DC Pandey Solutions: Magnetics- 3 | DC Pandey Solutions for JEE Physics PDF Download

 Page 1


14. W MB = - (cos cos ) q q
1 2
Here, q p
1
= , q p q
2
= -
W MB = - - (cos cos( )) p p q
= - - MB( cos ) 1 q
15. B
I
r
P
= ×
m
p
q
0
4
2 ( sin )
     r = - = 5 3 4
2 2
cm
= ´
-
4 10
2
 m
sin q =
3
5
\ B
P
=
´ ´ ´
´
-
-
10 50 2
3
5
4 10
7
2
            = ´
-
1.5 T 10
4
            =1.5 gauss.
16. Mag netic field on the axis of cur rent
car ry ing cir cu la r loop ,
B
M
r x
1
0
2 2 3 2
4
2
= ×
+
m
p ( )
/
…(i)
Magnetic field at the centre of current
carrying circular loop,
B
M
r
2
0
3
4
2
= ×
m
p
…(ii)
From Eqs. (i) and (ii),
       
B
B
r x
r
2
1
2 2 3 2
3
=
+ ( )
/
          =
+ ( )
/
3 4
3
2 2 3 2
3
          =
125
27
Þ    B
2
125
27
54 150 = ´ = mT
17. F l B ba B
®
=
®
´
®
=
®
´
®
I I ( ) ( )
=-
®
´
®
I( ) ab B =
®
´
®
I( ) B ab
18. Ki netic en ergy of elec tron,
K mv e V = =
1
2
2
Þ           v
eV
m
=
2
Magnetic force,
F evB
m
= sin q
F v
m
µ Þ F v
m
µ
Hence, if potential difference is doubled,
force will become 2 times.
19. Mag netic field at O due to P,
B
I
R
I
R
1
0 0
4
2
2
= × =
m
p
m
p /
 (inwards)
Magnetic field at O due to Q,
B
I
R
I
R
2
0 0
4
2
2
= × =
m
p
m
p /
 (inwards)
Net magnetic field at O,
B B B
I
R
= + =
1 2
0
2m
p
20. As solved in Ques tion 16,
B
B
x R
R
2
1
2 2
2
3 2
=
+ æ
è
ç
ç
ö
ø
÷
÷
/
 
Þ
x R
R
2 2
2
3 2
8
+ æ
è
ç
ç
ö
ø
÷
÷
=
/
Þ
x R
R
2 2
2
4
+
=
Þ x R = 3
21. Com po nent of ve loc ity of par ti cle along
mag netic field, i.e. ,
v
q E
m
t E t
y
= = a
is not constant, hence pitch is variable.
22. r
mv
qB
mK
qB
= =
2
Now, R
mK
e B
=
2
R
m K
e R
R ¢ = =
2 2
3
2
3
( )
( )
99 
R/2 R/2
I I
P Q
R
O
5 cm
P
5 cm
6 cm
3 cm 3 cm
I
r
q
q
Page 2


14. W MB = - (cos cos ) q q
1 2
Here, q p
1
= , q p q
2
= -
W MB = - - (cos cos( )) p p q
= - - MB( cos ) 1 q
15. B
I
r
P
= ×
m
p
q
0
4
2 ( sin )
     r = - = 5 3 4
2 2
cm
= ´
-
4 10
2
 m
sin q =
3
5
\ B
P
=
´ ´ ´
´
-
-
10 50 2
3
5
4 10
7
2
            = ´
-
1.5 T 10
4
            =1.5 gauss.
16. Mag netic field on the axis of cur rent
car ry ing cir cu la r loop ,
B
M
r x
1
0
2 2 3 2
4
2
= ×
+
m
p ( )
/
…(i)
Magnetic field at the centre of current
carrying circular loop,
B
M
r
2
0
3
4
2
= ×
m
p
…(ii)
From Eqs. (i) and (ii),
       
B
B
r x
r
2
1
2 2 3 2
3
=
+ ( )
/
          =
+ ( )
/
3 4
3
2 2 3 2
3
          =
125
27
Þ    B
2
125
27
54 150 = ´ = mT
17. F l B ba B
®
=
®
´
®
=
®
´
®
I I ( ) ( )
=-
®
´
®
I( ) ab B =
®
´
®
I( ) B ab
18. Ki netic en ergy of elec tron,
K mv e V = =
1
2
2
Þ           v
eV
m
=
2
Magnetic force,
F evB
m
= sin q
F v
m
µ Þ F v
m
µ
Hence, if potential difference is doubled,
force will become 2 times.
19. Mag netic field at O due to P,
B
I
R
I
R
1
0 0
4
2
2
= × =
m
p
m
p /
 (inwards)
Magnetic field at O due to Q,
B
I
R
I
R
2
0 0
4
2
2
= × =
m
p
m
p /
 (inwards)
Net magnetic field at O,
B B B
I
R
= + =
1 2
0
2m
p
20. As solved in Ques tion 16,
B
B
x R
R
2
1
2 2
2
3 2
=
+ æ
è
ç
ç
ö
ø
÷
÷
/
 
Þ
x R
R
2 2
2
3 2
8
+ æ
è
ç
ç
ö
ø
÷
÷
=
/
Þ
x R
R
2 2
2
4
+
=
Þ x R = 3
21. Com po nent of ve loc ity of par ti cle along
mag netic field, i.e. ,
v
q E
m
t E t
y
= = a
is not constant, hence pitch is variable.
22. r
mv
qB
mK
qB
= =
2
Now, R
mK
e B
=
2
R
m K
e R
R ¢ = =
2 2
3
2
3
( )
( )
99 
R/2 R/2
I I
P Q
R
O
5 cm
P
5 cm
6 cm
3 cm 3 cm
I
r
q
q
23. Same as ques tion 1. In tro duc tory ex er cise
23.6.
Note. Her di a gram is wrong cor rect di a gram 
should be
24. r
mv
qB
mK
qB
mqV
qB
= = =
2 2
 [K qV = ]
Þ r
mV
q B
=
æ
è
ç
ö
ø
÷
2 1
25. Mag netic field due to a con duc tor of fi nite
length.
B
I
r
= × +
m
p
a b
0
4
(sin sin )
Here, a q = -
2
, b q =
1
 and r a =
\ B
I
a
= -
m
q q
0
1 2
2
(sin sin )
26. In case C, mag netic field of con duc tor 1-2
and 2-3 at O is in ward while those of 3-4 and
4-1 at O is out ward, hence net mag netic field
at O in this case is zero.
27. dF dl B
®
=
®
´
®
I ( )
But B dl
® ®
| | at every point,
hence, dF
®
= 0.
28. B B
1 3
0 = = (Mag netic field on the axis of
cur rent car ry ing stra ight con duc t or is zero)
B k
2
0
1
4 2
®
=
-
æ
è
ç
ö
ø
÷
m I
b
^
 = -
m
0
8
I
b
k
^
,
B k
3
0
1
4 2
®
=
æ
è
ç
ö
ø
÷
m I
a
^
 =
m
0
8
I
a
k
^
B B B B B
®
=
®
+
®
+
®
+
®
1 2 3 4
= -
é
ë
ê
ù
û
ú
m
0
8
1 1 I
a b
k
^
29. Cur rent as so ci at ed with elec tron,
I
q
T
e f = =
B
I
R
e f
R
= =
m m
0 0
2 2
30. Same as ques tion 1(a). In tro duc tory Ex er cise 
23.5.
31. At point 1,
Magnetic field due to inner conductor is
non-zero, but due to outer conductor is zero.
Hence, B
1
0 ¹
At point 2,
Magnetic field due to both the conductors is
equal  and opposite.
Hence, B
2
0 =
32. Ap ply Flem ing’s left hand rule or right hand
thumb rule.
33. Mag netic field due to straight con duc tors at 
O is zero be cause O lies on axis of both the
con duc t ors.
Hence, B
I
x
I
x
=
f
× =
f
2 2 4
0 0
p
m m
p
34. In side a solid cyl in der hav ing uni form
cur rent den s ity,
 100
4
2
1
3
O x
b
y
a
I
2 3
4
O
1
I
A
B
x
x C
D
Page 3


14. W MB = - (cos cos ) q q
1 2
Here, q p
1
= , q p q
2
= -
W MB = - - (cos cos( )) p p q
= - - MB( cos ) 1 q
15. B
I
r
P
= ×
m
p
q
0
4
2 ( sin )
     r = - = 5 3 4
2 2
cm
= ´
-
4 10
2
 m
sin q =
3
5
\ B
P
=
´ ´ ´
´
-
-
10 50 2
3
5
4 10
7
2
            = ´
-
1.5 T 10
4
            =1.5 gauss.
16. Mag netic field on the axis of cur rent
car ry ing cir cu la r loop ,
B
M
r x
1
0
2 2 3 2
4
2
= ×
+
m
p ( )
/
…(i)
Magnetic field at the centre of current
carrying circular loop,
B
M
r
2
0
3
4
2
= ×
m
p
…(ii)
From Eqs. (i) and (ii),
       
B
B
r x
r
2
1
2 2 3 2
3
=
+ ( )
/
          =
+ ( )
/
3 4
3
2 2 3 2
3
          =
125
27
Þ    B
2
125
27
54 150 = ´ = mT
17. F l B ba B
®
=
®
´
®
=
®
´
®
I I ( ) ( )
=-
®
´
®
I( ) ab B =
®
´
®
I( ) B ab
18. Ki netic en ergy of elec tron,
K mv e V = =
1
2
2
Þ           v
eV
m
=
2
Magnetic force,
F evB
m
= sin q
F v
m
µ Þ F v
m
µ
Hence, if potential difference is doubled,
force will become 2 times.
19. Mag netic field at O due to P,
B
I
R
I
R
1
0 0
4
2
2
= × =
m
p
m
p /
 (inwards)
Magnetic field at O due to Q,
B
I
R
I
R
2
0 0
4
2
2
= × =
m
p
m
p /
 (inwards)
Net magnetic field at O,
B B B
I
R
= + =
1 2
0
2m
p
20. As solved in Ques tion 16,
B
B
x R
R
2
1
2 2
2
3 2
=
+ æ
è
ç
ç
ö
ø
÷
÷
/
 
Þ
x R
R
2 2
2
3 2
8
+ æ
è
ç
ç
ö
ø
÷
÷
=
/
Þ
x R
R
2 2
2
4
+
=
Þ x R = 3
21. Com po nent of ve loc ity of par ti cle along
mag netic field, i.e. ,
v
q E
m
t E t
y
= = a
is not constant, hence pitch is variable.
22. r
mv
qB
mK
qB
= =
2
Now, R
mK
e B
=
2
R
m K
e R
R ¢ = =
2 2
3
2
3
( )
( )
99 
R/2 R/2
I I
P Q
R
O
5 cm
P
5 cm
6 cm
3 cm 3 cm
I
r
q
q
23. Same as ques tion 1. In tro duc tory ex er cise
23.6.
Note. Her di a gram is wrong cor rect di a gram 
should be
24. r
mv
qB
mK
qB
mqV
qB
= = =
2 2
 [K qV = ]
Þ r
mV
q B
=
æ
è
ç
ö
ø
÷
2 1
25. Mag netic field due to a con duc tor of fi nite
length.
B
I
r
= × +
m
p
a b
0
4
(sin sin )
Here, a q = -
2
, b q =
1
 and r a =
\ B
I
a
= -
m
q q
0
1 2
2
(sin sin )
26. In case C, mag netic field of con duc tor 1-2
and 2-3 at O is in ward while those of 3-4 and
4-1 at O is out ward, hence net mag netic field
at O in this case is zero.
27. dF dl B
®
=
®
´
®
I ( )
But B dl
® ®
| | at every point,
hence, dF
®
= 0.
28. B B
1 3
0 = = (Mag netic field on the axis of
cur rent car ry ing stra ight con duc t or is zero)
B k
2
0
1
4 2
®
=
-
æ
è
ç
ö
ø
÷
m I
b
^
 = -
m
0
8
I
b
k
^
,
B k
3
0
1
4 2
®
=
æ
è
ç
ö
ø
÷
m I
a
^
 =
m
0
8
I
a
k
^
B B B B B
®
=
®
+
®
+
®
+
®
1 2 3 4
= -
é
ë
ê
ù
û
ú
m
0
8
1 1 I
a b
k
^
29. Cur rent as so ci at ed with elec tron,
I
q
T
e f = =
B
I
R
e f
R
= =
m m
0 0
2 2
30. Same as ques tion 1(a). In tro duc tory Ex er cise 
23.5.
31. At point 1,
Magnetic field due to inner conductor is
non-zero, but due to outer conductor is zero.
Hence, B
1
0 ¹
At point 2,
Magnetic field due to both the conductors is
equal  and opposite.
Hence, B
2
0 =
32. Ap ply Flem ing’s left hand rule or right hand
thumb rule.
33. Mag netic field due to straight con duc tors at 
O is zero be cause O lies on axis of both the
con duc t ors.
Hence, B
I
x
I
x
=
f
× =
f
2 2 4
0 0
p
m m
p
34. In side a solid cyl in der hav ing uni form
cur rent den s ity,
 100
4
2
1
3
O x
b
y
a
I
2 3
4
O
1
I
A
B
x
x C
D
B
Ir
R
=
m
p
0
2
2
Here, r R x = -
\ B
I R x
R
=
- m
p
0
2
2
( )
35. Mag netic force is act ing ra di ally out ward on
the loop.
JEE Corner
Assertion and Reason
1. For par a boli c path, ac ce l er a tio n must be
con stant and should not be par al lel or
antiparallel to ve loc ity.
2. By Flem ing’s left hand rule.
3. Magnetic force on up per wire must be in
up ward direction, hence cur rent should be in 
a di rec tion op po site to that of wire 1.
Reason is also correct but does not explain
Assertion.
4. t a = MBsin
a = ° 90
\ t = ¹ MB 0
5. F I lB x
O 2 1
=
F IlB x
4 0 2
=
\ F F
4 2
>   
Hence, net force is along X-axis.
6. Ra dii of both is dif fer ent be cause mass of
both is dif fer ent
r
mv
qB
meV
e B
= =
2
7. For equi lib riu m
F F
e m
®
+
®
= 0
Þ q q E v B
®
= -
®
´
®
( )
Þ      E v B B v
®
= -
®
´
®
=
®
´
®
8. P
m m
=
®
×
®
F v
As F
m
®
 is always perpendicular to v
®
,
P
m
®
= 0
Again, P
e e
=
®
×
®
F v, may or may not be zero.
9. Rea son cor rectly ex plains As ser tion.
10. Mag netic force can not change speed of
par ti c le as it is al way s per p en dic u lar to the
speed of the par ti cle.
11. a
v
R
=
2
but R also depends on v.
\ a
F
m
qvB
m
m
= =
Þ  a v µ
101 
x
y
F
2
F
4
1
2 4
3
x
1
x
2
B
®
R
r
Page 4


14. W MB = - (cos cos ) q q
1 2
Here, q p
1
= , q p q
2
= -
W MB = - - (cos cos( )) p p q
= - - MB( cos ) 1 q
15. B
I
r
P
= ×
m
p
q
0
4
2 ( sin )
     r = - = 5 3 4
2 2
cm
= ´
-
4 10
2
 m
sin q =
3
5
\ B
P
=
´ ´ ´
´
-
-
10 50 2
3
5
4 10
7
2
            = ´
-
1.5 T 10
4
            =1.5 gauss.
16. Mag netic field on the axis of cur rent
car ry ing cir cu la r loop ,
B
M
r x
1
0
2 2 3 2
4
2
= ×
+
m
p ( )
/
…(i)
Magnetic field at the centre of current
carrying circular loop,
B
M
r
2
0
3
4
2
= ×
m
p
…(ii)
From Eqs. (i) and (ii),
       
B
B
r x
r
2
1
2 2 3 2
3
=
+ ( )
/
          =
+ ( )
/
3 4
3
2 2 3 2
3
          =
125
27
Þ    B
2
125
27
54 150 = ´ = mT
17. F l B ba B
®
=
®
´
®
=
®
´
®
I I ( ) ( )
=-
®
´
®
I( ) ab B =
®
´
®
I( ) B ab
18. Ki netic en ergy of elec tron,
K mv e V = =
1
2
2
Þ           v
eV
m
=
2
Magnetic force,
F evB
m
= sin q
F v
m
µ Þ F v
m
µ
Hence, if potential difference is doubled,
force will become 2 times.
19. Mag netic field at O due to P,
B
I
R
I
R
1
0 0
4
2
2
= × =
m
p
m
p /
 (inwards)
Magnetic field at O due to Q,
B
I
R
I
R
2
0 0
4
2
2
= × =
m
p
m
p /
 (inwards)
Net magnetic field at O,
B B B
I
R
= + =
1 2
0
2m
p
20. As solved in Ques tion 16,
B
B
x R
R
2
1
2 2
2
3 2
=
+ æ
è
ç
ç
ö
ø
÷
÷
/
 
Þ
x R
R
2 2
2
3 2
8
+ æ
è
ç
ç
ö
ø
÷
÷
=
/
Þ
x R
R
2 2
2
4
+
=
Þ x R = 3
21. Com po nent of ve loc ity of par ti cle along
mag netic field, i.e. ,
v
q E
m
t E t
y
= = a
is not constant, hence pitch is variable.
22. r
mv
qB
mK
qB
= =
2
Now, R
mK
e B
=
2
R
m K
e R
R ¢ = =
2 2
3
2
3
( )
( )
99 
R/2 R/2
I I
P Q
R
O
5 cm
P
5 cm
6 cm
3 cm 3 cm
I
r
q
q
23. Same as ques tion 1. In tro duc tory ex er cise
23.6.
Note. Her di a gram is wrong cor rect di a gram 
should be
24. r
mv
qB
mK
qB
mqV
qB
= = =
2 2
 [K qV = ]
Þ r
mV
q B
=
æ
è
ç
ö
ø
÷
2 1
25. Mag netic field due to a con duc tor of fi nite
length.
B
I
r
= × +
m
p
a b
0
4
(sin sin )
Here, a q = -
2
, b q =
1
 and r a =
\ B
I
a
= -
m
q q
0
1 2
2
(sin sin )
26. In case C, mag netic field of con duc tor 1-2
and 2-3 at O is in ward while those of 3-4 and
4-1 at O is out ward, hence net mag netic field
at O in this case is zero.
27. dF dl B
®
=
®
´
®
I ( )
But B dl
® ®
| | at every point,
hence, dF
®
= 0.
28. B B
1 3
0 = = (Mag netic field on the axis of
cur rent car ry ing stra ight con duc t or is zero)
B k
2
0
1
4 2
®
=
-
æ
è
ç
ö
ø
÷
m I
b
^
 = -
m
0
8
I
b
k
^
,
B k
3
0
1
4 2
®
=
æ
è
ç
ö
ø
÷
m I
a
^
 =
m
0
8
I
a
k
^
B B B B B
®
=
®
+
®
+
®
+
®
1 2 3 4
= -
é
ë
ê
ù
û
ú
m
0
8
1 1 I
a b
k
^
29. Cur rent as so ci at ed with elec tron,
I
q
T
e f = =
B
I
R
e f
R
= =
m m
0 0
2 2
30. Same as ques tion 1(a). In tro duc tory Ex er cise 
23.5.
31. At point 1,
Magnetic field due to inner conductor is
non-zero, but due to outer conductor is zero.
Hence, B
1
0 ¹
At point 2,
Magnetic field due to both the conductors is
equal  and opposite.
Hence, B
2
0 =
32. Ap ply Flem ing’s left hand rule or right hand
thumb rule.
33. Mag netic field due to straight con duc tors at 
O is zero be cause O lies on axis of both the
con duc t ors.
Hence, B
I
x
I
x
=
f
× =
f
2 2 4
0 0
p
m m
p
34. In side a solid cyl in der hav ing uni form
cur rent den s ity,
 100
4
2
1
3
O x
b
y
a
I
2 3
4
O
1
I
A
B
x
x C
D
B
Ir
R
=
m
p
0
2
2
Here, r R x = -
\ B
I R x
R
=
- m
p
0
2
2
( )
35. Mag netic force is act ing ra di ally out ward on
the loop.
JEE Corner
Assertion and Reason
1. For par a boli c path, ac ce l er a tio n must be
con stant and should not be par al lel or
antiparallel to ve loc ity.
2. By Flem ing’s left hand rule.
3. Magnetic force on up per wire must be in
up ward direction, hence cur rent should be in 
a di rec tion op po site to that of wire 1.
Reason is also correct but does not explain
Assertion.
4. t a = MBsin
a = ° 90
\ t = ¹ MB 0
5. F I lB x
O 2 1
=
F IlB x
4 0 2
=
\ F F
4 2
>   
Hence, net force is along X-axis.
6. Ra dii of both is dif fer ent be cause mass of
both is dif fer ent
r
mv
qB
meV
e B
= =
2
7. For equi lib riu m
F F
e m
®
+
®
= 0
Þ q q E v B
®
= -
®
´
®
( )
Þ      E v B B v
®
= -
®
´
®
=
®
´
®
8. P
m m
=
®
×
®
F v
As F
m
®
 is always perpendicular to v
®
,
P
m
®
= 0
Again, P
e e
=
®
×
®
F v, may or may not be zero.
9. Rea son cor rectly ex plains As ser tion.
10. Mag netic force can not change speed of
par ti c le as it is al way s per p en dic u lar to the
speed of the par ti cle.
11. a
v
R
=
2
but R also depends on v.
\ a
F
m
qvB
m
m
= =
Þ  a v µ
101 
x
y
F
2
F
4
1
2 4
3
x
1
x
2
B
®
R
r
Objective Questions (Level 2)
1. For net torque to be zero.
        IAB mgR
0
=
I
mgR
AB
mgR
R B
= =
0
2
0
p
             =
mg
RB p
0
2. As it is clear from di a gram,
Effective length of wire,
l i
®
= ( )
^
4 m
    F l B
®
=
®
´
®
I ( )
      a
F
l B
®
=
®
=
®
´
®
m
I
m
( )
= ´ - =
2
4
0.1
0.02 1.6 ( ( ))
^ ^ ^
i k j m/s
2
3. Im pulse = Change in mo men tum
I lB dt mv
ò
= - 0
lB dq mv
ò
=
dq
mv
lB
m gh
l B
= =
2
4. Con sider the sphere to be made up of large
num ber of hol low, co ax ial cyl in der of
dif fer ent height and ra dius. Con sider one
such cyl in der of ra dius x, height y and
thickness.
Now, y R =2 cosq, x R = sinq, dx R d = cosq q
Charge on this cylinder,
dq
q
R
yx dx = ×
4
3
2
3
p
p ( )
  = 3
2
q d cos sin q q q
Current associated with this cylinder,
di
dq
T
dq q
d = = =
w
p
w
p
q q q
2
3
2
2
cos sin
Magnetic moment associated with this
cylinder,
dM diA
q
d x = = ´
3
2
2 2
w
p
q q q p cos sin
dM R qA d =
3
2
2 2 3
w q q q cos sin
M dM Rq d = =
ò ò
3
2
2 2
2
0
3
cos sin
p
q q q
= -
ò
3
2
1
2 2
2
0
2
R q d w q q q q
p
cos ( cos )sin
/
      = -
é
ë
ê
ê
ù
û
ú
ú
3
2 3 5
2
3 5
2
0
R q w
q q
p
cos cos
/
      =
1
5
2
R q w
5. As solved in ques tion 5(c). In tro duc tory
Ex er cise 23.2.
L
R
= sin q
Here, L d = , R
m V
q B
=
\
q B d
m V
= s in q
or
q
m
V
Bd
=
sin q
6. Force on por tion AC will more com pared to
that on por tion CB.
 102
q
R
x
y
O
l
(– 4,0) (2,0)
I
T
mg
Page 5


14. W MB = - (cos cos ) q q
1 2
Here, q p
1
= , q p q
2
= -
W MB = - - (cos cos( )) p p q
= - - MB( cos ) 1 q
15. B
I
r
P
= ×
m
p
q
0
4
2 ( sin )
     r = - = 5 3 4
2 2
cm
= ´
-
4 10
2
 m
sin q =
3
5
\ B
P
=
´ ´ ´
´
-
-
10 50 2
3
5
4 10
7
2
            = ´
-
1.5 T 10
4
            =1.5 gauss.
16. Mag netic field on the axis of cur rent
car ry ing cir cu la r loop ,
B
M
r x
1
0
2 2 3 2
4
2
= ×
+
m
p ( )
/
…(i)
Magnetic field at the centre of current
carrying circular loop,
B
M
r
2
0
3
4
2
= ×
m
p
…(ii)
From Eqs. (i) and (ii),
       
B
B
r x
r
2
1
2 2 3 2
3
=
+ ( )
/
          =
+ ( )
/
3 4
3
2 2 3 2
3
          =
125
27
Þ    B
2
125
27
54 150 = ´ = mT
17. F l B ba B
®
=
®
´
®
=
®
´
®
I I ( ) ( )
=-
®
´
®
I( ) ab B =
®
´
®
I( ) B ab
18. Ki netic en ergy of elec tron,
K mv e V = =
1
2
2
Þ           v
eV
m
=
2
Magnetic force,
F evB
m
= sin q
F v
m
µ Þ F v
m
µ
Hence, if potential difference is doubled,
force will become 2 times.
19. Mag netic field at O due to P,
B
I
R
I
R
1
0 0
4
2
2
= × =
m
p
m
p /
 (inwards)
Magnetic field at O due to Q,
B
I
R
I
R
2
0 0
4
2
2
= × =
m
p
m
p /
 (inwards)
Net magnetic field at O,
B B B
I
R
= + =
1 2
0
2m
p
20. As solved in Ques tion 16,
B
B
x R
R
2
1
2 2
2
3 2
=
+ æ
è
ç
ç
ö
ø
÷
÷
/
 
Þ
x R
R
2 2
2
3 2
8
+ æ
è
ç
ç
ö
ø
÷
÷
=
/
Þ
x R
R
2 2
2
4
+
=
Þ x R = 3
21. Com po nent of ve loc ity of par ti cle along
mag netic field, i.e. ,
v
q E
m
t E t
y
= = a
is not constant, hence pitch is variable.
22. r
mv
qB
mK
qB
= =
2
Now, R
mK
e B
=
2
R
m K
e R
R ¢ = =
2 2
3
2
3
( )
( )
99 
R/2 R/2
I I
P Q
R
O
5 cm
P
5 cm
6 cm
3 cm 3 cm
I
r
q
q
23. Same as ques tion 1. In tro duc tory ex er cise
23.6.
Note. Her di a gram is wrong cor rect di a gram 
should be
24. r
mv
qB
mK
qB
mqV
qB
= = =
2 2
 [K qV = ]
Þ r
mV
q B
=
æ
è
ç
ö
ø
÷
2 1
25. Mag netic field due to a con duc tor of fi nite
length.
B
I
r
= × +
m
p
a b
0
4
(sin sin )
Here, a q = -
2
, b q =
1
 and r a =
\ B
I
a
= -
m
q q
0
1 2
2
(sin sin )
26. In case C, mag netic field of con duc tor 1-2
and 2-3 at O is in ward while those of 3-4 and
4-1 at O is out ward, hence net mag netic field
at O in this case is zero.
27. dF dl B
®
=
®
´
®
I ( )
But B dl
® ®
| | at every point,
hence, dF
®
= 0.
28. B B
1 3
0 = = (Mag netic field on the axis of
cur rent car ry ing stra ight con duc t or is zero)
B k
2
0
1
4 2
®
=
-
æ
è
ç
ö
ø
÷
m I
b
^
 = -
m
0
8
I
b
k
^
,
B k
3
0
1
4 2
®
=
æ
è
ç
ö
ø
÷
m I
a
^
 =
m
0
8
I
a
k
^
B B B B B
®
=
®
+
®
+
®
+
®
1 2 3 4
= -
é
ë
ê
ù
û
ú
m
0
8
1 1 I
a b
k
^
29. Cur rent as so ci at ed with elec tron,
I
q
T
e f = =
B
I
R
e f
R
= =
m m
0 0
2 2
30. Same as ques tion 1(a). In tro duc tory Ex er cise 
23.5.
31. At point 1,
Magnetic field due to inner conductor is
non-zero, but due to outer conductor is zero.
Hence, B
1
0 ¹
At point 2,
Magnetic field due to both the conductors is
equal  and opposite.
Hence, B
2
0 =
32. Ap ply Flem ing’s left hand rule or right hand
thumb rule.
33. Mag netic field due to straight con duc tors at 
O is zero be cause O lies on axis of both the
con duc t ors.
Hence, B
I
x
I
x
=
f
× =
f
2 2 4
0 0
p
m m
p
34. In side a solid cyl in der hav ing uni form
cur rent den s ity,
 100
4
2
1
3
O x
b
y
a
I
2 3
4
O
1
I
A
B
x
x C
D
B
Ir
R
=
m
p
0
2
2
Here, r R x = -
\ B
I R x
R
=
- m
p
0
2
2
( )
35. Mag netic force is act ing ra di ally out ward on
the loop.
JEE Corner
Assertion and Reason
1. For par a boli c path, ac ce l er a tio n must be
con stant and should not be par al lel or
antiparallel to ve loc ity.
2. By Flem ing’s left hand rule.
3. Magnetic force on up per wire must be in
up ward direction, hence cur rent should be in 
a di rec tion op po site to that of wire 1.
Reason is also correct but does not explain
Assertion.
4. t a = MBsin
a = ° 90
\ t = ¹ MB 0
5. F I lB x
O 2 1
=
F IlB x
4 0 2
=
\ F F
4 2
>   
Hence, net force is along X-axis.
6. Ra dii of both is dif fer ent be cause mass of
both is dif fer ent
r
mv
qB
meV
e B
= =
2
7. For equi lib riu m
F F
e m
®
+
®
= 0
Þ q q E v B
®
= -
®
´
®
( )
Þ      E v B B v
®
= -
®
´
®
=
®
´
®
8. P
m m
=
®
×
®
F v
As F
m
®
 is always perpendicular to v
®
,
P
m
®
= 0
Again, P
e e
=
®
×
®
F v, may or may not be zero.
9. Rea son cor rectly ex plains As ser tion.
10. Mag netic force can not change speed of
par ti c le as it is al way s per p en dic u lar to the
speed of the par ti cle.
11. a
v
R
=
2
but R also depends on v.
\ a
F
m
qvB
m
m
= =
Þ  a v µ
101 
x
y
F
2
F
4
1
2 4
3
x
1
x
2
B
®
R
r
Objective Questions (Level 2)
1. For net torque to be zero.
        IAB mgR
0
=
I
mgR
AB
mgR
R B
= =
0
2
0
p
             =
mg
RB p
0
2. As it is clear from di a gram,
Effective length of wire,
l i
®
= ( )
^
4 m
    F l B
®
=
®
´
®
I ( )
      a
F
l B
®
=
®
=
®
´
®
m
I
m
( )
= ´ - =
2
4
0.1
0.02 1.6 ( ( ))
^ ^ ^
i k j m/s
2
3. Im pulse = Change in mo men tum
I lB dt mv
ò
= - 0
lB dq mv
ò
=
dq
mv
lB
m gh
l B
= =
2
4. Con sider the sphere to be made up of large
num ber of hol low, co ax ial cyl in der of
dif fer ent height and ra dius. Con sider one
such cyl in der of ra dius x, height y and
thickness.
Now, y R =2 cosq, x R = sinq, dx R d = cosq q
Charge on this cylinder,
dq
q
R
yx dx = ×
4
3
2
3
p
p ( )
  = 3
2
q d cos sin q q q
Current associated with this cylinder,
di
dq
T
dq q
d = = =
w
p
w
p
q q q
2
3
2
2
cos sin
Magnetic moment associated with this
cylinder,
dM diA
q
d x = = ´
3
2
2 2
w
p
q q q p cos sin
dM R qA d =
3
2
2 2 3
w q q q cos sin
M dM Rq d = =
ò ò
3
2
2 2
2
0
3
cos sin
p
q q q
= -
ò
3
2
1
2 2
2
0
2
R q d w q q q q
p
cos ( cos )sin
/
      = -
é
ë
ê
ê
ù
û
ú
ú
3
2 3 5
2
3 5
2
0
R q w
q q
p
cos cos
/
      =
1
5
2
R q w
5. As solved in ques tion 5(c). In tro duc tory
Ex er cise 23.2.
L
R
= sin q
Here, L d = , R
m V
q B
=
\
q B d
m V
= s in q
or
q
m
V
Bd
=
sin q
6. Force on por tion AC will more com pared to
that on por tion CB.
 102
q
R
x
y
O
l
(– 4,0) (2,0)
I
T
mg
7. Con sider an el e men tary por tion of the wire
car ry ing cur rent I
1
 of length dx at a dis tance 
x from end B.
Force on this portion
          dF I dxB =
1
= ×
+
m
p
0 1 2
4
2I I
a x
dx
Total force on wire AB
F dF I I
dx
a x
a
a
= = ×
+
ò ò
m
p
0
1 2
2
4
2
=
m
p
0 1 2
2
3
I I
ln
8. Mag netic field line due to cur rent car ry ing
con duc tor is shown in fig ure.
9.   B
IA
x r
1
0 1
1
2
1
2 3 2
4
2
= ×
+
m
p ( )
/
     = ×
×
+
m
p
p
0 1
2
1
2
1
2 3 2
4
2I r
x r ( )
/
   B
I r
x r
2
0 2
2
2
2
2
2 3 2
4
2
= ×
×
+
m
p
p
( )
/
  
B
B
r x r
r x r
1
2
1
2
2
2
2
2 3 2
2
2
1
2
1
2 3 2
=
+
+
( )
( )
/
/
But,      r x
1 1
= tanq
and       r x
2 2
= tanq
\       
B
B
1
2
2 =
10. b a - must be less than or equal to ra dius of
cir cu la r pat h,
i.e., b a
m v
qB
- £
or v
qB b a
m
³
- ( )
11. Con sider an el e men tary por tion of length d x
at a dis tance x from the piv oted end.
Charge on this portion
dq
q
l
dx =
Current associated with this portion
di
dq
T
qf
l
dx = =
Magnetic moment of this portion
dM x di
qf
l
x dx = = p
p
2 2
    M
qf
l
x dx qfl
l
= =
ò
p
p
0
2 2
1
3
12. At x =0, y= ±2 m
Effective length of wire
l = ( )
^
4 m j
\    F l B j k
m
I
®
=
®
´
®
= ´ ( ) ( )
^ ^
3 4 5
= 60 i
^
N
13. Ef fec tive length of wire,
l ST a = = ´ ´ ° 2
3
4
60 cot
=
a
2
For equilibrium, I lB Mg = 
Þ   I
Mg
lB
=
2
103 
× × × × × × × × × × × × × × × ×
× × × × × × × × × × × × × × × ×
× × × × × × × × × × × × × × × ×
60° T
g
P Q
I
S
R
d
3
4
x
dxe
z
y
x
d
x
I
1
I
2
B
a
2a
x