The magnetic moment of a bar magnet is $0.5 \text{ A m}^2$. It is suspended in a uniform magnetic field of $8 \times 10^{-2} \text{ T}$. The work done in rotating it from its most stable to most unstable position is:

At a point on the equatorial line (right bisector) of a magnetic dipole,the magnetic:

The dipole moment of a short bar magnet is $1.25 \, A-m^2$. The magnetic field on its axis at a distance of $0.5 \, m$ from the centre of the magnet is:

Some equipotential surfaces of the magnetic scalar potential are shown in the figure. The magnetic field at a point in the region is

Two short bar magnets have their magnetic moments $1.2 \text{ A m}^2$ and $1.0 \text{ A m}^2$. They are placed on a horizontal table parallel to each other at a distance of $20 \text{ cm}$ between their centres,such that their north poles point towards geographic south. They have a common magnetic equatorial line. The horizontal component of the Earth's magnetic field is $3.6 \times 10^{-5} \text{ T}$. The resultant horizontal magnetic induction at the midpoint of the line joining their centers is $\left(\frac{\mu_0}{4 \pi} = 10^{-7} \text{ N/A}^2\right)$.

$A$ magnet of magnetic moment $2 \,J \,T^{-1}$ is aligned in the direction of a magnetic field of $0.1 \,T$. What is the net work done to bring the magnet normal to the magnetic field (in $\,J$)?