A square loop OABCO of side l carries a current ‘I’. It is placed as shown in figure. The magnetic moment of the loop.

$$\begin{gathered} M=i(BC\times CO) \\ Here, BC=-l\widehat{k}, CO=l\cos {60}^{\circ }\widehat{i}-l\sin {60}^{\circ }\widehat{j}=-\frac{l}{2}\widehat{i}-\frac{\sqrt{3}l}{2}\widehat{i} \\ \therefore M=i\left[(-l\widehat{k})\times (-\frac{l}{2}\widehat{i}-\frac{\sqrt{3}l}{2}\widehat{i})\right] \\ or M=\frac{i{l}^2}{2}(\widehat{j}-\sqrt{3}i) \end{gathered}$$