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 :

Solution

Physics sir · 4 steps

Treat the loop like a compass needle: field torque pulls it back, ring inertia resists.

1. Restoring torque. The loop has magnetic moment M = Iπ a², and in equilibrium it lines up with B. Tilt it by a small angle θ and the torque is MB sin θ ≈ MBθ, pulling it back. That is the condition for SHM.
2. Moment of inertia. It rotates about a diameter. For a thin ring that moment of inertia is half the value about the central axis, so J = 1/2ma². The letter J is used here so it is not confused with the current I.

   T = 2π√(J/MB) = 2π√((1/2ma²)/(Iπ a² B)) = 2π√(m/(2π IB))

3. Tidy up. The a² cancels. Take 2π inside the root as 4π²: T = √((4π² m)/(2π IB)) = √((2π m)/IB).
4. So. The time period is T = √((2π m)/IB), which is option (d).

The idea to keep. Any magnetic dipole oscillating in a uniform field has T = 2π√(J/(MB)). Find M, pick the moment of inertia about the actual axis of rotation, and you are done.

Answer from the NTA final answer key. Solution by PrepWiser.