$A$ 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 is placed in a uniform magnetic field of $2 \ T$ such that the axis of the magnet makes an angle of $45^{\circ}$ with the direction of the magnetic field. If the torque acting on the magnet is $0.36 \sqrt{2} \ Nm$,then the magnetic moment of the magnet is: (in $J \ T^{-1}$)

$A$ bar magnet is held perpendicular to a uniform magnetic field. If the couple acting on the magnet is to be halved by rotating it,then the angle by which it is to be rotated is....$^o$

$A$ short bar magnet placed with its axis at $30^{\circ}$ to a uniform external magnetic field of $0.5 \ T$ experiences a torque of magnitude equal to $4.5 \times 10^{-2} \ J$. The magnitude of the magnetic moment is . . . . . . .

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

$A$ magnetic dipole of moment $2.5 \text{ A m}^2$ is free to rotate about a vertical axis passing through its centre. It is released from the East-West direction. What is its kinetic energy at the moment it takes the North-South position (in $\mu\text{J}$)? (Given: $B_H = 3 \times 10^{-5} \text{ T}$)