Consider a bar magnet of magnetic moment μ, suspended horizontally by a light twistless fibre in a region where the horizontal component of the Earth’s magnetic field is Bh. The bar magnet is free to rotate in a horizontal plane. It comes to rest in approximately the North-South direction, along Bh. If it is rotated in the horizontal plane by a small
displacement θ from its rest position (θ = 0), the suspension fibre is twisted. When the magnet is released, it oscillates about the rest position in angular or torsional oscillation.
The bar magnet experiences a torque \(\tau\) due to the field Bh. Which tends to restore it to its original orientation parallel to Bh. For small θ, this restoring torque is
\(\tau\) = – μBh sin θ = – μBhμ …. (1)
where the minus sign indicates that the torque is opposite in direction to the angular displacement θ. Equation (1) shows that the torque (and hence the angular acceleration) is directly proportional in 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 is simple harmonic.
