A small compass needle of magnetic moment 'm' and moment of inertia 'l' is free to oscillate in a magnetic field 'B'. It is slightly disturbed from its equilibrium position and then released. Show that it executes simple harmonic motion. Hence, write the expression for its time period.

Explanation

Let a small magnetic needle of magnetic moment `vecm` be freely suspended in a uniform magnetic field `vecB` so that in equilibrium positive magnet comes to rest along the direction of `vecB`.
If the magnetic needle is rotated by a small angle `theta` from its equilibrium position magnet comes to rest along the direction of `vecB`.
if the magnetic needle is rotated by a small angle `theta` from its equilibrium and then released , a restoring torque acts on the magnet, where
Restoring torque `vectau = vecm xx vecB`
or `tau = - m B sin theta`
If I be the moment of inertia of magnetic needle about the axis of suspension, then
`tau = I alpha = I (d^2 theta)/(dt^2)`
Hence, in equilibrium state, we have
`I = (d^2 theta)/(dt^2) = - m B sin theta`
If `theta ` is small then `sin theta to theta` and we get ,br> `I (d^2 theta)/(dt) = - mB theta ` or `(d^2 theta)/(dt^2) = - (mB)/(I) theta`
As here angular acceleration is directly proportional to angular displacement and direction towards the equilibrium position, motion of the magnetic needle is simple harmonic motion, and Angular frequency of SHM `omega = sqrt((mB)/(I))`
`therefore ` Time period of oscillation `T= (2pi)/(omega) = 2pi sqrt((I)/(mB))`.