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

Hint:

Split the loop into two rectangular loops that share the side AD; the shared side carries opposite currents and cancels. Add the two magnetic moment vectors.

Answer 1

The Correct Option is C

Side AD is traversed in opposite senses in the loops ABCDA and ADEFA, so the currents in AD cancel. The given loop ABCDEFA is therefore equivalent to two loops, each carrying current \(I\).
Loop ABCDA lies in the XY plane with area \(ab\). By the right-hand rule (A to B to C to D) its moment is \(\vec{m}_1 = Iab\,\hat{k}\).
Loop ADEFA lies in the vertical plane containing the X and Z directions (perpendicular to the first loop) with area \(ab\). By the right-hand rule its moment is \(\vec{m}_2 = Iab\,\hat{j}\).
Total moment: \(\vec{m} = Iab(\hat{j}+\hat{k})\).
Magnitude: \(|\vec{m}| = \sqrt{2}\,abI\).
Direction: \(\frac{\hat{j}+\hat{k}}{\sqrt{2}} = \frac{\hat{j}}{\sqrt{2}}+\frac{\hat{k}}{\sqrt{2}}\).
Answer: (3) \(\sqrt{2}\,abI\), along \(\left(\frac{\hat{j}}{\sqrt{2}}+\frac{\hat{k}}{\sqrt{2}}\right)\)