$A$ small bar magnet of magnetic moment $M$ is placed in a uniform magnetic field $H$. If the magnet makes an angle of $30^{\circ}$ with the field,the torque acting on the magnet is:

Two identical magnetic dipoles of magnetic moments $1.0 \, A-m^2$ each are placed at a separation of $2 \, m$ with their axes perpendicular to each other. The resultant magnetic field at a point $P$ midway between the dipoles is:

$A$ 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 $\text{ s}$)

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 of magnetic moment $10^4 \,J \,T^{-1}$ is free to rotate in a horizontal plane. The work done in rotating the magnet slowly from the direction parallel to a horizontal magnetic field of $4 \times 10^{-5} \,T$ to a direction $60^{\circ}$ to the direction of the field is (in $J$)

$A$ small bar magnet placed with its axis at $30^{\circ}$ with an external field of $0.06\, T$ experiences a torque of $0.018\, Nm$. The minimum work required to rotate it from its stable to unstable equilibrium position is