$A$ short bar magnet placed with its axis at $45^{\circ}$ with a uniform external magnetic field of $28.3 \times 10^{-3} \,T$ experiences a torque of magnitude equal to $3.6 \times 10^{-5} \,J$. The magnitude of the magnetic moment of the magnet is nearly:

$A$ 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 $(ii)?$

$A$ bar magnet of magnetic moment $\overrightarrow{M}$ is placed in a magnetic field of induction $\overrightarrow{B}$. The torque exerted on it is

$A$ magnet of magnetic moment $4 \, A-m^2$ is held in a uniform magnetic field $5 \times 10^{-4} \, T$ with the magnetic moment vector making an angle $30^{\circ}$ with the field. Calculate the work done in increasing the angle from $30^{\circ}$ to $45^{\circ}$.

Two short bar magnets have their magnetic moments $1.2 \text{ A m}^2$ and $1.0 \text{ A m}^2$. They are placed on a horizontal table parallel to each other at a distance of $20 \text{ cm}$ between their centres,such that their north poles point towards geographic south. They have a common magnetic equatorial line. The horizontal component of the Earth's magnetic field is $3.6 \times 10^{-5} \text{ T}$. The resultant horizontal magnetic induction at the midpoint of the line joining their centers is $\left(\frac{\mu_0}{4 \pi} = 10^{-7} \text{ N/A}^2\right)$.

Two short magnets are placed along the same axis with their like poles facing each other. They repel each other with a force which varies inversely as: