The intensity of the magnetic field is $H$ and the magnetic moment of the magnet is $M$. The maximum potential energy is $..... MH$.

$A$ short bar magnet produces a magnetic field of $6.4 \times 10^{-5} \,T$ at a distance of $20 \,cm$ from the centre of the magnet on the normal bisector of the magnet. The magnetic field produced by this magnet at a distance of $40 \,cm$ from the centre of the magnet on the axis,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 $\left(\frac{\mu_0}{4 \pi} = 10^{-7} \text{ N/A}^2\right)$.

$A$ 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 $\text{ J}$)

The midpoints of two small magnetic dipoles of length $d$ in end-on positions are separated by a distance $x$ $(x >> d)$. The force between them is proportional to $x^{-n}$,where $n$ is:

$A$ magnetic needle lying parallel to a magnetic field requires $W$ units of work to turn it through $60^{\circ}$. The torque required to maintain the needle in this position will be