Open App

NEET Exam > NEET Notes > Physics Class 11 > HC Verma Solutions: Chapter 36 - Permanent Magnets

HC Verma Solutions: Chapter 36 - Permanent Magnets | Physics Class 11 - NEET PDF Download

Download, print and study this document offline

Please wait while the PDF view is loading

 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

	Physics Class 11 97 videos\|378 docs\|103 tests

Physics Class 11

97 videos|378 docs|103 tests

Join Course for Free

Top Courses for NEET

View all

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.

Related Exams

NEET JEE Class 12 Grade 12 Class 11

About this Document

	1.1K Views
	4.89/5 Rating
	Dec 18, 2024 Last updated

Document Description: HC Verma Solutions: Chapter 36 - Permanent Magnets for NEET 2024 is part of Physics Class 11 preparation. The notes and questions for HC Verma Solutions: Chapter 36 - Permanent Magnets have been prepared according to the NEET exam syllabus. Information about HC Verma Solutions: Chapter 36 - Permanent Magnets covers topics like and HC Verma Solutions: Chapter 36 - Permanent Magnets Example, for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for HC Verma Solutions: Chapter 36 - Permanent Magnets.

Introduction of HC Verma Solutions: Chapter 36 - Permanent Magnets in English is available as part of our Physics Class 11 for NEET & HC Verma Solutions: Chapter 36 - Permanent Magnets 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 36 - Permanent Magnets | Physics Class 11 - NEET

Description

Full syllabus notes, lecture & questions for HC Verma Solutions: Chapter 36 - Permanent Magnets | 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 36 - Permanent Magnets

In this doc you can find the meaning of HC Verma Solutions: Chapter 36 - Permanent Magnets defined & explained in the simplest way possible. Besides explaining types of HC Verma Solutions: Chapter 36 - Permanent Magnets theory, EduRev gives you an ample number of questions to practice HC Verma Solutions: Chapter 36 - Permanent Magnets tests, examples and also practice NEET tests

	Physics Class 11 97 videos\|378 docs\|103 tests

Physics Class 11

97 videos|378 docs|103 tests

Join Course for Free

Download as PDF

Explore Courses for NEET exam

Top Courses for NEET

Explore Courses

Signup for Free!

Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.

Start learning for Free

10M+ students study on EduRev

Extra Questions

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

Summary

pdf

study material

ppt

Exam

Previous Year Questions with Solutions

mock tests for examination

past year papers

Objective type Questions

Important questions

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

practice quizzes

Semester Notes

shortcuts and tricks

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

video lectures

Free

Sample Paper

Viva Questions

MCQs

;

Additional Information about HC Verma Solutions: Chapter 36 - Permanent Magnets for NEET Preparation

HC Verma Solutions: Chapter 36 - Permanent Magnets Free PDF Download

The HC Verma Solutions: Chapter 36 - Permanent Magnets 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 36 - Permanent Magnets now and kickstart your journey towards success in the NEET exam.

Importance of HC Verma Solutions: Chapter 36 - Permanent Magnets

The importance of HC Verma Solutions: Chapter 36 - Permanent Magnets 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 36 - Permanent Magnets Notes

HC Verma Solutions: Chapter 36 - Permanent Magnets 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 36 - Permanent Magnets. 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 36 - Permanent Magnets Notes on EduRev are your ultimate resource for success.

HC Verma Solutions: Chapter 36 - Permanent Magnets NEET Questions

The "HC Verma Solutions: Chapter 36 - Permanent Magnets 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 36 - Permanent Magnets on the App

Students of NEET can study HC Verma Solutions: Chapter 36 - Permanent Magnets alongwith tests & analysis from the EduRev app, which will help them while preparing for their exam. Apart from the HC Verma Solutions: Chapter 36 - Permanent Magnets, 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 36 - Permanent Magnets is prepared as per the latest NEET syllabus.

Education Revolution