A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to $4.5\times {10}^{-2} J.$ What is the magnitude of the magnetic moment of the magnet?

(1) 0.36 J/T

(2) 0.21 J/T

(3) 0.01 J/T

(4) 0.12 J/T

A circular coil of 16 turns and a radius of 10 cm carrying a current of 0.75 A rests with its plane normal to an external field of magnitude $5.0\times {10}^{-2} T$ . The coil is free to turn about an axis in its plane perpendicular to the field direction. When the coil is turned slightly and released, it oscillates about its stable equilibrium with a frequency of 2.0$s^{-1}.$The moment of inertia of the coil about its axis of rotation is:

$$\begin{gathered} \left(1\right) 1.39\times {10}^{-4} kg m^2 \\ \left(2\right) 2.19\times {10}^{-4} kg m^2 \\ \left(3\right) 2.39\times {10}^{-4} kg m^2 \\ \left(4\right) 1.19\times {10}^{-4} kg m^2 \end{gathered}$$

A short bar magnet has a magnetic moment of $0.48 J T^{-1}.$ The direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on the equatorial lines (normal bisector) of the magnet is:

(1) 0.38 G along the N-S direction.

(2) 0.48 G along the N-S direction.

(3) 0.38 G along the S-N direction.

(4) 0.48 G along the S-N direction.