$A$ magnetic needle suspended parallel to a magnetic field requires $\sqrt{3} \text{ J}$ of work to turn it through $60^{\circ}$. The torque needed to maintain the needle in this position will be

$A$ bar magnet has a magnetic moment of $200 \text{ A m}^2$. The magnet is suspended in a magnetic field of $0.30 \text{ N A}^{-1} \text{ m}^{-1}$. The torque required to rotate the magnet from its equilibrium position through an angle of $30^{\circ}$ will be:

Two short bar magnets of magnetic moments $m$ each are arranged at the opposite corners of a square of side $d$ such that their centres coincide with the corners and their axes are parallel. If the like poles are in the same direction,the magnetic induction at any of the other corners of the square is

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

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

$A$ short bar magnet is placed in a uniform magnetic field of $2 \ T$ such that the axis of the magnet makes an angle of $45^{\circ}$ with the direction of the magnetic field. If the torque acting on the magnet is $0.36 \sqrt{2} \ Nm$,then the magnetic moment of the magnet is: (in $J \ T^{-1}$)