$A$ bar magnet is held perpendicular to a uniform magnetic field. The couple acting on the magnet is to be halved by rotating it. Through what angle should it be rotated?

$A$ bar magnet of magnetic moment $M$ and moment of inertia $I$ is freely suspended such that the magnetic axial line is in the direction of the magnetic meridian. If the magnet is displaced by a very small angle $\theta$, the angular acceleration is (Magnetic induction of earth's horizontal field $= B_H$)

$A$ magnet of magnetic moment $20$ $C.G.S.$ units is freely suspended in a uniform magnetic field of intensity $0.3$ $C.G.S.$ units. The amount of work done in deflecting it by an angle of $30^{\circ}$ in $C.G.S.$ units is:

The magnetic field due to a bar magnet at a distance $R$ from the center of the magnet (where $R$ is much larger than the length of the magnet) is proportional to:

Two short bar magnets have magnetic moments $1.2 \text{ Am}^2$ and $1.0 \text{ Am}^2$. 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 the geographic south. They share a common magnetic equatorial line. The horizontal component of the Earth's magnetic field is $3.6 \times 10^{-5} \text{ T}$. Calculate the resultant horizontal magnetic induction at the midpoint of the line joining their centres. (Given: $\frac{\mu_0}{4 \pi} = 10^{-7} \text{ N/A}^2$)

If there is no torsion in the suspension thread, then the time period of a magnet executing $SHM$ is