$A$ 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.

(N/A) The torque is given by $\tau = mB \sin \theta$. Given $\theta = 30^{\circ}$,$\sin 30^{\circ} = 0.5$,$B = 800 \; G = 0.08 \; T$,and $\tau = 0.016 \; Nm$.
$0.016 = m \times 0.08 \times 0.5$
$m = 0.016 / 0.04 = 0.40 \; Am^{2}$.
$(b)$ The most stable position is $\theta = 0^{\circ}$ and the most unstable position is $\theta = 180^{\circ}$. The work done is $W = U(\theta = 180^{\circ}) - U(\theta = 0^{\circ}) = -mB \cos 180^{\circ} - (-mB \cos 0^{\circ}) = mB + mB = 2mB$.
$W = 2 \times 0.40 \times 0.08 = 0.064 \; J$.
$(c)$ For a solenoid,$m = NIA$. Given $m = 0.40 \; Am^{2}$,$N = 1000$,and $A = 2 \times 10^{-4} \; m^{2}$.
$0.40 = 1000 \times I \times 2 \times 10^{-4}$
$0.40 = 0.2 \times I$
$I = 0.40 / 0.2 = 2 \; A$.