A wire carrying current $I$ is bent in the shape $ABCDEFA$ as shown, where rectangles $ABCDA$ and $ADEFA$ are perpendicular to each other. If the sides of the rectangles are of lengths $a$ and $b$, then the magnitude and direction of magnetic moment of the loop $ABCDEFA$ is:
Answer: (A) $\sqrt2\,abI$ along $\left(\dfrac{\hat j}{\sqrt2} + \dfrac{\hat k}{\sqrt2}\right)$
Add and subtract a current along $AD$: the loop is equivalent to two rectangular loops $ABCDA$ (in the $xy$-plane) and $ADEFA$ (in the $xz$-plane), each of area $ab$ and current $I$.
By the right-hand rule, $ABCDA$ has moment $abI\,\hat k$ and $ADEFA$ has moment $abI\,\hat j$.
$$\vec M = abI(\hat j + \hat k) = \sqrt2\,abI\left(\frac{\hat j}{\sqrt2} + \frac{\hat k}{\sqrt2}\right)$$
Solution by Sreeraj P, M.Sc Physics