Q.A bar magnet of dipole moment $0.9\ \text{A}\cdot\text{m}^2$ is placed in a uniform magnetic field of $0.25\ \text{T}$, with its axis making an angle of $30^\circ$ with the field. Calculate the torque acting on the magnet.

A short bar magnet placed in a uniform external magnetic field $\vec B$ experiences a torque $\vec\tau=\vec m\times\vec B$, of magnitude $\tau=mB\sin\theta$, where $\vec m$ is the magnet's dipole moment and $\theta$ is the angle between $\vec m$ and $\vec B$. This is exactly the same result derived earlier for a current-carrying coil in a uniform field, applied here to a bar magnet: the forces on its two poles are equal in magnitude, act along different (parallel) lines through the two poles, and so form a couple. Because the two pole forces are equal and opposite, they cancel to give zero NET force in a uniform field -- only a pure rotational torque results, which tends to rotate the magnet until $\vec m$ is aligned with $\vec B$. …