$A$ 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?

The magnetic field due to a small magnetic dipole of magnetic moment $M$,at a distance $r$ from the centre on the equatorial line is given by (in $M.K.S.$ system):

$A$ 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 $\mu_0 = 4\pi \times 10^{-7} \text{ Hm}^{-1}$)

$A$ 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: $B_H = 3 \times 10^{-5} \text{ T}$)

$A$ short bar magnet placed with its axis at $30^{\circ}$ with an external field of $800 \; G$ experiences a torque of $0.016 \; Nm.$
$(a)$ What is the magnetic moment of the magnet?
$(b)$ What is the work done in moving it from its most stable to most unstable position?
$(c)$ The bar magnet is replaced by a solenoid of cross-sectional area $2 \times 10^{-4} \; m^{2}$ and $1000$ turns,but of the same magnetic moment. Determine the current flowing through the solenoid.

$A$ short bar magnet placed with its axis at $30^{\circ}$ with an external field of $800 G$ experiences a torque of $0.016 Nm$. The magnetic moment of the bar magnet is (in $Am^2$)