A small compass needle of magnetic moment ‘M’ and moment of inertial ‘I’ 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

If a small bar magnet placed in uniform magnetic field `vecB` in equilibrium, is rotated through a small angle θ, then it experiences a restoring torque, which tends to align it in the direction of magnetic field, given by `tau_(R)=-Mbsintheta |tau_(R)|=I(d^(2)theta)/(dt^(2))= MB sin theta` or `(d^(2) theta)/(dt^(2))=(MB)/I sin theta`
For small angle `theta, sin theta~~theta` so this represents SHM. So small bar magnet executes SHM in uniform magnetic field of time period
`T=(2 pi)/(omega)=(2pi)/(sqrt(MB)/I)[ :' omega^(2)=(MB)I]` or `T=2pi sqrt(I/(MB))`
where I is moment of inertia of bar magnet.