$A$ bar magnet of magnetic moment $1.5 \, J/T$ lies aligned with the direction of a uniform magnetic field of $0.22 \, T$. What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment perpendicular to the field direction? (in $J$)

The magnetic moment of a magnet is $2.5 \, J \, T^{-1}$. How much work in $J$ must be done to rotate it from its stable equilibrium position to an unstable equilibrium position in a magnetic field of $0.2 \, T$?

$A$ 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 effect due to a uniform magnetic field on a freely suspended magnetic needle is as follows:

Two identical small bar magnets, each of dipole moment $3\sqrt{5} \text{ J/T}$, are placed at a center-to-center separation of $10 \text{ cm}$, with their axes perpendicular to each other as shown in the figure. The value of the magnetic field at the point $P$ midway between the magnets is $\alpha \times 10^{-3} \text{ T}$. The value of $\alpha$ is . . . . . . .

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