$A$ magnet of magnetic moment $M$ is rotated through $360^{\circ}$ in a magnetic field $H$. The work done will be:

$A$ small bar magnet of magnetic moment $M$ is placed in a uniform magnetic field $H$. If the magnet makes an angle of $30^{\circ}$ with the field,the torque acting on the magnet is:

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:

$A$ planet has a magnetic dipole moment of $27 \times 10^{22} \ A \ m^2$. If the radius of the planet is $300 \ km$, what would be the magnetic field at its equator (in $T$)? (Take $\frac{\mu_0}{4 \pi} = 10^{-7} \ T \ m/A$)

The magnetic moment of a magnet is $2.5 \, J \, T^{-1}$. How much work in $J$ must be done to rotate it from its stable equilibrium position to an unstable equilibrium position in a magnetic field of $0.2 \, T$?

$A$ 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 $, J$)?