The work done in rotating a bar magnet of magnetic moment $M$ from its unstable equilibrium position to its stable equilibrium position in a uniform magnetic field $B$ 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 $\left(\frac{\mu_0}{4 \pi} = 10^{-7} \text{ N/A}^2\right)$.

$A$ magnetic dipole of magnetic moment $M$ is freely suspended in a magnetic field of induction $B$. The minimum and maximum values of the potential energy of the dipole,respectively,are

$A$ short bar magnet placed with its axis at $30^{\circ}$ with an external field of $800 \; G$ experiences a torque of $0.016 \; Nm.$
$(a)$ What is the magnetic moment of the magnet?
$(b)$ What is the work done in moving it from its most stable to most unstable position?
$(c)$ The bar magnet is replaced by a solenoid of cross-sectional area $2 \times 10^{-4} \; m^{2}$ and $1000$ turns,but of the same magnetic moment. Determine the current flowing through the solenoid.

$A$ 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

The magnetic field due to a small magnetic dipole of magnetic moment $M$,at a distance $r$ from the centre on the equatorial line is given by (in $M.K.S.$ system):