Chapter 36 : Permanent Magnets - HC Verma Solution, Physics Class 11 Notes | EduRev

HC Verma and Irodov Solutions

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JEE : Chapter 36 : Permanent Magnets - HC Verma Solution, Physics Class 11 Notes | EduRev

 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
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