NEET  >  HC Verma Solutions: Chapter 38 - Electromagnetic Induction

# HC Verma Solutions: Chapter 38 - Electromagnetic Induction - Physics Class 11 - NEET

``` 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
?
?
?
16
10 2
R
e
6
1.25 × 10
–7
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
?
?
= 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
?
```

## FAQs on HC Verma Solutions: Chapter 38 - Electromagnetic Induction - Physics Class 11 - NEET

 1. What is electromagnetic induction?
Ans. Electromagnetic induction refers to the phenomenon where a changing magnetic field induces an electric current in a conductor. This process is governed by Faraday's law of electromagnetic induction.
 2. How can electromagnetic induction be demonstrated?
Ans. Electromagnetic induction can be demonstrated by performing simple experiments. For example, if a magnet is moved in and out of a coil of wire, an electric current is induced in the wire. This can be observed by connecting the coil to a galvanometer or a light bulb, which will show the presence of an electric current.
 3. What are the applications of electromagnetic induction?
Ans. Electromagnetic induction has various practical applications. It is used in generators to convert mechanical energy into electrical energy. It is also employed in transformers to change the voltage of an alternating current. Induction cooktops, magnetic levitation trains, and metal detectors are some other examples of applications of electromagnetic induction.
 4. How does electromagnetic induction work in a generator?
Ans. In a generator, electromagnetic induction is used to convert mechanical energy into electrical energy. This is achieved by rotating a coil of wire within a magnetic field. As the coil moves, the changing magnetic field induces an electric current in the wire. This current can be collected and used as an electrical output.
 5. What factors affect the magnitude of the induced current in electromagnetic induction?
Ans. The magnitude of the induced current in electromagnetic induction depends on several factors. These include the rate of change of the magnetic field, the number of turns in the coil, the strength of the magnetic field, and the resistance of the circuit. Higher rates of change of the magnetic field, more turns in the coil, stronger magnetic fields, and lower resistance will result in a larger induced current.

## Physics Class 11

130 videos|483 docs|210 tests

## Physics Class 11

130 videos|483 docs|210 tests

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