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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.
? ? ? ? ?
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FAQs on HC Verma Solutions: Chapter 37 - Magnetic Properties of Matter - Physics Class 11 - NEET

1. What are magnetic properties of matter?
Ans. Magnetic properties of matter refer to the ability of a substance to interact with magnetic fields. This includes the ability to be attracted or repelled by magnets, as well as the ability to generate a magnetic field of its own.
2. How are magnetic properties of matter measured?
Ans. Magnetic properties of matter are measured using various techniques. One common method is to use a magnetometer, which can measure the magnetic flux density or magnetic field strength of a substance. Other techniques include magnetic susceptibility measurements and hysteresis loop measurements.
3. What is paramagnetism?
Ans. Paramagnetism is a type of magnetic property exhibited by certain substances. In paramagnetic materials, the individual atoms or molecules have magnetic moments that align with an external magnetic field, causing the material to be weakly attracted to the magnet. Examples of paramagnetic materials include aluminum, oxygen, and platinum.
4. What is diamagnetism?
Ans. Diamagnetism is another type of magnetic property exhibited by certain substances. In diamagnetic materials, the individual atoms or molecules have magnetic moments that align opposite to an external magnetic field, causing the material to be weakly repelled by the magnet. Examples of diamagnetic materials include water, copper, and gold.
5. What is ferromagnetism?
Ans. Ferromagnetism is a strong magnetic property exhibited by certain substances. In ferromagnetic materials, the individual atoms or molecules have magnetic moments that align parallel to an external magnetic field, causing the material to be strongly attracted to the magnet and retain its magnetism even after the external field is removed. Examples of ferromagnetic materials include iron, nickel, and cobalt.
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