Two short magnets are placed along the same axis with their like poles facing each other. They repel each other with a force which varies inversely as:

$A$ 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:

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 $\left(\frac{\mu_0}{4 \pi} = 10^{-7} \text{ N/A}^2\right)$.

The magnetic field due to a short magnet at a point on its axis at a distance $X \, cm$ from the center of the magnet is $200 \, G$. The magnetic field at a point on the equatorial axis at a distance $X \, cm$ from the center of the magnet is.....$G$.

$A$ 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?

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