$A$ magnetic needle lying parallel to a magnetic field requires $W$ units of work to turn it through $60^{\circ}$. The torque required to maintain the needle in this position will be

The magnetic field due to a bar magnet at a distance $R$ from the center of the magnet (where $R$ is much larger than the length of the magnet) is proportional to:

$A$ bar magnet placed in a uniform magnetic field making an angle $\theta$ with the field experiences a torque. If the angle made by the magnet with the field is doubled,the torque experienced by the magnet increases by $41.4 \%$. The initial angle made by the magnet with the magnetic field is (in $^{\circ}$)

$A$ bar magnet of magnetic moment $\vec{M}$ is placed in a magnetic field of induction $\vec{B}$. The torque exerted on it 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 $\frac{\mu_0}{4 \pi} = 10^{-7} \ T \ m/A$)

$A$ 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