The magnetic moment of a magnet is $2.5 \, J \, T^{-1}$. How much work in $J$ must be done to rotate it from its stable equilibrium position to an unstable equilibrium position in a magnetic field of $0.2 \, T$?

$A$ bar magnet of magnetic moment $M$ and moment of inertia $I$ is freely suspended such that the magnetic axial line is in the direction of the magnetic meridian. If the magnet is displaced by a very small angle $\theta$,the angular acceleration is (Magnetic induction of earth's horizontal field $= B_H$)

$A$ 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

If there is no torsion in the suspension thread, then the time period of a magnet executing $SHM$ is

$A$ bar magnet of magnetic moment $2 \text{ A m}^2$ lies aligned with the direction of a uniform magnetic field of $0.3 \text{ T}$. The amount of work required by an external torque to turn the magnet so as to align its magnetic moment normal to the field direction is (in $\text{ J}$)

The magnetic field due to a small magnetic dipole of magnetic moment $M$,at a distance $r$ from the centre on the equatorial line is given by (in $M.K.S.$ system):