$A$ bar magnet of magnetic moment $10^4\,J/T$ is free to rotate in a horizontal plane. The work done in rotating the magnet slowly from a direction parallel to a horizontal magnetic field of $4 \times 10^{-5}\,T$ to a direction $60^\circ$ from the field will be... $J$.

The magnetic potential due to a magnetic dipole at a point on its axis situated at a distance of $20 \ cm$ from its center is $1.5 \times 10^{-5} \ T \cdot m$. The magnetic moment of the dipole is . . . . . . $A \cdot m^2$. (Given: $\frac{\mu_0}{4 \pi} = 10^{-7} \ T \cdot m \cdot A^{-1}$)

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 . . . . . . .

The work done in rotating a bar magnet of magnetic moment $M$ from its unstable equilibrium position to its stable equilibrium position in a uniform magnetic field $B$ is .........

$A$ 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?

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