$A$ magnet of magnetic moment $M$ is lying in a uniform magnetic field $B$. $W_1$ is the work done in turning it from $0^o$ to $60^o$ and $W_2$ is the work done in turning it from $30^o$ to $90^o$. Then:

$A$ bar magnet has a magnetic moment of $200 \text{ A m}^2$. The magnet is suspended in a magnetic field of $0.30 \text{ N A}^{-1} \text{ m}^{-1}$. The torque required to rotate the magnet from its equilibrium position through an angle of $30^{\circ}$ will be:

$A$ negative charge is given to a nonconducting loop and the loop is rotated in the plane of paper about its centre as shown in the figure. The magnetic field produced by the ring affects a small magnet placed above the ring in the same plane:

Two identical short bar magnets,each having a magnetic moment of $10 \, Am^2$,are arranged such that their axial lines are perpendicular to each other and their centres lie along the same straight line in a horizontal plane. If the distance between their centres is $0.2 \, m$,find the resultant magnetic induction at a point midway between them. (Given: $\mu_0 = 4\pi \times 10^{-7} \, Hm^{-1}$)

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

Derive the equation of torque on a magnetic needle in a uniform magnetic field.