$A$ 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?

Two short bar magnets have their magnetic moments $1.2 \text{ A m}^2$ and $1.0 \text{ A m}^2$. They are placed on a horizontal table parallel to each other at a distance of $20 \text{ cm}$ between their centres,such that their north poles point towards geographic south. They have a common magnetic equatorial line. The horizontal component of the Earth's magnetic field is $3.6 \times 10^{-5} \text{ T}$. The resultant horizontal magnetic induction at the midpoint of the line joining their centers is $\left(\frac{\mu_0}{4 \pi} = 10^{-7} \text{ N/A}^2\right)$.

At a point on the equatorial line (right bisector) of a magnetic dipole,the magnetic:

$A$ magnet of magnetic moment $M$ is situated with its axis along the direction of a magnetic field of strength $B$. The work done in rotating it by an angle of $180^{\circ}$ will be

$A$ magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is $60^{\circ},$ and one of the fields has a magnitude of $1.2 \times 10^{-2}\;T$. If the dipole comes to stable equilibrium at an angle of $15^{\circ}$ with this field,what is the magnitude of the other field?

$A$ 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 $\frac{\mu_0}{4 \pi} = 10^{-7} \ T \ m/A$)