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 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
& along ADC = 
i 3
2
Now, 
B
?
due to ADC = 
?
?
?
?
?
?
?
?
?
? ? ? ?
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
? ? ? ?
?? ? ?
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FAQs on HC Verma Solutions: Chapter 35 - Magnetic Field due to a Current - Physics Class 11 - NEET

1. What is the formula to calculate the magnetic field due to a current?
Ans. The formula to calculate the magnetic field due to a current is given by the Biot-Savart law, which states that the magnetic field at a point due to a small element of current is directly proportional to the current, the length of the element, the sine of the angle between the element and the line joining the element to the point, and inversely proportional to the square of the distance between the element and the point.
2. How does the direction of the magnetic field due to a current change with respect to the direction of the current?
Ans. The direction of the magnetic field due to a current can be determined using the right-hand rule. If the current flows in the direction of the thumb of the right hand, then the curling of the fingers gives the direction of the magnetic field lines around the current-carrying wire. The magnetic field forms concentric circles around the wire.
3. What is the difference between the magnetic field due to a current in a straight wire and a circular loop?
Ans. The magnetic field due to a current in a straight wire forms concentric circles around the wire, with the magnetic field lines being parallel to each other. On the other hand, the magnetic field due to a current in a circular loop forms a pattern similar to that of a bar magnet, with the magnetic field lines emerging from one end of the loop and converging towards the other end.
4. How does the strength of the magnetic field due to a current change with the distance from the wire?
Ans. The strength of the magnetic field due to a current decreases with increasing distance from the wire. According to the inverse square law, the magnetic field strength is inversely proportional to the square of the distance from the wire. This means that as the distance from the wire doubles, the magnetic field strength decreases by a factor of four.
5. Can the magnetic field due to a current be zero at any point?
Ans. No, the magnetic field due to a current cannot be zero at any point. According to the Biot-Savart law, the magnetic field at a point due to a current-carrying wire is non-zero for all points in space. However, the magnitude of the magnetic field can be very small at points far away from the wire, approaching zero as the distance increases.
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Document Description: HC Verma Solutions: Chapter 35 - Magnetic Field due to a Current for NEET 2024 is part of Physics Class 11 preparation. The notes and questions for HC Verma Solutions: Chapter 35 - Magnetic Field due to a Current have been prepared according to the NEET exam syllabus. Information about HC Verma Solutions: Chapter 35 - Magnetic Field due to a Current covers topics like and HC Verma Solutions: Chapter 35 - Magnetic Field due to a Current Example, for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for HC Verma Solutions: Chapter 35 - Magnetic Field due to a Current.
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