$A$ bar magnet of magnetic moment $1.5 \, A \cdot m^2$ lies aligned with the direction of a uniform magnetic field of $2 \, T$. What is the amount of work required to turn the magnet so as to align its magnetic moment normal to the field direction?

$A$ magnetic needle is placed in a uniform magnetic field and is aligned with the field. The needle is now rotated by an angle of $60^{\circ}$ and the work done is $W$. The torque on the magnetic needle at this position is

$A$ bar magnet having a magnetic moment of $2.0 \times 10^{5} \; J T^{-1}$ is placed along the direction of a uniform magnetic field of magnitude $B = 14 \times 10^{-5} \; T$. The work done in rotating the magnet slowly through $60^{\circ}$ from the direction of the field is .............. $J$.

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:

The magnetic field at a point $P$ on the axis of a short bar magnet of magnetic moment $M$ is $B$. If another short bar magnet of magnetic moment $2M$ is placed on the first magnet such that their axes are perpendicular and their centres coincide. The resultant magnetic field at the point $P$ due to both the magnets is

Deduce the expression for the potential energy of a bar magnet in a uniform magnetic field and discuss special cases.