Question

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. \(T = \sqrt{\frac{2m}{IB}}\)
• B. \(T = \sqrt{\frac{\pi m}{IB}}\)
• C. \(T = \sqrt{\frac{\pi m}{2IB}}\)
• D. \(T = \sqrt{\frac{2\pi m}{IB}}\)

Hint:

For a small angular displacement \(\theta\) the restoring torque is \(MB\theta\) with \(M = I\pi a^2\). The moment of inertia of a ring about its diameter is \(\tfrac12 ma^2\).

Answer 1

The Correct Option is D

Magnetic moment: \(M = I\pi a^2\). Rotating the loop by a small angle \(\theta\) about a diameter gives a restoring torque \(\tau = -MB\sin\theta \approx -I\pi a^2 B\,\theta\).
Moment of inertia of the ring about a diameter: \(I_d = \tfrac12 ma^2\).
Equation of motion: \(I_d\ddot\theta = -I\pi a^2 B\,\theta\), so \(\omega^2 = \frac{I\pi a^2 B}{\tfrac12 ma^2} = \frac{2\pi IB}{m}\).
\(T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{m}{2\pi IB}} = \sqrt{\frac{4\pi^2 m}{2\pi IB}} = \sqrt{\frac{2\pi m}{IB}}\).
Answer: (4) \(T = \sqrt{\frac{2\pi m}{IB}}\)