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 $A B C D$
$\vec{ M }_{1}=a bI \hat{ K }$
For $DEFA$
$\vec{ M }_{2}= abI \hat{ j }$
$\vec{ M }=\vec{ M }_{1}+\vec{ M }_{2}$
$=a b I(\hat{k}+\hat{j}) $
$=a b I \sqrt{2}\left(\frac{\hat{j}}{\sqrt{2}}+\frac{\hat{k}}{\sqrt{2}}\right)$