$A$ bar magnet of magnetic moment $1.5 \, J/T$ lies aligned with the direction of a uniform magnetic field of $0.22 \, T$. What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment perpendicular to the field direction? (in $J$)

$A$ magnetic dipole of magnetic moment $ 6 \times 10^{-2} \text{ A m}^2 $ and moment of inertia $ 12 \times 10^{-6} \text{ kg m}^2 $ performs oscillations in a magnetic field of $ 2 \times 10^{-2} \text{ T} $. The time taken by the dipole to complete $ 20 $ oscillations is (assume $ \pi \simeq 3 $). (in $\text{ s}$)

If a bar magnet of moment $10^{-4} Am^2$ is kept in a uniform magnetic field of $12 \times 10^{-3} T$ such that it makes an angle of $30^{\circ}$ with the direction of the magnetic field, then the torque acting on the magnet is:

Write an expression for the magnitude of the magnetic field at a point lying on the equatorial line of a bar magnet.

The work done in rotating a bar magnet from its equilibrium position by $60^{\circ}$ in a magnetic field is $W$. What is the torque required to hold it in this position?

$A$ small magnetized needle $P$ is placed at point $O$,and the arrow shows the direction of its magnetic moment. The other arrows show different positions (and orientations) of another identical magnet $Q$. In which configuration is the system $NOT$ in equilibrium?