$A$ magnetic dipole is placed in a uniform magnetic field of intensity $B$,oriented along the direction of the field. If the magnetic dipole moment is $M$,then the maximum work an external agent can perform in rotating the dipole will be

$A$ 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 $(ii)?$

$A$ short bar magnet of magnetic moment $0.21 \ A \cdot m^2$ is placed with its axis perpendicular to the direction of the horizontal component of the earth's magnetic field. The distance of the point on the axis of the magnet from the centre of the magnet where the resultant magnetic field is inclined at $45^{\circ}$ with the horizontal component of the earth's field direction is (horizontal component of the earth's magnetic field $= 4.2 \times 10^{-5} \ T$). (in $cm$)

$A$ magnet of length $0.1 \, m$ and pole strength $10^{-4} \, A \cdot m$ is kept in a magnetic field of $30 \, Wb/m^2$ at an angle of $30^\circ$. The couple (torque) acting on it is $...... \times 10^{-4} \, N \cdot m$.

Deduce the expression for the potential energy of a bar magnet in a uniform magnetic field and discuss special cases.

$A$ magnetic needle suspended parallel to a magnetic field requires $\sqrt{3} \text{ J}$ of work to turn it through $60^{\circ}$. The torque needed to maintain the needle in this position will be