Answer in brief.
Prove that under certain conditions a magnet vibrating in a uniform magnetic field performs angular S.H.M.
Consider a bar magnet of the magnetic moment μ, suspended horizontally by a light twistless fiber in a region where the horizontal component of the Earth's magnetic field is B. The bar magnet is free to rotate in a horizontal plane. It comes to rest in approximately the North-South direction, along with B. If it is rotated in the horizontal plane by a small
displacement θ from its rest position (θ = 0), the suspension fiber is twisted. When the magnet is released, it oscillates about the rest position in angular or torsional oscillation.
The bar magnet experiences a torque t due to field B. This tends to restore it to its original orientation parallel to B. For small θ, this restoring torque is Γ = μBθ ............(1)
where the minus sign indicates that the torque is opposite in direction to the angular displacement e. Equation (1) shows that the torque (and hence the angular acceleration) is directly proportional to the magnitude of the angular displacement but opposite in direction. Hence, for small angular displacement, the oscillations of the bar magnet in a uniform magnetic field are simple harmonic.