$A$ short bar magnet placed with its axis at $30^{\circ}$ to a uniform external magnetic field of $0.5 \ T$ experiences a torque of magnitude equal to $4.5 \times 10^{-2} \ J$. The magnitude of the magnetic moment is . . . . . . .

$A$ 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 $(ii)?$

If the dipole moment of a short bar magnet is $1.25 \ A-m^2$,find the magnetic field on its axis at a distance of $0.5 \ m$ from the centre of the magnet.

$A$ magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is $60^{\circ},$ and one of the fields has a magnitude of $1.2 \times 10^{-2}\;T$. If the dipole comes to stable equilibrium at an angle of $15^{\circ}$ with this field,what is the magnitude of the other field?

$A$ magnetic needle lying parallel to a magnetic field requires $W$ units of work to turn it through $60^{\circ}$. The torque required to maintain the needle in this position will be

$A$ 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: