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 $\cos 90^{\circ} = 0, \cos 60^{\circ} = 0.5$).

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

$A$ bar magnet having a magnetic moment of $2 \times 10^4 \, JT^{-1}$ is free to rotate in a horizontal plane. $A$ horizontal magnetic field $B = 6 \times 10^{-4} \, T$ exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction $60^{\circ}$ from the field is.....$J$

If a magnet of length $10 \, cm$ and pole strength $40 \, A-m$ is placed at an angle of $45^\circ$ in a uniform magnetic field of intensity $2 \times 10^{-4} \, T$,the torque acting on it is:

$A$ magnetic dipole is under the influence of two orthogonal magnetic fields, $B_1 = 0.5 \times 10^{-3} \ T$ and $B_2 = 0.866 \times 10^{-3} \ T$. If the dipole comes to stable equilibrium at an angle $\theta$ with respect to the $B_2$ field, then the value of $\theta$ is (in $^{\circ}$)

Figure shows a small magnetised needle $P$ placed at a point $O$. The arrow shows the direction of its magnetic moment. The other arrows show different positions (and orientations of the magnetic moment) of another identical magnetised needle $Q$.
$(a)$ In which configuration the system is not in equilibrium?
$(b)$ In which configuration is the system in $(i)$ stable,and $(ii)$ unstable equilibrium?
$(c)$ Which configuration corresponds to the lowest potential energy among all the configurations shown?