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JEE Main Physics Class 12 Magnetism and Matter Questions
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© examsnet.com
Question : 33
Total: 80
A bar magnet of length
14
c
m
is placed in the magnetic meridian with its North pole pointing towards the geographic North pole A neutral point is obtained at a distance of
18
c
m
from the centre of the magnet. If
B
H
=
0.4
G
, then the magnetic moment of the magnet is
(
1
G
=
10
−
4
T
)
[16 Mar 2021 Shift 1]
2.88
×
10
3
J
T
−
1
2.88
×
10
2
J
T
−
1
2.88
J
T
−
1
28.8
J
T
−
1
Validate
Solution:
The given situation can be shown as below
From the above figure, it is clear that the neutral point will lie on equitorial plane.
B
0
=
µ
0
m
4
π
1
r
2
B
H
=
2
B
0
cos
θ
⇒
0.4
×
10
−
4
=
2
µ
0
m
4
π
r
2
⋅
7
×
10
−
2
r
(
∵
B
H
=
0.4
G
=
0.4
×
10
−
4
T
)
[
∵
cos
θ
=
7
r
and
r
=
(
7
2
+
18
2
)
1
∕
2
]
⇒
0.4
×
10
−
4
=
2
×
10
−
7
×
m
×
7
(
7
2
+
18
2
)
3
/
2
×
10
4
⇒
m
=
4
×
10
−
2
×
(
373
)
3
/
2
14
...(i)
∵
Magnetic moment,
M
=
m
×
2
l
=
m
×
14
100
...(ii)
From Eqs. (i) & (ii), we get
M
=
4
×
10
−
2
×
(
373
)
3
/
2
14
×
14
100
=
4
×
10
−
4
×
7203.82
=
2.88
J
/
T
© examsnet.com
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