$A$ bar magnet of magnetic moment $200 \, A-m^2$ is suspended in a magnetic field of intensity $0.25 \, N/A-m$. The torque required to deflect it through $30^\circ$ is .... $N-m$.

$A$ 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 $20$ $C.G.S.$ units is freely suspended in a uniform magnetic field of intensity $0.3$ $C.G.S.$ units. The amount of work done in deflecting it by an angle of $30^{\circ}$ in $C.G.S.$ units is:

$A$ 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

$A$ short bar magnet of magnetic moment $M=0.32 \; J \, T^{-1}$ is placed in a uniform magnetic field of $0.15 \; T$. If the bar is free to rotate in the plane of the field,which orientation would correspond to its $(a)$ stable,and $(b)$ unstable equilibrium? What is the potential energy of the magnet in each case?

The work done in rotating a magnet of magnetic moment $2 \, A-m^2$ in a magnetic field of $5 \times 10^{-3} \, T$ from the direction along the magnetic field to the opposite direction to the magnetic field is: