$A$ magnetic needle suspended parallel to a magnetic field requires $\sqrt{3} \, J$ of work to turn it through $60^{\circ}$. The torque needed to maintain the needle in this position will be.....

The magnitude of the axial field due to a bar magnet at a distance of $1 \ m$ is found to be $5 \times 10^{-8} \ T$. The magnetic moment of the bar magnet is $\left(\mu_0 = 4 \pi \times 10^{-7} \ T \ m/A\right)$. (in $A \ m^2$)

The figure shows a bar magnet and a long straight wire $W$,carrying current into the plane of the paper. Point $P$ is the point of intersection of the axis of the magnet and the line of shortest distance between the magnet and the wire. If $P$ is the midpoint of the magnet,then which of the following statements is correct?

Two identical magnetic dipoles of magnetic moments $1.0 \, A-m^2$ each are placed at a separation of $2 \, m$ with their axes perpendicular to each other. The resultant magnetic field at a point $P$ midway between the dipoles is:

$A$ bar magnet of magnetic moment $1.5 \, J \, T^{-1}$ lies aligned with the direction of a uniform magnetic field of $0.22 \, T$.
$(a)$ What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment: $(i)$ normal to the field direction,$(ii)$ opposite to the field direction?
$(b)$ What is the torque on the magnet in cases $(i)$ and $(ii)?$

$A$ 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 $= B_H$)