JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Elasticity & SHM

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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:

(A) $T = \frac{π m}{IB}$
(B) $T = \frac{π m}{2 I B}$
(C) $T = \frac{2 π m}{I B}$ ✓ Correct
(D) $T = \frac{2 m}{I B}$

Answer: (C)

Solution

Tilting the current loop by a small angle

θ

from the field produces a restoring torque

τ = - M B sin θ \approx - M B θ

, with magnetic moment

M = I π a^{2}

. About a diameter the loop's moment of inertia is

\frac{m a ^{2}}{2}

, so the equation

\frac{m a ^{2}}{2} \ddot{θ} = - I π a^{2} B θ

describes simple harmonic motion with

ω^{2} = \frac{2 π I B}{m}

. The period is

T = \frac{2 π}{ω} = \frac{2 π m}{I B}

JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Single correct option

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JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Elasticity & SHM

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