$A$ bar magnet having a magnetic moment of $2 \times 10^4 \, JT^{-1}$ is free to rotate in a horizontal plane. $A$ horizontal magnetic field $B = 6 \times 10^{-4} \, T$ exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction $60^{\circ}$ from the field is.....$J$

The work done in turning a magnet of magnetic moment $M$ by an angle $\theta$ from its initial position parallel to the magnetic meridian is:

If the dipole moment of a short bar magnet is $1.25 \ A-m^2$,find the magnetic field on its axis at a distance of $0.5 \ m$ from the centre of the magnet.

Two magnets $A$ and $B$ are identical and are arranged as shown in the figure. Their length is negligible in comparison to the separation between them. $A$ magnetic needle is placed between the magnets at point $P$,which gets deflected through an angle $\theta$ under the influence of the magnets. The ratio of distances $d_1$ and $d_2$ will be:

The magnetic moment of a bar magnet is $0.5 \text{ A m}^2$. It is suspended in a uniform magnetic field of $8 \times 10^{-2} \text{ T}$. The work done in rotating it from its most stable to most unstable position is:

$A$ magnetic needle lying parallel to a magnetic field requires $W$ units of work to turn it through $60^{\circ}$. The torque required to maintain the needle in this position will be