The work done in turning a magnet of magnetic moment $M$ by an angle of $90^{\circ}$ from the magnetic meridian is $n$ times the corresponding work done to turn it through an angle of $60^{\circ}$. Find the value of $n$ (given $\cos 90^{\circ} = 0, \cos 60^{\circ} = 0.5$).

$A$ small magnetized needle $P$ is placed at point $O$,and the arrow shows the direction of its magnetic moment. The other arrows show different positions (and orientations) of another identical magnet $Q$. In which configuration is the system $NOT$ in equilibrium?

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: $\mu_0 = 4\pi \times 10^{-7} \, Hm^{-1}$)

$A$ magnet of length $10 \text{ cm}$ and magnetic moment $1 \text{ Am}^2$ is placed along side $AB$ of an equilateral triangle $ABC$. If the length of the side $AB$ is $10 \text{ cm}$, find the magnetic induction at point $C$. (Given $\mu_0 = 4\pi \times 10^{-7} \text{ Hm}^{-1}$)

$A$ bar magnet of length $10 \text{ cm}$ and having the pole strength equal to $10^{-3} \text{ A-m}$ is kept in a magnetic field having magnetic induction $B$ equal to $4 \pi \times 10^{-3} \text{ T}$. It makes an angle of $30^{\circ}$ with the direction of magnetic induction. The value of the torque acting on the magnet is

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 .........