$A$ 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 $45^{\circ}$.

$A$ magnetic dipole experiences a torque of $80 \sqrt{3} \ N \ m$ when placed in a uniform magnetic field in such a way that the dipole moment makes an angle of $60^{\circ}$ with the magnetic field. The potential energy of the dipole is

$A$ bar magnet of magnetic moment $200 \, A-m^2$ is suspended in a magnetic field of intensity $0.25 \, N/A-m$. The torque required to deflect it through $30^\circ$ is .... $N-m$.

Two magnets $A$ and $B$ are identical and are arranged as shown in the figure. Their length is negligible in comparison to the separation between them. $A$ magnetic needle is placed between the magnets at point $P$,which gets deflected through an angle $\theta$ under the influence of the magnets. The ratio of distances $d_1$ and $d_2$ will be:

$A$ 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 $(ii)?$