Question

A small compass needle of magnetic moment ‘m’ is free to turn about an axis perpendicular to the direction of uniform magnetic field ‘B’. The moment of inertia of the needle about the axis is ‘I’. The needle is slightly disturbed from its stable position and then released. Prove that it executes simple harmonic motion. Hence deduce the expression for its time period.

Answer 1

The torque on the needle is τ = m × B

In magnitude τ = mB sinθ

Here τ is restoring torque and θ is the angle between m and B.

Therefore, in equilibrium `I (d^2θ)/dt^2 =-mBsinθ`

Negative sign with mB sinθ implies that restoring torque is in opposition to deflecting torque. For small values of θ in radians, we approximate sinθ= θ and get

`I (d^2θ)/dt^2 =-mBθ`

or

`(d^2θ)/dt^2 =-(mB)/Iθ`

This represents a simple harmonic motion. The square of the angular frequency is ω² = mB/J and the time period is,`T = 2πsqrtI/(mB)`