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JEE Main Physics Class 12 Moving Charges and Magnetism Part 2 Questions
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Question : 95
Total: 100
A small circular loop of conducting wire has radius a and carries current I. It is placed in a uniform magnetic field B perpendicular to its plane such that when rotated slightly about its diameter and released, it starts performing simple harmonic motion of time period T. If the mass of the loop is m then :
[9 Jan 2020, II]
T
=
√
2
m
I
B
T
=
√
π
m
2
I
B
T
=
√
2
π
m
I
B
T
=
√
π
m
I
B
Validate
Solution:
Torque on circular loop,
τ
=
M
B
sin
θ
where,
M
=
magnetic moment
B
=
magnetic field
Now, using
τ
=
I
α
∴
τ
=
M
B
sin
θ
=
I
α
⇒
π
R
2
I
B
θ
=
m
R
2
α
2
(
∵
m
=
I
A
and moment of inertia of circular loop,
I
=
m
R
2
2
)
⇒
π
R
2
I
B
θ
=
m
R
2
2
ω
θ
⇒
ω
=
√
2
π
I
B
m
⇒
2
π
T
=
√
2
π
I
B
m
⇒
T
=
√
2
π
m
I
B
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