DC Pandey Solutions: Electromagnetic Induction- 1 | DC Pandey Solutions for JEE Physics PDF Download

 Page 1


In tro duc tory Ex er cise 24.1
1. Mag netic field in side the loop due to cur rent
car ry ing con duc t or is in ward s.
As the current in the conductor
increases, magnetic flux linked
with the loop increases as a
result of which, induced current
will produce in the loop to
produce an outward magnetic field, i.e.,
induced current will be anti-clockwise.
2. No.
Emf is in duced if the field is time vary ing.
3.
d
dt
B
f
= induced emf
\
d
dt
V
B
f é
ë
ê
ù
û
ú
= =
- -
[ ] [ML T I ]
2 3 1
Introductory Exercise 24.2
1. If the out ward mag netic flux in creases,
in duced cur rent will be in such a way that it
pro duces in wards mag net ic flux, i.e., it will
be clock wise.
2. Mag netic flux linked with the coil will not
change, hence in duced cur rent will be zero.
3. If the cur rent in coil 1 (clockwise) in creases,
out ward mag netic flux linked with the coil 2
in creases and the coil 2 will pro duce in duced
cur rent in clock wise di rec tion to op pose the
change in mag netic flux linked with it.
Hence, if the current in coil 1 increases,
induced current will be in same sense and
vice-versa.
In tro duc tory Ex er cise 24.3
1. f = =
-
B
at
BS B S e
0
e
d
dt
a B S e
B at
= -
f
=
-
0
2. No.
As, F i lB
m
= = 0
Because, i = 0 as the circuit is not closed. As
net force acting on the bar is zero, no
external force is required to move the bar
with constant velocity.
Electromagnetic
Induction
24
i
1
2 i'
i increasing
i
1
2 i'
i decreasing
Page 2


In tro duc tory Ex er cise 24.1
1. Mag netic field in side the loop due to cur rent
car ry ing con duc t or is in ward s.
As the current in the conductor
increases, magnetic flux linked
with the loop increases as a
result of which, induced current
will produce in the loop to
produce an outward magnetic field, i.e.,
induced current will be anti-clockwise.
2. No.
Emf is in duced if the field is time vary ing.
3.
d
dt
B
f
= induced emf
\
d
dt
V
B
f é
ë
ê
ù
û
ú
= =
- -
[ ] [ML T I ]
2 3 1
Introductory Exercise 24.2
1. If the out ward mag netic flux in creases,
in duced cur rent will be in such a way that it
pro duces in wards mag net ic flux, i.e., it will
be clock wise.
2. Mag netic flux linked with the coil will not
change, hence in duced cur rent will be zero.
3. If the cur rent in coil 1 (clockwise) in creases,
out ward mag netic flux linked with the coil 2
in creases and the coil 2 will pro duce in duced
cur rent in clock wise di rec tion to op pose the
change in mag netic flux linked with it.
Hence, if the current in coil 1 increases,
induced current will be in same sense and
vice-versa.
In tro duc tory Ex er cise 24.3
1. f = =
-
B
at
BS B S e
0
e
d
dt
a B S e
B at
= -
f
=
-
0
2. No.
As, F i lB
m
= = 0
Because, i = 0 as the circuit is not closed. As
net force acting on the bar is zero, no
external force is required to move the bar
with constant velocity.
Electromagnetic
Induction
24
i
1
2 i'
i increasing
i
1
2 i'
i decreasing
3. | | e
t
=
f - f
2 1
But, f =
1 1
NB Acos q, f =
2 2
NB Acos q
\ | |
cos ( )
e
NA B B
t
=
- q
2 1
Þ A
e t
N B B
=
-
| |
( )cos
2 1
q
 =
´ ´
´ ´ - ´ ´
-
- -
80.0 0.4 10
50 600 10 200 10
3
2
3
6 6
( )
   =1.85m
2
Side of square, a A = = 1.36 m
Total length of wire = ´ 50 4a
         = ´ ´ 50 4 272 1.36= m
4. (a) Consider an el e men tary por tion of length 
dx of the bar at a dis tance x from end a.
Magnetic field at this point,
B
i
a x
= ×
+
m
p
0
4
2
Induced emf in this portion,
de B dxv
vi
d x
dx = = ×
+
m
p
0
4
2
5. (a) EMF in duced in the bar ab,
   e de vi
dx
d x
l
= = ×
+
ò ò
m
p
0
0
4
2
= +
m
p
0
0
4
2 vi d x
l
[ln ( )]
          =
+ m
p
0
2
vi d l
d
ln
        = +
æ
è
ç
ö
ø
÷
m
p
0
2
1
vi l
d
ln
(b)Magnetic field in the region ab is
inwards, hence by Fleming’s left hand
rule, positive charge will move up and a
will be at higher potential.
Or
Use Fleming’s right hand rule.
(c) No.
As flux linked with the square loop will
remain same.
In tro duc tory Ex er cise 24.4
1. Po ten tial dif fer ence across an in duc tor,
                        V L
di
dt
L
d
dt
t t = = ( sin ) 3
                              = + 3L t t t [sin cos ]
In tro duc tory Ex er cise 24.5
1. (a) To tal num ber of turns on the so le noid,
         N
l
d
= =
´
´
-
-
40 10
10
2
2
0.10
               =400
L
N A
l
=
m
0
2
  =
´ ´ ´ ´
´
- -
-
4 10 400 10
40 10
7 2 4
2
p ( ) 0.90
   = ´
-
4.5 10
5
 H
(b)     e L
di
dt
= -
        = - ´ ´
-
-
4.5
0.10
10
0 10
5
        = ´
-
4.5 V 10
3
        =4.5 mV
109 
d
l
x
v
a
i
b
dx
Page 3


In tro duc tory Ex er cise 24.1
1. Mag netic field in side the loop due to cur rent
car ry ing con duc t or is in ward s.
As the current in the conductor
increases, magnetic flux linked
with the loop increases as a
result of which, induced current
will produce in the loop to
produce an outward magnetic field, i.e.,
induced current will be anti-clockwise.
2. No.
Emf is in duced if the field is time vary ing.
3.
d
dt
B
f
= induced emf
\
d
dt
V
B
f é
ë
ê
ù
û
ú
= =
- -
[ ] [ML T I ]
2 3 1
Introductory Exercise 24.2
1. If the out ward mag netic flux in creases,
in duced cur rent will be in such a way that it
pro duces in wards mag net ic flux, i.e., it will
be clock wise.
2. Mag netic flux linked with the coil will not
change, hence in duced cur rent will be zero.
3. If the cur rent in coil 1 (clockwise) in creases,
out ward mag netic flux linked with the coil 2
in creases and the coil 2 will pro duce in duced
cur rent in clock wise di rec tion to op pose the
change in mag netic flux linked with it.
Hence, if the current in coil 1 increases,
induced current will be in same sense and
vice-versa.
In tro duc tory Ex er cise 24.3
1. f = =
-
B
at
BS B S e
0
e
d
dt
a B S e
B at
= -
f
=
-
0
2. No.
As, F i lB
m
= = 0
Because, i = 0 as the circuit is not closed. As
net force acting on the bar is zero, no
external force is required to move the bar
with constant velocity.
Electromagnetic
Induction
24
i
1
2 i'
i increasing
i
1
2 i'
i decreasing
3. | | e
t
=
f - f
2 1
But, f =
1 1
NB Acos q, f =
2 2
NB Acos q
\ | |
cos ( )
e
NA B B
t
=
- q
2 1
Þ A
e t
N B B
=
-
| |
( )cos
2 1
q
 =
´ ´
´ ´ - ´ ´
-
- -
80.0 0.4 10
50 600 10 200 10
3
2
3
6 6
( )
   =1.85m
2
Side of square, a A = = 1.36 m
Total length of wire = ´ 50 4a
         = ´ ´ 50 4 272 1.36= m
4. (a) Consider an el e men tary por tion of length 
dx of the bar at a dis tance x from end a.
Magnetic field at this point,
B
i
a x
= ×
+
m
p
0
4
2
Induced emf in this portion,
de B dxv
vi
d x
dx = = ×
+
m
p
0
4
2
5. (a) EMF in duced in the bar ab,
   e de vi
dx
d x
l
= = ×
+
ò ò
m
p
0
0
4
2
= +
m
p
0
0
4
2 vi d x
l
[ln ( )]
          =
+ m
p
0
2
vi d l
d
ln
        = +
æ
è
ç
ö
ø
÷
m
p
0
2
1
vi l
d
ln
(b)Magnetic field in the region ab is
inwards, hence by Fleming’s left hand
rule, positive charge will move up and a
will be at higher potential.
Or
Use Fleming’s right hand rule.
(c) No.
As flux linked with the square loop will
remain same.
In tro duc tory Ex er cise 24.4
1. Po ten tial dif fer ence across an in duc tor,
                        V L
di
dt
L
d
dt
t t = = ( sin ) 3
                              = + 3L t t t [sin cos ]
In tro duc tory Ex er cise 24.5
1. (a) To tal num ber of turns on the so le noid,
         N
l
d
= =
´
´
-
-
40 10
10
2
2
0.10
               =400
L
N A
l
=
m
0
2
  =
´ ´ ´ ´
´
- -
-
4 10 400 10
40 10
7 2 4
2
p ( ) 0.90
   = ´
-
4.5 10
5
 H
(b)     e L
di
dt
= -
        = - ´ ´
-
-
4.5
0.10
10
0 10
5
        = ´
-
4.5 V 10
3
        =4.5 mV
109 
d
l
x
v
a
i
b
dx
In tro duc tory Ex er cise 24.6
1. Con sider a cur rent i is flow ing in the outer
loop.
Magnetic field at the centre of the loop.
B
i
R
=
m
0
2
As R r >> , magnetic field inside smaller loop
may assumed to be constant.
Hence, magnetic flux linked with the
smaller loop,
f = ´ =
m
B r
r i
R
p
m p
2 0
2
2
M
i
r
R
m
=
f
=
pm
0
2
2
   
In tro duc tory Ex er cise 24.7
1. (a) V i R
0 0
3
36 10 175 = = ´ ´ =
-
6.3 V
(b) i i e
t
= -
-
0
1 ( )
/ t
where, t =
L
R
Now, at t = 58 ms 
       i=4.9 mA
\  4.9 = -
-
36 1
58
( )
/
e
t
Þ e
-
=
58
36
/ t
31.1
Þ t m = 397 s
L
R
= 397 ms 
Þ                L = ´ ´
-
175 397 10
6
    = 69 mH
(c)          t m =397 s
2.        [ ]
[ ] [ ][ ]
[ ]
L
e
di
dt
V t
i
=
é
ë
ê
ù
û
ú
=
and     [ ]
[ ]
[ ]
R
V
i
=
\      
L
R
L
R
é
ë
ê
ù
û
ú
= =
[ ]
T
[ ]
[ ]
3. (a) Ini tially
       E L
di
dt
=
Þ     
di
dt
E
L
=
         = =
12.0
3.00
4 A/s
(b)    E V V
L R
= +
Þ    E L
di
dt
iR = +
Þ        
di
dt L
E iR = -
1
[ ]
 = ´ - ´
1
300
12 1 7
.
[ ]
Þ        
di
dt
= =
5
3
1.67 A/s
(c)   t = =
L
R
3
7
     i i e
t
= -
-
0
1 ( )
/t
= - = -
- -
E
R
e e
t
( ) ( )
/
1
12
7
1
3 t 1.4/
Þ    i=0.639 A
(d) i
E
R
0
12
7
= = = 1.71 A
4. (a) P Ei
E
R
e
t
= = -
-
2
1 ( )
/ t
   = - = -
- -
( )
( ) ( )
/
12
7
1 1
2
7 3
e e
t t
20.6
2.33
 W
(b) Rate of dissipation of energy,
      P i R i R e
R
t R
= = -
- 2
0
2 7 2
1 ( )
/
= -
-
20.6
2.33
( ) 1
2
e
t
 W    
(c) Rate of increase of magnetic energy
      P ei L
di
dt
i
L
= =
= -
- -
20.6
2.33 4.67
( ) e e
t t
 W
(d) Clearly, P P P
R L
= +
5. No.
 E V V
L R
= + and V
R
 cannot be negative in
RL circuit.
 110
R
r
i
Page 4


In tro duc tory Ex er cise 24.1
1. Mag netic field in side the loop due to cur rent
car ry ing con duc t or is in ward s.
As the current in the conductor
increases, magnetic flux linked
with the loop increases as a
result of which, induced current
will produce in the loop to
produce an outward magnetic field, i.e.,
induced current will be anti-clockwise.
2. No.
Emf is in duced if the field is time vary ing.
3.
d
dt
B
f
= induced emf
\
d
dt
V
B
f é
ë
ê
ù
û
ú
= =
- -
[ ] [ML T I ]
2 3 1
Introductory Exercise 24.2
1. If the out ward mag netic flux in creases,
in duced cur rent will be in such a way that it
pro duces in wards mag net ic flux, i.e., it will
be clock wise.
2. Mag netic flux linked with the coil will not
change, hence in duced cur rent will be zero.
3. If the cur rent in coil 1 (clockwise) in creases,
out ward mag netic flux linked with the coil 2
in creases and the coil 2 will pro duce in duced
cur rent in clock wise di rec tion to op pose the
change in mag netic flux linked with it.
Hence, if the current in coil 1 increases,
induced current will be in same sense and
vice-versa.
In tro duc tory Ex er cise 24.3
1. f = =
-
B
at
BS B S e
0
e
d
dt
a B S e
B at
= -
f
=
-
0
2. No.
As, F i lB
m
= = 0
Because, i = 0 as the circuit is not closed. As
net force acting on the bar is zero, no
external force is required to move the bar
with constant velocity.
Electromagnetic
Induction
24
i
1
2 i'
i increasing
i
1
2 i'
i decreasing
3. | | e
t
=
f - f
2 1
But, f =
1 1
NB Acos q, f =
2 2
NB Acos q
\ | |
cos ( )
e
NA B B
t
=
- q
2 1
Þ A
e t
N B B
=
-
| |
( )cos
2 1
q
 =
´ ´
´ ´ - ´ ´
-
- -
80.0 0.4 10
50 600 10 200 10
3
2
3
6 6
( )
   =1.85m
2
Side of square, a A = = 1.36 m
Total length of wire = ´ 50 4a
         = ´ ´ 50 4 272 1.36= m
4. (a) Consider an el e men tary por tion of length 
dx of the bar at a dis tance x from end a.
Magnetic field at this point,
B
i
a x
= ×
+
m
p
0
4
2
Induced emf in this portion,
de B dxv
vi
d x
dx = = ×
+
m
p
0
4
2
5. (a) EMF in duced in the bar ab,
   e de vi
dx
d x
l
= = ×
+
ò ò
m
p
0
0
4
2
= +
m
p
0
0
4
2 vi d x
l
[ln ( )]
          =
+ m
p
0
2
vi d l
d
ln
        = +
æ
è
ç
ö
ø
÷
m
p
0
2
1
vi l
d
ln
(b)Magnetic field in the region ab is
inwards, hence by Fleming’s left hand
rule, positive charge will move up and a
will be at higher potential.
Or
Use Fleming’s right hand rule.
(c) No.
As flux linked with the square loop will
remain same.
In tro duc tory Ex er cise 24.4
1. Po ten tial dif fer ence across an in duc tor,
                        V L
di
dt
L
d
dt
t t = = ( sin ) 3
                              = + 3L t t t [sin cos ]
In tro duc tory Ex er cise 24.5
1. (a) To tal num ber of turns on the so le noid,
         N
l
d
= =
´
´
-
-
40 10
10
2
2
0.10
               =400
L
N A
l
=
m
0
2
  =
´ ´ ´ ´
´
- -
-
4 10 400 10
40 10
7 2 4
2
p ( ) 0.90
   = ´
-
4.5 10
5
 H
(b)     e L
di
dt
= -
        = - ´ ´
-
-
4.5
0.10
10
0 10
5
        = ´
-
4.5 V 10
3
        =4.5 mV
109 
d
l
x
v
a
i
b
dx
In tro duc tory Ex er cise 24.6
1. Con sider a cur rent i is flow ing in the outer
loop.
Magnetic field at the centre of the loop.
B
i
R
=
m
0
2
As R r >> , magnetic field inside smaller loop
may assumed to be constant.
Hence, magnetic flux linked with the
smaller loop,
f = ´ =
m
B r
r i
R
p
m p
2 0
2
2
M
i
r
R
m
=
f
=
pm
0
2
2
   
In tro duc tory Ex er cise 24.7
1. (a) V i R
0 0
3
36 10 175 = = ´ ´ =
-
6.3 V
(b) i i e
t
= -
-
0
1 ( )
/ t
where, t =
L
R
Now, at t = 58 ms 
       i=4.9 mA
\  4.9 = -
-
36 1
58
( )
/
e
t
Þ e
-
=
58
36
/ t
31.1
Þ t m = 397 s
L
R
= 397 ms 
Þ                L = ´ ´
-
175 397 10
6
    = 69 mH
(c)          t m =397 s
2.        [ ]
[ ] [ ][ ]
[ ]
L
e
di
dt
V t
i
=
é
ë
ê
ù
û
ú
=
and     [ ]
[ ]
[ ]
R
V
i
=
\      
L
R
L
R
é
ë
ê
ù
û
ú
= =
[ ]
T
[ ]
[ ]
3. (a) Ini tially
       E L
di
dt
=
Þ     
di
dt
E
L
=
         = =
12.0
3.00
4 A/s
(b)    E V V
L R
= +
Þ    E L
di
dt
iR = +
Þ        
di
dt L
E iR = -
1
[ ]
 = ´ - ´
1
300
12 1 7
.
[ ]
Þ        
di
dt
= =
5
3
1.67 A/s
(c)   t = =
L
R
3
7
     i i e
t
= -
-
0
1 ( )
/t
= - = -
- -
E
R
e e
t
( ) ( )
/
1
12
7
1
3 t 1.4/
Þ    i=0.639 A
(d) i
E
R
0
12
7
= = = 1.71 A
4. (a) P Ei
E
R
e
t
= = -
-
2
1 ( )
/ t
   = - = -
- -
( )
( ) ( )
/
12
7
1 1
2
7 3
e e
t t
20.6
2.33
 W
(b) Rate of dissipation of energy,
      P i R i R e
R
t R
= = -
- 2
0
2 7 2
1 ( )
/
= -
-
20.6
2.33
( ) 1
2
e
t
 W    
(c) Rate of increase of magnetic energy
      P ei L
di
dt
i
L
= =
= -
- -
20.6
2.33 4.67
( ) e e
t t
 W
(d) Clearly, P P P
R L
= +
5. No.
 E V V
L R
= + and V
R
 cannot be negative in
RL circuit.
 110
R
r
i
6. Con sider the sys tem as a com bi na tion of two
bat ter i es ( E
1
10 = V and E
2
0 = ) as shown
E
E R E R
R R
=
+
+
=
1 2 2 1
1 2
5 V
R
R R
R R
=
+
=
1 2
1 2
2W         
i
E
R
0
8
5
10
=
+
= = 0.5 A
t =
+
=
L
R 8
1
10
             
i i e
t
= -
-
0
1 ( )
/ t
            
 i e
t
= -
-
0.5( ) 1
10
  A       
\ Current through inductor
i e
t
= -
-
2.5 A ( ) 1
10
In loop ABCDA
i R i L
di
dt
E
1 1 1
8 0 + + - =
i e e
t t
1
10 10
4 8 1 1 5 10 0 ´ + ´ - + - =
- -
0.5( ) ( )
i e
t
1
10
= -
-
( ) 1.5 0.25 A
In tro duc tory Ex er cise 24.8
1.   [ ]
[ ]
[ ]
[ ][
[ ]
C
q
V
i T
V
= =
]
]
   [ ]
[ ] [ ][
[ ]
L
e
di
dt
V T
i
=
é
ë
ê
ù
û
ú
=
]
Þ [ ] [ ] [ ] LC L C = = T
2. In LC os cil la tions, mag netic en er gy is
equiv a lent to ki netic en ergy in spring block
system.
i
dq
dt
= Þ v
dx
dt
=
Also L is equivalent to inertia (m) in
electricity, hence
Magnetic energy =
1
2
2
Li is equivalent to
kinetic energy =
1
2
2
mv .
3. In LC os cil la tions, 
(a) 
d i
d t L C
q = -
1
 Þ q L C
d i
d t
= -
Þ   | | q = ´ ´ ´
-
18 10
6
0.75 3.40
       = ´
-
46.5 10
6
 C
       =46.5 C m
(b) e L
di
dt
L
LC
q = - = - -
æ
è
ç
ö
ø
÷
1
    = =
´
´
=
-
-
q
C
4.8
23.3
10
18 10
4
6
 V
4. i q
0 0
= w
where, w =
1
LC
Þ    V
q
C
i
C
0
0 0
= =
w
     V i
L
C
0 0
3
6
20 10
10
= = ´
´
´
-
-
0.1
0.5
        =20 V
In tro duc tory Ex er cise 24.9
1. (a)  B ni = m
0
    f = =
m
NBA n NAi m
0
      e
d
dt
nNA
di
dt
m
= -
f
= - m
0
= - ´ ´ ´ ´ ´ ´ -
- -
4 10
25
10 10
7 4
p
0.01
5.0 0.2 ( )
111 
R = 4W
1 8W
A B
R = 4W
2
i
1
E = 10V
1
D C
E = 0
2
ß
8W
E
R
i
2
1H
1H
i
Page 5


In tro duc tory Ex er cise 24.1
1. Mag netic field in side the loop due to cur rent
car ry ing con duc t or is in ward s.
As the current in the conductor
increases, magnetic flux linked
with the loop increases as a
result of which, induced current
will produce in the loop to
produce an outward magnetic field, i.e.,
induced current will be anti-clockwise.
2. No.
Emf is in duced if the field is time vary ing.
3.
d
dt
B
f
= induced emf
\
d
dt
V
B
f é
ë
ê
ù
û
ú
= =
- -
[ ] [ML T I ]
2 3 1
Introductory Exercise 24.2
1. If the out ward mag netic flux in creases,
in duced cur rent will be in such a way that it
pro duces in wards mag net ic flux, i.e., it will
be clock wise.
2. Mag netic flux linked with the coil will not
change, hence in duced cur rent will be zero.
3. If the cur rent in coil 1 (clockwise) in creases,
out ward mag netic flux linked with the coil 2
in creases and the coil 2 will pro duce in duced
cur rent in clock wise di rec tion to op pose the
change in mag netic flux linked with it.
Hence, if the current in coil 1 increases,
induced current will be in same sense and
vice-versa.
In tro duc tory Ex er cise 24.3
1. f = =
-
B
at
BS B S e
0
e
d
dt
a B S e
B at
= -
f
=
-
0
2. No.
As, F i lB
m
= = 0
Because, i = 0 as the circuit is not closed. As
net force acting on the bar is zero, no
external force is required to move the bar
with constant velocity.
Electromagnetic
Induction
24
i
1
2 i'
i increasing
i
1
2 i'
i decreasing
3. | | e
t
=
f - f
2 1
But, f =
1 1
NB Acos q, f =
2 2
NB Acos q
\ | |
cos ( )
e
NA B B
t
=
- q
2 1
Þ A
e t
N B B
=
-
| |
( )cos
2 1
q
 =
´ ´
´ ´ - ´ ´
-
- -
80.0 0.4 10
50 600 10 200 10
3
2
3
6 6
( )
   =1.85m
2
Side of square, a A = = 1.36 m
Total length of wire = ´ 50 4a
         = ´ ´ 50 4 272 1.36= m
4. (a) Consider an el e men tary por tion of length 
dx of the bar at a dis tance x from end a.
Magnetic field at this point,
B
i
a x
= ×
+
m
p
0
4
2
Induced emf in this portion,
de B dxv
vi
d x
dx = = ×
+
m
p
0
4
2
5. (a) EMF in duced in the bar ab,
   e de vi
dx
d x
l
= = ×
+
ò ò
m
p
0
0
4
2
= +
m
p
0
0
4
2 vi d x
l
[ln ( )]
          =
+ m
p
0
2
vi d l
d
ln
        = +
æ
è
ç
ö
ø
÷
m
p
0
2
1
vi l
d
ln
(b)Magnetic field in the region ab is
inwards, hence by Fleming’s left hand
rule, positive charge will move up and a
will be at higher potential.
Or
Use Fleming’s right hand rule.
(c) No.
As flux linked with the square loop will
remain same.
In tro duc tory Ex er cise 24.4
1. Po ten tial dif fer ence across an in duc tor,
                        V L
di
dt
L
d
dt
t t = = ( sin ) 3
                              = + 3L t t t [sin cos ]
In tro duc tory Ex er cise 24.5
1. (a) To tal num ber of turns on the so le noid,
         N
l
d
= =
´
´
-
-
40 10
10
2
2
0.10
               =400
L
N A
l
=
m
0
2
  =
´ ´ ´ ´
´
- -
-
4 10 400 10
40 10
7 2 4
2
p ( ) 0.90
   = ´
-
4.5 10
5
 H
(b)     e L
di
dt
= -
        = - ´ ´
-
-
4.5
0.10
10
0 10
5
        = ´
-
4.5 V 10
3
        =4.5 mV
109 
d
l
x
v
a
i
b
dx
In tro duc tory Ex er cise 24.6
1. Con sider a cur rent i is flow ing in the outer
loop.
Magnetic field at the centre of the loop.
B
i
R
=
m
0
2
As R r >> , magnetic field inside smaller loop
may assumed to be constant.
Hence, magnetic flux linked with the
smaller loop,
f = ´ =
m
B r
r i
R
p
m p
2 0
2
2
M
i
r
R
m
=
f
=
pm
0
2
2
   
In tro duc tory Ex er cise 24.7
1. (a) V i R
0 0
3
36 10 175 = = ´ ´ =
-
6.3 V
(b) i i e
t
= -
-
0
1 ( )
/ t
where, t =
L
R
Now, at t = 58 ms 
       i=4.9 mA
\  4.9 = -
-
36 1
58
( )
/
e
t
Þ e
-
=
58
36
/ t
31.1
Þ t m = 397 s
L
R
= 397 ms 
Þ                L = ´ ´
-
175 397 10
6
    = 69 mH
(c)          t m =397 s
2.        [ ]
[ ] [ ][ ]
[ ]
L
e
di
dt
V t
i
=
é
ë
ê
ù
û
ú
=
and     [ ]
[ ]
[ ]
R
V
i
=
\      
L
R
L
R
é
ë
ê
ù
û
ú
= =
[ ]
T
[ ]
[ ]
3. (a) Ini tially
       E L
di
dt
=
Þ     
di
dt
E
L
=
         = =
12.0
3.00
4 A/s
(b)    E V V
L R
= +
Þ    E L
di
dt
iR = +
Þ        
di
dt L
E iR = -
1
[ ]
 = ´ - ´
1
300
12 1 7
.
[ ]
Þ        
di
dt
= =
5
3
1.67 A/s
(c)   t = =
L
R
3
7
     i i e
t
= -
-
0
1 ( )
/t
= - = -
- -
E
R
e e
t
( ) ( )
/
1
12
7
1
3 t 1.4/
Þ    i=0.639 A
(d) i
E
R
0
12
7
= = = 1.71 A
4. (a) P Ei
E
R
e
t
= = -
-
2
1 ( )
/ t
   = - = -
- -
( )
( ) ( )
/
12
7
1 1
2
7 3
e e
t t
20.6
2.33
 W
(b) Rate of dissipation of energy,
      P i R i R e
R
t R
= = -
- 2
0
2 7 2
1 ( )
/
= -
-
20.6
2.33
( ) 1
2
e
t
 W    
(c) Rate of increase of magnetic energy
      P ei L
di
dt
i
L
= =
= -
- -
20.6
2.33 4.67
( ) e e
t t
 W
(d) Clearly, P P P
R L
= +
5. No.
 E V V
L R
= + and V
R
 cannot be negative in
RL circuit.
 110
R
r
i
6. Con sider the sys tem as a com bi na tion of two
bat ter i es ( E
1
10 = V and E
2
0 = ) as shown
E
E R E R
R R
=
+
+
=
1 2 2 1
1 2
5 V
R
R R
R R
=
+
=
1 2
1 2
2W         
i
E
R
0
8
5
10
=
+
= = 0.5 A
t =
+
=
L
R 8
1
10
             
i i e
t
= -
-
0
1 ( )
/ t
            
 i e
t
= -
-
0.5( ) 1
10
  A       
\ Current through inductor
i e
t
= -
-
2.5 A ( ) 1
10
In loop ABCDA
i R i L
di
dt
E
1 1 1
8 0 + + - =
i e e
t t
1
10 10
4 8 1 1 5 10 0 ´ + ´ - + - =
- -
0.5( ) ( )
i e
t
1
10
= -
-
( ) 1.5 0.25 A
In tro duc tory Ex er cise 24.8
1.   [ ]
[ ]
[ ]
[ ][
[ ]
C
q
V
i T
V
= =
]
]
   [ ]
[ ] [ ][
[ ]
L
e
di
dt
V T
i
=
é
ë
ê
ù
û
ú
=
]
Þ [ ] [ ] [ ] LC L C = = T
2. In LC os cil la tions, mag netic en er gy is
equiv a lent to ki netic en ergy in spring block
system.
i
dq
dt
= Þ v
dx
dt
=
Also L is equivalent to inertia (m) in
electricity, hence
Magnetic energy =
1
2
2
Li is equivalent to
kinetic energy =
1
2
2
mv .
3. In LC os cil la tions, 
(a) 
d i
d t L C
q = -
1
 Þ q L C
d i
d t
= -
Þ   | | q = ´ ´ ´
-
18 10
6
0.75 3.40
       = ´
-
46.5 10
6
 C
       =46.5 C m
(b) e L
di
dt
L
LC
q = - = - -
æ
è
ç
ö
ø
÷
1
    = =
´
´
=
-
-
q
C
4.8
23.3
10
18 10
4
6
 V
4. i q
0 0
= w
where, w =
1
LC
Þ    V
q
C
i
C
0
0 0
= =
w
     V i
L
C
0 0
3
6
20 10
10
= = ´
´
´
-
-
0.1
0.5
        =20 V
In tro duc tory Ex er cise 24.9
1. (a)  B ni = m
0
    f = =
m
NBA n NAi m
0
      e
d
dt
nNA
di
dt
m
= -
f
= - m
0
= - ´ ´ ´ ´ ´ ´ -
- -
4 10
25
10 10
7 4
p
0.01
5.0 0.2 ( )
111 
R = 4W
1 8W
A B
R = 4W
2
i
1
E = 10V
1
D C
E = 0
2
ß
8W
E
R
i
2
1H
1H
i
     = ´
-
3.14 V 10
6
     = 3.14 V m
(b) E
e
R
= =
´
´ ´ ´ ´
-
-
2
10
2 25 10 10
6
2
p
3.14
3.14
        = ´
-
2 10
7
 V/m
2. B t t = - + ( ) 2.00 4.00 0.8
3 2
 T
dB
dt
t t = - ( ) 6.00 8.00
2
 T/s
From, t = 0 to t =1.33 s, 
dB
dt
 is negative,
hence B is decreasing in that interval.
For t > 1.33 s, 
dB
dt
 is positive, hence B is
increasing for t > 133 . s.
(a) For point P
2
,
induced emf, V
d
dt
R
dB
dt
m
2
2 2
= -
f
= - p
Induced electric field at P
2
,
E
V
r
R
r
dB
dt
= = - ×
2
2
2
2
2 2 p
              = - -
R
r
t t
2
2
2
2
( ) 6.00 8.00
      F eE
R
r
t t = - = -
2
2
2
2
( ) 6.00 8.00
       = ´
-
8.0 10
21
 N
As magnetic field is increasing in this region, 
induced electric field will be anti-clockwise
and hence, electron will experience force in
clockwise sense, i.e., downward at P
2
.
(b) For point P
1
,
Induced emf, V
d
dt
r
dB
dt
m
1 1
2 1
= -
f
= - p
Induced electric field at P
1
,
 E
V
r
r
dB
dt
= - = -
1
1
1
2
1
2 p
   = - -
1
2
1
2
r t t ( ) 6.00 8.00 =0.36 V/m
At, t = 2.00 s
magnetic field is increasing, hence, induced
electric field will be anti-clockwise, i.e.,
upward at P
1
 and perpendicular to r
1
.
AIEEE Cor ner
Subjective Questions (Level 1)
1. < > = -
f - f
= -
-
e
t
B A A
t
2 1 2 1
( )
              A r
1
2 2
= = ´ p 3.14 0.1 ( )
= ´ =
-
3.14 0.0314 10
2
              A a
r
2
2
2
2
4
= =
æ
è
ç
ö
ø
÷
p
      =
´ ´ æ
è
ç
ö
ø
÷ =
2
4
2
3.14 0.1
0.025
\         < > = -
-
e
100(0.025 0.0314)
0.1
                   = 6.4 V
2. f = = ´ ´ ´
-
1
4
500 4 10 NBA 0.2
= 0.04 Wb
f = - = -
2
NBA 0.04 Wb
Average induced emf,
< > = -
f - f
e
t
( )
2 1
Average induced current,
< > =
< >
= -
f - f
i
e
R Rt
( )
2 1
Charge flowing through the coil
q i t = < >
Þ q
R
= -
f - f
= -
- - ( ) (
2 1
50
0.04 0.04)
= = ´
-
0.08
1.6
50
10
3
 C
= = 1.6 mC C 1600 m
3. f =
1
NBS, f = -
2
NBS
Induced emf,
< > = -
f - f
= e
t
NBS
t
( )
2 1
2
Induced current
< > =
< >
= i
e
R
NBS
Rt
2
Charge flowing through the coil,
q i t
NBS
R
= < > =
2
Þ B
qR
NS
= =
´ ´
´ ´ ´
-
-
2
10 40
2 60 3 10
6
6
4.5
 112