The magnetic moment of a short magnet is $16 \ Am^2$. The magnetic induction at a point $20 \ cm$ away from its midpoint on $(i)$ axial point and $(ii)$ equatorial point respectively,will be:

The ratio of magnetic intensities for an axial point and a point on the broad side-on (equatorial) position at an equal distance $d$ from the centre of a short bar magnet is:

$A$ short bar magnet placed with its axis at $30^{\circ}$ with an external field of $800 \; G$ experiences a torque of $0.016 \; Nm.$
$(a)$ What is the magnetic moment of the magnet?
$(b)$ What is the work done in moving it from its most stable to most unstable position?
$(c)$ The bar magnet is replaced by a solenoid of cross-sectional area $2 \times 10^{-4} \; m^{2}$ and $1000$ turns,but of the same magnetic moment. Determine the current flowing through the solenoid.

Two short magnetic dipoles $m_{1}$ and $m_{2}$, each having a magnetic moment of $1\, \text{Am}^{2}$, are placed at points $O$ and $P$ respectively. The distance between $O$ and $P$ is $1\, \text{m}$. The torque experienced by the magnetic dipole $m_{2}$ due to the presence of $m_{1}$ is ...... $\times 10^{-7}\, \text{Nm}$.

$A$ bar magnet of magnetic moment $1.5 \, J \, T^{-1}$ lies aligned with the direction of a uniform magnetic field of $0.22 \, T$.
$(a)$ What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment: $(i)$ normal to the field direction,$(ii)$ opposite to the field direction?
$(b)$ What is the torque on the magnet in cases $(i)$ and $(ii)?$

Derive the equation of torque on a magnetic needle in a uniform magnetic field.