$A$ magnetic needle lying parallel to a magnetic field requires $W$ units of work to turn it through $60^{\circ}$. The torque required to maintain the needle in this position will be

$A$ short bar magnet placed in a uniform magnetic field making an angle with the field experiences a torque. If the angle made by the magnet with the field is changed from $30^{\circ}$ to $45^{\circ}$,the torque of the magnet

Two short magnets of equal dipole moments $M$ are fastened perpendicularly at their centre (figure). The magnitude of the magnetic field at a distance $d$ from the centre on the bisector of the right angle is

$A$ magnetic dipole of moment $2.5 \text{ A m}^2$ is free to rotate about a vertical axis passing through its centre. It is released from the East-West direction. What is its kinetic energy at the moment it takes the North-South position (in $\mu\text{J}$)? (Given: $B_H = 3 \times 10^{-5} \text{ T}$)

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

The magnetic moment of a short magnet is $16 \ Am^2$. The magnetic induction at a point $20 \ cm$ away from its midpoint on $(i)$ axial point and $(ii)$ equatorial point respectively,will be: