Write the equation of torque acting on a bar magnet placed in a uniform magnetic field.

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 . . . . . . .

Two small bar magnets are placed in a line with like poles facing each other at a certain distance $d$ apart. If the length of each magnet is negligible as compared to $d$,the force between them will be inversely proportional to

$A$ magnetic needle suspended parallel to a magnetic field requires $\sqrt{3} \, J$ of work to turn it through $60^{\circ}$. The torque needed to maintain the needle in this position will be.....

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: $\frac{\mu_0}{4 \pi} = 10^{-7} \ T \cdot m \cdot A^{-1}$)

$A$ 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... $J$.