The work done in rotating a magnet of magnetic moment $2 \, A-m^2$ in a magnetic field of $5 \times 10^{-3} \, T$ from the direction along the magnetic field to the opposite direction to the magnetic field is:

The ratio of magnetic fields due to a bar magnet at two axial points $P_1$ and $P_2$,which are separated from each other by $10 \ cm$,is $25: 2$. Point $P_1$ is situated at $10 \ cm$ from the center of the magnet. (Points $P_1$ and $P_2$ are on the same side of the magnet,and the distances of $P_1$ and $P_2$ from the center are much greater than the distance of the two ends of the magnet from the center). The magnetic length of the bar magnet is: (in $cm$)

$A$ short bar magnet placed with its axis at $30^{\circ}$ with a uniform external magnetic field of $0.25 \ T$ experiences a torque of magnitude $4.5 \times 10^{-2} \ J$. The magnitude of the magnetic moment of the magnet will be . . . . . . $J \ T^{-1}$.

The magnitude of the axial field due to a short bar magnet at a distance of $50 \,cm$ from its mid-point is (The magnetic moment of the bar magnet is $0.4 \,Am^2$)

$A$ 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?

$A$ short magnetic needle is placed in a magnetic field $B \hat{i}$. The needle is oriented in the direction $(\sqrt{3} \hat{i} + \hat{j})$. The needle experiences a torque of $0.06 \ N-m$. If the same magnetic needle is placed in a magnetic field $2B \hat{j}$ and is oriented in the direction $(\hat{i} + \sqrt{3} \hat{j})$,the torque experienced by it is: (in $N-m$)