The magnetic moment of a bar magnet is $0.5 \text{ A m}^2$. It is suspended in a uniform magnetic field of $8 \times 10^{-2} \text{ T}$. The work done in rotating it from its most stable to most unstable position is:

$A$ magnetic dipole is acted upon by two magnetic fields which are inclined to each other at an angle of $75^{\circ}$. One of the fields has a magnitude of $15 \, mT$. The dipole attains stable equilibrium at an angle of $30^{\circ}$ with this field. The magnitude of the other field (in $mT$) is close to:

$A$ 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).

$A$ short bar magnet of magnetic moment $M=0.32 \; J \, T^{-1}$ is placed in a uniform magnetic field of $0.15 \; T$. If the bar is free to rotate in the plane of the field,which orientation would correspond to its $(a)$ stable,and $(b)$ unstable equilibrium? What is the potential energy of the magnet in each case?

The work done in turning a magnet of magnetic moment $M$ by an angle of $90^{\circ}$ from the magnetic meridian is $n$ times the corresponding work done to turn it through an angle of $60^{\circ}$. The value of $n$ is:

$A$ bar magnet of magnetic moment $M$ and moment of inertia $I$ is freely suspended such that the magnetic axial line is in the direction of the magnetic meridian. If the magnet is displaced by a very small angle $\theta$, the angular acceleration is (Magnetic induction of earth's horizontal field $= B_H$)