Magnetism and Matter- 1 Practice Questions - DPP for NEET

 Page 1


1. (a). Force between magnetic poles in air is given by
F = 
0 12
2
mm
4
r
m
´
p
Given that m
1
 = 50 Am, m
2
 = 100Am,
r = 10 cm = 0.1 m and
µ
0
 = permeability of air = 4 p × 10
–7
 Hm
–1
.
\ F = 
7
4 10 50 100
.
4 0.1 0.1
-
p´´
p´
 = 50 × 10
–3
 N
2. (a) Strength of a magnetic field due to a pole of strength m
is given by
H = 
2
1m
.
4
r
p
Given that m = 40 Am, r = 20 cm = 20 × 10
–2
 m.
\ H = 
22
1 40
4
(20 10 )
-
´
p
´
 = 79.57 Am
–1
Now , magnetic induction at the same point :
B = µ
0
 H = 4p × 10
–7
 × 79.57  = 10
–4
 wb/m
2
3. (c) Couple acting on a bar magnet of dipole moment M
when placed in a magnetic field, is given by
t = MB sin q
where q is the angle made by the axis of magnet with
the direction of field.
Given that m = 5 Am, 2 l = 0.2 m, q = 30°
and B = 15 Wbm
–2
\ t = MB sin q  = (m × 2 l) B sin q
  = 5 × 0.2 × 15 × 
1
2
 = 7.5 Nm.
4. (a) F = mB = 
0
3
2m'
m
4
x
m
p
l
= 
7
3
10 2 200 0.05 100
8 10
-
-
´´ ´´
´
= 2.5 × 10
-2
 N
5. (d). W = MB (cosq
1
 - cosq
2
)
\ W
1
 = MB (cos 0º – cos 60º) = 
MB
2
W
2
 = MB (cos 30º – cos 90º) = 
3 MB
2
\ W
2
  = 
3
 WW
1
.
6. (b). Loss in P .E. = gain in K.E.
\ E
k
 = U
i
 – U
j
 = – MB cos 90º – (– MB cos 0º)
      = 4 × 25 × 10
-6
 = 10
-4
 J
7. (a). t = MB sin q = m l B sin q
       = 10
-3
 × 0.1 × 4p × 10
-3
 × 0.5
       = 2p × 10
-7
 N-m
8. (c)
9. ( d) Here, d = 10 cm  = 0.1 m ,
H = 0.4 gauss = 0.4 x 10
-4
 T,  M = ?
Neutral points in this case, lie on axial line of magnet,
such that
0
3
2M
H
4
d
m
=
p
\ 10
-7
 × 
3
2M
(0.1)
 = 0.4 × 10
-4
M = 0.2 A m
2
10. (d). t = F × r = MB sin 30
F = 
mBsin30
r
´ l
= 
2
2
25 10 24 0.25 1
2
12 10
-
-
´ ´´
´
´
 = 6.25 N
11. (b). B = 
0
3
2M
4
x
m
´
p
 = constant
\ x
2
 = x
1
 
1/3
2
1
M
M
æö
ç÷
èø
 = 20 × 
1/3
1
24
æö
ç÷
´ èø
 = 10 cm
12. (d). B =
00
223/23
2M 2M
44
(x)x
mm
==
pp
+l
\ 
3
12
21
Bx
Bx
æö
=
ç÷
èø
 = 8 : 1 approximately . .
13. (c). According to tangent law
B
A
 = B
B
 tan q
or  
00
33
11
2MM
44
dd
mm
=
pp
 tanq
\ 
1
2
d
d
 = (2 cot q)
1/3
14. (a).  T = 2p 
I
MB
 = 2p 
2
p
m
12 mB ´
l
l
 = 4 sec
T' = 2p 
2
p
m
2
m
12B
2
´
l
l
 = 4 sec
Page 2


1. (a). Force between magnetic poles in air is given by
F = 
0 12
2
mm
4
r
m
´
p
Given that m
1
 = 50 Am, m
2
 = 100Am,
r = 10 cm = 0.1 m and
µ
0
 = permeability of air = 4 p × 10
–7
 Hm
–1
.
\ F = 
7
4 10 50 100
.
4 0.1 0.1
-
p´´
p´
 = 50 × 10
–3
 N
2. (a) Strength of a magnetic field due to a pole of strength m
is given by
H = 
2
1m
.
4
r
p
Given that m = 40 Am, r = 20 cm = 20 × 10
–2
 m.
\ H = 
22
1 40
4
(20 10 )
-
´
p
´
 = 79.57 Am
–1
Now , magnetic induction at the same point :
B = µ
0
 H = 4p × 10
–7
 × 79.57  = 10
–4
 wb/m
2
3. (c) Couple acting on a bar magnet of dipole moment M
when placed in a magnetic field, is given by
t = MB sin q
where q is the angle made by the axis of magnet with
the direction of field.
Given that m = 5 Am, 2 l = 0.2 m, q = 30°
and B = 15 Wbm
–2
\ t = MB sin q  = (m × 2 l) B sin q
  = 5 × 0.2 × 15 × 
1
2
 = 7.5 Nm.
4. (a) F = mB = 
0
3
2m'
m
4
x
m
p
l
= 
7
3
10 2 200 0.05 100
8 10
-
-
´´ ´´
´
= 2.5 × 10
-2
 N
5. (d). W = MB (cosq
1
 - cosq
2
)
\ W
1
 = MB (cos 0º – cos 60º) = 
MB
2
W
2
 = MB (cos 30º – cos 90º) = 
3 MB
2
\ W
2
  = 
3
 WW
1
.
6. (b). Loss in P .E. = gain in K.E.
\ E
k
 = U
i
 – U
j
 = – MB cos 90º – (– MB cos 0º)
      = 4 × 25 × 10
-6
 = 10
-4
 J
7. (a). t = MB sin q = m l B sin q
       = 10
-3
 × 0.1 × 4p × 10
-3
 × 0.5
       = 2p × 10
-7
 N-m
8. (c)
9. ( d) Here, d = 10 cm  = 0.1 m ,
H = 0.4 gauss = 0.4 x 10
-4
 T,  M = ?
Neutral points in this case, lie on axial line of magnet,
such that
0
3
2M
H
4
d
m
=
p
\ 10
-7
 × 
3
2M
(0.1)
 = 0.4 × 10
-4
M = 0.2 A m
2
10. (d). t = F × r = MB sin 30
F = 
mBsin30
r
´ l
= 
2
2
25 10 24 0.25 1
2
12 10
-
-
´ ´´
´
´
 = 6.25 N
11. (b). B = 
0
3
2M
4
x
m
´
p
 = constant
\ x
2
 = x
1
 
1/3
2
1
M
M
æö
ç÷
èø
 = 20 × 
1/3
1
24
æö
ç÷
´ èø
 = 10 cm
12. (d). B =
00
223/23
2M 2M
44
(x)x
mm
==
pp
+l
\ 
3
12
21
Bx
Bx
æö
=
ç÷
èø
 = 8 : 1 approximately . .
13. (c). According to tangent law
B
A
 = B
B
 tan q
or  
00
33
11
2MM
44
dd
mm
=
pp
 tanq
\ 
1
2
d
d
 = (2 cot q)
1/3
14. (a).  T = 2p 
I
MB
 = 2p 
2
p
m
12 mB ´
l
l
 = 4 sec
T' = 2p 
2
p
m
2
m
12B
2
´
l
l
 = 4 sec
DPP/ P 42
118
15. (c).  T = 2p 
I
MB
  or    T µ 
1
M
1
2
T 3M 2M
T 3M 2M
-
=
+
  or  T T
2
 = 5 5s
16. (b).  T = 2p 
I
MB
         = 2p 
2
p
m /12
mB
l
l
  or T  µ ml
1/2
m
T'
nn
T
m
æö
ç÷
=
ç÷
ç÷
èø
l
l
  or  T' = 
T
n
17. (a). The volume of the bar magnet is
V  = 
mass
density
   = 
3
33
6.6 10 kg
7.9 10 kg / m
-
´
´
 = 8.3 × 10
-7
 m
3
.
The intensity of magnetization is
I = 
M
V
 = 
2
72
2.5Am
8.3 10 m
-
-
´
            = 3.0 × 10
6
 A/m
18. (d). The compass box will be on the axial line of the magnet,
Hence,  
33
1 2M 1 2 2m
..
44
rr
´
=
pp
l
 = H tan q
Given that H = horizontal component of the earth’s
magnetic field = 30 Am
–1
, q = 45°,
r = 20 cm = 0.02 m,
M = 2 l m = 4 × 10
–2
 m
Hence, 
2
3
2 4 10m
4 (0.2)
-
´´
p
 = 30 × tan 45°=30 × 1;
\ m = 
3
2
30 4 (0.2)
2 4 10
-
´ p´
´´
 = 37.7 Am
19. (c). As the magnet is placed with its south pole pointing
south, hence the neutral point lies on the equatorial
line. At the neutral point, the magnetic field B due to
the magnet becomes equal and opposite to horizontal
component of earth’s magnetic field i.e., B
H
.
Hence, if M be magnetic dipole moment of the magnet
of length 2l and r the distance of the neutral point from
its centre, then
B = 
0
2 2 3/2
µ M
4
(r)
p
+ l
 = B
H
Given that µ
0
 = 4p × 10
–1
 T mA
–1
,
M = 13.4 Am
2
,
r = 15 cm = 0.15 m and l = 5.0 cm = 0.05 m
\ B
H
 = 10
–7
 × 
2 2 3/2
1.34
[(0.15) (0.5) ] +
  = 10
–7
 × 
1.34
0.025 0.025
 = 0.34 × 10
–4
 TT
20. (a). As the magnet is placed with its north pole pointing
south, the neutral points are obtained on the axial line.
At the neutral points the magnetic field B due to the
magnet becomes equal and opposite to the horizontal
component of earth’s magnetic field i.e., B
H
.
Hence, if M be the magnetic dipole moment of the
magnet of length 2l and r the distance of neutral point
from the centre of the magnet, then we have
B = 
0
H
2 22
µ 2Mr
.B
4
(r)
=
p
-l
Given that µ
0
 = 4p × 10
–7
 TmA
–1
 ,
r = 40 cm = 0.40 m, l = 15 cm = 0.15 m and
H
H
 = 0.34 Gauss = 0.34 × 10
–4
 T
\ M = 
2 22
H
0
B(r) 4
.
µ 2r
- p l
   = 10
–7
× 
4 2 22
(0.34 10 ) [(0.40) (0.15) ]
2 0.40
-
´ --
´
   = 8.0 Am
2
The pole strength of the magnet is,
m = 
M 8.0
2 0.30
=
l
 = 26.7 Am
21. (a). The situation is shown in figure. The horizontal
component of earth’s magnetic field at the location of
the cable (angle of dip q = 0 is)
10°
10°
Magnetic 
North
North
Geographic 
Meridian
Cable
Geographic
East
Geographic
West
Magnetic
Meridian
Current
Magnetic
South
South
I
B
H 
= B cos q = B cos 0 = B = 0.33 Gauss
    = 0.33 × 10
–4
 Tesla
B
H
 is directed horizontally in the magnetic meridian.
The magnetic field produced by the cable at a distance
of R meter is given by
Page 3


1. (a). Force between magnetic poles in air is given by
F = 
0 12
2
mm
4
r
m
´
p
Given that m
1
 = 50 Am, m
2
 = 100Am,
r = 10 cm = 0.1 m and
µ
0
 = permeability of air = 4 p × 10
–7
 Hm
–1
.
\ F = 
7
4 10 50 100
.
4 0.1 0.1
-
p´´
p´
 = 50 × 10
–3
 N
2. (a) Strength of a magnetic field due to a pole of strength m
is given by
H = 
2
1m
.
4
r
p
Given that m = 40 Am, r = 20 cm = 20 × 10
–2
 m.
\ H = 
22
1 40
4
(20 10 )
-
´
p
´
 = 79.57 Am
–1
Now , magnetic induction at the same point :
B = µ
0
 H = 4p × 10
–7
 × 79.57  = 10
–4
 wb/m
2
3. (c) Couple acting on a bar magnet of dipole moment M
when placed in a magnetic field, is given by
t = MB sin q
where q is the angle made by the axis of magnet with
the direction of field.
Given that m = 5 Am, 2 l = 0.2 m, q = 30°
and B = 15 Wbm
–2
\ t = MB sin q  = (m × 2 l) B sin q
  = 5 × 0.2 × 15 × 
1
2
 = 7.5 Nm.
4. (a) F = mB = 
0
3
2m'
m
4
x
m
p
l
= 
7
3
10 2 200 0.05 100
8 10
-
-
´´ ´´
´
= 2.5 × 10
-2
 N
5. (d). W = MB (cosq
1
 - cosq
2
)
\ W
1
 = MB (cos 0º – cos 60º) = 
MB
2
W
2
 = MB (cos 30º – cos 90º) = 
3 MB
2
\ W
2
  = 
3
 WW
1
.
6. (b). Loss in P .E. = gain in K.E.
\ E
k
 = U
i
 – U
j
 = – MB cos 90º – (– MB cos 0º)
      = 4 × 25 × 10
-6
 = 10
-4
 J
7. (a). t = MB sin q = m l B sin q
       = 10
-3
 × 0.1 × 4p × 10
-3
 × 0.5
       = 2p × 10
-7
 N-m
8. (c)
9. ( d) Here, d = 10 cm  = 0.1 m ,
H = 0.4 gauss = 0.4 x 10
-4
 T,  M = ?
Neutral points in this case, lie on axial line of magnet,
such that
0
3
2M
H
4
d
m
=
p
\ 10
-7
 × 
3
2M
(0.1)
 = 0.4 × 10
-4
M = 0.2 A m
2
10. (d). t = F × r = MB sin 30
F = 
mBsin30
r
´ l
= 
2
2
25 10 24 0.25 1
2
12 10
-
-
´ ´´
´
´
 = 6.25 N
11. (b). B = 
0
3
2M
4
x
m
´
p
 = constant
\ x
2
 = x
1
 
1/3
2
1
M
M
æö
ç÷
èø
 = 20 × 
1/3
1
24
æö
ç÷
´ èø
 = 10 cm
12. (d). B =
00
223/23
2M 2M
44
(x)x
mm
==
pp
+l
\ 
3
12
21
Bx
Bx
æö
=
ç÷
èø
 = 8 : 1 approximately . .
13. (c). According to tangent law
B
A
 = B
B
 tan q
or  
00
33
11
2MM
44
dd
mm
=
pp
 tanq
\ 
1
2
d
d
 = (2 cot q)
1/3
14. (a).  T = 2p 
I
MB
 = 2p 
2
p
m
12 mB ´
l
l
 = 4 sec
T' = 2p 
2
p
m
2
m
12B
2
´
l
l
 = 4 sec
DPP/ P 42
118
15. (c).  T = 2p 
I
MB
  or    T µ 
1
M
1
2
T 3M 2M
T 3M 2M
-
=
+
  or  T T
2
 = 5 5s
16. (b).  T = 2p 
I
MB
         = 2p 
2
p
m /12
mB
l
l
  or T  µ ml
1/2
m
T'
nn
T
m
æö
ç÷
=
ç÷
ç÷
èø
l
l
  or  T' = 
T
n
17. (a). The volume of the bar magnet is
V  = 
mass
density
   = 
3
33
6.6 10 kg
7.9 10 kg / m
-
´
´
 = 8.3 × 10
-7
 m
3
.
The intensity of magnetization is
I = 
M
V
 = 
2
72
2.5Am
8.3 10 m
-
-
´
            = 3.0 × 10
6
 A/m
18. (d). The compass box will be on the axial line of the magnet,
Hence,  
33
1 2M 1 2 2m
..
44
rr
´
=
pp
l
 = H tan q
Given that H = horizontal component of the earth’s
magnetic field = 30 Am
–1
, q = 45°,
r = 20 cm = 0.02 m,
M = 2 l m = 4 × 10
–2
 m
Hence, 
2
3
2 4 10m
4 (0.2)
-
´´
p
 = 30 × tan 45°=30 × 1;
\ m = 
3
2
30 4 (0.2)
2 4 10
-
´ p´
´´
 = 37.7 Am
19. (c). As the magnet is placed with its south pole pointing
south, hence the neutral point lies on the equatorial
line. At the neutral point, the magnetic field B due to
the magnet becomes equal and opposite to horizontal
component of earth’s magnetic field i.e., B
H
.
Hence, if M be magnetic dipole moment of the magnet
of length 2l and r the distance of the neutral point from
its centre, then
B = 
0
2 2 3/2
µ M
4
(r)
p
+ l
 = B
H
Given that µ
0
 = 4p × 10
–1
 T mA
–1
,
M = 13.4 Am
2
,
r = 15 cm = 0.15 m and l = 5.0 cm = 0.05 m
\ B
H
 = 10
–7
 × 
2 2 3/2
1.34
[(0.15) (0.5) ] +
  = 10
–7
 × 
1.34
0.025 0.025
 = 0.34 × 10
–4
 TT
20. (a). As the magnet is placed with its north pole pointing
south, the neutral points are obtained on the axial line.
At the neutral points the magnetic field B due to the
magnet becomes equal and opposite to the horizontal
component of earth’s magnetic field i.e., B
H
.
Hence, if M be the magnetic dipole moment of the
magnet of length 2l and r the distance of neutral point
from the centre of the magnet, then we have
B = 
0
H
2 22
µ 2Mr
.B
4
(r)
=
p
-l
Given that µ
0
 = 4p × 10
–7
 TmA
–1
 ,
r = 40 cm = 0.40 m, l = 15 cm = 0.15 m and
H
H
 = 0.34 Gauss = 0.34 × 10
–4
 T
\ M = 
2 22
H
0
B(r) 4
.
µ 2r
- p l
   = 10
–7
× 
4 2 22
(0.34 10 ) [(0.40) (0.15) ]
2 0.40
-
´ --
´
   = 8.0 Am
2
The pole strength of the magnet is,
m = 
M 8.0
2 0.30
=
l
 = 26.7 Am
21. (a). The situation is shown in figure. The horizontal
component of earth’s magnetic field at the location of
the cable (angle of dip q = 0 is)
10°
10°
Magnetic 
North
North
Geographic 
Meridian
Cable
Geographic
East
Geographic
West
Magnetic
Meridian
Current
Magnetic
South
South
I
B
H 
= B cos q = B cos 0 = B = 0.33 Gauss
    = 0.33 × 10
–4
 Tesla
B
H
 is directed horizontally in the magnetic meridian.
The magnetic field produced by the cable at a distance
of R meter is given by
DPP/ P 42
119
B = 
0
µI
2R p
   = 2 × 10
–7
 × 
7
2.5 5.0 10
T
RR
-
´
=
According to right-hand-palm rule no, 1, the field B is
directed horizontally along B
H
 at a point below the
cable, and opposite to B
H
 at a point above the cable.
Therefore, neutral points will be obtained above the
cable. At these points,  will be equal and opposite to
B
H
. Thus
   
7
5.0 10
R
-
´
 = B
H
 = 0.33 × 10
–4
 .
or  R = 
7
4
5.0 10
0.33 10
-
-
´
´
 = 15 × 10
–3
 m = 1.5 cm
Thus, the line of neutral points lies above and parallel
to the cable at a distance of 1.5 cm from it.
22. (a) The magnetic lines of force are in the form of closed
curves whereas electric lines of force are open curves.
23. (a) Inside a magnet, magnetic lines of force move from
south pole to north pole.
24. (b) Given that pole strength, m = 5.25 × 10
–2
 JT
–1
, q =
45°
and  B = 0.42 G = 0.42 × 10
–4
 T.
45°
B
eq.
B
P
N S
r
Figure shows a point P on the normal bisector of a
short bar magnet lying at a distance r from the
centre O of the magnet. It is given that resultant
magnetic field at P makes an angle of 45° w .r.t. earth’s
field B.
From the figure, it is clear that
B
eq
 = B tan 45° = B.
Now, for a short bar magnet in this position
B
eq
 = B = 
0
3
µ M
.
4
r
p
or r
3
 = 
0
µ M
.
4B p
  = 10
–7
 × 
2
4
5.25 10
0.42 10
-
-
´
´
 = 125 × 10
–7
.
\ r  = 5 × 10
–2
 m = 5 cm.
N
S O
r
Q
B
45°
B
a
Figure shows a point Q on the axis of a short bar magnet
lying at a distance r from the centre of the magnet. It is
given that resultant magnetic field at Q is inclined at
an angle of 45° w.r.t. earth’s field B.
From the figure, it is clear that
B
a
 = B tan 45° = B
Now for a short bar magnet in this position,
B
a
 = B = 
0
3
2M
.
4
r
m
p
or   r
3
 = 
0
2M
.
4B
m
p
= 2 × 125 × 10
–6
\   r = (2 × 125 × 10
–6
)
1/3
 = 6.3 cm
25-27
Given that earth’s magnetic field, B = 0.39 G and angle
of dip, q = 35°
Horizontal and vertical components of earth’s magnetic
field B at the location of the cable are
B
H
 = B cos q = 0.39 cos 35°
= 0.39 × 0.82 = 0.32 Gauss and
B
V
 = B sin q = 0.39 sin 35°
= 0.39 × 0.57 = 0.22 Gauss
The magnetic field produced by four current carrying
straight cable wires at a distance R is
B¢ = 
R 2
µ
0
p
I
× 4 = 2× 10
–7
 × 
1.0
0.04
 × 4
= 0.2 × 10
–4
 T = 0.2 Gauss
Resultant magnetic field below the cable
According to right - hand, palm rule no. 1, the direction
of B¢ below the cable will be opposite to that of
B
H
.Therefore, at a point 4 cm below the cable, resultant
horizontal component of earth’s magnetic field
R
H
 = B
H
 – B¢ = 0.32 – 0.2 = 0.12 Gauss.
Resultant vertical component of earth’s magnetic field
 R
V
 = B
V
 = 0.22 Gauss (unchanged)
\ Resultant magnetic field below the cable is
R = 
22
HV
[R R] +
        = 
22
[(0.12) (0.22) ] +
   = 0.25 Gauss
The angle that R makes with the horizontal is given by
q = tan
–1
 
V
H
R
R
     = tan
–1
 
0.22
0.12
 = tan
–1
(1.8) @ 62°
Resultant magnetic field above the cable
Again, according to right - hand - palm no. 1, the
direction of B¢ at a point above the cable is the same as
that of B
H
.
Therefore, at a point 4 cm below the cable, the
horizontal component of resultant magnetic field will
be
R
H
 = B
H
 + B¢ = 0.32 + 0.20 = 0.52 Gauss
Page 4


1. (a). Force between magnetic poles in air is given by
F = 
0 12
2
mm
4
r
m
´
p
Given that m
1
 = 50 Am, m
2
 = 100Am,
r = 10 cm = 0.1 m and
µ
0
 = permeability of air = 4 p × 10
–7
 Hm
–1
.
\ F = 
7
4 10 50 100
.
4 0.1 0.1
-
p´´
p´
 = 50 × 10
–3
 N
2. (a) Strength of a magnetic field due to a pole of strength m
is given by
H = 
2
1m
.
4
r
p
Given that m = 40 Am, r = 20 cm = 20 × 10
–2
 m.
\ H = 
22
1 40
4
(20 10 )
-
´
p
´
 = 79.57 Am
–1
Now , magnetic induction at the same point :
B = µ
0
 H = 4p × 10
–7
 × 79.57  = 10
–4
 wb/m
2
3. (c) Couple acting on a bar magnet of dipole moment M
when placed in a magnetic field, is given by
t = MB sin q
where q is the angle made by the axis of magnet with
the direction of field.
Given that m = 5 Am, 2 l = 0.2 m, q = 30°
and B = 15 Wbm
–2
\ t = MB sin q  = (m × 2 l) B sin q
  = 5 × 0.2 × 15 × 
1
2
 = 7.5 Nm.
4. (a) F = mB = 
0
3
2m'
m
4
x
m
p
l
= 
7
3
10 2 200 0.05 100
8 10
-
-
´´ ´´
´
= 2.5 × 10
-2
 N
5. (d). W = MB (cosq
1
 - cosq
2
)
\ W
1
 = MB (cos 0º – cos 60º) = 
MB
2
W
2
 = MB (cos 30º – cos 90º) = 
3 MB
2
\ W
2
  = 
3
 WW
1
.
6. (b). Loss in P .E. = gain in K.E.
\ E
k
 = U
i
 – U
j
 = – MB cos 90º – (– MB cos 0º)
      = 4 × 25 × 10
-6
 = 10
-4
 J
7. (a). t = MB sin q = m l B sin q
       = 10
-3
 × 0.1 × 4p × 10
-3
 × 0.5
       = 2p × 10
-7
 N-m
8. (c)
9. ( d) Here, d = 10 cm  = 0.1 m ,
H = 0.4 gauss = 0.4 x 10
-4
 T,  M = ?
Neutral points in this case, lie on axial line of magnet,
such that
0
3
2M
H
4
d
m
=
p
\ 10
-7
 × 
3
2M
(0.1)
 = 0.4 × 10
-4
M = 0.2 A m
2
10. (d). t = F × r = MB sin 30
F = 
mBsin30
r
´ l
= 
2
2
25 10 24 0.25 1
2
12 10
-
-
´ ´´
´
´
 = 6.25 N
11. (b). B = 
0
3
2M
4
x
m
´
p
 = constant
\ x
2
 = x
1
 
1/3
2
1
M
M
æö
ç÷
èø
 = 20 × 
1/3
1
24
æö
ç÷
´ èø
 = 10 cm
12. (d). B =
00
223/23
2M 2M
44
(x)x
mm
==
pp
+l
\ 
3
12
21
Bx
Bx
æö
=
ç÷
èø
 = 8 : 1 approximately . .
13. (c). According to tangent law
B
A
 = B
B
 tan q
or  
00
33
11
2MM
44
dd
mm
=
pp
 tanq
\ 
1
2
d
d
 = (2 cot q)
1/3
14. (a).  T = 2p 
I
MB
 = 2p 
2
p
m
12 mB ´
l
l
 = 4 sec
T' = 2p 
2
p
m
2
m
12B
2
´
l
l
 = 4 sec
DPP/ P 42
118
15. (c).  T = 2p 
I
MB
  or    T µ 
1
M
1
2
T 3M 2M
T 3M 2M
-
=
+
  or  T T
2
 = 5 5s
16. (b).  T = 2p 
I
MB
         = 2p 
2
p
m /12
mB
l
l
  or T  µ ml
1/2
m
T'
nn
T
m
æö
ç÷
=
ç÷
ç÷
èø
l
l
  or  T' = 
T
n
17. (a). The volume of the bar magnet is
V  = 
mass
density
   = 
3
33
6.6 10 kg
7.9 10 kg / m
-
´
´
 = 8.3 × 10
-7
 m
3
.
The intensity of magnetization is
I = 
M
V
 = 
2
72
2.5Am
8.3 10 m
-
-
´
            = 3.0 × 10
6
 A/m
18. (d). The compass box will be on the axial line of the magnet,
Hence,  
33
1 2M 1 2 2m
..
44
rr
´
=
pp
l
 = H tan q
Given that H = horizontal component of the earth’s
magnetic field = 30 Am
–1
, q = 45°,
r = 20 cm = 0.02 m,
M = 2 l m = 4 × 10
–2
 m
Hence, 
2
3
2 4 10m
4 (0.2)
-
´´
p
 = 30 × tan 45°=30 × 1;
\ m = 
3
2
30 4 (0.2)
2 4 10
-
´ p´
´´
 = 37.7 Am
19. (c). As the magnet is placed with its south pole pointing
south, hence the neutral point lies on the equatorial
line. At the neutral point, the magnetic field B due to
the magnet becomes equal and opposite to horizontal
component of earth’s magnetic field i.e., B
H
.
Hence, if M be magnetic dipole moment of the magnet
of length 2l and r the distance of the neutral point from
its centre, then
B = 
0
2 2 3/2
µ M
4
(r)
p
+ l
 = B
H
Given that µ
0
 = 4p × 10
–1
 T mA
–1
,
M = 13.4 Am
2
,
r = 15 cm = 0.15 m and l = 5.0 cm = 0.05 m
\ B
H
 = 10
–7
 × 
2 2 3/2
1.34
[(0.15) (0.5) ] +
  = 10
–7
 × 
1.34
0.025 0.025
 = 0.34 × 10
–4
 TT
20. (a). As the magnet is placed with its north pole pointing
south, the neutral points are obtained on the axial line.
At the neutral points the magnetic field B due to the
magnet becomes equal and opposite to the horizontal
component of earth’s magnetic field i.e., B
H
.
Hence, if M be the magnetic dipole moment of the
magnet of length 2l and r the distance of neutral point
from the centre of the magnet, then we have
B = 
0
H
2 22
µ 2Mr
.B
4
(r)
=
p
-l
Given that µ
0
 = 4p × 10
–7
 TmA
–1
 ,
r = 40 cm = 0.40 m, l = 15 cm = 0.15 m and
H
H
 = 0.34 Gauss = 0.34 × 10
–4
 T
\ M = 
2 22
H
0
B(r) 4
.
µ 2r
- p l
   = 10
–7
× 
4 2 22
(0.34 10 ) [(0.40) (0.15) ]
2 0.40
-
´ --
´
   = 8.0 Am
2
The pole strength of the magnet is,
m = 
M 8.0
2 0.30
=
l
 = 26.7 Am
21. (a). The situation is shown in figure. The horizontal
component of earth’s magnetic field at the location of
the cable (angle of dip q = 0 is)
10°
10°
Magnetic 
North
North
Geographic 
Meridian
Cable
Geographic
East
Geographic
West
Magnetic
Meridian
Current
Magnetic
South
South
I
B
H 
= B cos q = B cos 0 = B = 0.33 Gauss
    = 0.33 × 10
–4
 Tesla
B
H
 is directed horizontally in the magnetic meridian.
The magnetic field produced by the cable at a distance
of R meter is given by
DPP/ P 42
119
B = 
0
µI
2R p
   = 2 × 10
–7
 × 
7
2.5 5.0 10
T
RR
-
´
=
According to right-hand-palm rule no, 1, the field B is
directed horizontally along B
H
 at a point below the
cable, and opposite to B
H
 at a point above the cable.
Therefore, neutral points will be obtained above the
cable. At these points,  will be equal and opposite to
B
H
. Thus
   
7
5.0 10
R
-
´
 = B
H
 = 0.33 × 10
–4
 .
or  R = 
7
4
5.0 10
0.33 10
-
-
´
´
 = 15 × 10
–3
 m = 1.5 cm
Thus, the line of neutral points lies above and parallel
to the cable at a distance of 1.5 cm from it.
22. (a) The magnetic lines of force are in the form of closed
curves whereas electric lines of force are open curves.
23. (a) Inside a magnet, magnetic lines of force move from
south pole to north pole.
24. (b) Given that pole strength, m = 5.25 × 10
–2
 JT
–1
, q =
45°
and  B = 0.42 G = 0.42 × 10
–4
 T.
45°
B
eq.
B
P
N S
r
Figure shows a point P on the normal bisector of a
short bar magnet lying at a distance r from the
centre O of the magnet. It is given that resultant
magnetic field at P makes an angle of 45° w .r.t. earth’s
field B.
From the figure, it is clear that
B
eq
 = B tan 45° = B.
Now, for a short bar magnet in this position
B
eq
 = B = 
0
3
µ M
.
4
r
p
or r
3
 = 
0
µ M
.
4B p
  = 10
–7
 × 
2
4
5.25 10
0.42 10
-
-
´
´
 = 125 × 10
–7
.
\ r  = 5 × 10
–2
 m = 5 cm.
N
S O
r
Q
B
45°
B
a
Figure shows a point Q on the axis of a short bar magnet
lying at a distance r from the centre of the magnet. It is
given that resultant magnetic field at Q is inclined at
an angle of 45° w.r.t. earth’s field B.
From the figure, it is clear that
B
a
 = B tan 45° = B
Now for a short bar magnet in this position,
B
a
 = B = 
0
3
2M
.
4
r
m
p
or   r
3
 = 
0
2M
.
4B
m
p
= 2 × 125 × 10
–6
\   r = (2 × 125 × 10
–6
)
1/3
 = 6.3 cm
25-27
Given that earth’s magnetic field, B = 0.39 G and angle
of dip, q = 35°
Horizontal and vertical components of earth’s magnetic
field B at the location of the cable are
B
H
 = B cos q = 0.39 cos 35°
= 0.39 × 0.82 = 0.32 Gauss and
B
V
 = B sin q = 0.39 sin 35°
= 0.39 × 0.57 = 0.22 Gauss
The magnetic field produced by four current carrying
straight cable wires at a distance R is
B¢ = 
R 2
µ
0
p
I
× 4 = 2× 10
–7
 × 
1.0
0.04
 × 4
= 0.2 × 10
–4
 T = 0.2 Gauss
Resultant magnetic field below the cable
According to right - hand, palm rule no. 1, the direction
of B¢ below the cable will be opposite to that of
B
H
.Therefore, at a point 4 cm below the cable, resultant
horizontal component of earth’s magnetic field
R
H
 = B
H
 – B¢ = 0.32 – 0.2 = 0.12 Gauss.
Resultant vertical component of earth’s magnetic field
 R
V
 = B
V
 = 0.22 Gauss (unchanged)
\ Resultant magnetic field below the cable is
R = 
22
HV
[R R] +
        = 
22
[(0.12) (0.22) ] +
   = 0.25 Gauss
The angle that R makes with the horizontal is given by
q = tan
–1
 
V
H
R
R
     = tan
–1
 
0.22
0.12
 = tan
–1
(1.8) @ 62°
Resultant magnetic field above the cable
Again, according to right - hand - palm no. 1, the
direction of B¢ at a point above the cable is the same as
that of B
H
.
Therefore, at a point 4 cm below the cable, the
horizontal component of resultant magnetic field will
be
R
H
 = B
H
 + B¢ = 0.32 + 0.20 = 0.52 Gauss
DPP/ P 42
120
V ertical component of resultant magnetic field will be
R
V
 = B
V
 – 0.22 Gauss ( unchanged)
Hence, magnitude of resultant magnetic field below
the cable
R  =  
22
HV
[R R] +
    = 
22
[(0.52) (0.52) ] +
     = 0.56 Gauss
The angle that R makes with the horizontal is given by
q = tan
–1
 
V
H
R
R
æö
ç÷
èø
= tan
–1
 
0.22
0.52
æö
ç÷
èø
   = tan
–1
 (0.43) @ 23°
25. (a)         26. (a)        27. (b)
28. (a)
29. (d) The earth has only vertical component of its magnetic
field at the magnetic poles. Since compass needle is
only free to rotate in horizontal plane. At north pole
the vertical component of earth's field will exert torque
on the magnetic needle so as to aligh it along its direc-
tion. As the compass needle can not rotate in vertical
plane, it will rest horizontally , when placed ont he mag-
netic pole of the earth.
30. (c) It is quite clear that magnetic poles always exists in
pairs. Since, one can imagine magnetic field configu-
ration with three poles. When north poles or south poles
of two magnets are glued together. They provide a three
pole field configuration. It is also known that a bar
magnet does not exert a torque on itself due to own its
field.