Q.A short bar magnet is placed in an external magnetic field of 700 gauss. When its axis makes an angle of $30^\circ$ with the external magnetic field, it experiences a torque of 0.014 N m. Find the magnetic moment of the magnet, and the work done in moving it from its most stable to its most unstable position.

[!TLDR] $\tau=mB\sin\theta\Rightarrow m=\tau/(B\sin\theta)=0.4$ A m$^2$; work from stable to unstable is $W=2mB= …

Converting the field, $B=700$ gauss $=700\times10^{-4}=0.07$ T. From $\tau=mB\sin\theta$: $m=\dfrac{\tau}{B\sin\theta}=\dfrac{0.014}{0.07\times\sin30^\circ}=\dfrac{0.014}{0.07\times0.5}=\dfrac{0.014}{0.035}=0.4$ A m$^2$. The work done moving the magnet from its most stable ($\theta=0^\circ$) to its most unstable ($\th …

Find m from $\tau=mB\sin\theta$ using the given angle and torque, then use $W=2mB$ for the full stable-to …

Forgetting to convert the field from gauss to tesla (1 G $=10^{-4}$ T) before using it in SI-unit formulas; or using $W=mB$ instead of the correct $W=2mB$ for a full $180^\circ$ reversal (the work for a $1 …