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# Chapter 24 - Electromagnetic Induction (Part - 2) - Physics, Solution by D C Pandey, NEET NEET Notes | EduRev

## DC Pandey (Questions & Solutions) of Physics: NEET

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## NEET : Chapter 24 - Electromagnetic Induction (Part - 2) - Physics, Solution by D C Pandey, NEET NEET Notes | EduRev

``` Page 1

27. At any in stant of time,
L
di
dt
L
di
dt
1
1
2
2
=
Þ  L i L i
1 1 2 2
=
i i
1 2
2 = â€¦(i)
inductors offer zero resistance, hence
i= =
20
5
4 A
But      i i i
1 2
+ =
i
2
4
3
= A, i
1
8
3
= A
28. When the switch is closed,
i
E
R
e
R t L
2
2
1
2
= -
-
( )
/
di
dt
E
L
e
R t L 2
2
=
- /

Potential difference across L
V L
di
dt
E e
R t L
+ =
- 2
2
/
=
-
( ) 12
5
e
t
V
When the switch S is open, current  i
2
flows
in the circuit in clockwise direction and is
given by
i i e
t
2 0
=
- / t
Here, i
E
R
2
2
=
t =
+
L
R R
1 2
i
E
R
e
R R
L
t
2
2
1 2
=
-
+ æ
è
ç
ç
ö
ø
÷
÷
= =
- -
12
2
6
10 10
e e
t t
( ) A
29. For cur ren t through gal va nom e ter to be
zero,
V V
P Q
=
L
d i
d t
i R L
d i
d t
i R
1
1
1 1 2
2
2 2
+ = + â€¦(i )
Also, i R i R
1 3 2 4
= â€¦(ii)
From Eqs.(i) and (ii),
L
di
dt
i R
i R
L
di
dt
i R
i R
1
1
1 1
1 3
2
2
2 2
2 4
+
=
+
â€¦(iii)
In the steady state,
di
dt
di
dt
1 2
0 = =
\
R
R
R
R
1
3
2
4
= Þ
R
R
R
R
1
2
3
4
=
Again as current through galvanometer is
always zero.

i
i
1
2
= constant
or
di dt
di dt
1
2
/
/
= constant
or
di
dt
di
dt
i
i
1
2
1
2
= â€¦(iv)
From Eqs. (iii) and (iv),
117
L ,R
11
i
1
i
2
P
R
3
i
1
G
L,R
22
i
2
Q
K
E
R
4
E
ii
2
i
1L
R
1
S
R
2
E
i
2
i
1
R
1
S
R
2
i
2
L
i
i
1
5 mH
10 mH
20 V
5W
Page 2

27. At any in stant of time,
L
di
dt
L
di
dt
1
1
2
2
=
Þ  L i L i
1 1 2 2
=
i i
1 2
2 = â€¦(i)
inductors offer zero resistance, hence
i= =
20
5
4 A
But      i i i
1 2
+ =
i
2
4
3
= A, i
1
8
3
= A
28. When the switch is closed,
i
E
R
e
R t L
2
2
1
2
= -
-
( )
/
di
dt
E
L
e
R t L 2
2
=
- /

Potential difference across L
V L
di
dt
E e
R t L
+ =
- 2
2
/
=
-
( ) 12
5
e
t
V
When the switch S is open, current  i
2
flows
in the circuit in clockwise direction and is
given by
i i e
t
2 0
=
- / t
Here, i
E
R
2
2
=
t =
+
L
R R
1 2
i
E
R
e
R R
L
t
2
2
1 2
=
-
+ æ
è
ç
ç
ö
ø
÷
÷
= =
- -
12
2
6
10 10
e e
t t
( ) A
29. For cur ren t through gal va nom e ter to be
zero,
V V
P Q
=
L
d i
d t
i R L
d i
d t
i R
1
1
1 1 2
2
2 2
+ = + â€¦(i )
Also, i R i R
1 3 2 4
= â€¦(ii)
From Eqs.(i) and (ii),
L
di
dt
i R
i R
L
di
dt
i R
i R
1
1
1 1
1 3
2
2
2 2
2 4
+
=
+
â€¦(iii)
In the steady state,
di
dt
di
dt
1 2
0 = =
\
R
R
R
R
1
3
2
4
= Þ
R
R
R
R
1
2
3
4
=
Again as current through galvanometer is
always zero.

i
i
1
2
= constant
or
di dt
di dt
1
2
/
/
= constant
or
di
dt
di
dt
i
i
1
2
1
2
= â€¦(iv)
From Eqs. (iii) and (iv),
117
L ,R
11
i
1
i
2
P
R
3
i
1
G
L,R
22
i
2
Q
K
E
R
4
E
ii
2
i
1L
R
1
S
R
2
E
i
2
i
1
R
1
S
R
2
i
2
L
i
i
1
5 mH
10 mH
20 V
5W
L
L
R
R
R
R
1
2
3
4
1
2
= =
30. (a) In LC cir cuit
Maximum electrical energy = Maximum
magnetic energy
Þ
1
2
1
2
0
2
0
2
CV Li =
L C
V
i
=
æ
è
ç
ç
ö
ø
÷
÷
= ´
´
æ
è
ç
ç
ö
ø
÷
÷
-
-
0
0
2
6
3
2
4 10
50 10
1.50
= ´
-
3.6 10
3
H
Þ L = 3.6 mH
(b) f
LC
=
1
2p
=
´ ´ ´ ´
- -
1
2 10 4 10
3 6
3.14 3.6

= ´ 1.33 10
3
Hz
=1.33 kHz
(c) Time taken to rise from zero to maximum
value,
t
T
f
= = =
´ ´ 4
1
4
1
10
3
4 1.33
= ´
-
3 10
3
s = 3 ms.
31. (a) w p = = ´ ´ 2 2 10
3
f 3.14
T
f
= = =
-
1 1
10
10
3
3
s = 1 ms
(b) As initially charge is maximum, (i.e.., it is
extreme position for charge).
q q t =
0
cosw
q CV
0 0
6
1 10 100 = = ´ ´
-
=
-
10
4
\ q t = ´
-
[ cos( ) ] 10 10
4 3
6.28 C.
(c) w =
1
LC

Þ L
C
= =
´ ´
-
1 1
10 10
2 3 2 6
w ( ) 6.28
= ´
-
2.53 10
3
Þ      L =2.53 mH
(d) In one quarter cycle, entire charge of the
capacitor flows out.
< > = = i
q
t
CV
T
4
=
´ ´
=
-
-
4 10 100
10
6
3
0.4 A
32. (a) V
q
C
0
0
6
4
10
4 10
= =
´
´
-
-
5.00
= ´
-
1.25 10
2
V= 12.5  mV
(b) Maximum magnetic energy = Maximum
electric energy
1
2 2
0
2 0
2
Li
q
C
=
Þ       i
q
LC
0
0
=
Þ i
0
6
4
4
10
4 10
10 =
´
´ ´
= ´
-
-
-
5.00
0.090
8.33 A
(c) Maximum energy stored in inductor,
=
1
2
0
2
Li
= ´ ´ ´
-
1
2
10
4 2
0.0900 8.33 ( )
= ´
-
3.125 10
8
J
(d) By conservation of energy,
q
C
Li Li
2
2
0
2
2
1
2
1
2
+ =
But i
i
=
0
2
q
C
Li
2
0
2
2
3
8
=
q
i
LC q = =
0
0
2
3
3
2
= ´ ´
-
1.732
5.00
2
10
6
= ´
-
4.33 10
6
C
U Li Li
m
= =
æ
è
ç
ö
ø
÷
1
2
1
4
1
2
2
0
2
= ´
-
7.8 10
9
J
33. (a) w = =
´ ´ ´
- -
1 1
10 10
3 6
LC
2.0 5.0
=10
4

di
dt
Q
½
½
½
½
½
½
= w
2
= ´ ´ =
-
( ) 10 100 10 10
4 2 6 4
A/s
(b) i Q Q = - w
0
2 2
= ´ - ´ =
- -
10 200 10 200 10 0
4 6 2 6 2
( ) ( )
(c) i Q
0 0
4 6
10 200 10 = = ´ ´
-
w = 2 A
(d) i Q Q = - w
0
2 2
118
Page 3

27. At any in stant of time,
L
di
dt
L
di
dt
1
1
2
2
=
Þ  L i L i
1 1 2 2
=
i i
1 2
2 = â€¦(i)
inductors offer zero resistance, hence
i= =
20
5
4 A
But      i i i
1 2
+ =
i
2
4
3
= A, i
1
8
3
= A
28. When the switch is closed,
i
E
R
e
R t L
2
2
1
2
= -
-
( )
/
di
dt
E
L
e
R t L 2
2
=
- /

Potential difference across L
V L
di
dt
E e
R t L
+ =
- 2
2
/
=
-
( ) 12
5
e
t
V
When the switch S is open, current  i
2
flows
in the circuit in clockwise direction and is
given by
i i e
t
2 0
=
- / t
Here, i
E
R
2
2
=
t =
+
L
R R
1 2
i
E
R
e
R R
L
t
2
2
1 2
=
-
+ æ
è
ç
ç
ö
ø
÷
÷
= =
- -
12
2
6
10 10
e e
t t
( ) A
29. For cur ren t through gal va nom e ter to be
zero,
V V
P Q
=
L
d i
d t
i R L
d i
d t
i R
1
1
1 1 2
2
2 2
+ = + â€¦(i )
Also, i R i R
1 3 2 4
= â€¦(ii)
From Eqs.(i) and (ii),
L
di
dt
i R
i R
L
di
dt
i R
i R
1
1
1 1
1 3
2
2
2 2
2 4
+
=
+
â€¦(iii)
In the steady state,
di
dt
di
dt
1 2
0 = =
\
R
R
R
R
1
3
2
4
= Þ
R
R
R
R
1
2
3
4
=
Again as current through galvanometer is
always zero.

i
i
1
2
= constant
or
di dt
di dt
1
2
/
/
= constant
or
di
dt
di
dt
i
i
1
2
1
2
= â€¦(iv)
From Eqs. (iii) and (iv),
117
L ,R
11
i
1
i
2
P
R
3
i
1
G
L,R
22
i
2
Q
K
E
R
4
E
ii
2
i
1L
R
1
S
R
2
E
i
2
i
1
R
1
S
R
2
i
2
L
i
i
1
5 mH
10 mH
20 V
5W
L
L
R
R
R
R
1
2
3
4
1
2
= =
30. (a) In LC cir cuit
Maximum electrical energy = Maximum
magnetic energy
Þ
1
2
1
2
0
2
0
2
CV Li =
L C
V
i
=
æ
è
ç
ç
ö
ø
÷
÷
= ´
´
æ
è
ç
ç
ö
ø
÷
÷
-
-
0
0
2
6
3
2
4 10
50 10
1.50
= ´
-
3.6 10
3
H
Þ L = 3.6 mH
(b) f
LC
=
1
2p
=
´ ´ ´ ´
- -
1
2 10 4 10
3 6
3.14 3.6

= ´ 1.33 10
3
Hz
=1.33 kHz
(c) Time taken to rise from zero to maximum
value,
t
T
f
= = =
´ ´ 4
1
4
1
10
3
4 1.33
= ´
-
3 10
3
s = 3 ms.
31. (a) w p = = ´ ´ 2 2 10
3
f 3.14
T
f
= = =
-
1 1
10
10
3
3
s = 1 ms
(b) As initially charge is maximum, (i.e.., it is
extreme position for charge).
q q t =
0
cosw
q CV
0 0
6
1 10 100 = = ´ ´
-
=
-
10
4
\ q t = ´
-
[ cos( ) ] 10 10
4 3
6.28 C.
(c) w =
1
LC

Þ L
C
= =
´ ´
-
1 1
10 10
2 3 2 6
w ( ) 6.28
= ´
-
2.53 10
3
Þ      L =2.53 mH
(d) In one quarter cycle, entire charge of the
capacitor flows out.
< > = = i
q
t
CV
T
4
=
´ ´
=
-
-
4 10 100
10
6
3
0.4 A
32. (a) V
q
C
0
0
6
4
10
4 10
= =
´
´
-
-
5.00
= ´
-
1.25 10
2
V= 12.5  mV
(b) Maximum magnetic energy = Maximum
electric energy
1
2 2
0
2 0
2
Li
q
C
=
Þ       i
q
LC
0
0
=
Þ i
0
6
4
4
10
4 10
10 =
´
´ ´
= ´
-
-
-
5.00
0.090
8.33 A
(c) Maximum energy stored in inductor,
=
1
2
0
2
Li
= ´ ´ ´
-
1
2
10
4 2
0.0900 8.33 ( )
= ´
-
3.125 10
8
J
(d) By conservation of energy,
q
C
Li Li
2
2
0
2
2
1
2
1
2
+ =
But i
i
=
0
2
q
C
Li
2
0
2
2
3
8
=
q
i
LC q = =
0
0
2
3
3
2
= ´ ´
-
1.732
5.00
2
10
6
= ´
-
4.33 10
6
C
U Li Li
m
= =
æ
è
ç
ö
ø
÷
1
2
1
4
1
2
2
0
2
= ´
-
7.8 10
9
J
33. (a) w = =
´ ´ ´
- -
1 1
10 10
3 6
LC
2.0 5.0
=10
4

di
dt
Q
½
½
½
½
½
½
= w
2
= ´ ´ =
-
( ) 10 100 10 10
4 2 6 4
A/s
(b) i Q Q = - w
0
2 2
= ´ - ´ =
- -
10 200 10 200 10 0
4 6 2 6 2
( ) ( )
(c) i Q
0 0
4 6
10 200 10 = = ´ ´
-
w = 2 A
(d) i Q Q = - w
0
2 2
118
Þ
i
Q Q
0
0
2 2
2
= - w
Þ
w
w
Q
Q Q
0
0
2 2
2
= -
Q Q = =
´ ´
-
3
2
200 10
2
0
6
1.73
=173 mC
34. As ini tially charge is max i mum
q q t =
0
cos w
and | | sin i i t =
0
w
where, w = =
´ ´
-
1 1
840 10
6
LC
3.3
i q
0 0
6
19 105 10 = = ´ ´
-
w
» ´
-
2.0 10
3
A = 2.0 mA
At t = 2.00 ms
(a) U
q
C
q
C
t
e
= =
2
0
2
2
2 2
(cos ) w
=
´
´ ´
-
-
( )
[cos ( ]
105 10
2 840 10
38
6 2
6
2
U
e
= ´
-
6.55 10
6
J = 6.55 mJ
(b) U Li Li t
m
= =
1
2
1
2
2
0
2
(sin ) w
= ´ ´ ´
-
1
2
2 10 38
3 2 2
3.3 rad ( ) sin ( )
= ´
-
0.009 10
6
J = 0.009 mj
(c) U
q
C
Li = =
0
2
0
2
2
1
2
= ´
-
6.56 10
6
J = 6.56 mJ
35. As the in ward mag netic field is in creas ing,
in duced elec tric field will be anticlockwise.
At a distance x from centre of the region,
Magnetic flux linked with the imaginary
loop of radius x
f =
m
x B p
2
e
d
dt
x
dB
dt
m
=
- f
= - p
2
Induced electric field,
E
e
x
x
dB
dt
= =
2
1
2 p
At a,
E r
dB
dt
=
1
4
, towards left.
At b ,
E r
dB
dt
=
1
2
, upwards.
At c,
E = 0
36. In side the so le noid,
B n i = m
0
d B
d t
n
d i
d t
= m
0
Inside the region of varying magnetic field
E r
dB
dt
nr
di
dt
= =
1
2
1
2
0
m
(a) r = 0.5 cm = ´
-
5.0 10
3
m
E rn
di
dt
=
1
2
0
m
= ´ ´ ´ ´ ´ ´
- -
1
2
4 10 10 900 60
7 3
p 5.0
= ´
-
1.7 10
4
V/m
(b) r = 10 . cm = ´
-
1.0 10
2
m
E rn
di
dt
=
1
0
m
= ´ ´ ´ ´ ´ ´
- -
1
2
4 10 10 900 60
3 3
p 5.0
= ´
-
3.4 10
4
V/m
119
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E
x
Page 4

27. At any in stant of time,
L
di
dt
L
di
dt
1
1
2
2
=
Þ  L i L i
1 1 2 2
=
i i
1 2
2 = â€¦(i)
inductors offer zero resistance, hence
i= =
20
5
4 A
But      i i i
1 2
+ =
i
2
4
3
= A, i
1
8
3
= A
28. When the switch is closed,
i
E
R
e
R t L
2
2
1
2
= -
-
( )
/
di
dt
E
L
e
R t L 2
2
=
- /

Potential difference across L
V L
di
dt
E e
R t L
+ =
- 2
2
/
=
-
( ) 12
5
e
t
V
When the switch S is open, current  i
2
flows
in the circuit in clockwise direction and is
given by
i i e
t
2 0
=
- / t
Here, i
E
R
2
2
=
t =
+
L
R R
1 2
i
E
R
e
R R
L
t
2
2
1 2
=
-
+ æ
è
ç
ç
ö
ø
÷
÷
= =
- -
12
2
6
10 10
e e
t t
( ) A
29. For cur ren t through gal va nom e ter to be
zero,
V V
P Q
=
L
d i
d t
i R L
d i
d t
i R
1
1
1 1 2
2
2 2
+ = + â€¦(i )
Also, i R i R
1 3 2 4
= â€¦(ii)
From Eqs.(i) and (ii),
L
di
dt
i R
i R
L
di
dt
i R
i R
1
1
1 1
1 3
2
2
2 2
2 4
+
=
+
â€¦(iii)
In the steady state,
di
dt
di
dt
1 2
0 = =
\
R
R
R
R
1
3
2
4
= Þ
R
R
R
R
1
2
3
4
=
Again as current through galvanometer is
always zero.

i
i
1
2
= constant
or
di dt
di dt
1
2
/
/
= constant
or
di
dt
di
dt
i
i
1
2
1
2
= â€¦(iv)
From Eqs. (iii) and (iv),
117
L ,R
11
i
1
i
2
P
R
3
i
1
G
L,R
22
i
2
Q
K
E
R
4
E
ii
2
i
1L
R
1
S
R
2
E
i
2
i
1
R
1
S
R
2
i
2
L
i
i
1
5 mH
10 mH
20 V
5W
L
L
R
R
R
R
1
2
3
4
1
2
= =
30. (a) In LC cir cuit
Maximum electrical energy = Maximum
magnetic energy
Þ
1
2
1
2
0
2
0
2
CV Li =
L C
V
i
=
æ
è
ç
ç
ö
ø
÷
÷
= ´
´
æ
è
ç
ç
ö
ø
÷
÷
-
-
0
0
2
6
3
2
4 10
50 10
1.50
= ´
-
3.6 10
3
H
Þ L = 3.6 mH
(b) f
LC
=
1
2p
=
´ ´ ´ ´
- -
1
2 10 4 10
3 6
3.14 3.6

= ´ 1.33 10
3
Hz
=1.33 kHz
(c) Time taken to rise from zero to maximum
value,
t
T
f
= = =
´ ´ 4
1
4
1
10
3
4 1.33
= ´
-
3 10
3
s = 3 ms.
31. (a) w p = = ´ ´ 2 2 10
3
f 3.14
T
f
= = =
-
1 1
10
10
3
3
s = 1 ms
(b) As initially charge is maximum, (i.e.., it is
extreme position for charge).
q q t =
0
cosw
q CV
0 0
6
1 10 100 = = ´ ´
-
=
-
10
4
\ q t = ´
-
[ cos( ) ] 10 10
4 3
6.28 C.
(c) w =
1
LC

Þ L
C
= =
´ ´
-
1 1
10 10
2 3 2 6
w ( ) 6.28
= ´
-
2.53 10
3
Þ      L =2.53 mH
(d) In one quarter cycle, entire charge of the
capacitor flows out.
< > = = i
q
t
CV
T
4
=
´ ´
=
-
-
4 10 100
10
6
3
0.4 A
32. (a) V
q
C
0
0
6
4
10
4 10
= =
´
´
-
-
5.00
= ´
-
1.25 10
2
V= 12.5  mV
(b) Maximum magnetic energy = Maximum
electric energy
1
2 2
0
2 0
2
Li
q
C
=
Þ       i
q
LC
0
0
=
Þ i
0
6
4
4
10
4 10
10 =
´
´ ´
= ´
-
-
-
5.00
0.090
8.33 A
(c) Maximum energy stored in inductor,
=
1
2
0
2
Li
= ´ ´ ´
-
1
2
10
4 2
0.0900 8.33 ( )
= ´
-
3.125 10
8
J
(d) By conservation of energy,
q
C
Li Li
2
2
0
2
2
1
2
1
2
+ =
But i
i
=
0
2
q
C
Li
2
0
2
2
3
8
=
q
i
LC q = =
0
0
2
3
3
2
= ´ ´
-
1.732
5.00
2
10
6
= ´
-
4.33 10
6
C
U Li Li
m
= =
æ
è
ç
ö
ø
÷
1
2
1
4
1
2
2
0
2
= ´
-
7.8 10
9
J
33. (a) w = =
´ ´ ´
- -
1 1
10 10
3 6
LC
2.0 5.0
=10
4

di
dt
Q
½
½
½
½
½
½
= w
2
= ´ ´ =
-
( ) 10 100 10 10
4 2 6 4
A/s
(b) i Q Q = - w
0
2 2
= ´ - ´ =
- -
10 200 10 200 10 0
4 6 2 6 2
( ) ( )
(c) i Q
0 0
4 6
10 200 10 = = ´ ´
-
w = 2 A
(d) i Q Q = - w
0
2 2
118
Þ
i
Q Q
0
0
2 2
2
= - w
Þ
w
w
Q
Q Q
0
0
2 2
2
= -
Q Q = =
´ ´
-
3
2
200 10
2
0
6
1.73
=173 mC
34. As ini tially charge is max i mum
q q t =
0
cos w
and | | sin i i t =
0
w
where, w = =
´ ´
-
1 1
840 10
6
LC
3.3
i q
0 0
6
19 105 10 = = ´ ´
-
w
» ´
-
2.0 10
3
A = 2.0 mA
At t = 2.00 ms
(a) U
q
C
q
C
t
e
= =
2
0
2
2
2 2
(cos ) w
=
´
´ ´
-
-
( )
[cos ( ]
105 10
2 840 10
38
6 2
6
2
U
e
= ´
-
6.55 10
6
J = 6.55 mJ
(b) U Li Li t
m
= =
1
2
1
2
2
0
2
(sin ) w
= ´ ´ ´
-
1
2
2 10 38
3 2 2
3.3 rad ( ) sin ( )
= ´
-
0.009 10
6
J = 0.009 mj
(c) U
q
C
Li = =
0
2
0
2
2
1
2
= ´
-
6.56 10
6
J = 6.56 mJ
35. As the in ward mag netic field is in creas ing,
in duced elec tric field will be anticlockwise.
At a distance x from centre of the region,
Magnetic flux linked with the imaginary
loop of radius x
f =
m
x B p
2
e
d
dt
x
dB
dt
m
=
- f
= - p
2
Induced electric field,
E
e
x
x
dB
dt
= =
2
1
2 p
At a,
E r
dB
dt
=
1
4
, towards left.
At b ,
E r
dB
dt
=
1
2
, upwards.
At c,
E = 0
36. In side the so le noid,
B n i = m
0
d B
d t
n
d i
d t
= m
0
Inside the region of varying magnetic field
E r
dB
dt
nr
di
dt
= =
1
2
1
2
0
m
(a) r = 0.5 cm = ´
-
5.0 10
3
m
E rn
di
dt
=
1
2
0
m
= ´ ´ ´ ´ ´ ´
- -
1
2
4 10 10 900 60
7 3
p 5.0
= ´
-
1.7 10
4
V/m
(b) r = 10 . cm = ´
-
1.0 10
2
m
E rn
di
dt
=
1
0
m
= ´ ´ ´ ´ ´ ´
- -
1
2
4 10 10 900 60
3 3
p 5.0
= ´
-
3.4 10
4
V/m
119
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E
x
AIEEE Corner
Ob jec ti ve Ques tions (Level 1)
1. V L
di
dt
=
[ ]
[ ] [ ]
[ ]
[ ] [ ]
[ ]
L
V T
i
= =
- -
M L T A T
A
2 3 1
=
- -
[ ] M L T A
2 2 2
2. M n n µ
1 2
3. Both will tend to op pose the mag netic flux
chang ing with them by in creas ing cur rent in
op po site di rec tion.
4. Mov ing charged par ti cle will pro duced
mag netic field par al lel to ring, Hence
f =
m
0
Velocity of particle increases continuously
due to gravity.
5. In duced elec tric field can ex ist at a point
where mag netic field is not pres ent, i.e.,
out side the re gion oc cu py ing the mag netic
field.
6. At, t = 1 s
q t = = 4 4
2
C
i
dq
dt
t = = = 8 8 A

di
dt
= 8 A/s
As,
di
dt
d q
dt
= =
2
Positive
Charge in capacitor is increasing, current i
must be towards left.
V I L
di
dt
q
C
ab
= - + - - 2 4
=- ´ + - ´ =- 2 8 4 2 8
4
2
30 _ V
7. | | ( sin ) e M
di
dt
M
d
dt
i t = =
0
w
= w w Mi t
0
cos
Maximum induced emf = w Mi
0
= ´ ´ 100 10 p 0.005
=5p
8.
1
2
1
2
0
2
0
2
Li CV = Þ i
L
C
0 6
2
2
4 10
= ´
´
-
= ´ 2 10
3
V
9. e B l =
1
2
2
w , is in de p end ent of t .
10. | | e
d
d t t
=
f
=
f D
D f = | | e t = iRt
= ´ ´ ´
-
10 10 5
3
0.5
= ´
-
25 10
3
Wb
=25 mWb.
11. As in ward mag netic field is in creas ing,
in duced elec tric field must be anti-clockwise.
Hence, direction of induced electric field at P
will be towards and electron will experience
force towards right (opposite to electric
field).
12. f= - = - at t a t at ( ) t t
2
|| e
d
dt
a at =
f
= - t 2
i
e
R
a at
R
= =
- || t 2
H i Rdt
a at
R
dt = =
-
ò ò
2
0
2
0
2 t t t ( )
=
-
´ -
é
ë
ê
ê
ù
û
ú
ú
1 2
3 2
3
0
R
a at
a
( )
( )
t
t
=
-
- -
1
6
3 3 3 3
Ra
a a [ ] t t
=
a
R
2 3
3
t
13. E L
di
dt
= -
14. V L
di
dt
iR
BA
=- + - 15
= - ´ - + - ´
-
5 10 10 15 5 1
3 3
( )
=15 V
15.
di
dt
= 10 A/s, at t = 0, i = 5A
di
dt
= 10 A/s
120
2W
4 V 2 H 2F
b a
i â€“
a
+
Page 5

27. At any in stant of time,
L
di
dt
L
di
dt
1
1
2
2
=
Þ  L i L i
1 1 2 2
=
i i
1 2
2 = â€¦(i)
inductors offer zero resistance, hence
i= =
20
5
4 A
But      i i i
1 2
+ =
i
2
4
3
= A, i
1
8
3
= A
28. When the switch is closed,
i
E
R
e
R t L
2
2
1
2
= -
-
( )
/
di
dt
E
L
e
R t L 2
2
=
- /

Potential difference across L
V L
di
dt
E e
R t L
+ =
- 2
2
/
=
-
( ) 12
5
e
t
V
When the switch S is open, current  i
2
flows
in the circuit in clockwise direction and is
given by
i i e
t
2 0
=
- / t
Here, i
E
R
2
2
=
t =
+
L
R R
1 2
i
E
R
e
R R
L
t
2
2
1 2
=
-
+ æ
è
ç
ç
ö
ø
÷
÷
= =
- -
12
2
6
10 10
e e
t t
( ) A
29. For cur ren t through gal va nom e ter to be
zero,
V V
P Q
=
L
d i
d t
i R L
d i
d t
i R
1
1
1 1 2
2
2 2
+ = + â€¦(i )
Also, i R i R
1 3 2 4
= â€¦(ii)
From Eqs.(i) and (ii),
L
di
dt
i R
i R
L
di
dt
i R
i R
1
1
1 1
1 3
2
2
2 2
2 4
+
=
+
â€¦(iii)
In the steady state,
di
dt
di
dt
1 2
0 = =
\
R
R
R
R
1
3
2
4
= Þ
R
R
R
R
1
2
3
4
=
Again as current through galvanometer is
always zero.

i
i
1
2
= constant
or
di dt
di dt
1
2
/
/
= constant
or
di
dt
di
dt
i
i
1
2
1
2
= â€¦(iv)
From Eqs. (iii) and (iv),
117
L ,R
11
i
1
i
2
P
R
3
i
1
G
L,R
22
i
2
Q
K
E
R
4
E
ii
2
i
1L
R
1
S
R
2
E
i
2
i
1
R
1
S
R
2
i
2
L
i
i
1
5 mH
10 mH
20 V
5W
L
L
R
R
R
R
1
2
3
4
1
2
= =
30. (a) In LC cir cuit
Maximum electrical energy = Maximum
magnetic energy
Þ
1
2
1
2
0
2
0
2
CV Li =
L C
V
i
=
æ
è
ç
ç
ö
ø
÷
÷
= ´
´
æ
è
ç
ç
ö
ø
÷
÷
-
-
0
0
2
6
3
2
4 10
50 10
1.50
= ´
-
3.6 10
3
H
Þ L = 3.6 mH
(b) f
LC
=
1
2p
=
´ ´ ´ ´
- -
1
2 10 4 10
3 6
3.14 3.6

= ´ 1.33 10
3
Hz
=1.33 kHz
(c) Time taken to rise from zero to maximum
value,
t
T
f
= = =
´ ´ 4
1
4
1
10
3
4 1.33
= ´
-
3 10
3
s = 3 ms.
31. (a) w p = = ´ ´ 2 2 10
3
f 3.14
T
f
= = =
-
1 1
10
10
3
3
s = 1 ms
(b) As initially charge is maximum, (i.e.., it is
extreme position for charge).
q q t =
0
cosw
q CV
0 0
6
1 10 100 = = ´ ´
-
=
-
10
4
\ q t = ´
-
[ cos( ) ] 10 10
4 3
6.28 C.
(c) w =
1
LC

Þ L
C
= =
´ ´
-
1 1
10 10
2 3 2 6
w ( ) 6.28
= ´
-
2.53 10
3
Þ      L =2.53 mH
(d) In one quarter cycle, entire charge of the
capacitor flows out.
< > = = i
q
t
CV
T
4
=
´ ´
=
-
-
4 10 100
10
6
3
0.4 A
32. (a) V
q
C
0
0
6
4
10
4 10
= =
´
´
-
-
5.00
= ´
-
1.25 10
2
V= 12.5  mV
(b) Maximum magnetic energy = Maximum
electric energy
1
2 2
0
2 0
2
Li
q
C
=
Þ       i
q
LC
0
0
=
Þ i
0
6
4
4
10
4 10
10 =
´
´ ´
= ´
-
-
-
5.00
0.090
8.33 A
(c) Maximum energy stored in inductor,
=
1
2
0
2
Li
= ´ ´ ´
-
1
2
10
4 2
0.0900 8.33 ( )
= ´
-
3.125 10
8
J
(d) By conservation of energy,
q
C
Li Li
2
2
0
2
2
1
2
1
2
+ =
But i
i
=
0
2
q
C
Li
2
0
2
2
3
8
=
q
i
LC q = =
0
0
2
3
3
2
= ´ ´
-
1.732
5.00
2
10
6
= ´
-
4.33 10
6
C
U Li Li
m
= =
æ
è
ç
ö
ø
÷
1
2
1
4
1
2
2
0
2
= ´
-
7.8 10
9
J
33. (a) w = =
´ ´ ´
- -
1 1
10 10
3 6
LC
2.0 5.0
=10
4

di
dt
Q
½
½
½
½
½
½
= w
2
= ´ ´ =
-
( ) 10 100 10 10
4 2 6 4
A/s
(b) i Q Q = - w
0
2 2
= ´ - ´ =
- -
10 200 10 200 10 0
4 6 2 6 2
( ) ( )
(c) i Q
0 0
4 6
10 200 10 = = ´ ´
-
w = 2 A
(d) i Q Q = - w
0
2 2
118
Þ
i
Q Q
0
0
2 2
2
= - w
Þ
w
w
Q
Q Q
0
0
2 2
2
= -
Q Q = =
´ ´
-
3
2
200 10
2
0
6
1.73
=173 mC
34. As ini tially charge is max i mum
q q t =
0
cos w
and | | sin i i t =
0
w
where, w = =
´ ´
-
1 1
840 10
6
LC
3.3
i q
0 0
6
19 105 10 = = ´ ´
-
w
» ´
-
2.0 10
3
A = 2.0 mA
At t = 2.00 ms
(a) U
q
C
q
C
t
e
= =
2
0
2
2
2 2
(cos ) w
=
´
´ ´
-
-
( )
[cos ( ]
105 10
2 840 10
38
6 2
6
2
U
e
= ´
-
6.55 10
6
J = 6.55 mJ
(b) U Li Li t
m
= =
1
2
1
2
2
0
2
(sin ) w
= ´ ´ ´
-
1
2
2 10 38
3 2 2
3.3 rad ( ) sin ( )
= ´
-
0.009 10
6
J = 0.009 mj
(c) U
q
C
Li = =
0
2
0
2
2
1
2
= ´
-
6.56 10
6
J = 6.56 mJ
35. As the in ward mag netic field is in creas ing,
in duced elec tric field will be anticlockwise.
At a distance x from centre of the region,
Magnetic flux linked with the imaginary
loop of radius x
f =
m
x B p
2
e
d
dt
x
dB
dt
m
=
- f
= - p
2
Induced electric field,
E
e
x
x
dB
dt
= =
2
1
2 p
At a,
E r
dB
dt
=
1
4
, towards left.
At b ,
E r
dB
dt
=
1
2
, upwards.
At c,
E = 0
36. In side the so le noid,
B n i = m
0
d B
d t
n
d i
d t
= m
0
Inside the region of varying magnetic field
E r
dB
dt
nr
di
dt
= =
1
2
1
2
0
m
(a) r = 0.5 cm = ´
-
5.0 10
3
m
E rn
di
dt
=
1
2
0
m
= ´ ´ ´ ´ ´ ´
- -
1
2
4 10 10 900 60
7 3
p 5.0
= ´
-
1.7 10
4
V/m
(b) r = 10 . cm = ´
-
1.0 10
2
m
E rn
di
dt
=
1
0
m
= ´ ´ ´ ´ ´ ´
- -
1
2
4 10 10 900 60
3 3
p 5.0
= ´
-
3.4 10
4
V/m
119
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E
x
AIEEE Corner
Ob jec ti ve Ques tions (Level 1)
1. V L
di
dt
=
[ ]
[ ] [ ]
[ ]
[ ] [ ]
[ ]
L
V T
i
= =
- -
M L T A T
A
2 3 1
=
- -
[ ] M L T A
2 2 2
2. M n n µ
1 2
3. Both will tend to op pose the mag netic flux
chang ing with them by in creas ing cur rent in
op po site di rec tion.
4. Mov ing charged par ti cle will pro duced
mag netic field par al lel to ring, Hence
f =
m
0
Velocity of particle increases continuously
due to gravity.
5. In duced elec tric field can ex ist at a point
where mag netic field is not pres ent, i.e.,
out side the re gion oc cu py ing the mag netic
field.
6. At, t = 1 s
q t = = 4 4
2
C
i
dq
dt
t = = = 8 8 A

di
dt
= 8 A/s
As,
di
dt
d q
dt
= =
2
Positive
Charge in capacitor is increasing, current i
must be towards left.
V I L
di
dt
q
C
ab
= - + - - 2 4
=- ´ + - ´ =- 2 8 4 2 8
4
2
30 _ V
7. | | ( sin ) e M
di
dt
M
d
dt
i t = =
0
w
= w w Mi t
0
cos
Maximum induced emf = w Mi
0
= ´ ´ 100 10 p 0.005
=5p
8.
1
2
1
2
0
2
0
2
Li CV = Þ i
L
C
0 6
2
2
4 10
= ´
´
-
= ´ 2 10
3
V
9. e B l =
1
2
2
w , is in de p end ent of t .
10. | | e
d
d t t
=
f
=
f D
D f = | | e t = iRt
= ´ ´ ´
-
10 10 5
3
0.5
= ´
-
25 10
3
Wb
=25 mWb.
11. As in ward mag netic field is in creas ing,
in duced elec tric field must be anti-clockwise.
Hence, direction of induced electric field at P
will be towards and electron will experience
force towards right (opposite to electric
field).
12. f= - = - at t a t at ( ) t t
2
|| e
d
dt
a at =
f
= - t 2
i
e
R
a at
R
= =
- || t 2
H i Rdt
a at
R
dt = =
-
ò ò
2
0
2
0
2 t t t ( )
=
-
´ -
é
ë
ê
ê
ù
û
ú
ú
1 2
3 2
3
0
R
a at
a
( )
( )
t
t
=
-
- -
1
6
3 3 3 3
Ra
a a [ ] t t
=
a
R
2 3
3
t
13. E L
di
dt
= -
14. V L
di
dt
iR
BA
=- + - 15
= - ´ - + - ´
-
5 10 10 15 5 1
3 3
( )
=15 V
15.
di
dt
= 10 A/s, at t = 0, i = 5A
di
dt
= 10 A/s
120
2W
4 V 2 H 2F
b a
i â€“
a
+
V V iR L
di
dt
E
A B
- = + - = 0
= ´ + ´ - = 5 3 1 10 10 15 V
16.
di
dt
d q
dt
q
q
LC
æ
è
ç
ö
ø
÷ =
æ
è
ç
ç
ö
ø
÷
÷
= =
max
max
2
2
0
0
w
17. V L
di
dt
=
18. f =
m
BAcos q
Þ e
d
dt
BA
d
dt
m
= -
f
= sin q
q
Þ            iR BA
d
dt
= sin q
q
Þ
dq
dt
R BA
d
dt
= sin q
q
Þ           dq
BA
R
d = sin q q
Þ             q
BA
R
d = =
ò
sin
/
/
q q
p
p
2
3 2
0
19. A k
®
= ab
^
, B i j k
®
= + + 20 10 50
2
t t
^ ^ ^

f =
®
×
®
=
m
ab B A 50
e
d
dt
m
= -
f
= 0
20. E V iR
b
= +
V E iR
b
= - = - ´ = 200 20 1 5 170 . V
21.
V
V
N
N
s
p
s
p
= Þ V
s
= ´ =
1
2
290 10 V
i
i
N
N
p
s
s
p
=
Þ i
N
N
i
s
p
s
p
= = ´ = 2 4 8 A
22. V
r
= 0, hence mag netic flux linked with the
coil re main same.
\  e
d
dt
=
- f
= 0
23. s at =
1
2
2
Due to change in magnetic flux linked with
the ring, magnet experiences an upward
force, hence,
a g <
s gt <
1
2
2
Þ s < 5 m
24. V V L
di
dt
t
A B
- = = - a
25. i
E
R
0
12
40 = = =
0.3
A
U Li
0 0
2 3 2
1
2
1
2
50 10 40 = = ´ ´ ´
-
( )
=40 J
26. i i e
E
R
e
t Rt
L
= -
æ
è
ç
ç
ö
ø
÷
÷
= -
æ
è
ç
ç
ö
ø
÷
÷
- -
0
1 1
t
di
dt
E
L
e
Rt
L
=
-
V L
di
dt
Ee
L
Rt
L
= =
-
at t = 0
V E
L
= =20 V
at t = 20 ms
V Ee
L
R
L
=
-
´ ´
-
20 10
3
Þ 5 20
50
=
-
e
R
L
R
L 50
4 = ln Þ R = ( ln ) 100 4 W
27. | |
| |
i
e
R R
d
dt R
NA
dB
dt
= = ×
f
=
1 1
=
´ ´
´ ´
-
-
10 10 10
20
10 10
4
8 4
= 5 A
28. In the steady state, in duc tor be haves as
short cir cuit, hence en tire cur rent flows
through it.
29.   f =
m
ABcosq
But, q= ° 90
\  f =
m
0
30. i
e
R R
d
dt
m
= = -
f | | 1
Þ
dq
dt
nBA
R
d
dt
=
-
(cos ) q
=
nBA
R
d
dt
sinq
q
Þ         dq
nBA
R
d = sinq q
Q
nBA
R
d
nBA
R
1
0
2
= =
ò
sin
p
q q
Q
nBA
R
d
2
0
2
0 = =
ò
sin
p
q q
\
Q
Q
2
1
0 =
31. Ac cord i ng to Lenzâ€™s law, in duced cur rent
al ways op poses the cause pro duc ing it.
121
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