NEET  >  HC Verma Solutions: Chapter 35 - Magnetic Field due to a Current

# HC Verma Solutions: Chapter 35 - Magnetic Field due to a Current | Physics Class 11 - NEET

``` Page 1

35.1
CHAPTER – 35
MAGNETIC FIELD DUE TO CURRENT
1. F = B q
?
?
? ? or, B =
? q
F
=
? ?T
F
=
. sec / . sec . A
N
=
m A
N
?
B =
r 2
0
?
? ?
or, ?
0
=
?
?rB 2
=
A m A
N m
? ?
?
=
2
A
N
2. i = 10 A, d = 1 m
B =
r 2
i
0
?
?
=
1 2
10 4 10
7
? ?
? ? ?
?
= 20 × 10
–6
T = 2 ?T
Along +ve Y direction. ?
3. d = 1.6 mm
So, r = 0.8 mm = 0.0008 m
i = 20 A
B
?
=
r 2
i
0
?
?
=
4
7
10 8 2
20 10 4
?
?
? ? ? ?
? ? ?
= 5 × 10
–3
T = 5 mT
4. i = 100 A, d = 8 m
B =
r 2
i
0
?
?
=
8 2
100 10 4
7
? ? ?
? ? ?
?
= 2.5 ?T ?
5. ?
0
= 4 ? × 10
–7
T-m/A
r = 2 cm = 0.02 m, ? = 1 A, B
?
= 1 × 10
–5
T
We know: Magnetic field due to a long straight wire carrying current =
r 2
0
?
? ?
B
?
at P =
02 . 0 2
1 10 4
7
? ?
? ? ?
?
= 1 × 10
–5
T upward
net B = 2 × 1 × 10
–7
T = 20 ?T
B at Q = 1 × 10
–5
T downwards
Hence net B
?
= 0 ?
6. (a) The maximum magnetic field is
r 2
B
0
?
? ?
? which are along the left keeping the sense along the
direction of traveling current.
(b)The minimum
r 2
B
0
?
? ?
?
If r =
B 2
0
?
? ?
B net = 0
r <
B 2
0
?
? ?
B net = 0
r >
B 2
0
?
? ?
B net =
r 2
B
0
?
? ?
?
7. ?
0
= 4 ? × 10
–7
T-m/A, ? = 30 A, B = 4.0 × 10
–4
T Parallel to current.
B
?
due to wore at a pt. 2 cm
=
r 2
0
?
? ?
=
02 . 0 2
30 10 4
7
? ?
? ? ?
?
= 3 × 10
–4
T
net field = ? ? ? ?
2
4
2
4
10 4 10 3
? ?
? ? ? = 5 × 10
–4
T
1 m
X axis
Z axis
r
8 m
100 A
i
?
Q
P
2 cm
2 cm
r
i
r 2
i
0
?
?
30 A
B
?
= 40 × 10
–4
T
– – – – –
– – – – –
– – – – –
– – – – –
Page 2

35.1
CHAPTER – 35
MAGNETIC FIELD DUE TO CURRENT
1. F = B q
?
?
? ? or, B =
? q
F
=
? ?T
F
=
. sec / . sec . A
N
=
m A
N
?
B =
r 2
0
?
? ?
or, ?
0
=
?
?rB 2
=
A m A
N m
? ?
?
=
2
A
N
2. i = 10 A, d = 1 m
B =
r 2
i
0
?
?
=
1 2
10 4 10
7
? ?
? ? ?
?
= 20 × 10
–6
T = 2 ?T
Along +ve Y direction. ?
3. d = 1.6 mm
So, r = 0.8 mm = 0.0008 m
i = 20 A
B
?
=
r 2
i
0
?
?
=
4
7
10 8 2
20 10 4
?
?
? ? ? ?
? ? ?
= 5 × 10
–3
T = 5 mT
4. i = 100 A, d = 8 m
B =
r 2
i
0
?
?
=
8 2
100 10 4
7
? ? ?
? ? ?
?
= 2.5 ?T ?
5. ?
0
= 4 ? × 10
–7
T-m/A
r = 2 cm = 0.02 m, ? = 1 A, B
?
= 1 × 10
–5
T
We know: Magnetic field due to a long straight wire carrying current =
r 2
0
?
? ?
B
?
at P =
02 . 0 2
1 10 4
7
? ?
? ? ?
?
= 1 × 10
–5
T upward
net B = 2 × 1 × 10
–7
T = 20 ?T
B at Q = 1 × 10
–5
T downwards
Hence net B
?
= 0 ?
6. (a) The maximum magnetic field is
r 2
B
0
?
? ?
? which are along the left keeping the sense along the
direction of traveling current.
(b)The minimum
r 2
B
0
?
? ?
?
If r =
B 2
0
?
? ?
B net = 0
r <
B 2
0
?
? ?
B net = 0
r >
B 2
0
?
? ?
B net =
r 2
B
0
?
? ?
?
7. ?
0
= 4 ? × 10
–7
T-m/A, ? = 30 A, B = 4.0 × 10
–4
T Parallel to current.
B
?
due to wore at a pt. 2 cm
=
r 2
0
?
? ?
=
02 . 0 2
30 10 4
7
? ?
? ? ?
?
= 3 × 10
–4
T
net field = ? ? ? ?
2
4
2
4
10 4 10 3
? ?
? ? ? = 5 × 10
–4
T
1 m
X axis
Z axis
r
8 m
100 A
i
?
Q
P
2 cm
2 cm
r
i
r 2
i
0
?
?
30 A
B
?
= 40 × 10
–4
T
– – – – –
– – – – –
– – – – –
– – – – –
Magnetic Field due to Current
35.2
8. i = 10 A.  ( K
ˆ
)
B = 2 × 10
–3
T South to North ( J
ˆ
)
To cancel the magnetic field the point should be choosen so that the net magnetic field is along - J
ˆ
direction.
? The point is along - i
ˆ
direction or along west of the wire.
B =
r 2
0
?
? ?
? 2 × 10
–3
=
r 2
10 10 4
7
? ?
? ? ?
?
? r =
3
7
10 2
10 2
?
?
?
?
= 10
–3
m = 1 mm.
9. Let the tow wires be positioned at O & P
R = OA, =
2 2
) 02 . 0 ( ) 02 . 0 ( ? =
4
10 8
?
? = 2.828 × 10
–2
m
(a) B
?
due to Q, at A
1
=
02 . 0 2
10 10 4
7
? ?
? ? ?
?
= 1 × 10
–4
T ( ?r towards up the line)
B
?
due to P, at A
1
=
06 . 0 2
10 10 4
7
? ?
? ? ?
?
= 0.33 × 10
–4
T ( ?r towards down the line)
net B
?
= 1 × 10
–4
– 0.33 × 10
–4
= 0.67 × 10
–4
T
(b) B
?
due to O at A
2
=
01 . 0
10 10 2
7
? ?
?
= 2 × 10
–4
T ?r down the line
B
?
due to P at A
2
=
03 . 0
10 10 2
7
? ?
?
= 0.67 × 10
–4
T ?r down the line
net  B
?
at A
2
= 2 × 10
–4
+ 0.67 × 10
–4
= 2.67 × 10
–4
T
(c) B
?
at A
3
due to O = 1 × 10
–4
T ?r towards down the line
B
?
at A
3
due to P = 1 × 10
–4
T ?r towards down the line
Net B
?
at A
3
= 2 × 10
–4
T
(d) B
?
at A
4
due to O =
2
7
10 828 . 2
10 10 2
?
?
?
? ?
= 0.7 × 10
–4
T towards SE
B
?
at A
4
due to P = 0.7 × 10
–4
T towards SW
Net B
?
= ? ? ? ?
2
4 -
2
4 -
10 0.7 10 0.7 ? ? ? = 0.989 ×10
–4
˜ 1 × 10
–4
T
10. Cos ? = ½ , ? = 60° & ?AOB = 60°
B =
r 2
0
?
? ?
=
2
7
10 2
10 2 10
?
?
?
? ?
= 10
–4
T
So net is [(10
–4
)
2
+ (10
–4
)
2
+ 2(10
–8
) Cos 60°]
1/2
= 10
–4
[1 + 1 + 2 × ½ ]
1/2
= 10
-4
× 3 T = 1.732 × 10
–4
T
11. (a) B
?
for X = B
?
for Y
Both are oppositely directed hence net B
?
= 0
(b) B
?
due to X = B
?
due to X both directed along Z–axis
Net B
?
=
1
5 2 10 2
7
? ? ?
?
= 2 × 10
–6
T = 2 ?T ?
(c) B
?
due to X = B
?
due to Y both directed opposite to each other.
Hence Net B
?
= 0
(d) B
?
due to X = B
?
due to Y = 1 × 10
–6
T both directed along (–) ve Z–axis
Hence Net B
?
= 2 × 1.0 × 10
–6
= 2 ?T ?
A 1
?
O
?
A 4
A 3 A 2
2 cm
(1, 1)
(–1, 1)
(–1, –1)
(1, –1)
? ?
B A
2 cm
O
2 cm
2 cm
Page 3

35.1
CHAPTER – 35
MAGNETIC FIELD DUE TO CURRENT
1. F = B q
?
?
? ? or, B =
? q
F
=
? ?T
F
=
. sec / . sec . A
N
=
m A
N
?
B =
r 2
0
?
? ?
or, ?
0
=
?
?rB 2
=
A m A
N m
? ?
?
=
2
A
N
2. i = 10 A, d = 1 m
B =
r 2
i
0
?
?
=
1 2
10 4 10
7
? ?
? ? ?
?
= 20 × 10
–6
T = 2 ?T
Along +ve Y direction. ?
3. d = 1.6 mm
So, r = 0.8 mm = 0.0008 m
i = 20 A
B
?
=
r 2
i
0
?
?
=
4
7
10 8 2
20 10 4
?
?
? ? ? ?
? ? ?
= 5 × 10
–3
T = 5 mT
4. i = 100 A, d = 8 m
B =
r 2
i
0
?
?
=
8 2
100 10 4
7
? ? ?
? ? ?
?
= 2.5 ?T ?
5. ?
0
= 4 ? × 10
–7
T-m/A
r = 2 cm = 0.02 m, ? = 1 A, B
?
= 1 × 10
–5
T
We know: Magnetic field due to a long straight wire carrying current =
r 2
0
?
? ?
B
?
at P =
02 . 0 2
1 10 4
7
? ?
? ? ?
?
= 1 × 10
–5
T upward
net B = 2 × 1 × 10
–7
T = 20 ?T
B at Q = 1 × 10
–5
T downwards
Hence net B
?
= 0 ?
6. (a) The maximum magnetic field is
r 2
B
0
?
? ?
? which are along the left keeping the sense along the
direction of traveling current.
(b)The minimum
r 2
B
0
?
? ?
?
If r =
B 2
0
?
? ?
B net = 0
r <
B 2
0
?
? ?
B net = 0
r >
B 2
0
?
? ?
B net =
r 2
B
0
?
? ?
?
7. ?
0
= 4 ? × 10
–7
T-m/A, ? = 30 A, B = 4.0 × 10
–4
T Parallel to current.
B
?
due to wore at a pt. 2 cm
=
r 2
0
?
? ?
=
02 . 0 2
30 10 4
7
? ?
? ? ?
?
= 3 × 10
–4
T
net field = ? ? ? ?
2
4
2
4
10 4 10 3
? ?
? ? ? = 5 × 10
–4
T
1 m
X axis
Z axis
r
8 m
100 A
i
?
Q
P
2 cm
2 cm
r
i
r 2
i
0
?
?
30 A
B
?
= 40 × 10
–4
T
– – – – –
– – – – –
– – – – –
– – – – –
Magnetic Field due to Current
35.2
8. i = 10 A.  ( K
ˆ
)
B = 2 × 10
–3
T South to North ( J
ˆ
)
To cancel the magnetic field the point should be choosen so that the net magnetic field is along - J
ˆ
direction.
? The point is along - i
ˆ
direction or along west of the wire.
B =
r 2
0
?
? ?
? 2 × 10
–3
=
r 2
10 10 4
7
? ?
? ? ?
?
? r =
3
7
10 2
10 2
?
?
?
?
= 10
–3
m = 1 mm.
9. Let the tow wires be positioned at O & P
R = OA, =
2 2
) 02 . 0 ( ) 02 . 0 ( ? =
4
10 8
?
? = 2.828 × 10
–2
m
(a) B
?
due to Q, at A
1
=
02 . 0 2
10 10 4
7
? ?
? ? ?
?
= 1 × 10
–4
T ( ?r towards up the line)
B
?
due to P, at A
1
=
06 . 0 2
10 10 4
7
? ?
? ? ?
?
= 0.33 × 10
–4
T ( ?r towards down the line)
net B
?
= 1 × 10
–4
– 0.33 × 10
–4
= 0.67 × 10
–4
T
(b) B
?
due to O at A
2
=
01 . 0
10 10 2
7
? ?
?
= 2 × 10
–4
T ?r down the line
B
?
due to P at A
2
=
03 . 0
10 10 2
7
? ?
?
= 0.67 × 10
–4
T ?r down the line
net  B
?
at A
2
= 2 × 10
–4
+ 0.67 × 10
–4
= 2.67 × 10
–4
T
(c) B
?
at A
3
due to O = 1 × 10
–4
T ?r towards down the line
B
?
at A
3
due to P = 1 × 10
–4
T ?r towards down the line
Net B
?
at A
3
= 2 × 10
–4
T
(d) B
?
at A
4
due to O =
2
7
10 828 . 2
10 10 2
?
?
?
? ?
= 0.7 × 10
–4
T towards SE
B
?
at A
4
due to P = 0.7 × 10
–4
T towards SW
Net B
?
= ? ? ? ?
2
4 -
2
4 -
10 0.7 10 0.7 ? ? ? = 0.989 ×10
–4
˜ 1 × 10
–4
T
10. Cos ? = ½ , ? = 60° & ?AOB = 60°
B =
r 2
0
?
? ?
=
2
7
10 2
10 2 10
?
?
?
? ?
= 10
–4
T
So net is [(10
–4
)
2
+ (10
–4
)
2
+ 2(10
–8
) Cos 60°]
1/2
= 10
–4
[1 + 1 + 2 × ½ ]
1/2
= 10
-4
× 3 T = 1.732 × 10
–4
T
11. (a) B
?
for X = B
?
for Y
Both are oppositely directed hence net B
?
= 0
(b) B
?
due to X = B
?
due to X both directed along Z–axis
Net B
?
=
1
5 2 10 2
7
? ? ?
?
= 2 × 10
–6
T = 2 ?T ?
(c) B
?
due to X = B
?
due to Y both directed opposite to each other.
Hence Net B
?
= 0
(d) B
?
due to X = B
?
due to Y = 1 × 10
–6
T both directed along (–) ve Z–axis
Hence Net B
?
= 2 × 1.0 × 10
–6
= 2 ?T ?
A 1
?
O
?
A 4
A 3 A 2
2 cm
(1, 1)
(–1, 1)
(–1, –1)
(1, –1)
? ?
B A
2 cm
O
2 cm
2 cm
Magnetic Field due to Current
35.3
12. (a) For each of the wire
Magnitude of magnetic field
= ) 45 Sin 45 Sin (
r 4
i
0
? ? ?
?
?
=
? ?
2
2
2 / 5 4
5
0
? ?
? ?
For AB ? for BC ? For CD ? and for DA ?.
The two ? and 2 ? fields cancel each other. Thus B
net
= 0
(b) At point Q
1
due to (1) B =
2
0
10 5 . 2 2
i
?
? ? ?
?
=
2
7
10 5 2
10 2 5 4
?
?
? ? ?
? ? ? ?
= 4 × 10
–5
?
due to (2) B =
2
0
10 ) 2 / 15 ( 2
i
?
? ? ?
?
=
2
7
10 15 2
10 2 5 4
?
?
? ? ?
? ? ? ?
= (4/3) × 10
–5
?
due to (3) B =
2
0
10 ) 2 / 5 5 ( 2
i
?
? ? ? ?
?
=
2
7
10 15 2
10 2 5 4
?
?
? ? ?
? ? ? ?
= (4/3) × 10
–5
?
due to (4) B =
2
0
10 5 . 2 2
i
?
? ? ?
?
=
2
7
10 5 2
10 2 5 4
?
?
? ? ?
? ? ? ?
= 4 × 10
–5
?
B
net
= [4 + 4 + (4/3) + (4/3)] × 10
–5
=
3
32
× 10
–5
= 10.6 × 10
–5
˜ 1.1 × 10
–4
T
At point Q
2
due to (1)
2
o
10 ) 5 . 2 ( 2
i
?
? ? ?
?
?
due to (2)
2
o
10 ) 2 / 15 ( 2
i
?
? ? ?
?
?
due to (3)
2
o
10 ) 5 . 2 ( 2
i
?
? ? ?
?
?
due to (4)
2
o
10 ) 2 / 15 ( 2
i
?
? ? ?
?
?
B
net
= 0
At point Q
3
due to (1)
2
7
10 ) 2 / 15 ( 2
5 10 4
?
?
? ? ?
? ? ?
= 4/3 × 10
–5
?
due to (2)
2
7
10 ) 2 / 5 ( 2
5 10 4
?
?
? ? ?
? ? ?
= 4 × 10
–5
?
due to (3)
2
7
10 ) 2 / 5 ( 2
5 10 4
?
?
? ? ?
? ? ?
= 4 × 10
–5
?
due to (4)
2
7
10 ) 2 / 15 ( 2
5 10 4
?
?
? ? ?
? ? ?
= 4/3 × 10
–5
?
B
net
= [4 + 4 + (4/3) + (4/3)] × 10
–5
=
3
32
× 10
–5
= 10.6 × 10
–5
˜ 1.1 × 10
–4
T
For Q
4
due to (1) 4/3 × 10
–5
?
due to (2) 4 × 10
–5
?
due to (3) 4/3 × 10
–5
?
due to (4) 4 × 10
–5
?
B
net
= 0
P
D
C
4 3
B
A
5 cm
2 5
2 / 2 5
Q 1 Q 2
Q 3
Q 4
Page 4

35.1
CHAPTER – 35
MAGNETIC FIELD DUE TO CURRENT
1. F = B q
?
?
? ? or, B =
? q
F
=
? ?T
F
=
. sec / . sec . A
N
=
m A
N
?
B =
r 2
0
?
? ?
or, ?
0
=
?
?rB 2
=
A m A
N m
? ?
?
=
2
A
N
2. i = 10 A, d = 1 m
B =
r 2
i
0
?
?
=
1 2
10 4 10
7
? ?
? ? ?
?
= 20 × 10
–6
T = 2 ?T
Along +ve Y direction. ?
3. d = 1.6 mm
So, r = 0.8 mm = 0.0008 m
i = 20 A
B
?
=
r 2
i
0
?
?
=
4
7
10 8 2
20 10 4
?
?
? ? ? ?
? ? ?
= 5 × 10
–3
T = 5 mT
4. i = 100 A, d = 8 m
B =
r 2
i
0
?
?
=
8 2
100 10 4
7
? ? ?
? ? ?
?
= 2.5 ?T ?
5. ?
0
= 4 ? × 10
–7
T-m/A
r = 2 cm = 0.02 m, ? = 1 A, B
?
= 1 × 10
–5
T
We know: Magnetic field due to a long straight wire carrying current =
r 2
0
?
? ?
B
?
at P =
02 . 0 2
1 10 4
7
? ?
? ? ?
?
= 1 × 10
–5
T upward
net B = 2 × 1 × 10
–7
T = 20 ?T
B at Q = 1 × 10
–5
T downwards
Hence net B
?
= 0 ?
6. (a) The maximum magnetic field is
r 2
B
0
?
? ?
? which are along the left keeping the sense along the
direction of traveling current.
(b)The minimum
r 2
B
0
?
? ?
?
If r =
B 2
0
?
? ?
B net = 0
r <
B 2
0
?
? ?
B net = 0
r >
B 2
0
?
? ?
B net =
r 2
B
0
?
? ?
?
7. ?
0
= 4 ? × 10
–7
T-m/A, ? = 30 A, B = 4.0 × 10
–4
T Parallel to current.
B
?
due to wore at a pt. 2 cm
=
r 2
0
?
? ?
=
02 . 0 2
30 10 4
7
? ?
? ? ?
?
= 3 × 10
–4
T
net field = ? ? ? ?
2
4
2
4
10 4 10 3
? ?
? ? ? = 5 × 10
–4
T
1 m
X axis
Z axis
r
8 m
100 A
i
?
Q
P
2 cm
2 cm
r
i
r 2
i
0
?
?
30 A
B
?
= 40 × 10
–4
T
– – – – –
– – – – –
– – – – –
– – – – –
Magnetic Field due to Current
35.2
8. i = 10 A.  ( K
ˆ
)
B = 2 × 10
–3
T South to North ( J
ˆ
)
To cancel the magnetic field the point should be choosen so that the net magnetic field is along - J
ˆ
direction.
? The point is along - i
ˆ
direction or along west of the wire.
B =
r 2
0
?
? ?
? 2 × 10
–3
=
r 2
10 10 4
7
? ?
? ? ?
?
? r =
3
7
10 2
10 2
?
?
?
?
= 10
–3
m = 1 mm.
9. Let the tow wires be positioned at O & P
R = OA, =
2 2
) 02 . 0 ( ) 02 . 0 ( ? =
4
10 8
?
? = 2.828 × 10
–2
m
(a) B
?
due to Q, at A
1
=
02 . 0 2
10 10 4
7
? ?
? ? ?
?
= 1 × 10
–4
T ( ?r towards up the line)
B
?
due to P, at A
1
=
06 . 0 2
10 10 4
7
? ?
? ? ?
?
= 0.33 × 10
–4
T ( ?r towards down the line)
net B
?
= 1 × 10
–4
– 0.33 × 10
–4
= 0.67 × 10
–4
T
(b) B
?
due to O at A
2
=
01 . 0
10 10 2
7
? ?
?
= 2 × 10
–4
T ?r down the line
B
?
due to P at A
2
=
03 . 0
10 10 2
7
? ?
?
= 0.67 × 10
–4
T ?r down the line
net  B
?
at A
2
= 2 × 10
–4
+ 0.67 × 10
–4
= 2.67 × 10
–4
T
(c) B
?
at A
3
due to O = 1 × 10
–4
T ?r towards down the line
B
?
at A
3
due to P = 1 × 10
–4
T ?r towards down the line
Net B
?
at A
3
= 2 × 10
–4
T
(d) B
?
at A
4
due to O =
2
7
10 828 . 2
10 10 2
?
?
?
? ?
= 0.7 × 10
–4
T towards SE
B
?
at A
4
due to P = 0.7 × 10
–4
T towards SW
Net B
?
= ? ? ? ?
2
4 -
2
4 -
10 0.7 10 0.7 ? ? ? = 0.989 ×10
–4
˜ 1 × 10
–4
T
10. Cos ? = ½ , ? = 60° & ?AOB = 60°
B =
r 2
0
?
? ?
=
2
7
10 2
10 2 10
?
?
?
? ?
= 10
–4
T
So net is [(10
–4
)
2
+ (10
–4
)
2
+ 2(10
–8
) Cos 60°]
1/2
= 10
–4
[1 + 1 + 2 × ½ ]
1/2
= 10
-4
× 3 T = 1.732 × 10
–4
T
11. (a) B
?
for X = B
?
for Y
Both are oppositely directed hence net B
?
= 0
(b) B
?
due to X = B
?
due to X both directed along Z–axis
Net B
?
=
1
5 2 10 2
7
? ? ?
?
= 2 × 10
–6
T = 2 ?T ?
(c) B
?
due to X = B
?
due to Y both directed opposite to each other.
Hence Net B
?
= 0
(d) B
?
due to X = B
?
due to Y = 1 × 10
–6
T both directed along (–) ve Z–axis
Hence Net B
?
= 2 × 1.0 × 10
–6
= 2 ?T ?
A 1
?
O
?
A 4
A 3 A 2
2 cm
(1, 1)
(–1, 1)
(–1, –1)
(1, –1)
? ?
B A
2 cm
O
2 cm
2 cm
Magnetic Field due to Current
35.3
12. (a) For each of the wire
Magnitude of magnetic field
= ) 45 Sin 45 Sin (
r 4
i
0
? ? ?
?
?
=
? ?
2
2
2 / 5 4
5
0
? ?
? ?
For AB ? for BC ? For CD ? and for DA ?.
The two ? and 2 ? fields cancel each other. Thus B
net
= 0
(b) At point Q
1
due to (1) B =
2
0
10 5 . 2 2
i
?
? ? ?
?
=
2
7
10 5 2
10 2 5 4
?
?
? ? ?
? ? ? ?
= 4 × 10
–5
?
due to (2) B =
2
0
10 ) 2 / 15 ( 2
i
?
? ? ?
?
=
2
7
10 15 2
10 2 5 4
?
?
? ? ?
? ? ? ?
= (4/3) × 10
–5
?
due to (3) B =
2
0
10 ) 2 / 5 5 ( 2
i
?
? ? ? ?
?
=
2
7
10 15 2
10 2 5 4
?
?
? ? ?
? ? ? ?
= (4/3) × 10
–5
?
due to (4) B =
2
0
10 5 . 2 2
i
?
? ? ?
?
=
2
7
10 5 2
10 2 5 4
?
?
? ? ?
? ? ? ?
= 4 × 10
–5
?
B
net
= [4 + 4 + (4/3) + (4/3)] × 10
–5
=
3
32
× 10
–5
= 10.6 × 10
–5
˜ 1.1 × 10
–4
T
At point Q
2
due to (1)
2
o
10 ) 5 . 2 ( 2
i
?
? ? ?
?
?
due to (2)
2
o
10 ) 2 / 15 ( 2
i
?
? ? ?
?
?
due to (3)
2
o
10 ) 5 . 2 ( 2
i
?
? ? ?
?
?
due to (4)
2
o
10 ) 2 / 15 ( 2
i
?
? ? ?
?
?
B
net
= 0
At point Q
3
due to (1)
2
7
10 ) 2 / 15 ( 2
5 10 4
?
?
? ? ?
? ? ?
= 4/3 × 10
–5
?
due to (2)
2
7
10 ) 2 / 5 ( 2
5 10 4
?
?
? ? ?
? ? ?
= 4 × 10
–5
?
due to (3)
2
7
10 ) 2 / 5 ( 2
5 10 4
?
?
? ? ?
? ? ?
= 4 × 10
–5
?
due to (4)
2
7
10 ) 2 / 15 ( 2
5 10 4
?
?
? ? ?
? ? ?
= 4/3 × 10
–5
?
B
net
= [4 + 4 + (4/3) + (4/3)] × 10
–5
=
3
32
× 10
–5
= 10.6 × 10
–5
˜ 1.1 × 10
–4
T
For Q
4
due to (1) 4/3 × 10
–5
?
due to (2) 4 × 10
–5
?
due to (3) 4/3 × 10
–5
?
due to (4) 4 × 10
–5
?
B
net
= 0
P
D
C
4 3
B
A
5 cm
2 5
2 / 2 5
Q 1 Q 2
Q 3
Q 4
Magnetic Field due to Current
35.4
13. Since all the points lie along a circle with radius = ‘d’
Hence ‘R’ & ‘Q’ both at a distance ‘d’ from the wire.
So, magnetic field B
?
due to are same in magnitude.
As the wires can be treated as semi infinite straight current carrying
conductors. Hence magnetic field B
?
=
d 4
i
0
?
?
At P
B
1
due to 1 is 0
B
2
due to 2 is
d 4
i
0
?
?
At Q
B
1
due to 1 is
d 4
i
0
?
?
B
2
due to 2 is 0
At R
B
1
due to 1 is 0
B
2
due to 2 is
d 4
i
0
?
?
At S
B
1
due to 1 is
d 4
i
0
?
?
B
2
due to 2 is 0
14. B =
d 4
i
0
?
?
2 Sin ?
=
4
x
d 2
x 2
d 4
i
2
2
0
? ?
?
?
?
=
4
x
d d 4
ix
2
2
0
? ?
?
(a)  When d >> x
Neglecting x w.r.t. d
B =
2
0
d d
ix
??
?
=
2
0
d
ix
??
?
? B ?
2
d
1
(b) When x >> d, neglecting d w.r.t. x
B =
2 / dx 4
ix
0
?
?
=
d 4
i 2
0
?
?
? B ?
d
1
15. ? = 10 A, a = 10 cm = 0.1 m
r = OP = 1 . 0
2
3
? m
B = ) Sin Sin (
r 4
2 1
0
? ? ?
?
? ?
=
1 . 0
2
3
1 10 10
7
?
? ?
?
=
732 . 1
10 2
5 ?
?
= 1.154 × 10
–5
T = 11.54 ?T
P
i
1
i
Q
R
S
d
2
d
i
? ?
? ?
x
x/2
O
10 A
P
B
A
10 cm
Q 1
Q 2
30
30
P
Page 5

35.1
CHAPTER – 35
MAGNETIC FIELD DUE TO CURRENT
1. F = B q
?
?
? ? or, B =
? q
F
=
? ?T
F
=
. sec / . sec . A
N
=
m A
N
?
B =
r 2
0
?
? ?
or, ?
0
=
?
?rB 2
=
A m A
N m
? ?
?
=
2
A
N
2. i = 10 A, d = 1 m
B =
r 2
i
0
?
?
=
1 2
10 4 10
7
? ?
? ? ?
?
= 20 × 10
–6
T = 2 ?T
Along +ve Y direction. ?
3. d = 1.6 mm
So, r = 0.8 mm = 0.0008 m
i = 20 A
B
?
=
r 2
i
0
?
?
=
4
7
10 8 2
20 10 4
?
?
? ? ? ?
? ? ?
= 5 × 10
–3
T = 5 mT
4. i = 100 A, d = 8 m
B =
r 2
i
0
?
?
=
8 2
100 10 4
7
? ? ?
? ? ?
?
= 2.5 ?T ?
5. ?
0
= 4 ? × 10
–7
T-m/A
r = 2 cm = 0.02 m, ? = 1 A, B
?
= 1 × 10
–5
T
We know: Magnetic field due to a long straight wire carrying current =
r 2
0
?
? ?
B
?
at P =
02 . 0 2
1 10 4
7
? ?
? ? ?
?
= 1 × 10
–5
T upward
net B = 2 × 1 × 10
–7
T = 20 ?T
B at Q = 1 × 10
–5
T downwards
Hence net B
?
= 0 ?
6. (a) The maximum magnetic field is
r 2
B
0
?
? ?
? which are along the left keeping the sense along the
direction of traveling current.
(b)The minimum
r 2
B
0
?
? ?
?
If r =
B 2
0
?
? ?
B net = 0
r <
B 2
0
?
? ?
B net = 0
r >
B 2
0
?
? ?
B net =
r 2
B
0
?
? ?
?
7. ?
0
= 4 ? × 10
–7
T-m/A, ? = 30 A, B = 4.0 × 10
–4
T Parallel to current.
B
?
due to wore at a pt. 2 cm
=
r 2
0
?
? ?
=
02 . 0 2
30 10 4
7
? ?
? ? ?
?
= 3 × 10
–4
T
net field = ? ? ? ?
2
4
2
4
10 4 10 3
? ?
? ? ? = 5 × 10
–4
T
1 m
X axis
Z axis
r
8 m
100 A
i
?
Q
P
2 cm
2 cm
r
i
r 2
i
0
?
?
30 A
B
?
= 40 × 10
–4
T
– – – – –
– – – – –
– – – – –
– – – – –
Magnetic Field due to Current
35.2
8. i = 10 A.  ( K
ˆ
)
B = 2 × 10
–3
T South to North ( J
ˆ
)
To cancel the magnetic field the point should be choosen so that the net magnetic field is along - J
ˆ
direction.
? The point is along - i
ˆ
direction or along west of the wire.
B =
r 2
0
?
? ?
? 2 × 10
–3
=
r 2
10 10 4
7
? ?
? ? ?
?
? r =
3
7
10 2
10 2
?
?
?
?
= 10
–3
m = 1 mm.
9. Let the tow wires be positioned at O & P
R = OA, =
2 2
) 02 . 0 ( ) 02 . 0 ( ? =
4
10 8
?
? = 2.828 × 10
–2
m
(a) B
?
due to Q, at A
1
=
02 . 0 2
10 10 4
7
? ?
? ? ?
?
= 1 × 10
–4
T ( ?r towards up the line)
B
?
due to P, at A
1
=
06 . 0 2
10 10 4
7
? ?
? ? ?
?
= 0.33 × 10
–4
T ( ?r towards down the line)
net B
?
= 1 × 10
–4
– 0.33 × 10
–4
= 0.67 × 10
–4
T
(b) B
?
due to O at A
2
=
01 . 0
10 10 2
7
? ?
?
= 2 × 10
–4
T ?r down the line
B
?
due to P at A
2
=
03 . 0
10 10 2
7
? ?
?
= 0.67 × 10
–4
T ?r down the line
net  B
?
at A
2
= 2 × 10
–4
+ 0.67 × 10
–4
= 2.67 × 10
–4
T
(c) B
?
at A
3
due to O = 1 × 10
–4
T ?r towards down the line
B
?
at A
3
due to P = 1 × 10
–4
T ?r towards down the line
Net B
?
at A
3
= 2 × 10
–4
T
(d) B
?
at A
4
due to O =
2
7
10 828 . 2
10 10 2
?
?
?
? ?
= 0.7 × 10
–4
T towards SE
B
?
at A
4
due to P = 0.7 × 10
–4
T towards SW
Net B
?
= ? ? ? ?
2
4 -
2
4 -
10 0.7 10 0.7 ? ? ? = 0.989 ×10
–4
˜ 1 × 10
–4
T
10. Cos ? = ½ , ? = 60° & ?AOB = 60°
B =
r 2
0
?
? ?
=
2
7
10 2
10 2 10
?
?
?
? ?
= 10
–4
T
So net is [(10
–4
)
2
+ (10
–4
)
2
+ 2(10
–8
) Cos 60°]
1/2
= 10
–4
[1 + 1 + 2 × ½ ]
1/2
= 10
-4
× 3 T = 1.732 × 10
–4
T
11. (a) B
?
for X = B
?
for Y
Both are oppositely directed hence net B
?
= 0
(b) B
?
due to X = B
?
due to X both directed along Z–axis
Net B
?
=
1
5 2 10 2
7
? ? ?
?
= 2 × 10
–6
T = 2 ?T ?
(c) B
?
due to X = B
?
due to Y both directed opposite to each other.
Hence Net B
?
= 0
(d) B
?
due to X = B
?
due to Y = 1 × 10
–6
T both directed along (–) ve Z–axis
Hence Net B
?
= 2 × 1.0 × 10
–6
= 2 ?T ?
A 1
?
O
?
A 4
A 3 A 2
2 cm
(1, 1)
(–1, 1)
(–1, –1)
(1, –1)
? ?
B A
2 cm
O
2 cm
2 cm
Magnetic Field due to Current
35.3
12. (a) For each of the wire
Magnitude of magnetic field
= ) 45 Sin 45 Sin (
r 4
i
0
? ? ?
?
?
=
? ?
2
2
2 / 5 4
5
0
? ?
? ?
For AB ? for BC ? For CD ? and for DA ?.
The two ? and 2 ? fields cancel each other. Thus B
net
= 0
(b) At point Q
1
due to (1) B =
2
0
10 5 . 2 2
i
?
? ? ?
?
=
2
7
10 5 2
10 2 5 4
?
?
? ? ?
? ? ? ?
= 4 × 10
–5
?
due to (2) B =
2
0
10 ) 2 / 15 ( 2
i
?
? ? ?
?
=
2
7
10 15 2
10 2 5 4
?
?
? ? ?
? ? ? ?
= (4/3) × 10
–5
?
due to (3) B =
2
0
10 ) 2 / 5 5 ( 2
i
?
? ? ? ?
?
=
2
7
10 15 2
10 2 5 4
?
?
? ? ?
? ? ? ?
= (4/3) × 10
–5
?
due to (4) B =
2
0
10 5 . 2 2
i
?
? ? ?
?
=
2
7
10 5 2
10 2 5 4
?
?
? ? ?
? ? ? ?
= 4 × 10
–5
?
B
net
= [4 + 4 + (4/3) + (4/3)] × 10
–5
=
3
32
× 10
–5
= 10.6 × 10
–5
˜ 1.1 × 10
–4
T
At point Q
2
due to (1)
2
o
10 ) 5 . 2 ( 2
i
?
? ? ?
?
?
due to (2)
2
o
10 ) 2 / 15 ( 2
i
?
? ? ?
?
?
due to (3)
2
o
10 ) 5 . 2 ( 2
i
?
? ? ?
?
?
due to (4)
2
o
10 ) 2 / 15 ( 2
i
?
? ? ?
?
?
B
net
= 0
At point Q
3
due to (1)
2
7
10 ) 2 / 15 ( 2
5 10 4
?
?
? ? ?
? ? ?
= 4/3 × 10
–5
?
due to (2)
2
7
10 ) 2 / 5 ( 2
5 10 4
?
?
? ? ?
? ? ?
= 4 × 10
–5
?
due to (3)
2
7
10 ) 2 / 5 ( 2
5 10 4
?
?
? ? ?
? ? ?
= 4 × 10
–5
?
due to (4)
2
7
10 ) 2 / 15 ( 2
5 10 4
?
?
? ? ?
? ? ?
= 4/3 × 10
–5
?
B
net
= [4 + 4 + (4/3) + (4/3)] × 10
–5
=
3
32
× 10
–5
= 10.6 × 10
–5
˜ 1.1 × 10
–4
T
For Q
4
due to (1) 4/3 × 10
–5
?
due to (2) 4 × 10
–5
?
due to (3) 4/3 × 10
–5
?
due to (4) 4 × 10
–5
?
B
net
= 0
P
D
C
4 3
B
A
5 cm
2 5
2 / 2 5
Q 1 Q 2
Q 3
Q 4
Magnetic Field due to Current
35.4
13. Since all the points lie along a circle with radius = ‘d’
Hence ‘R’ & ‘Q’ both at a distance ‘d’ from the wire.
So, magnetic field B
?
due to are same in magnitude.
As the wires can be treated as semi infinite straight current carrying
conductors. Hence magnetic field B
?
=
d 4
i
0
?
?
At P
B
1
due to 1 is 0
B
2
due to 2 is
d 4
i
0
?
?
At Q
B
1
due to 1 is
d 4
i
0
?
?
B
2
due to 2 is 0
At R
B
1
due to 1 is 0
B
2
due to 2 is
d 4
i
0
?
?
At S
B
1
due to 1 is
d 4
i
0
?
?
B
2
due to 2 is 0
14. B =
d 4
i
0
?
?
2 Sin ?
=
4
x
d 2
x 2
d 4
i
2
2
0
? ?
?
?
?
=
4
x
d d 4
ix
2
2
0
? ?
?
(a)  When d >> x
Neglecting x w.r.t. d
B =
2
0
d d
ix
??
?
=
2
0
d
ix
??
?
? B ?
2
d
1
(b) When x >> d, neglecting d w.r.t. x
B =
2 / dx 4
ix
0
?
?
=
d 4
i 2
0
?
?
? B ?
d
1
15. ? = 10 A, a = 10 cm = 0.1 m
r = OP = 1 . 0
2
3
? m
B = ) Sin Sin (
r 4
2 1
0
? ? ?
?
? ?
=
1 . 0
2
3
1 10 10
7
?
? ?
?
=
732 . 1
10 2
5 ?
?
= 1.154 × 10
–5
T = 11.54 ?T
P
i
1
i
Q
R
S
d
2
d
i
? ?
? ?
x
x/2
O
10 A
P
B
A
10 cm
Q 1
Q 2
30
30
P
Magnetic Field due to Current
35.5
16. B
1
=
d 2
i
0
?
?
, B
2
= ) Sin 2 (
d 4
i
0
? ?
?
?
=
4
d 2
2
d 4
i
2
2
0
?
?
?
?
?
?
=
4
d d 4
i
2
2
0
?
?
? ?
?
B
1
– B
2
=
100
1
B
2
?
4
d d 4
i
d 2
i
2
2
0 0
?
?
? ?
?
?
?
?
=
d 200
i
0
?
?
?
4
d d 4
i
2
2
0
?
?
? ?
?
= ?
?
?
?
?
?
?
?
?
200
1
2
1
d
i
0
?
4
d 4
2
2
?
?
?
=
200
99
?
4
d
2
2
2
?
?
?
=
2
200
4 99
?
?
?
?
?
? ?
=
40000
156816
= 3.92
? l
2
= 3.92 d
2
+
2
4
92 . 3
?
2
4
92 . 3 1
? ?
?
?
?
?
? ?
= 3.92 d
2
? 0.02 l
2
= 3.92 d
2
?
2
2
d
?
=
92 . 3
02 . 0
=
?
d
=
92 . 3
02 . 0
= 0.07
17. As resistances vary as r & 2r
Hence Current along ABC =
3
i
i 3
2
Now,
B
?
?
?
?
?
?
?
?
?
?
? ? ? ?
a 3 4
2 2 2 i
2
0
=
a 3
i 2 2
0
?
?
B
?
due to ABC =
?
?
?
?
?
?
?
?
?
? ? ?
a 3 4
2 2 i
2
0
=
a 6
i 2 2
0
?
?
Now B
?
=
a 3
i 2 2
0
?
?
–
a 6
i 2 2
0
?
?
=
a 3
i 2
0
?
?
?
18. A
0
=
4
a
16
a
2 2
? =
16
a 5
2
=
4
5 a
D
0
=
2 2
2
a
4
a 3
?
?
?
?
?
?
? ?
?
?
?
?
?
=
4
a
16
a 9
2 2
? =
16
a 13
2
=
4
13 a
Magnetic field due to AB
B
AB
=
? ? 4 / a 2
i
4
0
?
?
?
(Sin (90 – i) + Sin (90 – ?))
= ?
?
? ?
Cos 2
a 4
i 2
0
=
? ?
) 4 / 5 ( a
2 / a
2
a 4
i 2
0
? ?
?
? ?
=
5
i 2
0
?
?
Magnetic field due to DC
B
DC
=
? ? 4 / a 3 2
i
4
0
?
?
?
2Sin (90° – B)
= ?
? ?
? ? ?
Cos
a 3 4
2 4 i
0
=
? ?
) 4 / a 13 (
2 / a
a 3
i
0
?
? ?
?
=
13 3 a
i 2
0
?
?
The magnetic field due to AD & BC are equal and appropriate hence cancle each other.
Hence, net magnetic field is
5
i 2
0
?
?
–
13 3 a
i 2
0
?
?
=
?
?
?
?
?
?
?
?
?
13 3
1
5
1
a
i 2
0
d
i
? ?
l
C
D
B
A
i/3
2i/3a
a/2
2
a
i
D
C
B A
i
i
3a/4
a/4
O
a
a/2 a/2
? ? ? ?
?? ? ?
```

## Physics Class 11

124 videos|464 docs|210 tests

## Physics Class 11

124 videos|464 docs|210 tests

### How to Prepare for NEET

Read our guide to prepare for NEET which is created by Toppers & the best Teachers

Track your progress, build streaks, highlight & save important lessons and more! (Scan QR code)

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;