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Test Index
JEE Main Physics Class 12 Magnetism and Matter Questions
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© examsnet.com
Question : 40
Total: 80
A circular coil has moment of inertia
0.8
k
g
m
2
around any diameter and is carrying current to produce a magnetic moment of
20
A
m
2
. The coil is kept initially in a vertical position and it can rotate freely around a horizontal diameter. When a uniform magnetic field of
4
T
is applied along the vertical, it starts rotating around its horizontal diameter. The angular speed the coil acquires after rotating by
60
°
will be:
[Sep. 04, 2020 (II)]
10
r
a
d
s
−
1
10
π
r
a
d
s
−
1
20
π
r
a
d
s
−
1
20
r
a
d
s
−
1
Validate
Solution:
Given,
Moment of inertia of circular coil,
I
=
0.8
k
g
m
2
Magnetic moment of circular coil,
M
=
20
A
m
2
Rotational kinetic energy of circular coil,
K
E
=
1
2
I
ω
2
Here,
ω
=
angular speed of coil
Potential energy of bar magnet
=
−
M
B
cos
ϕ
From energy conservation
1
2
I
ω
2
=
U
in
−
U
f
=
−
M
B
cos
60
°
−
(
−
M
B
)
⇒
M
B
2
=
1
2
I
ω
2
⇒
20
×
4
2
=
1
2
(
0.8
)
ω
2
⇒
100
=
ω
2
⇒
ω
=
10
r
a
d
© examsnet.com
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