HC Verma Solutions: Chapter 37 - Magnetic Properties of Matter | Physics Class 11 - NEET PDF Download

 Page 1


37.1
CHAPTER – 37
MAGNETIC PROPERTIES OF MATTER
1. B = ?
0
ni, H = 
0
B
?
? H = ni
? 1500 A/m = n× 2
? n = 750 turns/meter
? n = 7.5 turns/cm
2. (a) H = 1500 A/m
As the solenoid and the rod are long and we are interested in the magnetic intensity at the centre, 
the end effects may be neglected. There is no effect of the rod on the magnetic intensity at the 
centre.
(b) ? = 0.12 A/m
We know ?
?
= H X
?
  X = Susceptibility
? X = 
H
?
= 
1500
12 . 0
= 0.00008 = 8 × 10
–5
(c)  The material is paramagnetic 
3. B
1
= 2.5 × 10
–3
, B
2
= 2.5
A = 4 × 10
–4
m
2
, n = 50 turns/cm = 5000 turns/m
(a) B = ?
0
ni, 
? 2.5 × 10
–3
= 4 ? × 10
–7
× 5000 × i
? i = 
5000 10 4
10 5 . 2
7
3
? ? ?
?
?
?
= 0.398 A ˜ 0.4 A ?
(b) ? = H
B
0
2
?
?
= ) B B (
10 4
5 . 2
1 2
7
? ?
? ?
?
= 497 . 2
10 4
5 . 2
7
?
? ?
?
= 1.99 × 10
6
˜ 2 × 10
6
(c) ? = 
V
M
? ? = 
?
?
A
m
= 
A
m
? m = ?A = 2 × 10
6
× 4 × 10
–4
= 800 A-m ?
4. (a) Given d = 15 cm = 0.15 m
l = 1 cm = 0.01 m
A = 1.0 cm
2
= 1 × 10
–4
m
2
B = 1.5 × 10
–4
T
M = ?
We Know B
?
= 
2 2 2
0
) d (
Md 2
4
? ?
?
?
?
? 1.5 × 10
–4
= 
2
7
) 0001 . 0 0225 . 0 (
15 . 0 M 2 10
?
? ? ?
?
  = 
4
8
10 01 . 5
M 10 3
?
?
?
?
? M = 
8
4 4
10 3
10 01 . 5 10 5 . 1
?
? ?
?
? ? ?
= 2.5 A
(b) Magnetisation ? = 
V
M
= 
2 4
10 10
5 . 2
? ?
?
= 2.5 × 10
6
A/m
(c) H = 
2
d 4
m
?
= 
2
d 4
M
? ?
= 
2
) 15 . 0 ( 01 . 0 14 . 3 4
5 . 2
? ? ?
net H = H
N
+ H
?
= 2 × 884.6 = 8.846 × 10
2
B
?
= ?
0
(–H + ?) = 4 ? × 10
–7
(2.5×10
6
– 2 × 884.6) ˜ 3,14 T
?
Page 2


37.1
CHAPTER – 37
MAGNETIC PROPERTIES OF MATTER
1. B = ?
0
ni, H = 
0
B
?
? H = ni
? 1500 A/m = n× 2
? n = 750 turns/meter
? n = 7.5 turns/cm
2. (a) H = 1500 A/m
As the solenoid and the rod are long and we are interested in the magnetic intensity at the centre, 
the end effects may be neglected. There is no effect of the rod on the magnetic intensity at the 
centre.
(b) ? = 0.12 A/m
We know ?
?
= H X
?
  X = Susceptibility
? X = 
H
?
= 
1500
12 . 0
= 0.00008 = 8 × 10
–5
(c)  The material is paramagnetic 
3. B
1
= 2.5 × 10
–3
, B
2
= 2.5
A = 4 × 10
–4
m
2
, n = 50 turns/cm = 5000 turns/m
(a) B = ?
0
ni, 
? 2.5 × 10
–3
= 4 ? × 10
–7
× 5000 × i
? i = 
5000 10 4
10 5 . 2
7
3
? ? ?
?
?
?
= 0.398 A ˜ 0.4 A ?
(b) ? = H
B
0
2
?
?
= ) B B (
10 4
5 . 2
1 2
7
? ?
? ?
?
= 497 . 2
10 4
5 . 2
7
?
? ?
?
= 1.99 × 10
6
˜ 2 × 10
6
(c) ? = 
V
M
? ? = 
?
?
A
m
= 
A
m
? m = ?A = 2 × 10
6
× 4 × 10
–4
= 800 A-m ?
4. (a) Given d = 15 cm = 0.15 m
l = 1 cm = 0.01 m
A = 1.0 cm
2
= 1 × 10
–4
m
2
B = 1.5 × 10
–4
T
M = ?
We Know B
?
= 
2 2 2
0
) d (
Md 2
4
? ?
?
?
?
? 1.5 × 10
–4
= 
2
7
) 0001 . 0 0225 . 0 (
15 . 0 M 2 10
?
? ? ?
?
  = 
4
8
10 01 . 5
M 10 3
?
?
?
?
? M = 
8
4 4
10 3
10 01 . 5 10 5 . 1
?
? ?
?
? ? ?
= 2.5 A
(b) Magnetisation ? = 
V
M
= 
2 4
10 10
5 . 2
? ?
?
= 2.5 × 10
6
A/m
(c) H = 
2
d 4
m
?
= 
2
d 4
M
? ?
= 
2
) 15 . 0 ( 01 . 0 14 . 3 4
5 . 2
? ? ?
net H = H
N
+ H
?
= 2 × 884.6 = 8.846 × 10
2
B
?
= ?
0
(–H + ?) = 4 ? × 10
–7
(2.5×10
6
– 2 × 884.6) ˜ 3,14 T
?
Magnetic Properties of Matter
37.2
5. Permiability ( ?) = ?
0
(1 + x)
Given susceptibility = 5500
? = 4 × 10
–7
(1 + 5500) 
= 4 × 3.14 × 10
–7
× 5501 6909.56 × 10
–7
˜ 6.9 × 10
–3
6. B = 1.6 T, H = 1000 A/m
? = Permeability of material
? = 
H
B
= 
1000
6 . 1
= 1.6 × 10
–3
?r = 
0
?
?
= 
7
3
10 4
10 6 . 1
?
?
? ?
?
= 0.127 × 10
4
˜ 1.3 × 10
3
? = ?
0
(1 + x)
? x = 1
0
?
?
?
= ?
r
– 1 = 1.3 × 10
3
– 1 = 1300 – 1 = 1299 ˜ 1.3 × 10
3
  ?
7. x = 
T
C
= ?
2
1
x
x
= 
1
2
T
T
?
5
5
10 8 . 1
10 2 . 1
?
?
?
?
= 
300
T
2
  
? T
2
= 300
18
12
? = 200 K.
8. ? = 8.52 × 10
28
atoms/m
3
For maximum ‘ ?’, Let us consider the no. of atoms present in 1 m
3
of volume. 
Given: m per atom = 2 × 9.27 × 10
–24
A–m
2
? = 
V
m net
= 2 × 9.27 × 10
–24
× 8.52 × 10
28
˜ 1.58 × 10
6
A/m
B = ?
0
(H + ?) = ?
0
? [ ? H = 0 in this case]
= 4 ? × 10
–7
× 1.58 × 10
6
= 1.98 × 10
–1
˜ 2.0 T ?
9. B = ?
0
ni, H = 
0
B
?
Given n = 40 turn/cm = 4000 turns/m
? H = ni
H = 4 × 10
4
A/m
? i = 
n
H
= 
4000
10 4
4
?
= 10 A.
? ? ? ? ?