The magnetic moment of a magnet is $20 \, C.G.S$ units. How much work is required to rotate it by $30^{\circ}$ from its equilibrium position in a magnetic field of $0.3 \, C.G.S$ units?

$A$ bar magnet having a magnetic moment of $2 \times 10^4 \, JT^{-1}$ is free to rotate in a horizontal plane. $A$ horizontal magnetic field $B = 6 \times 10^{-4} \, T$ exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction $60^{\circ}$ from the field is.....$J$

The magnetic field at a point $P$ on the axis of a short bar magnet of magnetic moment $M$ is $B$. If another short bar magnet of magnetic moment $2M$ is placed on the first magnet such that their axes are perpendicular and their centres coincide. The resultant magnetic field at the point $P$ due to both the magnets is

$A$ short bar magnet of magnetic moment $0.4 \, J T^{-1}$ is placed in a uniform magnetic field of $0.16 \, T$. The magnet is in stable equilibrium when the potential energy is ....... $J$.

Verify Ampere's law for the magnetic field of a point dipole with dipole moment $\vec{M} = M\hat{k}$. Take $C$ as the closed curve running clockwise along the quarter circle of radius $a$ and center at the origin in the first quadrant of the $x-z$ plane,closed by segments along the $x$ and $z$ axes.

$A$ 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: