$A$ magnet of magnetic moment $M$ is rotated through $360^{\circ}$ in a magnetic field $H$,the work done will be

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

Write the equation of torque acting on a bar magnet placed in a uniform magnetic field.

Points $A$ and $B$ are situated perpendicular to the axis of a small bar magnet at large distances $x$ and $3x$ from its centre on opposite sides. The ratio of the magnetic fields at $A$ and $B$ will be approximately equal to

Torques $\tau_1$ and $\tau_2$ are required for a magnetic needle to remain perpendicular to the magnetic fields of $B_1$ and $B_2$ at two different places. The ratio of $B_1: B_2$ is equal to

The magnetic field at a point $P$ on the axis of a short bar magnet of magnetic moment $M$ is $B$. If another short bar magnet of magnetic moment $2M$ is placed on the first magnet such that their axes are perpendicular and their centres coincide. The resultant magnetic field at the point $P$ due to both the magnets is