$A$ magnet is parallel to a uniform magnetic field. If it is rotated by $60^\circ$,the work done is $0.8\, J$. How much work is done in moving it $30^\circ$ further?

$A$ short bar magnet placed with its axis at $45^{\circ}$ with a uniform external magnetic field of $28.3 \times 10^{-3} \,T$ experiences a torque of magnitude equal to $3.6 \times 10^{-5} \,J$. The magnitude of the magnetic moment of the magnet is nearly:

$A$ 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... $J$.

$A$ short bar magnet placed with its axis at $30^{\circ}$ with a uniform external magnetic field of $0.25 \ T$ experiences a torque of magnitude $4.5 \times 10^{-2} \ J$. The magnitude of the magnetic moment of the magnet will be . . . . . . $J \ T^{-1}$.

$A$ bar magnet is held perpendicular to a uniform magnetic field. The couple acting on the magnet is to be halved by rotating it. Through what angle should it be rotated?

Write the equation of torque on a magnetic needle placed in a uniform magnetic field and derive the expression for its time period $T = 2\pi \sqrt{\frac{I}{mB}}$.