$A$ magnet is parallel to a uniform magnetic field. If it is rotated by $60^o$,the work done is $0.8\, J$. How much work is done in moving it $30^o$ further (in $, J$)?

$A$ short bar magnet placed with its axis at $30^{\circ}$ with a uniform external magnetic field of $0.25 \; T$ experiences a torque of magnitude equal to $4.5 \times 10^{-2} \; J$. What is the magnitude of magnetic moment (in $J \; T^{-1}$) of the magnet?

The magnetic moment of a current-carrying loop is $2.1 \times 10^{-25} \text{ A m}^2$. The magnetic field at a point on its axis at a distance of $1 \text{ Å}$ is:

The rate of change of torque $\tau$ with respect to deflection $\theta$ is maximum for a magnet suspended freely in a uniform magnetic field of induction $B$ when $\theta = ........ ^\circ$.

$A$ bar magnet of magnetic moment $2 \text{ A m}^2$ lies aligned with the direction of a uniform magnetic field of $0.3 \text{ T}$. The amount of work required by an external torque to turn the magnet so as to align its magnetic moment normal to the field direction is (in $\text{ J}$)

$A$ bar magnet of length $6\,cm$ has a magnetic moment of $4\,J\,T^{-1}$. Find the strength of magnetic field at a distance of $200\,cm$ from the centre of the magnet along its equatorial line.