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

What is the magnitude of the equatorial magnetic field due to a bar magnet of length $5.0 \ cm$ at a distance $75 \ cm$ from its midpoint? The magnetic moment of the bar magnet is $0.75 \ A \ m^2$.

In which direction would the magnetic field on the axis at a distance $Z$ from the centre of the bar magnet be?

$A$ 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 $= B_H$)

The work done in rotating a bar magnet from its equilibrium position by $60^{\circ}$ in a magnetic field is $W$. What is the torque required to hold it in this position?

$A$ magnet of magnetic moment $50 \hat{i} \text{ Am}^2$ is placed along the $x$-axis in a magnetic field $\vec{B} = (0.5 \hat{i} + 3.0 \hat{j}) \text{ T}$. The torque acting on the magnet is: