All Courses
Get the App Signup / Login
Sign in Open App
NEET Exam  >  NEET Notes  >  Physics Class 11  >  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 PDF Download

Download, print and study this document offline
Please wait while the PDF view is loading
 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
? ? ? ?
?? ? ?
Read More
98 videos|388 docs|105 tests
Join Course for Free

Top Courses for NEET

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.
Related Exams
About this Document
2K Views
4.97/5 Rating
Nov 15, 2024 Last updated
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.
Introduction of HC Verma Solutions: Chapter 35 - Magnetic Field due to a Current in English is available as part of our Physics Class 11 for NEET & HC Verma Solutions: Chapter 35 - Magnetic Field due to a Current in Hindi for Physics Class 11 course. Download more important topics related with notes, lectures and mock test series for NEET Exam by signing up for free. NEET: HC Verma Solutions: Chapter 35 - Magnetic Field due to a Current | Physics Class 11 - NEET
Description
Full syllabus notes, lecture & questions for HC Verma Solutions: Chapter 35 - Magnetic Field due to a Current | Physics Class 11 - NEET - NEET | Plus excerises question with solution to help you revise complete syllabus for Physics Class 11 | Best notes, free PDF download
Information about HC Verma Solutions: Chapter 35 - Magnetic Field due to a Current
In this doc you can find the meaning of HC Verma Solutions: Chapter 35 - Magnetic Field due to a Current defined & explained in the simplest way possible. Besides explaining types of HC Verma Solutions: Chapter 35 - Magnetic Field due to a Current theory, EduRev gives you an ample number of questions to practice HC Verma Solutions: Chapter 35 - Magnetic Field due to a Current tests, examples and also practice NEET tests
98 videos|388 docs|105 tests
Join Course for Free
Download as PDF
Explore Courses for NEET exam

Top Courses for NEET

Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

pdf

,

ppt

,

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

,

video lectures

,

shortcuts and tricks

,

Important questions

,

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

,

Viva Questions

,

practice quizzes

,

Objective type Questions

,

past year papers

,

mock tests for examination

,

MCQs

,

Semester Notes

,

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

,

Exam

,

Extra Questions

,

Previous Year Questions with Solutions

,

study material

,

Free

,

Summary

,

Sample Paper

;
Additional Information about HC Verma Solutions: Chapter 35 - Magnetic Field due to a Current for NEET Preparation

HC Verma Solutions: Chapter 35 - Magnetic Field due to a Current Free PDF Download

The HC Verma Solutions: Chapter 35 - Magnetic Field due to a Current is an invaluable resource that delves deep into the core of the NEET exam. These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective. With the help of these notes, you can grasp complex subjects quickly, revise important points easily, and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner, allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material, or toppers' notes, this PDF has got you covered. Download the HC Verma Solutions: Chapter 35 - Magnetic Field due to a Current now and kickstart your journey towards success in the NEET exam.

Importance of HC Verma Solutions: Chapter 35 - Magnetic Field due to a Current

The importance of HC Verma Solutions: Chapter 35 - Magnetic Field due to a Current cannot be overstated, especially for NEET aspirants. This document holds the key to success in the NEET exam. It offers a detailed understanding of the concept, providing invaluable insights into the topic. By knowing the concepts well in advance, students can plan their preparation effectively. Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.

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

HC Verma Solutions: Chapter 35 - Magnetic Field due to a Current Notes offer in-depth insights into the specific topic to help you master it with ease. This comprehensive document covers all aspects related to HC Verma Solutions: Chapter 35 - Magnetic Field due to a Current. It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation. Practice papers and question papers enable you to assess your progress effectively. Additionally, the paper analysis provides valuable tips for tackling the exam strategically. Access to Toppers' notes gives you an edge in understanding complex concepts. Whether you're a beginner or aiming for advanced proficiency, HC Verma Solutions: Chapter 35 - Magnetic Field due to a Current Notes on EduRev are your ultimate resource for success.

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

The "HC Verma Solutions: Chapter 35 - Magnetic Field due to a Current NEET Questions" guide is a valuable resource for all aspiring students preparing for the NEET exam. It focuses on providing a wide range of practice questions to help students gauge their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation. The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level. Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.

Study HC Verma Solutions: Chapter 35 - Magnetic Field due to a Current on the App

Students of NEET can study HC Verma Solutions: Chapter 35 - Magnetic Field due to a Current alongwith tests & analysis from the EduRev app, which will help them while preparing for their exam. Apart from the HC Verma Solutions: Chapter 35 - Magnetic Field due to a Current, students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc. The EduRev App will make your learning easier as you can access it from anywhere you want. The content of HC Verma Solutions: Chapter 35 - Magnetic Field due to a Current is prepared as per the latest NEET syllabus.
© EduRev
Education Revolution
Follow Us
Signup to see your scores go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup