$A$ 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?

Verify Ampere's law for the magnetic field of a point dipole with dipole moment $\vec{M} = M\hat{k}$. Take $C$ as the closed curve running clockwise along the quarter circle of radius $a$ and center at the origin in the first quadrant of the $x-z$ plane,closed by segments along the $x$ and $z$ axes.

Two small magnets, each of magnetic moment $10 \, A \cdot m^2$, are placed in an end-on position at a distance of $0.1 \, m$ apart from their centers. The force acting between them is .... $N$.

The work done in rotating a bar magnet from its equilibrium position by $60^{\circ}$ in a magnetic field is $W$. What is the torque required to hold it in this position?

Two short magnets of equal dipole moments $M$ are fastened perpendicularly at their centres, which lie at the origin. Let the two magnets lie along the $X$-axis and $Y$-axis, respectively. The magnitude of the magnetic field at a distance $R$ from the centre on the $Y$-axis is $\frac{\mu_0}{4 \pi} \frac{M_0}{R^3}$. Assuming $R >> l$ (magnet length), the magnitude of $M$ is

$A$ planet has a magnetic dipole moment of $27 \times 10^{22} \ A \ m^2$. If the radius of the planet is $300 \ km$, what would be the magnetic field at its equator (in $T$)? (Take $\frac{\mu_0}{4 \pi} = 10^{-7} \ T \ m/A$)