If a magnet of length $10 \, cm$ and pole strength $40 \, A-m$ is placed at an angle of $45^\circ$ in a uniform magnetic field of intensity $2 \times 10^{-4} \, T$,the torque acting on it is:

$A$ bar magnet of length $6\,cm$ has a magnetic moment of $4\,J\,T^{-1}$. Find the strength of magnetic field at a distance of $200\,cm$ from the centre of the magnet along its equatorial line.

$A$ magnet of magnetic moment $2 \, J \, T^{-1}$ is aligned in the direction of a magnetic field of $0.1 \, T$. What is the net work done to bring the magnet normal to the magnetic field?

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

$A$ magnet is parallel to a uniform magnetic field. If it is rotated by $60^o$,the work done is $0.8\, J$. How much work is done in moving it $30^o$ further (in $, J$)?

The torque acting on a magnetic dipole placed in a uniform magnetic field is zero,when the angle between the dipole axis and the magnetic field is