$A$ magnetized wire of magnetic moment $M$ and length $l$ is bent in the form of a semicircle of radius $r$. Then its new magnetic moment is:

The ultimate individual unit of magnetism in any magnet is called

$A$ bar magnet with a magnetic moment of $3.0 \, A \cdot m^2$ is placed in a magnetic field of $2 \times 10^{-5} \, T$. If the force acting on each pole is $6 \times 10^{-4} \, N$,what is the length of the magnet in $m$?

The magnetic field at a point $x$ on the axis of a small bar magnet is equal to the field at a point $y$ on the equator of the same magnet. The ratio of the distances of $x$ and $y$ from the centre of the magnet is

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 $x$-axis from $x = R$ to $x = a$,then along a circular arc of radius $r$ in the $xz$-plane from $\theta = 0$ to $\theta = \pi$,and finally back to the starting point.

$A$ short bar magnet has a magnetic moment of $0.48 \; J \; T^{-1}$. Give 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
$(a)$ the axis,
$(b)$ the equatorial lines (normal bisector) of the magnet.