$A$ 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... $J$.

$A$ magnetic dipole of magnetic moment $ 6 \times 10^{-2} \text{ A m}^2 $ and moment of inertia $ 12 \times 10^{-6} \text{ kg m}^2 $ performs oscillations in a magnetic field of $ 2 \times 10^{-2} \text{ T} $. The time taken by the dipole to complete $ 20 $ oscillations is (assume $ \pi \simeq 3 $). (in $\text{ s}$)

$A$ magnet of length $10 \text{ cm}$ and magnetic moment $1 \text{ Am}^2$ is placed along side $AB$ of an equilateral triangle $ABC$. If the length of the side $AB$ is $10 \text{ cm}$,find the magnetic induction at point $C$. (Given $\mu_0 = 4\pi \times 10^{-7} \text{ Hm}^{-1}$)

$A$ bar magnet of length $10 \text{ cm}$ and having the pole strength equal to $10^{-3} \text{ A-m}$ is kept in a magnetic field having magnetic induction $B$ equal to $4 \pi \times 10^{-3} \text{ T}$. It makes an angle of $30^{\circ}$ with the direction of magnetic induction. The value of the torque acting on the magnet is

The work done in rotating a magnet of magnetic moment $2 \, A-m^2$ in a magnetic field of $5 \times 10^{-3} \, T$ from the direction along the magnetic field to the opposite direction to the magnetic field is:

$A$ 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?