If a bar magnet of magnetic moment $M$ is freely suspended in a uniform magnetic field of strength $B$,the work done in rotating the magnet through an angle $\theta$ is

Give the stability position of a bar magnet for $\theta = 0^{\circ}$ and $\theta = 180^{\circ}$.

$A$ bar magnet of magnetic moment $1.5 \, A \cdot m^2$ lies aligned with the direction of a uniform magnetic field of $2 \, T$. What is the amount of work required to turn the magnet so as to align its magnetic moment normal to the field direction?

$A$ magnetic needle is placed in a uniform magnetic field and is aligned with the field. The needle is now rotated by an angle of $60^{\circ}$ and the work done is $W$. The torque on the magnetic needle at this position is

The magnetic field due to a short magnet at a point on its axis at a distance $X \, cm$ from the center of the magnet is $200 \, G$. The magnetic field at a point on the equatorial axis at a distance $X \, cm$ from the center of the magnet is.....$G$.

$A$ 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 $(ii)?$