DC Pandey Solutions: Electromagnetic Induction- 1

# DC Pandey Solutions: Electromagnetic Induction- 1 | DC Pandey Solutions for NEET 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
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
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
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
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
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
```

122 docs

## FAQs on DC Pandey Solutions: Electromagnetic Induction- 1 - DC Pandey Solutions for NEET Physics

 1. What is electromagnetic induction?
Ans. Electromagnetic induction is the process of generating an electromotive force (EMF) or voltage in a conductor when it is exposed to a changing magnetic field. This phenomenon was first discovered by Michael Faraday in the 1830s and is the basis for the functioning of various electrical devices such as generators and transformers.
 2. How does electromagnetic induction work?
Ans. Electromagnetic induction works on the principle that when a conductor, such as a wire, is exposed to a changing magnetic field, an EMF or voltage is induced in the conductor. This induced voltage causes the movement of charge carriers, creating an electric current in the conductor. The magnitude of the induced voltage depends on factors such as the rate of change of the magnetic field and the number of turns in the conductor.
 3. What is Faraday's law of electromagnetic induction?
Ans. Faraday's law of electromagnetic induction states that the magnitude of the induced EMF in a conductor is directly proportional to the rate of change of magnetic flux through the conductor. Mathematically, it can be expressed as E = -dΦ/dt, where E is the induced EMF, Φ is the magnetic flux, and dt represents the time interval during which the change in magnetic flux occurs.
 4. What are the applications of electromagnetic induction?
Ans. Electromagnetic induction has numerous applications in our daily lives. Some of the common applications include: 1. Electric generators: Electromagnetic induction is used to generate electricity in power plants through the rotation of conductors in a magnetic field. 2. Transformers: Transformers utilize electromagnetic induction to transfer electrical energy from one circuit to another, typically at different voltage levels. 3. Induction cooktops: These cooktops use electromagnetic induction to heat the cooking vessel directly, resulting in faster and more efficient cooking. 4. Wireless charging: Inductive charging allows for the wireless transfer of energy between two objects, commonly used in smartphones and electric vehicles.
 5. What is Lenz's law in electromagnetic induction?
Ans. Lenz's law is a fundamental principle in electromagnetic induction that states that the direction of the induced current in a conductor will always oppose the change in magnetic field that produced it. This law is based on the conservation of energy and the principle of electromagnetic induction. Lenz's law helps in determining the direction of the induced current and is widely used in various applications, including the design of electromagnetic devices and circuits.

## DC Pandey Solutions for NEET Physics

122 docs

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