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

The ratio of magnetic fields due to a bar magnet at two axial points $P_1$ and $P_2$,which are separated from each other by $10 \ cm$,is $25: 2$. Point $P_1$ is situated at $10 \ cm$ from the center of the magnet. (Points $P_1$ and $P_2$ are on the same side of the magnet,and the distances of $P_1$ and $P_2$ from the center are much greater than the distance of the two ends of the magnet from the center). The magnetic length of the bar magnet is: (in $cm$)

Derive the equation of torque on a magnetic needle in a uniform magnetic field.

The magnetic potential due to a magnetic dipole at a point on its axis situated at a distance of $20 \ cm$ from its center is $1.5 \times 10^{-5} \ T \cdot m$. The magnetic moment of the dipole is . . . . . . $A \cdot m^2$. (Given: $\frac{\mu_0}{4 \pi} = 10^{-7} \ T \cdot m \cdot A^{-1}$)

$A$ short bar magnet placed with its axis at $30^{\circ}$ with a uniform external magnetic field of $0.16 \ T$ experiences a torque of magnitude $0.032 \ J$. The magnetic moment of the bar magnet will be......$J/T$.

$A$ bar magnet of length $3 \, cm$ has points $A$ and $B$ along its axis at distances of $24 \, cm$ and $48 \, cm$ on the opposite sides. The ratio of the magnetic fields at these points is