$A$ short bar magnet of magnetic moment $10^4 \,J \,T^{-1}$ is free to rotate in a horizontal plane. The work done in rotating the magnet slowly from the direction parallel to a horizontal magnetic field of $4 \times 10^{-5} \,T$ to a direction $60^{\circ}$ to the direction of the field is (in $J$)

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

Two short bar magnets, each with a magnetic moment of $9 \text{ Am}^2$, are placed such that one is at $x = -3 \text{ cm}$ and the other is at $y = -3 \text{ cm}$. If their magnetic moments are directed along the positive and negative $X$-directions respectively, then the resultant magnetic field at the origin is: (in $\text{ T}$)

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

$A$ 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?

The magnetic moment of a current-carrying loop is $2.1 \times 10^{-25} \text{ A m}^2$. The magnetic field at a point on its axis at a distance of $1 \text{ Å}$ is: