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

$A$ 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 (in $\,J$)?

The magnetic moment of a short magnet is $16 \ Am^2$. The magnetic induction at a point $20 \ cm$ away from its midpoint on $(i)$ axial point and $(ii)$ equatorial point respectively,will be:

$A$ bar magnet is held perpendicular to a uniform magnetic field. The couple acting on the magnet is to be halved by rotating it. Through what angle should it be rotated?

$A$ bar magnet of magnetic moment $200 \, A-m^2$ is suspended in a magnetic field of intensity $0.25 \, N/A-m$. The torque required to deflect it through $30^\circ$ is .... $N-m$.

$A$ hoop and a solid cylinder of same mass and radius are made of a permanent magnetic material with their magnetic moment parallel to their respective axes. But the magnetic moment of the hoop is twice that of the solid cylinder. They are placed in a uniform magnetic field in such a manner that their magnetic moments make a small angle with the field. If the oscillation periods of the hoop and cylinder are $T_h$ and $T_c$ respectively,then the relation between them is....