HC Verma Solutions: Chapter 34 - Magnetic Field

# HC Verma Solutions: Chapter 34 - Magnetic Field | Physics Class 11 - NEET PDF Download

``` Page 1

34.1
CHAPTER – 34
MAGNETIC FIELD
1. q = 2 ×1.6 × 10
–19
C, ? = 3 × 10
4
km/s  = 3 × 10
7
m/s
B = 1 T, F = qB ? = 2 × 1.6 × 10
–19
× 3 × 10
7
× 1 = 9.6 10
–12
N. towards west. ?
2. KE = 10 Kev = 1.6 × 10
–15
J, B
?
= 1 × 10
–7
T
(a) The electron will be deflected towards left
(b) (1/2) mv
2
= KE ? V =
m
2 KE ?
F = qVB & accln =
e
m
qVB
Applying s = ut + (1/2) at
2
=
2
2
e
V
x
m
qVB
2
1
? ? =
V m 2
qBx
e
2
=
m
2 KE
m 2
qBx
e
2
?
=
31
15
31
2 7 19
10 1 . 9
2 10 6 . 1
10 1 . 9
1 10 1 10 6 . 1
2
1
?
?
?
? ?
?
? ?
? ?
? ? ? ?
?
By solving we get, s = 0.0148 ˜ 1.5 × 10
–2
cm
3. B = 4 × 10
–3
T  ) K
ˆ
(
F = [4 i
ˆ
+ 3 j
ˆ
× 10
–10
] N. F
X
= 4 × 10
–10
N F
Y
= 3 × 10
–10
N
Q = 1 × 10
–9
C.
Considering the motion along x-axis :–
F
X
= quV
Y
B ? V
Y
=
qB
F
=
3 9
10
10 4 10 1
10 4
? ?
?
? ? ?
?
= 100 m/s
Along y-axis
F
Y
= qV
X
B ? V
X
=
qB
F
=
3 9
10
10 4 10 1
10 3
? ?
?
? ? ?
?
= 75 m/s
Velocity = (–75 i
ˆ
+ 100 j
ˆ
) m/s
4. B
?
= (7.0 i – 3.0 j) × 10
–3
T
a
?
= acceleration = (---i + 7j) × 10
–6
m/s
2
Let the gap be x.
Since B
?
and a
?
are always perpendicular
a B
?
?
? = 0
? (7x × 10
–3
× 10
–6
– 3 × 10
–3
7 × 10
–6
) = 0
? 7x – 21 = 0 ? x = 3
5. m = 10 g = 10 × 10
–3
kg
q = 400 mc = 400 × 10
–6
C
? = 270 m/s, B = 500 ?t = 500 × 10
–6
Tesla
Force on the particle = quB = 4 × 10
–6
× 270 × 500 × 10
–6
= 54 × 10
–8
(K)
Acceleration on the particle = 54 × 10
–6
m/s
2
(K)
Velocity along i
ˆ
and acceleration along k
ˆ
along x-axis the motion is uniform motion and
along y-axis it is accelerated motion.
Along – X axis 100 = 270 × t ? t =
27
10
Along – Z axis s = ut + (1/2) at
2

? s =
2
1
× 54 × 10
–6
×
27
10
×
27
10
= 3.7 × 10
–6
?
B
X
s
? ?
100
a
Page 2

34.1
CHAPTER – 34
MAGNETIC FIELD
1. q = 2 ×1.6 × 10
–19
C, ? = 3 × 10
4
km/s  = 3 × 10
7
m/s
B = 1 T, F = qB ? = 2 × 1.6 × 10
–19
× 3 × 10
7
× 1 = 9.6 10
–12
N. towards west. ?
2. KE = 10 Kev = 1.6 × 10
–15
J, B
?
= 1 × 10
–7
T
(a) The electron will be deflected towards left
(b) (1/2) mv
2
= KE ? V =
m
2 KE ?
F = qVB & accln =
e
m
qVB
Applying s = ut + (1/2) at
2
=
2
2
e
V
x
m
qVB
2
1
? ? =
V m 2
qBx
e
2
=
m
2 KE
m 2
qBx
e
2
?
=
31
15
31
2 7 19
10 1 . 9
2 10 6 . 1
10 1 . 9
1 10 1 10 6 . 1
2
1
?
?
?
? ?
?
? ?
? ?
? ? ? ?
?
By solving we get, s = 0.0148 ˜ 1.5 × 10
–2
cm
3. B = 4 × 10
–3
T  ) K
ˆ
(
F = [4 i
ˆ
+ 3 j
ˆ
× 10
–10
] N. F
X
= 4 × 10
–10
N F
Y
= 3 × 10
–10
N
Q = 1 × 10
–9
C.
Considering the motion along x-axis :–
F
X
= quV
Y
B ? V
Y
=
qB
F
=
3 9
10
10 4 10 1
10 4
? ?
?
? ? ?
?
= 100 m/s
Along y-axis
F
Y
= qV
X
B ? V
X
=
qB
F
=
3 9
10
10 4 10 1
10 3
? ?
?
? ? ?
?
= 75 m/s
Velocity = (–75 i
ˆ
+ 100 j
ˆ
) m/s
4. B
?
= (7.0 i – 3.0 j) × 10
–3
T
a
?
= acceleration = (---i + 7j) × 10
–6
m/s
2
Let the gap be x.
Since B
?
and a
?
are always perpendicular
a B
?
?
? = 0
? (7x × 10
–3
× 10
–6
– 3 × 10
–3
7 × 10
–6
) = 0
? 7x – 21 = 0 ? x = 3
5. m = 10 g = 10 × 10
–3
kg
q = 400 mc = 400 × 10
–6
C
? = 270 m/s, B = 500 ?t = 500 × 10
–6
Tesla
Force on the particle = quB = 4 × 10
–6
× 270 × 500 × 10
–6
= 54 × 10
–8
(K)
Acceleration on the particle = 54 × 10
–6
m/s
2
(K)
Velocity along i
ˆ
and acceleration along k
ˆ
along x-axis the motion is uniform motion and
along y-axis it is accelerated motion.
Along – X axis 100 = 270 × t ? t =
27
10
Along – Z axis s = ut + (1/2) at
2

? s =
2
1
× 54 × 10
–6
×
27
10
×
27
10
= 3.7 × 10
–6
?
B
X
s
? ?
100
a
Magnetic Field
34.2
6. q
P
= e, mp = m, F = q
P
× E
or ma
0
= eE or, E =
e
ma
0
towards west
The acceleration changes from a
0
to 3a
0
Hence net acceleration produced by magnetic field B
?
is 2a
0
.
Force due to magnetic field
=
B
F = m × 2a
0
= e × V
0
× B
? B =
0
0
eV
ma 2
downwards
7. l = 10 cm = 10 × 10
–3
m = 10
–1
m
i = 10 A, B = 0.1 T, ? = 53°
F = iL B Sin ? = 10 × 10
–1
×0.1 × 0.79 = 0.0798 ˜ 0.08
direction of F is along a direction ?r to both l and B.
8. F
?
= ilB = 1 × 0.20 × 0.1 = 0.02 N
For F
?
= il × B
So, For
da & cb ? l × B = l B sin 90° towards left
Hence F
?
0.02 N towards left
For
dc & ab ? F
?
= 0.02 N downward
9. F = ilB Sin ?
= ilB Sin 90°
= i 2RB
= 2 × (8 × 10
–2
) × 1
= 16 × 10
–2
= 0.16 N.
10. Length = l, Current = l i
ˆ
B
?
= B
0 T ) k
ˆ
j
ˆ
i
ˆ
( ? ?
=
T k
ˆ
B j
ˆ
B i
ˆ
B
0 0 0
? ?
F = ??l × B
?
= ??l i
ˆ
×
k
ˆ
B j
ˆ
B i
ˆ
B
0 0 0
? ?
= ??l B
0
i
ˆ
× i
ˆ
+ lB
0
i
ˆ
× j
ˆ
+ lB
0
i
ˆ
× k
ˆ
= ? l B
0K
ˆ
– ? l B
0 j
ˆ
or, F
?
=
2
0
2 2
B l 2 ? = 2 ? l B
0
11. i = 5 A, l = 50 cm = 0.5 m
B = 0.2 T,
F = ilB Sin ? = ilB Sin 90°
= 5 × 0.5 × 0.2
= 0.05 N
( j
ˆ
)
12. l = 2 ?a
Magnetic field = B
?
Current ? ‘i’
F = i l× B
= i × (2 ?a × B
?
)
? = 2 ?ai B perpendicular to the plane of the figure going inside.
a 0
W E
53°
d c
a
–lA
2A
b
N
2R
S
i
? ?
X
l
l =50 cm
0.2 T
5A
x x x x x x
x x x x x x
x x x x x x
x x x x x x
P Q
B
?
a
i
Page 3

34.1
CHAPTER – 34
MAGNETIC FIELD
1. q = 2 ×1.6 × 10
–19
C, ? = 3 × 10
4
km/s  = 3 × 10
7
m/s
B = 1 T, F = qB ? = 2 × 1.6 × 10
–19
× 3 × 10
7
× 1 = 9.6 10
–12
N. towards west. ?
2. KE = 10 Kev = 1.6 × 10
–15
J, B
?
= 1 × 10
–7
T
(a) The electron will be deflected towards left
(b) (1/2) mv
2
= KE ? V =
m
2 KE ?
F = qVB & accln =
e
m
qVB
Applying s = ut + (1/2) at
2
=
2
2
e
V
x
m
qVB
2
1
? ? =
V m 2
qBx
e
2
=
m
2 KE
m 2
qBx
e
2
?
=
31
15
31
2 7 19
10 1 . 9
2 10 6 . 1
10 1 . 9
1 10 1 10 6 . 1
2
1
?
?
?
? ?
?
? ?
? ?
? ? ? ?
?
By solving we get, s = 0.0148 ˜ 1.5 × 10
–2
cm
3. B = 4 × 10
–3
T  ) K
ˆ
(
F = [4 i
ˆ
+ 3 j
ˆ
× 10
–10
] N. F
X
= 4 × 10
–10
N F
Y
= 3 × 10
–10
N
Q = 1 × 10
–9
C.
Considering the motion along x-axis :–
F
X
= quV
Y
B ? V
Y
=
qB
F
=
3 9
10
10 4 10 1
10 4
? ?
?
? ? ?
?
= 100 m/s
Along y-axis
F
Y
= qV
X
B ? V
X
=
qB
F
=
3 9
10
10 4 10 1
10 3
? ?
?
? ? ?
?
= 75 m/s
Velocity = (–75 i
ˆ
+ 100 j
ˆ
) m/s
4. B
?
= (7.0 i – 3.0 j) × 10
–3
T
a
?
= acceleration = (---i + 7j) × 10
–6
m/s
2
Let the gap be x.
Since B
?
and a
?
are always perpendicular
a B
?
?
? = 0
? (7x × 10
–3
× 10
–6
– 3 × 10
–3
7 × 10
–6
) = 0
? 7x – 21 = 0 ? x = 3
5. m = 10 g = 10 × 10
–3
kg
q = 400 mc = 400 × 10
–6
C
? = 270 m/s, B = 500 ?t = 500 × 10
–6
Tesla
Force on the particle = quB = 4 × 10
–6
× 270 × 500 × 10
–6
= 54 × 10
–8
(K)
Acceleration on the particle = 54 × 10
–6
m/s
2
(K)
Velocity along i
ˆ
and acceleration along k
ˆ
along x-axis the motion is uniform motion and
along y-axis it is accelerated motion.
Along – X axis 100 = 270 × t ? t =
27
10
Along – Z axis s = ut + (1/2) at
2

? s =
2
1
× 54 × 10
–6
×
27
10
×
27
10
= 3.7 × 10
–6
?
B
X
s
? ?
100
a
Magnetic Field
34.2
6. q
P
= e, mp = m, F = q
P
× E
or ma
0
= eE or, E =
e
ma
0
towards west
The acceleration changes from a
0
to 3a
0
Hence net acceleration produced by magnetic field B
?
is 2a
0
.
Force due to magnetic field
=
B
F = m × 2a
0
= e × V
0
× B
? B =
0
0
eV
ma 2
downwards
7. l = 10 cm = 10 × 10
–3
m = 10
–1
m
i = 10 A, B = 0.1 T, ? = 53°
F = iL B Sin ? = 10 × 10
–1
×0.1 × 0.79 = 0.0798 ˜ 0.08
direction of F is along a direction ?r to both l and B.
8. F
?
= ilB = 1 × 0.20 × 0.1 = 0.02 N
For F
?
= il × B
So, For
da & cb ? l × B = l B sin 90° towards left
Hence F
?
0.02 N towards left
For
dc & ab ? F
?
= 0.02 N downward
9. F = ilB Sin ?
= ilB Sin 90°
= i 2RB
= 2 × (8 × 10
–2
) × 1
= 16 × 10
–2
= 0.16 N.
10. Length = l, Current = l i
ˆ
B
?
= B
0 T ) k
ˆ
j
ˆ
i
ˆ
( ? ?
=
T k
ˆ
B j
ˆ
B i
ˆ
B
0 0 0
? ?
F = ??l × B
?
= ??l i
ˆ
×
k
ˆ
B j
ˆ
B i
ˆ
B
0 0 0
? ?
= ??l B
0
i
ˆ
× i
ˆ
+ lB
0
i
ˆ
× j
ˆ
+ lB
0
i
ˆ
× k
ˆ
= ? l B
0K
ˆ
– ? l B
0 j
ˆ
or, F
?
=
2
0
2 2
B l 2 ? = 2 ? l B
0
11. i = 5 A, l = 50 cm = 0.5 m
B = 0.2 T,
F = ilB Sin ? = ilB Sin 90°
= 5 × 0.5 × 0.2
= 0.05 N
( j
ˆ
)
12. l = 2 ?a
Magnetic field = B
?
Current ? ‘i’
F = i l× B
= i × (2 ?a × B
?
)
? = 2 ?ai B perpendicular to the plane of the figure going inside.
a 0
W E
53°
d c
a
–lA
2A
b
N
2R
S
i
? ?
X
l
l =50 cm
0.2 T
5A
x x x x x x
x x x x x x
x x x x x x
x x x x x x
P Q
B
?
a
i
Magnetic Field
34.3
13. B
?
= B
0
r
e
r
e
= Unit  vector along radial direction
F = i( B I
? ?
? ) = ilB Sin ?
=
2 2
0
d a
a B ) a 2 ( i
?
?
=
2 2
0
2
d a
B a 2 i
?
?
14. Current anticlockwise
Since the horizontal Forces have no effect.
Let us check the forces for current along AD & BC [Since there is no B
?
]
For BC
F = iaB upward
Current clockwise
Similarly, F = – iaB downwards
Hence change in force = change in tension
= iaB – (–iaB) = 2 iaB
15. F
1
= Force on AD = ilB inwards
F
2
= Force on BC = ilB inwards
They cancel each other
F
3
= Force on CD = ilB inwards
F
4
= Force on AB = ilB inwards
They also cancel each other.
So the net force on the body is 0.
16. For force on a current carrying wire in an uniform magnetic field
We need, l ? length of wire
i ? Current
B ? Magnitude of magnetic field
Since F
?
= ilB
Now, since the length of the wire is fixed from A to B, so force is independent of the shape of the wire.
17. Force on a semicircular wire
= 2iRB
= 2 × 5 × 0.05 × 0.5
= 0.25 N
18. Here the displacement vector dI = ?
So magnetic for i ?t B dl
?
? = i × ?B
19. Force due to the wire AB and force due to wire CD are equal and opposite to each
other. Thus they cancel each other.
Net force is the force due to the semicircular loop = 2iRB
20. Mass = 10 mg = 10
–5
kg
Length = 1 m
? = 2 A, B = ?
Now, Mg = ilB
?B =
iI
mg
=
1 2
8 . 9 10
5
?
?
?
= 4.9 × 10
–5
T ?
21. (a) When switch S is open
2T Cos 30° = mg
? T =
? 30 Cos 2
mg
=
) 2 / 3 ( 2
8 . 9 10 200
3
? ?
?
= 1.13
dl
B
?
a
i
2 2
d a ?
d
? ?
B
?
l
? ?
C
D
B
A
x x x x x x
x x x x x x
x x x x x x
x x x x x x
B
?
a
B
A
D
?
C
? B
?  l
?
?
l
l
l
b
B
a
5 A
5 cm
? B = 0.5 T
D
C
B
X X X
X X X
X X X
X X X
A
R
O
/
20 cm
P Q
T T
Page 4

34.1
CHAPTER – 34
MAGNETIC FIELD
1. q = 2 ×1.6 × 10
–19
C, ? = 3 × 10
4
km/s  = 3 × 10
7
m/s
B = 1 T, F = qB ? = 2 × 1.6 × 10
–19
× 3 × 10
7
× 1 = 9.6 10
–12
N. towards west. ?
2. KE = 10 Kev = 1.6 × 10
–15
J, B
?
= 1 × 10
–7
T
(a) The electron will be deflected towards left
(b) (1/2) mv
2
= KE ? V =
m
2 KE ?
F = qVB & accln =
e
m
qVB
Applying s = ut + (1/2) at
2
=
2
2
e
V
x
m
qVB
2
1
? ? =
V m 2
qBx
e
2
=
m
2 KE
m 2
qBx
e
2
?
=
31
15
31
2 7 19
10 1 . 9
2 10 6 . 1
10 1 . 9
1 10 1 10 6 . 1
2
1
?
?
?
? ?
?
? ?
? ?
? ? ? ?
?
By solving we get, s = 0.0148 ˜ 1.5 × 10
–2
cm
3. B = 4 × 10
–3
T  ) K
ˆ
(
F = [4 i
ˆ
+ 3 j
ˆ
× 10
–10
] N. F
X
= 4 × 10
–10
N F
Y
= 3 × 10
–10
N
Q = 1 × 10
–9
C.
Considering the motion along x-axis :–
F
X
= quV
Y
B ? V
Y
=
qB
F
=
3 9
10
10 4 10 1
10 4
? ?
?
? ? ?
?
= 100 m/s
Along y-axis
F
Y
= qV
X
B ? V
X
=
qB
F
=
3 9
10
10 4 10 1
10 3
? ?
?
? ? ?
?
= 75 m/s
Velocity = (–75 i
ˆ
+ 100 j
ˆ
) m/s
4. B
?
= (7.0 i – 3.0 j) × 10
–3
T
a
?
= acceleration = (---i + 7j) × 10
–6
m/s
2
Let the gap be x.
Since B
?
and a
?
are always perpendicular
a B
?
?
? = 0
? (7x × 10
–3
× 10
–6
– 3 × 10
–3
7 × 10
–6
) = 0
? 7x – 21 = 0 ? x = 3
5. m = 10 g = 10 × 10
–3
kg
q = 400 mc = 400 × 10
–6
C
? = 270 m/s, B = 500 ?t = 500 × 10
–6
Tesla
Force on the particle = quB = 4 × 10
–6
× 270 × 500 × 10
–6
= 54 × 10
–8
(K)
Acceleration on the particle = 54 × 10
–6
m/s
2
(K)
Velocity along i
ˆ
and acceleration along k
ˆ
along x-axis the motion is uniform motion and
along y-axis it is accelerated motion.
Along – X axis 100 = 270 × t ? t =
27
10
Along – Z axis s = ut + (1/2) at
2

? s =
2
1
× 54 × 10
–6
×
27
10
×
27
10
= 3.7 × 10
–6
?
B
X
s
? ?
100
a
Magnetic Field
34.2
6. q
P
= e, mp = m, F = q
P
× E
or ma
0
= eE or, E =
e
ma
0
towards west
The acceleration changes from a
0
to 3a
0
Hence net acceleration produced by magnetic field B
?
is 2a
0
.
Force due to magnetic field
=
B
F = m × 2a
0
= e × V
0
× B
? B =
0
0
eV
ma 2
downwards
7. l = 10 cm = 10 × 10
–3
m = 10
–1
m
i = 10 A, B = 0.1 T, ? = 53°
F = iL B Sin ? = 10 × 10
–1
×0.1 × 0.79 = 0.0798 ˜ 0.08
direction of F is along a direction ?r to both l and B.
8. F
?
= ilB = 1 × 0.20 × 0.1 = 0.02 N
For F
?
= il × B
So, For
da & cb ? l × B = l B sin 90° towards left
Hence F
?
0.02 N towards left
For
dc & ab ? F
?
= 0.02 N downward
9. F = ilB Sin ?
= ilB Sin 90°
= i 2RB
= 2 × (8 × 10
–2
) × 1
= 16 × 10
–2
= 0.16 N.
10. Length = l, Current = l i
ˆ
B
?
= B
0 T ) k
ˆ
j
ˆ
i
ˆ
( ? ?
=
T k
ˆ
B j
ˆ
B i
ˆ
B
0 0 0
? ?
F = ??l × B
?
= ??l i
ˆ
×
k
ˆ
B j
ˆ
B i
ˆ
B
0 0 0
? ?
= ??l B
0
i
ˆ
× i
ˆ
+ lB
0
i
ˆ
× j
ˆ
+ lB
0
i
ˆ
× k
ˆ
= ? l B
0K
ˆ
– ? l B
0 j
ˆ
or, F
?
=
2
0
2 2
B l 2 ? = 2 ? l B
0
11. i = 5 A, l = 50 cm = 0.5 m
B = 0.2 T,
F = ilB Sin ? = ilB Sin 90°
= 5 × 0.5 × 0.2
= 0.05 N
( j
ˆ
)
12. l = 2 ?a
Magnetic field = B
?
Current ? ‘i’
F = i l× B
= i × (2 ?a × B
?
)
? = 2 ?ai B perpendicular to the plane of the figure going inside.
a 0
W E
53°
d c
a
–lA
2A
b
N
2R
S
i
? ?
X
l
l =50 cm
0.2 T
5A
x x x x x x
x x x x x x
x x x x x x
x x x x x x
P Q
B
?
a
i
Magnetic Field
34.3
13. B
?
= B
0
r
e
r
e
= Unit  vector along radial direction
F = i( B I
? ?
? ) = ilB Sin ?
=
2 2
0
d a
a B ) a 2 ( i
?
?
=
2 2
0
2
d a
B a 2 i
?
?
14. Current anticlockwise
Since the horizontal Forces have no effect.
Let us check the forces for current along AD & BC [Since there is no B
?
]
For BC
F = iaB upward
Current clockwise
Similarly, F = – iaB downwards
Hence change in force = change in tension
= iaB – (–iaB) = 2 iaB
15. F
1
= Force on AD = ilB inwards
F
2
= Force on BC = ilB inwards
They cancel each other
F
3
= Force on CD = ilB inwards
F
4
= Force on AB = ilB inwards
They also cancel each other.
So the net force on the body is 0.
16. For force on a current carrying wire in an uniform magnetic field
We need, l ? length of wire
i ? Current
B ? Magnitude of magnetic field
Since F
?
= ilB
Now, since the length of the wire is fixed from A to B, so force is independent of the shape of the wire.
17. Force on a semicircular wire
= 2iRB
= 2 × 5 × 0.05 × 0.5
= 0.25 N
18. Here the displacement vector dI = ?
So magnetic for i ?t B dl
?
? = i × ?B
19. Force due to the wire AB and force due to wire CD are equal and opposite to each
other. Thus they cancel each other.
Net force is the force due to the semicircular loop = 2iRB
20. Mass = 10 mg = 10
–5
kg
Length = 1 m
? = 2 A, B = ?
Now, Mg = ilB
?B =
iI
mg
=
1 2
8 . 9 10
5
?
?
?
= 4.9 × 10
–5
T ?
21. (a) When switch S is open
2T Cos 30° = mg
? T =
? 30 Cos 2
mg
=
) 2 / 3 ( 2
8 . 9 10 200
3
? ?
?
= 1.13
dl
B
?
a
i
2 2
d a ?
d
? ?
B
?
l
? ?
C
D
B
A
x x x x x x
x x x x x x
x x x x x x
x x x x x x
B
?
a
B
A
D
?
C
? B
?  l
?
?
l
l
l
b
B
a
5 A
5 cm
? B = 0.5 T
D
C
B
X X X
X X X
X X X
X X X
A
R
O
/
20 cm
P Q
T T
Magnetic Field
34.4
(b) When the switch is closed and a current passes through the circuit = 2 A
Then
? 2T Cos 30° = mg + ilB
= 200 × 10
–3
9.8 + 2 × 0.2 × 0.5 = 1.96 + 0.2 = 2.16
? 2T =
3
2 16 . 2 ?
= 2.49
? T =
2
49 . 2
= 1.245 ˜ 1.25
22. Let ‘F’ be the force applied due to magnetic field on the wire and ‘x’ be
the dist covered.
So, F × l = ?mg × x
? ibBl = ?mgx
? x =
mg
ibBl
?
?
23. ?R = F
? ? × m × g = ilB
? ? × 10 × 10
–3
× 9.8 =
20
6
× 4.9 × 10
–2
× 0.8
? ? =
2
2
10 2
10 8 . 0 3 . 0
?
?
?
? ?
= 0.12 ?
24. Mass = m
length = l
Current = i
Magnetic field = B = ?
friction Coefficient = ?
iBl = ?mg
? B =
il
mg ?
?
25. (a) F
dl
= i × dl × B towards centre. (By cross product rule)
(b) Let the length of subtends an small angle of 20 at the centre.
Here 2T sin ??= i × dl × B
? 2T ? = i × a × 2 ? × B [As ??? 0, Sin ? ˜ 0]
? T = i × a × B dl = a × 2 ?
Force of compression on the wire = i a B ?
26. Y =
Strain
Stress
=
?
?
?
?
?
?
?
?
?
?
?
?
?
L
dl
r
F
2
? Y
L
dl
=
2
r
F
?
? dl =
Y
L
r
F
2
?
?
=
Y
a 2
r
iaB
2
?
?
?
=
Y r
iB a 2
2
2
?
?
So, dp =
Y r
iB a 2
2
2
?
?
(for small cross sectional circle)
dr =
?
?
?
?
2
1
Y r
iB a 2
2
2
=
Y r
iB a
2
2
?
S
b
l
X
P
X X X X
X X X X X
X X X X X
X X X X X
6 V ?
Q
l
i
T
? ?
? ?
T
Page 5

34.1
CHAPTER – 34
MAGNETIC FIELD
1. q = 2 ×1.6 × 10
–19
C, ? = 3 × 10
4
km/s  = 3 × 10
7
m/s
B = 1 T, F = qB ? = 2 × 1.6 × 10
–19
× 3 × 10
7
× 1 = 9.6 10
–12
N. towards west. ?
2. KE = 10 Kev = 1.6 × 10
–15
J, B
?
= 1 × 10
–7
T
(a) The electron will be deflected towards left
(b) (1/2) mv
2
= KE ? V =
m
2 KE ?
F = qVB & accln =
e
m
qVB
Applying s = ut + (1/2) at
2
=
2
2
e
V
x
m
qVB
2
1
? ? =
V m 2
qBx
e
2
=
m
2 KE
m 2
qBx
e
2
?
=
31
15
31
2 7 19
10 1 . 9
2 10 6 . 1
10 1 . 9
1 10 1 10 6 . 1
2
1
?
?
?
? ?
?
? ?
? ?
? ? ? ?
?
By solving we get, s = 0.0148 ˜ 1.5 × 10
–2
cm
3. B = 4 × 10
–3
T  ) K
ˆ
(
F = [4 i
ˆ
+ 3 j
ˆ
× 10
–10
] N. F
X
= 4 × 10
–10
N F
Y
= 3 × 10
–10
N
Q = 1 × 10
–9
C.
Considering the motion along x-axis :–
F
X
= quV
Y
B ? V
Y
=
qB
F
=
3 9
10
10 4 10 1
10 4
? ?
?
? ? ?
?
= 100 m/s
Along y-axis
F
Y
= qV
X
B ? V
X
=
qB
F
=
3 9
10
10 4 10 1
10 3
? ?
?
? ? ?
?
= 75 m/s
Velocity = (–75 i
ˆ
+ 100 j
ˆ
) m/s
4. B
?
= (7.0 i – 3.0 j) × 10
–3
T
a
?
= acceleration = (---i + 7j) × 10
–6
m/s
2
Let the gap be x.
Since B
?
and a
?
are always perpendicular
a B
?
?
? = 0
? (7x × 10
–3
× 10
–6
– 3 × 10
–3
7 × 10
–6
) = 0
? 7x – 21 = 0 ? x = 3
5. m = 10 g = 10 × 10
–3
kg
q = 400 mc = 400 × 10
–6
C
? = 270 m/s, B = 500 ?t = 500 × 10
–6
Tesla
Force on the particle = quB = 4 × 10
–6
× 270 × 500 × 10
–6
= 54 × 10
–8
(K)
Acceleration on the particle = 54 × 10
–6
m/s
2
(K)
Velocity along i
ˆ
and acceleration along k
ˆ
along x-axis the motion is uniform motion and
along y-axis it is accelerated motion.
Along – X axis 100 = 270 × t ? t =
27
10
Along – Z axis s = ut + (1/2) at
2

? s =
2
1
× 54 × 10
–6
×
27
10
×
27
10
= 3.7 × 10
–6
?
B
X
s
? ?
100
a
Magnetic Field
34.2
6. q
P
= e, mp = m, F = q
P
× E
or ma
0
= eE or, E =
e
ma
0
towards west
The acceleration changes from a
0
to 3a
0
Hence net acceleration produced by magnetic field B
?
is 2a
0
.
Force due to magnetic field
=
B
F = m × 2a
0
= e × V
0
× B
? B =
0
0
eV
ma 2
downwards
7. l = 10 cm = 10 × 10
–3
m = 10
–1
m
i = 10 A, B = 0.1 T, ? = 53°
F = iL B Sin ? = 10 × 10
–1
×0.1 × 0.79 = 0.0798 ˜ 0.08
direction of F is along a direction ?r to both l and B.
8. F
?
= ilB = 1 × 0.20 × 0.1 = 0.02 N
For F
?
= il × B
So, For
da & cb ? l × B = l B sin 90° towards left
Hence F
?
0.02 N towards left
For
dc & ab ? F
?
= 0.02 N downward
9. F = ilB Sin ?
= ilB Sin 90°
= i 2RB
= 2 × (8 × 10
–2
) × 1
= 16 × 10
–2
= 0.16 N.
10. Length = l, Current = l i
ˆ
B
?
= B
0 T ) k
ˆ
j
ˆ
i
ˆ
( ? ?
=
T k
ˆ
B j
ˆ
B i
ˆ
B
0 0 0
? ?
F = ??l × B
?
= ??l i
ˆ
×
k
ˆ
B j
ˆ
B i
ˆ
B
0 0 0
? ?
= ??l B
0
i
ˆ
× i
ˆ
+ lB
0
i
ˆ
× j
ˆ
+ lB
0
i
ˆ
× k
ˆ
= ? l B
0K
ˆ
– ? l B
0 j
ˆ
or, F
?
=
2
0
2 2
B l 2 ? = 2 ? l B
0
11. i = 5 A, l = 50 cm = 0.5 m
B = 0.2 T,
F = ilB Sin ? = ilB Sin 90°
= 5 × 0.5 × 0.2
= 0.05 N
( j
ˆ
)
12. l = 2 ?a
Magnetic field = B
?
Current ? ‘i’
F = i l× B
= i × (2 ?a × B
?
)
? = 2 ?ai B perpendicular to the plane of the figure going inside.
a 0
W E
53°
d c
a
–lA
2A
b
N
2R
S
i
? ?
X
l
l =50 cm
0.2 T
5A
x x x x x x
x x x x x x
x x x x x x
x x x x x x
P Q
B
?
a
i
Magnetic Field
34.3
13. B
?
= B
0
r
e
r
e
= Unit  vector along radial direction
F = i( B I
? ?
? ) = ilB Sin ?
=
2 2
0
d a
a B ) a 2 ( i
?
?
=
2 2
0
2
d a
B a 2 i
?
?
14. Current anticlockwise
Since the horizontal Forces have no effect.
Let us check the forces for current along AD & BC [Since there is no B
?
]
For BC
F = iaB upward
Current clockwise
Similarly, F = – iaB downwards
Hence change in force = change in tension
= iaB – (–iaB) = 2 iaB
15. F
1
= Force on AD = ilB inwards
F
2
= Force on BC = ilB inwards
They cancel each other
F
3
= Force on CD = ilB inwards
F
4
= Force on AB = ilB inwards
They also cancel each other.
So the net force on the body is 0.
16. For force on a current carrying wire in an uniform magnetic field
We need, l ? length of wire
i ? Current
B ? Magnitude of magnetic field
Since F
?
= ilB
Now, since the length of the wire is fixed from A to B, so force is independent of the shape of the wire.
17. Force on a semicircular wire
= 2iRB
= 2 × 5 × 0.05 × 0.5
= 0.25 N
18. Here the displacement vector dI = ?
So magnetic for i ?t B dl
?
? = i × ?B
19. Force due to the wire AB and force due to wire CD are equal and opposite to each
other. Thus they cancel each other.
Net force is the force due to the semicircular loop = 2iRB
20. Mass = 10 mg = 10
–5
kg
Length = 1 m
? = 2 A, B = ?
Now, Mg = ilB
?B =
iI
mg
=
1 2
8 . 9 10
5
?
?
?
= 4.9 × 10
–5
T ?
21. (a) When switch S is open
2T Cos 30° = mg
? T =
? 30 Cos 2
mg
=
) 2 / 3 ( 2
8 . 9 10 200
3
? ?
?
= 1.13
dl
B
?
a
i
2 2
d a ?
d
? ?
B
?
l
? ?
C
D
B
A
x x x x x x
x x x x x x
x x x x x x
x x x x x x
B
?
a
B
A
D
?
C
? B
?  l
?
?
l
l
l
b
B
a
5 A
5 cm
? B = 0.5 T
D
C
B
X X X
X X X
X X X
X X X
A
R
O
/
20 cm
P Q
T T
Magnetic Field
34.4
(b) When the switch is closed and a current passes through the circuit = 2 A
Then
? 2T Cos 30° = mg + ilB
= 200 × 10
–3
9.8 + 2 × 0.2 × 0.5 = 1.96 + 0.2 = 2.16
? 2T =
3
2 16 . 2 ?
= 2.49
? T =
2
49 . 2
= 1.245 ˜ 1.25
22. Let ‘F’ be the force applied due to magnetic field on the wire and ‘x’ be
the dist covered.
So, F × l = ?mg × x
? ibBl = ?mgx
? x =
mg
ibBl
?
?
23. ?R = F
? ? × m × g = ilB
? ? × 10 × 10
–3
× 9.8 =
20
6
× 4.9 × 10
–2
× 0.8
? ? =
2
2
10 2
10 8 . 0 3 . 0
?
?
?
? ?
= 0.12 ?
24. Mass = m
length = l
Current = i
Magnetic field = B = ?
friction Coefficient = ?
iBl = ?mg
? B =
il
mg ?
?
25. (a) F
dl
= i × dl × B towards centre. (By cross product rule)
(b) Let the length of subtends an small angle of 20 at the centre.
Here 2T sin ??= i × dl × B
? 2T ? = i × a × 2 ? × B [As ??? 0, Sin ? ˜ 0]
? T = i × a × B dl = a × 2 ?
Force of compression on the wire = i a B ?
26. Y =
Strain
Stress
=
?
?
?
?
?
?
?
?
?
?
?
?
?
L
dl
r
F
2
? Y
L
dl
=
2
r
F
?
? dl =
Y
L
r
F
2
?
?
=
Y
a 2
r
iaB
2
?
?
?
=
Y r
iB a 2
2
2
?
?
So, dp =
Y r
iB a 2
2
2
?
?
(for small cross sectional circle)
dr =
?
?
?
?
2
1
Y r
iB a 2
2
2
=
Y r
iB a
2
2
?
S
b
l
X
P
X X X X
X X X X X
X X X X X
X X X X X
6 V ?
Q
l
i
T
? ?
? ?
T
Magnetic Field
34.5
27. B
?
= B
0 K
ˆ
l
x
1 ?
?
?
?
?
?
?
f
1
= force on AB = iB
0
[1 + 0]l = iB
0
l
f
2
= force on CD = iB
0
[1 + 0]l = iB
0
l
f
3
= force on AD = iB
0
[1 + 0/1]l = iB
0
l
f
4
= force on AB = iB
0
[1 + 1/1]l = 2iB
0
l
Net horizontal force = F
1
– F
2
= 0
Net vertical force = F
4
– F
3
= iB
0
l
28. (a) Velocity of electron = ?
Magnetic force on electron
F = e ?B
(b) F = qE; F = e ?B
or, qE = e ?B
? eE = e ?B or, E
?
= ?B
(c) E =
dr
dV
=
l
V
? V = lE = l ?B ?
29. (a) i = V
0
nAe
? V
0
=
nae
i
(b) F = ilB =
nA
iBl
=
nA
iB
(upwards)
(c) Let the electric field be E
Ee =
An
iB
? E =
Aen
iB
(d)
dr
dv
= E ? dV = Edr
= E×d = d
Aen
iB
30. q = 2.0 × 10
–8
C B
?
= 0.10 T
m = 2.0 × 10
–10
g = 2 × 10
–13
g
? = 2.0 × 10
3
m/ ?
R =
qB
m ?
=
1 8
3 13
10 10 2
10 2 10 2
? ?
?
? ?
? ? ?
= 0.2 m = 20 cm
T =
qB
m 2 ?
=
1 8
13
10 10 2
10 2 14 . 3 2
? ?
?
? ?
? ? ?
= 6.28 × 10
–4
s
31. r =
qB
mv
0.01 =
e0.1
mv
…(1)
r =
0.1 2e
V m 4
?
?
…(2)
(2) ÷ (1)
?
01 . 0
r
=
mv 1 . 0 e 2
0.1 mVe 4
? ?
?
=
2
4
= 2 ? r = 0.02 m = 2 cm.
32. KE = 100ev = 1.6 × 10
–17
J
(1/2) × 9.1 × 10
–31
× V
2
= 1.6 × 10
–17
J
? V
2
=
31
17
10 1 . 9
2 10 6 . 1
?
?
?
? ?
= 0.35 × 10
14
l
l
C
D
B
A
V l
X X X X X X
X X X X X X
X X X X X X
X X X X X X
X X X X X X
X X X X X X
X X X X X X
```

## Physics Class 11

102 videos|411 docs|121 tests

## FAQs on HC Verma Solutions: Chapter 34 - Magnetic Field - Physics Class 11 - NEET

 1. What is a magnetic field?
Ans. A magnetic field is a region in which a magnetic force is experienced by a magnetic material or a moving electric charge.
 2. How is a magnetic field created?
Ans. A magnetic field is created by moving electric charges or by the intrinsic magnetic moments of elementary particles associated with their spin.
 3. What is the direction of a magnetic field around a current-carrying wire?
Ans. The magnetic field around a current-carrying wire forms concentric circles with the wire as the center, and the direction of the field is given by the right-hand rule, where the thumb points in the direction of the current and the curled fingers give the direction of the magnetic field.
 4. How does the strength of a magnetic field depend on the distance from the source?
Ans. The strength of a magnetic field decreases with increasing distance from the source. The magnetic field strength is inversely proportional to the square of the distance from the source.
 5. What is the unit of magnetic field strength?
Ans. The unit of magnetic field strength is Tesla (T). It can also be expressed in Gauss (G), where 1 T = 10,000 G.

## Physics Class 11

102 videos|411 docs|121 tests

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