NEET  >  HC Verma Solutions: Chapter 36 - Permanent Magnets

# HC Verma Solutions: Chapter 36 - Permanent Magnets - Physics Class 11 - NEET

``` Page 1

36.1
CHAPTER – 36
PERMANENT MAGNETS
1. m = 10 A-m, d = 5 cm = 0.05 m
B =
2
0
r
m
4 ?
?
=
? ?
2
2
7
10 5
10 10
?
?
?
?
=
25
10
2 ?
= 4 × 10
–4
Tesla
2. m
1
=m
2
= 10 A-m
r = 2 cm = 0.02 m
we know
Force exerted by tow magnetic poles on each other =
2
2 1 0
r
m m
4 ?
?
=
4
2 7
10 4 4
10 10 4
?
?
? ? ?
? ? ?
= 2.5 × 10
–2
N
3. B = –
? d
dv
? dv = –B dl = – 0.2 × 10
–3
× 0.5 = – 0.1 × 10
–3
T-m
Since the sigh is –ve therefore potential decreases.
4. Here dx = 10 sin 30° cm = 5 cm
dx
dV
= B =
m 10 5
m T 10 1 . 0
2
4
?
?
?
? ?
Since B is perpendicular to equipotential surface.
Here it is at angle 120° with (+ve) x-axis and B = 2 × 10
–4
T
5. B = 2 × 10
–4
T
d = 10 cm = 0.1 m
(a) if the point at end-on postion.
B =
3
0
d
M 2
4 ?
?
? 2 × 10
–4
=
3 1
7
) 10 (
M 2 10
?
?
?
?
2 10
10 10 2
7
3 4
?
? ?
?
? ?
= M ? M = 1 Am
2
(b) If the point is at broad-on position
3
0
d
M
4 ?
?
? 2 × 10
–4
=
3 1
7
) 10 (
M 10
?
?
?
?  M = 2 Am
2
6. Given :
? = tan
–1
2 ? tan ? = 2 ? 2 = tan
2
?
? tan ? = 2 cot ? ?
2
tan ?
= cot ?
We know
2
tan ?
= tan ?
Comparing we get, tan ? = cot ?
or, tan ? = tan(90 – ?) or ? = 90 – ? or ? + ? = 90
Hence magnetic field due to the dipole is ?r to the magnetic axis. ?
7. Magnetic field at the broad side on position :
B =
? ?
2 / 3
2 2
0
d
M
4
? ?
?
?
2l = 8 cm d = 3 cm
? 4 × 10
–6
=
? ?
2 / 3
4 4
2 7
10 16 10 9
10 8 m 10
? ?
? ?
? ? ?
? ? ?
? 4 × 10
–6
=
? ? ? ?
2 / 3
2 / 3
4
9
25 10
8 m 10
?
? ?
?
?
? m =
9
8 6
10 8
10 125 10 4
?
? ?
?
? ? ?
= 62.5 × 10
–5
A-m
N S
In cm
30°
X
V
0.1×10
–4
T-m
30°
0.2×10
–4
T-m
0.4×10
–4
T-m
0.3×10
–4
T-m
? ?
? ?
P
S
N
? ?
Page 2

36.1
CHAPTER – 36
PERMANENT MAGNETS
1. m = 10 A-m, d = 5 cm = 0.05 m
B =
2
0
r
m
4 ?
?
=
? ?
2
2
7
10 5
10 10
?
?
?
?
=
25
10
2 ?
= 4 × 10
–4
Tesla
2. m
1
=m
2
= 10 A-m
r = 2 cm = 0.02 m
we know
Force exerted by tow magnetic poles on each other =
2
2 1 0
r
m m
4 ?
?
=
4
2 7
10 4 4
10 10 4
?
?
? ? ?
? ? ?
= 2.5 × 10
–2
N
3. B = –
? d
dv
? dv = –B dl = – 0.2 × 10
–3
× 0.5 = – 0.1 × 10
–3
T-m
Since the sigh is –ve therefore potential decreases.
4. Here dx = 10 sin 30° cm = 5 cm
dx
dV
= B =
m 10 5
m T 10 1 . 0
2
4
?
?
?
? ?
Since B is perpendicular to equipotential surface.
Here it is at angle 120° with (+ve) x-axis and B = 2 × 10
–4
T
5. B = 2 × 10
–4
T
d = 10 cm = 0.1 m
(a) if the point at end-on postion.
B =
3
0
d
M 2
4 ?
?
? 2 × 10
–4
=
3 1
7
) 10 (
M 2 10
?
?
?
?
2 10
10 10 2
7
3 4
?
? ?
?
? ?
= M ? M = 1 Am
2
(b) If the point is at broad-on position
3
0
d
M
4 ?
?
? 2 × 10
–4
=
3 1
7
) 10 (
M 10
?
?
?
?  M = 2 Am
2
6. Given :
? = tan
–1
2 ? tan ? = 2 ? 2 = tan
2
?
? tan ? = 2 cot ? ?
2
tan ?
= cot ?
We know
2
tan ?
= tan ?
Comparing we get, tan ? = cot ?
or, tan ? = tan(90 – ?) or ? = 90 – ? or ? + ? = 90
Hence magnetic field due to the dipole is ?r to the magnetic axis. ?
7. Magnetic field at the broad side on position :
B =
? ?
2 / 3
2 2
0
d
M
4
? ?
?
?
2l = 8 cm d = 3 cm
? 4 × 10
–6
=
? ?
2 / 3
4 4
2 7
10 16 10 9
10 8 m 10
? ?
? ?
? ? ?
? ? ?
? 4 × 10
–6
=
? ? ? ?
2 / 3
2 / 3
4
9
25 10
8 m 10
?
? ?
?
?
? m =
9
8 6
10 8
10 125 10 4
?
? ?
?
? ? ?
= 62.5 × 10
–5
A-m
N S
In cm
30°
X
V
0.1×10
–4
T-m
30°
0.2×10
–4
T-m
0.4×10
–4
T-m
0.3×10
–4
T-m
? ?
? ?
P
S
N
? ?
Permanent Magnets
36.2
8. We know for a magnetic dipole with its north pointing the north, the neutral point in the broadside on
position.
Again B
?
in this case =
3
0
d 4
M
?
?
?
3
0
d 4
M
?
?
=
H
B due to earth
?
3
7
d
44 . 1 10 ?
?
= 18 ?T
?
3
7
d
44 . 1 10 ?
?
= 18 × 10
–6
? d
3
= 8 × 10
–3
? d = 2 × 10
–1
m = 20 cm
In the plane bisecting the dipole.
9. When the magnet is such that its North faces the geographic south of earth. The neutral point lies along
the axial line of the magnet.
3
0
d
M 2
4 ?
?
= 18 × 10
–6
?

3
7
d
72 . 0 2 10 ? ?
?
= 18 × 10
–6
? d
3
=
10 18
10 7 . 0 2
6 -
7
?
? ?
?
? d =
3 / 1
6
9
10
10 8
?
?
?
?
?
?
?
?
?
?
?
= 2 × 10
–1
m = 20 cm
10. Magnetic moment = 0.72 2 A-m
2
= M
B =
3
0
d
M
4 ?
?
B
H
= 18 ?T
?
3
7
d 4
2 72 . 0 10 4
? ?
? ? ?
?
= 18 × 10
–6
? d
3
=
6
7
10 18
10 414 . 1 72 . 0
?
?
?
? ?
= 0.005656
? d ˜ 0.2 m = 20 cm
11. The geomagnetic pole is at the end on position of the earth.
B =
3
0
d
M 2
4 ?
?
=
3 3
22 7
) 10 6400 (
10 8 2 10
?
? ? ?
?
˜ 60 × 10
–6
T = 60 ?T
12. B
?
= 3.4 × 10
–5
T
Given
3
0
R
M
4 ?
?
= 3.4 × 10
–5
? M =
7
3 5
10 4
4 R 10 4 . 3
?
?
? ?
? ? ? ?
= 3.4 × 10
2
R
3
B
?
at Poles =
3
0
R
M 2
4 ?
?
=  = 6.8 × 10
–5
T
13. ?(dip) = 60°
B
H
= B cos 60°
? B = 52 × 10
–6
= 52 ?T
B
V
= B sin ? = 52 × 10
–6
2
3
= 44.98 ?T ˜ 45 ?T ?
14. If ?
1
and ?
2
be the apparent dips shown by the dip circle in the 2 ?r positions, the true dip ? is given by
Cot
2
? = Cot
2
?
1
+ Cot
2
?
2
? Cot
2
? = Cot
2
45° + Cot
2
53°
? Cot
2
? = 1.56 ? ? = 38.6 ˜ 39° ?
N
S
P
d
B M
S N
B H
N
W N S
S
E
tan
–1
2
N 2
N 1
B H
d
Page 3

36.1
CHAPTER – 36
PERMANENT MAGNETS
1. m = 10 A-m, d = 5 cm = 0.05 m
B =
2
0
r
m
4 ?
?
=
? ?
2
2
7
10 5
10 10
?
?
?
?
=
25
10
2 ?
= 4 × 10
–4
Tesla
2. m
1
=m
2
= 10 A-m
r = 2 cm = 0.02 m
we know
Force exerted by tow magnetic poles on each other =
2
2 1 0
r
m m
4 ?
?
=
4
2 7
10 4 4
10 10 4
?
?
? ? ?
? ? ?
= 2.5 × 10
–2
N
3. B = –
? d
dv
? dv = –B dl = – 0.2 × 10
–3
× 0.5 = – 0.1 × 10
–3
T-m
Since the sigh is –ve therefore potential decreases.
4. Here dx = 10 sin 30° cm = 5 cm
dx
dV
= B =
m 10 5
m T 10 1 . 0
2
4
?
?
?
? ?
Since B is perpendicular to equipotential surface.
Here it is at angle 120° with (+ve) x-axis and B = 2 × 10
–4
T
5. B = 2 × 10
–4
T
d = 10 cm = 0.1 m
(a) if the point at end-on postion.
B =
3
0
d
M 2
4 ?
?
? 2 × 10
–4
=
3 1
7
) 10 (
M 2 10
?
?
?
?
2 10
10 10 2
7
3 4
?
? ?
?
? ?
= M ? M = 1 Am
2
(b) If the point is at broad-on position
3
0
d
M
4 ?
?
? 2 × 10
–4
=
3 1
7
) 10 (
M 10
?
?
?
?  M = 2 Am
2
6. Given :
? = tan
–1
2 ? tan ? = 2 ? 2 = tan
2
?
? tan ? = 2 cot ? ?
2
tan ?
= cot ?
We know
2
tan ?
= tan ?
Comparing we get, tan ? = cot ?
or, tan ? = tan(90 – ?) or ? = 90 – ? or ? + ? = 90
Hence magnetic field due to the dipole is ?r to the magnetic axis. ?
7. Magnetic field at the broad side on position :
B =
? ?
2 / 3
2 2
0
d
M
4
? ?
?
?
2l = 8 cm d = 3 cm
? 4 × 10
–6
=
? ?
2 / 3
4 4
2 7
10 16 10 9
10 8 m 10
? ?
? ?
? ? ?
? ? ?
? 4 × 10
–6
=
? ? ? ?
2 / 3
2 / 3
4
9
25 10
8 m 10
?
? ?
?
?
? m =
9
8 6
10 8
10 125 10 4
?
? ?
?
? ? ?
= 62.5 × 10
–5
A-m
N S
In cm
30°
X
V
0.1×10
–4
T-m
30°
0.2×10
–4
T-m
0.4×10
–4
T-m
0.3×10
–4
T-m
? ?
? ?
P
S
N
? ?
Permanent Magnets
36.2
8. We know for a magnetic dipole with its north pointing the north, the neutral point in the broadside on
position.
Again B
?
in this case =
3
0
d 4
M
?
?
?
3
0
d 4
M
?
?
=
H
B due to earth
?
3
7
d
44 . 1 10 ?
?
= 18 ?T
?
3
7
d
44 . 1 10 ?
?
= 18 × 10
–6
? d
3
= 8 × 10
–3
? d = 2 × 10
–1
m = 20 cm
In the plane bisecting the dipole.
9. When the magnet is such that its North faces the geographic south of earth. The neutral point lies along
the axial line of the magnet.
3
0
d
M 2
4 ?
?
= 18 × 10
–6
?

3
7
d
72 . 0 2 10 ? ?
?
= 18 × 10
–6
? d
3
=
10 18
10 7 . 0 2
6 -
7
?
? ?
?
? d =
3 / 1
6
9
10
10 8
?
?
?
?
?
?
?
?
?
?
?
= 2 × 10
–1
m = 20 cm
10. Magnetic moment = 0.72 2 A-m
2
= M
B =
3
0
d
M
4 ?
?
B
H
= 18 ?T
?
3
7
d 4
2 72 . 0 10 4
? ?
? ? ?
?
= 18 × 10
–6
? d
3
=
6
7
10 18
10 414 . 1 72 . 0
?
?
?
? ?
= 0.005656
? d ˜ 0.2 m = 20 cm
11. The geomagnetic pole is at the end on position of the earth.
B =
3
0
d
M 2
4 ?
?
=
3 3
22 7
) 10 6400 (
10 8 2 10
?
? ? ?
?
˜ 60 × 10
–6
T = 60 ?T
12. B
?
= 3.4 × 10
–5
T
Given
3
0
R
M
4 ?
?
= 3.4 × 10
–5
? M =
7
3 5
10 4
4 R 10 4 . 3
?
?
? ?
? ? ? ?
= 3.4 × 10
2
R
3
B
?
at Poles =
3
0
R
M 2
4 ?
?
=  = 6.8 × 10
–5
T
13. ?(dip) = 60°
B
H
= B cos 60°
? B = 52 × 10
–6
= 52 ?T
B
V
= B sin ? = 52 × 10
–6
2
3
= 44.98 ?T ˜ 45 ?T ?
14. If ?
1
and ?
2
be the apparent dips shown by the dip circle in the 2 ?r positions, the true dip ? is given by
Cot
2
? = Cot
2
?
1
+ Cot
2
?
2
? Cot
2
? = Cot
2
45° + Cot
2
53°
? Cot
2
? = 1.56 ? ? = 38.6 ˜ 39° ?
N
S
P
d
B M
S N
B H
N
W N S
S
E
tan
–1
2
N 2
N 1
B H
d
Permanent Magnets
36.3
15. We know B
H
=
r 2
in
0
?
Give : B
H
= 3.6 × 10
–5
T ? = 45° ?
i = 10 mA = 10
–2
A tan ? = 1
n = ? r = 10 cm = 0.1 m
n =
i
r 2 tan B
0
H
?
? ?
=
2 7
1 5
10 10 4
10 1 2 10 6 . 3
? ?
? ?
? ? ?
? ? ? ?
= 0.5732 × 10
3
˜ 573 turns ?
16. n = 50 A = 2 cm × 2 cm = 2 × 2 × 10
–4
m
2
i = 20 × 10
–3
A B = 0.5 T
? = ? ? B A ni
? ?
? = niAB Sin 90° = 50 × 20 × 10
–3
× 4 × 10
–4
× 0.5 = 2 × 10
–4
N-M
17. Given ? = 37° d = 10 cm = 0.1 m
We know
H
B
M
= ?
?
?
?
tan
d 2
) d ( 4
2 2 2
0
?
= ? ?
?
?
tan
d 2
d 4
4
0
[As the magnet is short]
= ? ? ?
? ?
?
?
37 tan
2
) 1 . 0 (
10 4
4
3
7
= 0.5 × 0.75 × 1 × 10
–3
× 10
7
= 0.375 × 10
4
= 3.75 ×10
3
A-m
2
T
–1
18.
H
B
M
(found in the previous problem) = 3.75 ×10
3
A-m
2
T
–1
? = 37°, d = ?
H
B
M
= ? ?
?
?
tan ) d (
4
2 / 3 2 2
0
?
l << d neglecting l w.r.t.d
??
H
B
M
= ?
?
?
Tan d
4
3
0
? 3.75 × 10
3
=
7
10
1
?
× d
3
× 0.75
? d
3
=
75 . 0
10 10 75 . 3
7 3 ?
? ?
= 5 × 10
–4
? d = 0.079 m = 7.9 cm ?
19. Given
H
B
M
= 40 A-m
2
/T
Since the magnet is short ‘l’ can be neglected
So,
H
B
M
=
2
d 4
3
0
?
?
?
= 40
? d
3
=
?
? ? ? ?
?
4
2 10 4 40
7
= 8 × 10
–6
? d = 2 × 10
–2
m = 2 cm
with the northpole pointing towards south.
20. According to oscillation magnetometer,
T =
H
MB
2
?
?
?
10
?
=
6
4
10 30 M
10 2 . 1
2
?
?
? ?
?
?
?
2
20
1
?
?
?
?
?
?
=
6
4
10 30 M
10 2 . 1
?
?
? ?
?
? M =
6
4
10 30
400 10 2 . 1
?
?
?
? ?
= 16 × 10
2
A-m
2
= 1600 A-m
2
N
S
Page 4

36.1
CHAPTER – 36
PERMANENT MAGNETS
1. m = 10 A-m, d = 5 cm = 0.05 m
B =
2
0
r
m
4 ?
?
=
? ?
2
2
7
10 5
10 10
?
?
?
?
=
25
10
2 ?
= 4 × 10
–4
Tesla
2. m
1
=m
2
= 10 A-m
r = 2 cm = 0.02 m
we know
Force exerted by tow magnetic poles on each other =
2
2 1 0
r
m m
4 ?
?
=
4
2 7
10 4 4
10 10 4
?
?
? ? ?
? ? ?
= 2.5 × 10
–2
N
3. B = –
? d
dv
? dv = –B dl = – 0.2 × 10
–3
× 0.5 = – 0.1 × 10
–3
T-m
Since the sigh is –ve therefore potential decreases.
4. Here dx = 10 sin 30° cm = 5 cm
dx
dV
= B =
m 10 5
m T 10 1 . 0
2
4
?
?
?
? ?
Since B is perpendicular to equipotential surface.
Here it is at angle 120° with (+ve) x-axis and B = 2 × 10
–4
T
5. B = 2 × 10
–4
T
d = 10 cm = 0.1 m
(a) if the point at end-on postion.
B =
3
0
d
M 2
4 ?
?
? 2 × 10
–4
=
3 1
7
) 10 (
M 2 10
?
?
?
?
2 10
10 10 2
7
3 4
?
? ?
?
? ?
= M ? M = 1 Am
2
(b) If the point is at broad-on position
3
0
d
M
4 ?
?
? 2 × 10
–4
=
3 1
7
) 10 (
M 10
?
?
?
?  M = 2 Am
2
6. Given :
? = tan
–1
2 ? tan ? = 2 ? 2 = tan
2
?
? tan ? = 2 cot ? ?
2
tan ?
= cot ?
We know
2
tan ?
= tan ?
Comparing we get, tan ? = cot ?
or, tan ? = tan(90 – ?) or ? = 90 – ? or ? + ? = 90
Hence magnetic field due to the dipole is ?r to the magnetic axis. ?
7. Magnetic field at the broad side on position :
B =
? ?
2 / 3
2 2
0
d
M
4
? ?
?
?
2l = 8 cm d = 3 cm
? 4 × 10
–6
=
? ?
2 / 3
4 4
2 7
10 16 10 9
10 8 m 10
? ?
? ?
? ? ?
? ? ?
? 4 × 10
–6
=
? ? ? ?
2 / 3
2 / 3
4
9
25 10
8 m 10
?
? ?
?
?
? m =
9
8 6
10 8
10 125 10 4
?
? ?
?
? ? ?
= 62.5 × 10
–5
A-m
N S
In cm
30°
X
V
0.1×10
–4
T-m
30°
0.2×10
–4
T-m
0.4×10
–4
T-m
0.3×10
–4
T-m
? ?
? ?
P
S
N
? ?
Permanent Magnets
36.2
8. We know for a magnetic dipole with its north pointing the north, the neutral point in the broadside on
position.
Again B
?
in this case =
3
0
d 4
M
?
?
?
3
0
d 4
M
?
?
=
H
B due to earth
?
3
7
d
44 . 1 10 ?
?
= 18 ?T
?
3
7
d
44 . 1 10 ?
?
= 18 × 10
–6
? d
3
= 8 × 10
–3
? d = 2 × 10
–1
m = 20 cm
In the plane bisecting the dipole.
9. When the magnet is such that its North faces the geographic south of earth. The neutral point lies along
the axial line of the magnet.
3
0
d
M 2
4 ?
?
= 18 × 10
–6
?

3
7
d
72 . 0 2 10 ? ?
?
= 18 × 10
–6
? d
3
=
10 18
10 7 . 0 2
6 -
7
?
? ?
?
? d =
3 / 1
6
9
10
10 8
?
?
?
?
?
?
?
?
?
?
?
= 2 × 10
–1
m = 20 cm
10. Magnetic moment = 0.72 2 A-m
2
= M
B =
3
0
d
M
4 ?
?
B
H
= 18 ?T
?
3
7
d 4
2 72 . 0 10 4
? ?
? ? ?
?
= 18 × 10
–6
? d
3
=
6
7
10 18
10 414 . 1 72 . 0
?
?
?
? ?
= 0.005656
? d ˜ 0.2 m = 20 cm
11. The geomagnetic pole is at the end on position of the earth.
B =
3
0
d
M 2
4 ?
?
=
3 3
22 7
) 10 6400 (
10 8 2 10
?
? ? ?
?
˜ 60 × 10
–6
T = 60 ?T
12. B
?
= 3.4 × 10
–5
T
Given
3
0
R
M
4 ?
?
= 3.4 × 10
–5
? M =
7
3 5
10 4
4 R 10 4 . 3
?
?
? ?
? ? ? ?
= 3.4 × 10
2
R
3
B
?
at Poles =
3
0
R
M 2
4 ?
?
=  = 6.8 × 10
–5
T
13. ?(dip) = 60°
B
H
= B cos 60°
? B = 52 × 10
–6
= 52 ?T
B
V
= B sin ? = 52 × 10
–6
2
3
= 44.98 ?T ˜ 45 ?T ?
14. If ?
1
and ?
2
be the apparent dips shown by the dip circle in the 2 ?r positions, the true dip ? is given by
Cot
2
? = Cot
2
?
1
+ Cot
2
?
2
? Cot
2
? = Cot
2
45° + Cot
2
53°
? Cot
2
? = 1.56 ? ? = 38.6 ˜ 39° ?
N
S
P
d
B M
S N
B H
N
W N S
S
E
tan
–1
2
N 2
N 1
B H
d
Permanent Magnets
36.3
15. We know B
H
=
r 2
in
0
?
Give : B
H
= 3.6 × 10
–5
T ? = 45° ?
i = 10 mA = 10
–2
A tan ? = 1
n = ? r = 10 cm = 0.1 m
n =
i
r 2 tan B
0
H
?
? ?
=
2 7
1 5
10 10 4
10 1 2 10 6 . 3
? ?
? ?
? ? ?
? ? ? ?
= 0.5732 × 10
3
˜ 573 turns ?
16. n = 50 A = 2 cm × 2 cm = 2 × 2 × 10
–4
m
2
i = 20 × 10
–3
A B = 0.5 T
? = ? ? B A ni
? ?
? = niAB Sin 90° = 50 × 20 × 10
–3
× 4 × 10
–4
× 0.5 = 2 × 10
–4
N-M
17. Given ? = 37° d = 10 cm = 0.1 m
We know
H
B
M
= ?
?
?
?
tan
d 2
) d ( 4
2 2 2
0
?
= ? ?
?
?
tan
d 2
d 4
4
0
[As the magnet is short]
= ? ? ?
? ?
?
?
37 tan
2
) 1 . 0 (
10 4
4
3
7
= 0.5 × 0.75 × 1 × 10
–3
× 10
7
= 0.375 × 10
4
= 3.75 ×10
3
A-m
2
T
–1
18.
H
B
M
(found in the previous problem) = 3.75 ×10
3
A-m
2
T
–1
? = 37°, d = ?
H
B
M
= ? ?
?
?
tan ) d (
4
2 / 3 2 2
0
?
l << d neglecting l w.r.t.d
??
H
B
M
= ?
?
?
Tan d
4
3
0
? 3.75 × 10
3
=
7
10
1
?
× d
3
× 0.75
? d
3
=
75 . 0
10 10 75 . 3
7 3 ?
? ?
= 5 × 10
–4
? d = 0.079 m = 7.9 cm ?
19. Given
H
B
M
= 40 A-m
2
/T
Since the magnet is short ‘l’ can be neglected
So,
H
B
M
=
2
d 4
3
0
?
?
?
= 40
? d
3
=
?
? ? ? ?
?
4
2 10 4 40
7
= 8 × 10
–6
? d = 2 × 10
–2
m = 2 cm
with the northpole pointing towards south.
20. According to oscillation magnetometer,
T =
H
MB
2
?
?
?
10
?
=
6
4
10 30 M
10 2 . 1
2
?
?
? ?
?
?
?
2
20
1
?
?
?
?
?
?
=
6
4
10 30 M
10 2 . 1
?
?
? ?
?
? M =
6
4
10 30
400 10 2 . 1
?
?
?
? ?
= 16 × 10
2
A-m
2
= 1600 A-m
2
N
S
Permanent Magnets
36.4
21. We know :  ? =
? ?
H
mB
2
1
For like poles tied together
M = M
1
– M
2
For unlike poles M ? = M
1
+ M
2
2
1
?
?
=
2 1
2 1
M M
M M
?
?
?
2
2
10
?
?
?
?
?
?
=
2 1
2 1
M M
M M
?
?
? 25 =
2 1
2 1
M M
M M
?
?
?
24
26
=
2
1
M 2
M 2
?
2
1
M
M
=
12
13
22. B
H
= 24 × 10
–6
T T
1
= 0.1 ?
B = B
H
– B
wire
= 2.4 × 10
–6
–
r
i
2
o
?
?
= 24 × 10
–6
–
2 . 0
18 10 2
7
? ?
?
= (24 –10) × 10
–6
= 14 × 10
–6
T =
H
MB
2
?
?
2
1
T
T
=
H
B
B
?
2
T
1 . 0
=
6
6
10 24
10 14
?
?
?
?
?
2
2
T
1 . 0
?
?
?
?
?
?
?
?
=
24
14
? T
2
2
=
24
14 01 . 0 ?
?T
2
= 0.076
23. T =
H
MB
2
?
? Here ?? = 2 ?
T
1
=
40
1
min T
2
= ?
2
1
T
T
=
? ?
?
?
2
T 40
1
=
2
1
?
2
2
T 1600
1
=
2
1
? T
2
2
=
800
1
? T
2
= 0.03536 min
For 1 oscillation Time taken = 0.03536 min.
For 40 Oscillation Time = 4 × 0.03536 = 1.414 = 2 min
24. ?
1
= 40 oscillations/minute
B
H
= 25 ?T
m of second magnet = 1.6 A-m
2
d = 20 cm = 0.2 m
(a) For north facing north
?
1
=
? ?
H
MB
2
1
?
2
=
? ?
?
?
?
B B M
2
1
H
B =
3
0
d
m
4 ?
?
???
3
7
10 8
6 . 1 10
?
?
?
?
?= 20 ?T ?
2
1
?
?
=
B B
B
H
?
?
2
40
?
=
5
25
? ?
2
=
5
40
= 17.88 ˜ 18 osci/min ?
(b) For north pole facing south
?
?
????
? ?
H
MB
2
1
?
2
=
? ?
?
?
?
B B M
2
1
H
2
1
?
?
=
B B
B
H
?
?
2
40
?
=
45
25
? ?
2
=
?
?
?
?
?
?
45
25
40
= 53.66 ˜ 54 osci/min
? ? ? ? ??
S N
S N
N S
S N
```

## FAQs on HC Verma Solutions: Chapter 36 - Permanent Magnets - Physics Class 11 - NEET

 1. What are permanent magnets?
Ans. Permanent magnets are objects made from materials that are magnetized and can create their own persistent magnetic fields. These magnets retain their magnetism for a long period, hence the name "permanent."
 2. How are permanent magnets different from temporary magnets?
Ans. Permanent magnets differ from temporary magnets in that they can retain their magnetism without the need for an external magnetic field. Temporary magnets, on the other hand, only exhibit magnetism when in the presence of a magnetic field.
 3. What are some common uses of permanent magnets?
Ans. Permanent magnets have various applications in everyday life. They are commonly used in electric motors, generators, speakers, hard disk drives, magnetic separators, and MRI machines, among others.
 4. How can you demagnetize a permanent magnet?
Ans. Permanent magnets can be demagnetized by subjecting them to high temperatures, dropping or hitting them, or by placing them in a strong alternating magnetic field. These processes disrupt the alignment of the magnetic domains in the material, causing the magnetism to diminish or disappear.
 5. What are some materials that can be used to make permanent magnets?
Ans. Permanent magnets are typically made from materials such as iron, cobalt, nickel, and certain alloys like neodymium-iron-boron (NdFeB) or samarium-cobalt (SmCo). These materials have high magnetic permeability and can retain their magnetism for extended periods.

## Physics Class 11

130 videos|483 docs|210 tests

## Physics Class 11

130 videos|483 docs|210 tests

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