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

$A$ magnet of magnetic moment $2 \,J \,T^{-1}$ is aligned in the direction of a magnetic field of $0.1 \,T$. What is the net work done to bring the magnet normal to the magnetic field (in $\,J$)?

Figure shows a small magnetised needle $P$ placed at a point $O$. The arrow shows the direction of its magnetic moment. The other arrows show different positions (and orientations of the magnetic moment) of another identical magnetised needle $Q$.
$(a)$ In which configuration the system is not in equilibrium?
$(b)$ In which configuration is the system in $(i)$ stable,and $(ii)$ unstable equilibrium?
$(c)$ Which configuration corresponds to the lowest potential energy among all the configurations shown?

$A$ magnetic dipole of magnetic moment $M$ is freely suspended in a magnetic field of induction $B$. The minimum and maximum values of the potential energy of the dipole,respectively,are

$A$ bar magnet of magnetic moment $M$ is placed at a distance $D$ with its axis along the positive $X$-axis. Likewise, a second bar magnet of magnetic moment $M$ is placed at a distance $2D$ on the positive $Y$-axis and perpendicular to it as shown in the figure. The magnitude of the magnetic field at the origin is $|\vec{B}| = \alpha \left[ \frac{\mu_0}{4 \pi} \frac{M}{D^3} \right]$. The value of $\alpha$ must be (Assume $D \gg l$, where $l$ is the length of the magnets).

The torque acting on a magnetic dipole placed in a uniform magnetic field is zero,when the angle between the dipole axis and the magnetic field is