If a bar magnet of magnetic moment $M$ is freely suspended in a uniform magnetic field of strength $B$,the work done in rotating the magnet through an angle $\theta$ is

$A$ short bar magnet is placed in a uniform magnetic field of $2 \ T$ such that the axis of the magnet makes an angle of $45^{\circ}$ with the direction of the magnetic field. If the torque acting on the magnet is $0.36 \sqrt{2} \ Nm$,then the magnetic moment of the magnet is: (in $J \ T^{-1}$)

$A$ short bar magnet experiences a torque of magnitude $0.64 \ J$ when it is placed in a uniform magnetic field of $0.32 \ T$. The magnetic moment of the magnet is . . . . . .

Two short bar magnets of magnetic moments $m$ each are arranged at the opposite corners of a square of side $d$ such that their centres coincide with the corners and their axes are parallel. If the like poles are in the same direction,the magnetic induction at any of the other corners of the square is

$A$ bar magnet of length $10 \text{ cm}$ and having the pole strength equal to $10^{-3} \text{ A-m}$ is kept in a magnetic field having magnetic induction $B$ equal to $4 \pi \times 10^{-3} \text{ T}$. It makes an angle of $30^{\circ}$ with the direction of magnetic induction. The value of the torque acting on the magnet is

The magnetic field due to a bar magnet at a distance $R$ from the center of the magnet (where $R$ is much larger than the length of the magnet) is proportional to: