Question 257 of 276

A wire carrying current $I$ is bent in the shape$ABCDEFA$ as shown, where rectangle $ABCDA$ 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 :

A$\sqrt{2}abIalong\left(\dfrac{\hat{j}}{\sqrt{2}}+\dfrac{\hat{k}}{\sqrt{2}}\right)$B$abI$ along $\left(\dfrac{\hat{j}}{\sqrt{2}}+\dfrac{\hat{k}}{\sqrt{2}}\right)$C$\sqrt{2}abI$ along $\left(\dfrac{\hat{j}}{\sqrt{5}}+\dfrac{2\hat{k}}{\sqrt{5}}\right)$D$abI$ along $\left(\dfrac{\hat{j}}{\sqrt{5}}+\dfrac{\hat{k}}{\sqrt{5}}\right)$

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Answer: (A) $\sqrt{2}abIalong\left(\dfrac{\hat{j}}{\sqrt{2}}+\dfrac{\hat{k}}{\sqrt{2}}\right)$

Loop ABCD
${\vec{M}}_{1}=(abI)(\hat{k})$
For Loop DEFA ${\vec{M}}_{2}=(abI)(\hat{j})$
$\vec{M}={\vec{M}}_{1}+{\vec{M}}_{2};\vec{M}=(abI)(\hat{j}+\hat{k})$
$\sqrt{2}abI$ along $\left(\dfrac{\hat{j}}{\sqrt{2}}+\dfrac{\hat{k}}{\sqrt{2}}\right)$.