weber
(b) E = 
d
dt
?
[a = 0.2, b = 0.4, c = 0.6, t = 2s]
= 2at + b
= 2 ? 0.2 ? 2 + 0.4 = 1.2 volt
3. (a) ?
2
= B.A. = 0.01 ? 2 ? 10
–3
= 2 ? 10
–5
. 
?
1
= 0
e = 
5
3
d 2 10
dt 10 10
?
?
? ? ?
? ?
?
= – 2 mV
?
3
= B.A. = 0.03 ? 2 ? 10
–3
= 6 ? 10
–5
d ? = 4 ? 10
–5
e =
d
dt
?
? = –4 mV
?
4
= B.A. = 0.01 ? 2 ? 10
–3
= 2 ? 10
–5
d ? = –4 ? 10
–5
e = 
d
dt
?
? = 4 mV
?
5
= B.A. = 0
d ? = –2 ? 10
–5
e = 
d
dt
?
? = 2 mV
(b) emf is not constant in case of ? 10 – 20 ms and 20 – 30 ms as –4 mV and 4 mV. ?
4. ?
1
= BA = 0.5 ? ?(5 ? 10
–2
)
2
= 5 ? 25 ? 10
–5
= 125 ? 10
–5
?
2
= 0 
E = 
5
1 2
1
125 10
t 5 10
?
?
? ? ? ? ?
?
?
= 25 ? ? 10
–4
= 7.8 ? 10
–3
. ?
5. A = 1 mm
2
; i = 10 A, d = 20 cm ; dt = 0.1 s
e = 
0
i d BA A
dt dt 2 d dt
? ?
? ? ?
?
= 
7 6
10
1 1
4 10 10 10
1 10 V
2 2 10 1 10
? ?
?
? ?
? ? ?
? ? ?
? ? ? ?
.
6. (a) During removal,
?
?
= B.A. = 1 ? 50 ? 0.5 ? 0.5 – 25 ? 0.5 = 12.5 Tesla-m
2
?
?
????? ??????????
(ms) 
10   20  30  40  50   t ?
0.01 ?
0.03 ?
0.02 ?
10A ?
20cm 
Electromagnetic Induction
38.2
e = 
1
2 1
2
d 12.5 125 10
50V
dt dt 0.25 25 10
?
?
? ? ? ? ?
? ? ? ? ?
?
(b) During its restoration 
?
1
= 0 ; ?
2
= 12.5 Tesla-m
2
; t = 0.25 s
E = 
12.5 0
0.25
?
= 50 V.
(c) During the motion
?
1
= 0, ?
2
= 0
E = 
d
0
dt
?
? ?
7. R = 25 ?
(a) e = 50 V, T = 0.25 s
i = e/R = 2A, H =  i
2
RT
= 4 ? 25 ? 0.25 = 25 J
(b) e = 50 V, T = 0.25 s
i = e/R = 2A, H =  i
2
RT = 25 J
(c) Since energy is a scalar quantity
Net thermal energy developed = 25 J + 25 J = 50 J.
8. A = 5 cm
2
= 5 ? 10
–4
m
2
B = B
0
sin ?t = 0.2 sin(300 t)
? = 60°
a) Max emf induced in the coil
E = 
d d
(BAcos )
dt dt
?
? ? ?
= 
4
0
d 1
(B sin t 5 10 )
dt 2
?
? ? ? ?
= 
4
0
5 d
B 10 (sin t)
2 dt
?
? ? ? = 
4 0
B 5
10 cos t
2
?
? ? ? ?
=
4 3
0.2 5
300 10 cos t 15 10 cos t
2
? ?
?
? ? ? ? ? ? ?
E
max
= 15 ? 10
–3
= 0.015 V
b) Induced emf at t = ( ?/900) s
E = 15 ? 10
–3
? cos ?t 
= 15 ? 10
–3
? cos (300 ? ?/900) = 15 ? 10
–3
? ½ 
= 0.015/2 = 0.0075 = 7.5 ? 10
–3
V
c) Induced emf at t = ?/600 s
E = 15 ? 10
–3
? cos (300 ? ?/600)
= 15 ? 10
–3
? 0 = 0 V. ?
9. B
?
= 0.10 T
A = 1 cm
2
= 10
–4
m
2
T = 1 s
? = B.A. = 10
–1
? 10
–4
= 10
–5
e = 
5
5
d 10
10
dt 1
?
?
?
? ? = 10 ?V
10. E = 20 mV = 20 ? 10
–3
V
A = (2 ? 10
–2
)
2
= 4 ? 10
–4
Dt = 0.2 s, ? = 180°
Page 3


38.1
ELECTROMAGNETIC INDUCTION
CHAPTER - 38
1. (a) 
3 1 2 1 3
E.dl MLT I L ML I T
? ? ? ?
? ? ?
?
(b) 
1 1 2 2 1 3
BI LT MI T L ML I T
? ? ? ? ?
? ? ? ? ?
(c) 
1 2 2 2 1 2
s
d / dt MI T L ML I T
? ? ? ?
? ? ? ?
2. ? = at
2
+ bt + c
(a) a = 
2
/ t Volt
t Sec t
? ? ? ? ? ?
? ?
? ? ? ?
? ? ? ?
b = 
t
? ? ?
? ?
? ?
= Volt
c = [ ?] = Weber
(b) E = 
d
dt
?
[a = 0.2, b = 0.4, c = 0.6, t = 2s]
= 2at + b
= 2 ? 0.2 ? 2 + 0.4 = 1.2 volt
3. (a) ?
2
= B.A. = 0.01 ? 2 ? 10
–3
= 2 ? 10
–5
. 
?
1
= 0
e = 
5
3
d 2 10
dt 10 10
?
?
? ? ?
? ?
?
= – 2 mV
?
3
= B.A. = 0.03 ? 2 ? 10
–3
= 6 ? 10
–5
d ? = 4 ? 10
–5
e =
d
dt
?
? = –4 mV
?
4
= B.A. = 0.01 ? 2 ? 10
–3
= 2 ? 10
–5
d ? = –4 ? 10
–5
e = 
d
dt
?
? = 4 mV
?
5
= B.A. = 0
d ? = –2 ? 10
–5
e = 
d
dt
?
? = 2 mV
(b) emf is not constant in case of ? 10 – 20 ms and 20 – 30 ms as –4 mV and 4 mV. ?
4. ?
1
= BA = 0.5 ? ?(5 ? 10
–2
)
2
= 5 ? 25 ? 10
–5
= 125 ? 10
–5
?
2
= 0 
E = 
5
1 2
1
125 10
t 5 10
?
?
? ? ? ? ?
?
?
= 25 ? ? 10
–4
= 7.8 ? 10
–3
. ?
5. A = 1 mm
2
; i = 10 A, d = 20 cm ; dt = 0.1 s
e = 
0
i d BA A
dt dt 2 d dt
? ?
? ? ?
?
= 
7 6
10
1 1
4 10 10 10
1 10 V
2 2 10 1 10
? ?
?
? ?
? ? ?
? ? ?
? ? ? ?
.
6. (a) During removal,
?
?
= B.A. = 1 ? 50 ? 0.5 ? 0.5 – 25 ? 0.5 = 12.5 Tesla-m
2
?
?
????? ??????????
(ms) 
10   20  30  40  50   t ?
0.01 ?
0.03 ?
0.02 ?
10A ?
20cm 
Electromagnetic Induction
38.2
e = 
1
2 1
2
d 12.5 125 10
50V
dt dt 0.25 25 10
?
?
? ? ? ? ?
? ? ? ? ?
?
(b) During its restoration 
?
1
= 0 ; ?
2
= 12.5 Tesla-m
2
; t = 0.25 s
E = 
12.5 0
0.25
?
= 50 V.
(c) During the motion
?
1
= 0, ?
2
= 0
E = 
d
0
dt
?
? ?
7. R = 25 ?
(a) e = 50 V, T = 0.25 s
i = e/R = 2A, H =  i
2
RT
= 4 ? 25 ? 0.25 = 25 J
(b) e = 50 V, T = 0.25 s
i = e/R = 2A, H =  i
2
RT = 25 J
(c) Since energy is a scalar quantity
Net thermal energy developed = 25 J + 25 J = 50 J.
8. A = 5 cm
2
= 5 ? 10
–4
m
2
B = B
0
sin ?t = 0.2 sin(300 t)
? = 60°
a) Max emf induced in the coil
E = 
d d
(BAcos )
dt dt
?
? ? ?
= 
4
0
d 1
(B sin t 5 10 )
dt 2
?
? ? ? ?
= 
4
0
5 d
B 10 (sin t)
2 dt
?
? ? ? = 
4 0
B 5
10 cos t
2
?
? ? ? ?
=
4 3
0.2 5
300 10 cos t 15 10 cos t
2
? ?
?
? ? ? ? ? ? ?
E
max
= 15 ? 10
–3
= 0.015 V
b) Induced emf at t = ( ?/900) s
E = 15 ? 10
–3
? cos ?t 
= 15 ? 10
–3
? cos (300 ? ?/900) = 15 ? 10
–3
? ½ 
= 0.015/2 = 0.0075 = 7.5 ? 10
–3
V
c) Induced emf at t = ?/600 s
E = 15 ? 10
–3
? cos (300 ? ?/600)
= 15 ? 10
–3
? 0 = 0 V. ?
9. B
?
= 0.10 T
A = 1 cm
2
= 10
–4
m
2
T = 1 s
? = B.A. = 10
–1
? 10
–4
= 10
–5
e = 
5
5
d 10
10
dt 1
?
?
?
? ? = 10 ?V
10. E = 20 mV = 20 ? 10
–3
V
A = (2 ? 10
–2
)
2
= 4 ? 10
–4
Dt = 0.2 s, ? = 180°
Electromagnetic Induction
38.3
?
1
= BA, ?
2
= –BA
d ? = 2BA
E = 
d 2BA
dt dt
?
?
? 20 ? 10
–3
= 
4
1
2 B 2 10
2 10
?
?
? ? ?
?
?? 20 ? 10
–3
= 4 ? B ? 10
–3
? B = 
3
3
20 10
42 10
?
?
?
?
= 5T
11. Area = A, Resistance = R, B = Magnetic field
? = BA = Ba cos 0° = BA
e = 
d BA
dt 1
?
? ; i = 
e BA
R R
?
? = iT = BA/R ?
12. r = 2 cm = 2 ? 10
–2
m
n = 100 turns / cm = 10000 turns/m
i = 5 A
B = ?
0
ni
= 4 ? ? 10
–7
? 10000 ? 5 = 20 ? ? 10
–3
= 62.8 ? 10
–3
T
n
2
= 100 turns
R = 20 ?
r = 1 cm = 10
–2
m
Flux linking per turn of the second coil = B ?r
2
= B ? ? 10
–4
?
1
= Total flux linking = Bn
2
?r
2
= 100 ? ? ? 10
–4
? 20 ? ? 10
–3
When current is reversed.
?
2
= – ?
1
d ? = ?
2
– ?
1
= 2 ? 100 ? ? ? 10
–4
? 20 ? ? 10
–3
E = 
2 4
d 4 10
dt dt
?
? ? ?
? ?
I = 
2 4
E 4 10
R dt 20
?
? ?
?
?
q = Idt = 
2 4
4 10
dt
dt 20
?
? ?
?
?
= 2 ? 10
–4
C. ?
13. Speed = u
Magnetic field = B
Side = a
a) The perpendicular component i.e. a sin ? is to be taken which is ?r to 
velocity.
So, l = a sin ? 30° = a/2.
Net ‘a’ charge = 4 ? a/2 = 2a
So, induced emf = B ?I = 2auB
b) Current = 
E 2auB
R R
? ?
14. ?
1
= 0.35 weber, ?
2
= 0.85 weber
D ? = ?
2
– ?
1
= (0.85 – 0.35) weber = 0.5 weber
dt = 0.5 sec
B ?
a ?
u ?
a ?
B ?
a sin ? ?
30° ?
a ?
Page 4


38.1
ELECTROMAGNETIC INDUCTION
CHAPTER - 38
1. (a) 
3 1 2 1 3
E.dl MLT I L ML I T
? ? ? ?
? ? ?
?
(b) 
1 1 2 2 1 3
BI LT MI T L ML I T
? ? ? ? ?
? ? ? ? ?
(c) 
1 2 2 2 1 2
s
d / dt MI T L ML I T
? ? ? ?
? ? ? ?
2. ? = at
2
+ bt + c
(a) a = 
2
/ t Volt
t Sec t
? ? ? ? ? ?
? ?
? ? ? ?
? ? ? ?
b = 
t
? ? ?
? ?
? ?
= Volt
c = [ ?] = Weber
(b) E = 
d
dt
?
[a = 0.2, b = 0.4, c = 0.6, t = 2s]
= 2at + b
= 2 ? 0.2 ? 2 + 0.4 = 1.2 volt
3. (a) ?
2
= B.A. = 0.01 ? 2 ? 10
–3
= 2 ? 10
–5
. 
?
1
= 0
e = 
5
3
d 2 10
dt 10 10
?
?
? ? ?
? ?
?
= – 2 mV
?
3
= B.A. = 0.03 ? 2 ? 10
–3
= 6 ? 10
–5
d ? = 4 ? 10
–5
e =
d
dt
?
? = –4 mV
?
4
= B.A. = 0.01 ? 2 ? 10
–3
= 2 ? 10
–5
d ? = –4 ? 10
–5
e = 
d
dt
?
? = 4 mV
?
5
= B.A. = 0
d ? = –2 ? 10
–5
e = 
d
dt
?
? = 2 mV
(b) emf is not constant in case of ? 10 – 20 ms and 20 – 30 ms as –4 mV and 4 mV. ?
4. ?
1
= BA = 0.5 ? ?(5 ? 10
–2
)
2
= 5 ? 25 ? 10
–5
= 125 ? 10
–5
?
2
= 0 
E = 
5
1 2
1
125 10
t 5 10
?
?
? ? ? ? ?
?
?
= 25 ? ? 10
–4
= 7.8 ? 10
–3
. ?
5. A = 1 mm
2
; i = 10 A, d = 20 cm ; dt = 0.1 s
e = 
0
i d BA A
dt dt 2 d dt
? ?
? ? ?
?
= 
7 6
10
1 1
4 10 10 10
1 10 V
2 2 10 1 10
? ?
?
? ?
? ? ?
? ? ?
? ? ? ?
.
6. (a) During removal,
?
?
= B.A. = 1 ? 50 ? 0.5 ? 0.5 – 25 ? 0.5 = 12.5 Tesla-m
2
?
?
????? ??????????
(ms) 
10   20  30  40  50   t ?
0.01 ?
0.03 ?
0.02 ?
10A ?
20cm 
Electromagnetic Induction
38.2
e = 
1
2 1
2
d 12.5 125 10
50V
dt dt 0.25 25 10
?
?
? ? ? ? ?
? ? ? ? ?
?
(b) During its restoration 
?
1
= 0 ; ?
2
= 12.5 Tesla-m
2
; t = 0.25 s
E = 
12.5 0
0.25
?
= 50 V.
(c) During the motion
?
1
= 0, ?
2
= 0
E = 
d
0
dt
?
? ?
7. R = 25 ?
(a) e = 50 V, T = 0.25 s
i = e/R = 2A, H =  i
2
RT
= 4 ? 25 ? 0.25 = 25 J
(b) e = 50 V, T = 0.25 s
i = e/R = 2A, H =  i
2
RT = 25 J
(c) Since energy is a scalar quantity
Net thermal energy developed = 25 J + 25 J = 50 J.
8. A = 5 cm
2
= 5 ? 10
–4
m
2
B = B
0
sin ?t = 0.2 sin(300 t)
? = 60°
a) Max emf induced in the coil
E = 
d d
(BAcos )
dt dt
?
? ? ?
= 
4
0
d 1
(B sin t 5 10 )
dt 2
?
? ? ? ?
= 
4
0
5 d
B 10 (sin t)
2 dt
?
? ? ? = 
4 0
B 5
10 cos t
2
?
? ? ? ?
=
4 3
0.2 5
300 10 cos t 15 10 cos t
2
? ?
?
? ? ? ? ? ? ?
E
max
= 15 ? 10
–3
= 0.015 V
b) Induced emf at t = ( ?/900) s
E = 15 ? 10
–3
? cos ?t 
= 15 ? 10
–3
? cos (300 ? ?/900) = 15 ? 10
–3
? ½ 
= 0.015/2 = 0.0075 = 7.5 ? 10
–3
V
c) Induced emf at t = ?/600 s
E = 15 ? 10
–3
? cos (300 ? ?/600)
= 15 ? 10
–3
? 0 = 0 V. ?
9. B
?
= 0.10 T
A = 1 cm
2
= 10
–4
m
2
T = 1 s
? = B.A. = 10
–1
? 10
–4
= 10
–5
e = 
5
5
d 10
10
dt 1
?
?
?
? ? = 10 ?V
10. E = 20 mV = 20 ? 10
–3
V
A = (2 ? 10
–2
)
2
= 4 ? 10
–4
Dt = 0.2 s, ? = 180°
Electromagnetic Induction
38.3
?
1
= BA, ?
2
= –BA
d ? = 2BA
E = 
d 2BA
dt dt
?
?
? 20 ? 10
–3
= 
4
1
2 B 2 10
2 10
?
?
? ? ?
?
?? 20 ? 10
–3
= 4 ? B ? 10
–3
? B = 
3
3
20 10
42 10
?
?
?
?
= 5T
11. Area = A, Resistance = R, B = Magnetic field
? = BA = Ba cos 0° = BA
e = 
d BA
dt 1
?
? ; i = 
e BA
R R
?
? = iT = BA/R ?
12. r = 2 cm = 2 ? 10
–2
m
n = 100 turns / cm = 10000 turns/m
i = 5 A
B = ?
0
ni
= 4 ? ? 10
–7
? 10000 ? 5 = 20 ? ? 10
–3
= 62.8 ? 10
–3
T
n
2
= 100 turns
R = 20 ?
r = 1 cm = 10
–2
m
Flux linking per turn of the second coil = B ?r
2
= B ? ? 10
–4
?
1
= Total flux linking = Bn
2
?r
2
= 100 ? ? ? 10
–4
? 20 ? ? 10
–3
When current is reversed.
?
2
= – ?
1
d ? = ?
2
– ?
1
= 2 ? 100 ? ? ? 10
–4
? 20 ? ? 10
–3
E = 
2 4
d 4 10
dt dt
?
? ? ?
? ?
I = 
2 4
E 4 10
R dt 20
?
? ?
?
?
q = Idt = 
2 4
4 10
dt
dt 20
?
? ?
?
?
= 2 ? 10
–4
C. ?
13. Speed = u
Magnetic field = B
Side = a
a) The perpendicular component i.e. a sin ? is to be taken which is ?r to 
velocity.
So, l = a sin ? 30° = a/2.
Net ‘a’ charge = 4 ? a/2 = 2a
So, induced emf = B ?I = 2auB
b) Current = 
E 2auB
R R
? ?
14. ?
1
= 0.35 weber, ?
2
= 0.85 weber
D ? = ?
2
– ?
1
= (0.85 – 0.35) weber = 0.5 weber
dt = 0.5 sec
B ?
a ?
u ?
a ?
B ?
a sin ? ?
30° ?
a ?
Electromagnetic Induction
38.4
E = 
5 . 0
5 . 0
t d
d
?
?
?
= 1 v.
The induced current is anticlockwise as seen from above. ?
15. i = v(B × l)
= v B l cos ?
? is angle between normal to plane and B
?
= 90°.
= v B l cos 90° = 0. ?
16. u = 1 cm/ ?, B = 0.6 T
a) At t = 2 sec, distance moved = 2 × 1 cm/s = 2 cm
E = 
2
10 ) 0 5 2 ( 6 . 0
dt
d
4 ?
? ? ? ?
?
?
= 3 ×10
–4
V
b) At t = 10 sec
distance moved = 10 × 1 = 10 cm
The flux linked does not change with time
? E = 0
c) At t = 22 sec
distance = 22 × 1 = 22 cm
The loop is moving out of the field and 2 cm outside.
E = 
dt
dA
B
dt
d
? ?
?
= 
2
) 10 5 2 ( 6 . 0
4 ?
? ? ?
= 3 × 10
–4
V
d) At t = 30 sec
The loop is total outside and flux linked = 0
? E = 0.
17. As heat produced is a scalar prop.
So, net heat produced = H
a
+ H
b
+ H
c
+ H
d
R = 4.5 m ? = 4.5 ×10
–3
?
a) e = 3 × 10
–4
V
i = ?
?
?
?
?
?
3
4
10 5 . 4
10 3
R
e
6.7 × 10
–2
Amp.
H
a
= (6.7 ×10
–2
)
2
× 4.5 × 10
–3
× 5
H
b
= H
d
= 0 [since emf is induced for 5 sec]
H
c
= (6.7 ×10
–2
)
2
× 4.5 × 10
–3
× 5
So Total heat = H
a 
+ H
c
= 2 × (6.7 ×10
–2
)
2
× 4.5 × 10
–3
× 5 = 2 × 10
–4
J.
18. r = 10 cm, R = 4 ?
010 . 0
dt
dB
? T/ ?, A
dt
dB
dt
d
?
?
E = 
?
?
?
?
?
?
?
?
? ?
? ? ?
?
2
r
01 . 0 A
dt
dB
dt
d
2
= 
4
10
2
14 . 3
2
01 . 0 14 . 3 01 . 0
?
? ?
? ?
= 1.57 × 10
–4
i = 
4
10 57 . 1
R
E
4 ?
?
? = 0.39 × 10
–4
= 3.9 × 10
–5
A
19. a) S
1
closed S
2
open
net R = 4 × 4 = 16 ?
B 
20cm
× × × ×
× × × ×
× × × ×
× × × ×
× × × ×
5cm
b
× × × ×
× × × ×
× × × ×
× × × ×
× × × ×
a d
c
r
Page 5


38.1
ELECTROMAGNETIC INDUCTION
CHAPTER - 38
1. (a) 
3 1 2 1 3
E.dl MLT I L ML I T
? ? ? ?
? ? ?
?
(b) 
1 1 2 2 1 3
BI LT MI T L ML I T
? ? ? ? ?
? ? ? ? ?
(c) 
1 2 2 2 1 2
s
d / dt MI T L ML I T
? ? ? ?
? ? ? ?
2. ? = at
2
+ bt + c
(a) a = 
2
/ t Volt
t Sec t
? ? ? ? ? ?
? ?
? ? ? ?
? ? ? ?
b = 
t
? ? ?
? ?
? ?
= Volt
c = [ ?] = Weber
(b) E = 
d
dt
?
[a = 0.2, b = 0.4, c = 0.6, t = 2s]
= 2at + b
= 2 ? 0.2 ? 2 + 0.4 = 1.2 volt
3. (a) ?
2
= B.A. = 0.01 ? 2 ? 10
–3
= 2 ? 10
–5
. 
?
1
= 0
e = 
5
3
d 2 10
dt 10 10
?
?
? ? ?
? ?
?
= – 2 mV
?
3
= B.A. = 0.03 ? 2 ? 10
–3
= 6 ? 10
–5
d ? = 4 ? 10
–5
e =
d
dt
?
? = –4 mV
?
4
= B.A. = 0.01 ? 2 ? 10
–3
= 2 ? 10
–5
d ? = –4 ? 10
–5
e = 
d
dt
?
? = 4 mV
?
5
= B.A. = 0
d ? = –2 ? 10
–5
e = 
d
dt
?
? = 2 mV
(b) emf is not constant in case of ? 10 – 20 ms and 20 – 30 ms as –4 mV and 4 mV. ?
4. ?
1
= BA = 0.5 ? ?(5 ? 10
–2
)
2
= 5 ? 25 ? 10
–5
= 125 ? 10
–5
?
2
= 0 
E = 
5
1 2
1
125 10
t 5 10
?
?
? ? ? ? ?
?
?
= 25 ? ? 10
–4
= 7.8 ? 10
–3
. ?
5. A = 1 mm
2
; i = 10 A, d = 20 cm ; dt = 0.1 s
e = 
0
i d BA A
dt dt 2 d dt
? ?
? ? ?
?
= 
7 6
10
1 1
4 10 10 10
1 10 V
2 2 10 1 10
? ?
?
? ?
? ? ?
? ? ?
? ? ? ?
.
6. (a) During removal,
?
?
= B.A. = 1 ? 50 ? 0.5 ? 0.5 – 25 ? 0.5 = 12.5 Tesla-m
2
?
?
????? ??????????
(ms) 
10   20  30  40  50   t ?
0.01 ?
0.03 ?
0.02 ?
10A ?
20cm 
Electromagnetic Induction
38.2
e = 
1
2 1
2
d 12.5 125 10
50V
dt dt 0.25 25 10
?
?
? ? ? ? ?
? ? ? ? ?
?
(b) During its restoration 
?
1
= 0 ; ?
2
= 12.5 Tesla-m
2
; t = 0.25 s
E = 
12.5 0
0.25
?
= 50 V.
(c) During the motion
?
1
= 0, ?
2
= 0
E = 
d
0
dt
?
? ?
7. R = 25 ?
(a) e = 50 V, T = 0.25 s
i = e/R = 2A, H =  i
2
RT
= 4 ? 25 ? 0.25 = 25 J
(b) e = 50 V, T = 0.25 s
i = e/R = 2A, H =  i
2
RT = 25 J
(c) Since energy is a scalar quantity
Net thermal energy developed = 25 J + 25 J = 50 J.
8. A = 5 cm
2
= 5 ? 10
–4
m
2
B = B
0
sin ?t = 0.2 sin(300 t)
? = 60°
a) Max emf induced in the coil
E = 
d d
(BAcos )
dt dt
?
? ? ?
= 
4
0
d 1
(B sin t 5 10 )
dt 2
?
? ? ? ?
= 
4
0
5 d
B 10 (sin t)
2 dt
?
? ? ? = 
4 0
B 5
10 cos t
2
?
? ? ? ?
=
4 3
0.2 5
300 10 cos t 15 10 cos t
2
? ?
?
? ? ? ? ? ? ?
E
max
= 15 ? 10
–3
= 0.015 V
b) Induced emf at t = ( ?/900) s
E = 15 ? 10
–3
? cos ?t 
= 15 ? 10
–3
? cos (300 ? ?/900) = 15 ? 10
–3
? ½ 
= 0.015/2 = 0.0075 = 7.5 ? 10
–3
V
c) Induced emf at t = ?/600 s
E = 15 ? 10
–3
? cos (300 ? ?/600)
= 15 ? 10
–3
? 0 = 0 V. ?
9. B
?
= 0.10 T
A = 1 cm
2
= 10
–4
m
2
T = 1 s
? = B.A. = 10
–1
? 10
–4
= 10
–5
e = 
5
5
d 10
10
dt 1
?
?
?
? ? = 10 ?V
10. E = 20 mV = 20 ? 10
–3
V
A = (2 ? 10
–2
)
2
= 4 ? 10
–4
Dt = 0.2 s, ? = 180°
Electromagnetic Induction
38.3
?
1
= BA, ?
2
= –BA
d ? = 2BA
E = 
d 2BA
dt dt
?
?
? 20 ? 10
–3
= 
4
1
2 B 2 10
2 10
?
?
? ? ?
?
?? 20 ? 10
–3
= 4 ? B ? 10
–3
? B = 
3
3
20 10
42 10
?
?
?
?
= 5T
11. Area = A, Resistance = R, B = Magnetic field
? = BA = Ba cos 0° = BA
e = 
d BA
dt 1
?
? ; i = 
e BA
R R
?
? = iT = BA/R ?
12. r = 2 cm = 2 ? 10
–2
m
n = 100 turns / cm = 10000 turns/m
i = 5 A
B = ?
0
ni
= 4 ? ? 10
–7
? 10000 ? 5 = 20 ? ? 10
–3
= 62.8 ? 10
–3
T
n
2
= 100 turns
R = 20 ?
r = 1 cm = 10
–2
m
Flux linking per turn of the second coil = B ?r
2
= B ? ? 10
–4
?
1
= Total flux linking = Bn
2
?r
2
= 100 ? ? ? 10
–4
? 20 ? ? 10
–3
When current is reversed.
?
2
= – ?
1
d ? = ?
2
– ?
1
= 2 ? 100 ? ? ? 10
–4
? 20 ? ? 10
–3
E = 
2 4
d 4 10
dt dt
?
? ? ?
? ?
I = 
2 4
E 4 10
R dt 20
?
? ?
?
?
q = Idt = 
2 4
4 10
dt
dt 20
?
? ?
?
?
= 2 ? 10
–4
C. ?
13. Speed = u
Magnetic field = B
Side = a
a) The perpendicular component i.e. a sin ? is to be taken which is ?r to 
velocity.
So, l = a sin ? 30° = a/2.
Net ‘a’ charge = 4 ? a/2 = 2a
So, induced emf = B ?I = 2auB
b) Current = 
E 2auB
R R
? ?
14. ?
1
= 0.35 weber, ?
2
= 0.85 weber
D ? = ?
2
– ?
1
= (0.85 – 0.35) weber = 0.5 weber
dt = 0.5 sec
B ?
a ?
u ?
a ?
B ?
a sin ? ?
30° ?
a ?
Electromagnetic Induction
38.4
E = 
5 . 0
5 . 0
t d
d
?
?
?
= 1 v.
The induced current is anticlockwise as seen from above. ?
15. i = v(B × l)
= v B l cos ?
? is angle between normal to plane and B
?
= 90°.
= v B l cos 90° = 0. ?
16. u = 1 cm/ ?, B = 0.6 T
a) At t = 2 sec, distance moved = 2 × 1 cm/s = 2 cm
E = 
2
10 ) 0 5 2 ( 6 . 0
dt
d
4 ?
? ? ? ?
?
?
= 3 ×10
–4
V
b) At t = 10 sec
distance moved = 10 × 1 = 10 cm
The flux linked does not change with time
? E = 0
c) At t = 22 sec
distance = 22 × 1 = 22 cm
The loop is moving out of the field and 2 cm outside.
E = 
dt
dA
B
dt
d
? ?
?
= 
2
) 10 5 2 ( 6 . 0
4 ?
? ? ?
= 3 × 10
–4
V
d) At t = 30 sec
The loop is total outside and flux linked = 0
? E = 0.
17. As heat produced is a scalar prop.
So, net heat produced = H
a
+ H
b
+ H
c
+ H
d
R = 4.5 m ? = 4.5 ×10
–3
?
a) e = 3 × 10
–4
V
i = ?
?
?
?
?
?
3
4
10 5 . 4
10 3
R
e
6.7 × 10
–2
Amp.
H
a
= (6.7 ×10
–2
)
2
× 4.5 × 10
–3
× 5
H
b
= H
d
= 0 [since emf is induced for 5 sec]
H
c
= (6.7 ×10
–2
)
2
× 4.5 × 10
–3
× 5
So Total heat = H
a 
+ H
c
= 2 × (6.7 ×10
–2
)
2
× 4.5 × 10
–3
× 5 = 2 × 10
–4
J.
18. r = 10 cm, R = 4 ?
010 . 0
dt
dB
? T/ ?, A
dt
dB
dt
d
?
?
E = 
?
?
?
?
?
?
?
?
? ?
? ? ?
?
2
r
01 . 0 A
dt
dB
dt
d
2
= 
4
10
2
14 . 3
2
01 . 0 14 . 3 01 . 0
?
? ?
? ?
= 1.57 × 10
–4
i = 
4
10 57 . 1
R
E
4 ?
?
? = 0.39 × 10
–4
= 3.9 × 10
–5
A
19. a) S
1
closed S
2
open
net R = 4 × 4 = 16 ?
B 
20cm
× × × ×
× × × ×
× × × ×
× × × ×
× × × ×
5cm
b
× × × ×
× × × ×
× × × ×
× × × ×
× × × ×
a d
c
r
Electromagnetic Induction
38.5
e = 
2 4
10 2 10
dt
dB
A
dt
d
? ?
? ? ? ?
?
= 2 × 10
–6
V
i through ad = ?
?
?
?
16
10 2
R
e
6
1.25 × 10
–7
A along ad
b) R = 16 ?
e = A × 
dt
dB
= 2 × 0
–5
V
i = 
16
10 2
6 ?
?
= 1.25 × 10
–7
A along d a
c) Since both S
1
and S
2
are open, no current is passed as circuit is open i.e. i = 0
d) Since both S
1
and S
2
are closed, the circuit forms a balanced wheat stone bridge and no current will 
flow along ad i.e. i = 0. ?
20. Magnetic field due to the coil (1) at the center of (2) is B = 
2 / 3 2 2
2
0
) x a ( 2
Nia
?
?
Flux linked with the second, 
= B.A 
(2)
= 
2 / 3 2 2
2
0
) x a ( 2
Nia
?
?
?a ?
2
E.m.f. induced 
dt
di
) x a ( 2
a Na
dt
d
2 / 3 2 2
2 2
0
?
? ? ?
?
?
= 
? ? r x ) L / R (
E
dt
d
) x a ( 2
a a N
2 / 3 2 2
2 2
0
? ?
? ? ?
= 
? ?
2 2 / 3 2 2
2 2
0
r x ) L / R (
v . L / R . 1
E
) x a ( 2
a a N
?
?
?
? ? ?
b) = 
2 2 / 3 2 2
2 2
0
) r 2 / R ( L
ERV
) x a ( 2
a a N
? ?
? ? ?
(for x = L/2, R/L x = R/2)
a) For x = L
E = 
2 2 / 3 2 2
2 2
0
) r R ( ) x a ( 2
RvE a a N
? ?
? ? ?
21. N = 50, B
?
= 0.200 T ; r = 2.00 cm = 0.02 m
? = 60°, t = 0.100 s
a) e = 
T
60 cos NBA
T
A . B N
dt
Nd ?
?
?
?
?
= 
1 . 0
) 02 . 0 ( 10 2 50
2 1
? ? ? ? ?
?
= 5 × 4 × 10
–3
× ?
= 2 ? × 10
–2
V = 6.28 × 10
–2
V
b) i = 
4
10 28 . 6
R
e
2 ?
?
? = 1.57 × 10
–2
A
Q = it = 1.57 × 10
–2
× 10
–1
= 1.57 × 10
–3
C
22. n = 100 turns, B = 4 × 10
–4
T
A = 25 cm
2
= 25 × 10
–4
m
2
a) When the coil is perpendicular to the field
? = nBA
When coil goes through half a turn
? = BA cos 18° = 0 – nBA 
d ? = 2nBA
b
× × × ×
× × × ×
× × × ×
× × × ×
× × × ×
a
d c e
d
(1)
a ?
a
(2)
B
?
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