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

$A$ short bar magnet placed with its axis at $30^{\circ}$ with an external field of $800 G$ experiences a torque of $0.016 Nm$. The magnetic moment of the bar magnet is (in $Am^2$)

$A$ magnetic dipole of magnetic moment $M$ is freely suspended in a magnetic field of induction $B$. The minimum and maximum values of the potential energy of the dipole,respectively,are

$A$ magnetic dipole of magnetic moment $ 6 \times 10^{-2} \text{ A m}^2 $ and moment of inertia $ 12 \times 10^{-6} \text{ kg m}^2 $ performs oscillations in a magnetic field of $ 2 \times 10^{-2} \text{ T} $. The time taken by the dipole to complete $ 20 $ oscillations is (assume $ \pi \simeq 3 $). (in $\text{ s}$)

The midpoints of two small magnetic dipoles of length $d$ in end-on positions are separated by a distance $x$ $(x >> d)$. The force between them is proportional to $x^{-n}$,where $n$ is:

Two short magnetic dipoles $m_{1}$ and $m_{2}$, each having a magnetic moment of $1\, \text{Am}^{2}$, are placed at points $O$ and $P$ respectively. The distance between $O$ and $P$ is $1\, \text{m}$. The torque experienced by the magnetic dipole $m_{2}$ due to the presence of $m_{1}$ is ...... $\times 10^{-7}\, \text{Nm}$.