Question

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. $\sqrt{2}abI,$ along $\left(\frac{\hat{j}}{\sqrt{2}}+\frac{\hat{k}}{\sqrt{2}}\right)$
• B. $\sqrt{2}abI,$ along $\left(\frac{\hat{j}}{\sqrt{5}}+\frac{2\hat{k}}{\sqrt{5}}\right)$
• C. $abI,along\left(\frac{\hat{j}}{\sqrt{2}}+\frac{\hat{k}}{\sqrt{2}}\right)$
• D. $abI,$ along $\left(\frac{\hat{j}}{\sqrt{5}}+\frac{2\hat{k}}{\sqrt{5}}\right)$

Answer 1

Answer: (A)

$M=NIA$
$N=1$
For $ABCD$
${\vec{M}}_1=abI\hat{K}$
For $DEFA$
${\vec{M}}_2=abI\hat{j}$
$\vec{M}={\vec{M}}_1+{\vec{M}}_2$
$=abI(\hat{k}+\hat{j})$
$=abI\sqrt{2}\left(\frac{\hat{j}}{\sqrt{2}}+\frac{\hat{k}}{\sqrt{2}}\right)$