$A$ magnet of magnetic moment $M$ is rotated through $360^{\circ}$ in a magnetic field $H$. The work done will be:

$A$ magnet of magnetic moment $50 \hat{i} \text{ Am}^2$ is placed along the $x$-axis in a magnetic field $\vec{B} = (0.5 \hat{i} + 3.0 \hat{j}) \text{ T}$. The torque acting on the magnet is:

$A$ short bar magnet placed with its axis at $30^{\circ}$ with a uniform external magnetic field of $0.16 \ T$ experiences a torque of magnitude $0.032 \ J$. The magnetic moment of the bar magnet will be......$J/T$.

Torques $\tau_1$ and $\tau_2$ are required for a magnetic needle to remain perpendicular to the magnetic fields of $B_1$ and $B_2$ at two different places. The ratio of $B_1: B_2$ is equal to

$A$ bar magnet is held perpendicular to a uniform magnetic field. If the couple acting on the magnet is to be halved by rotating it,then the angle by which it is to be rotated is....$^o$

The intensity of the magnetic field is $H$ and the magnetic moment of the magnet is $M$. The maximum potential energy is $..... MH$.