Show that the time period `(T)` of oscillations of a freely suspended magnetic dipole of magnetic moment (m) in a uniform magnetic field (B) is given by `T=2 pi sqrt((I)/(mB))` , where I is moment of inertia of the magnetic dipole.

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))`.