Question

A steady current $i$ flows in a small square loop of wire of side $L$ in a horizontal plane. The loop is now folded about its middle such that half of it lies in a vertical plane. Let $\overrightarrow{{\mu }_1}$ and $\overrightarrow{{\mu }_2}$ respectively denote the magnetic moments due to the current loop before and after folding. Then

• A. $\overrightarrow{{\mu }_2}=0$
• B. $\overrightarrow{{\mu }_1}$ and $\overrightarrow{{\mu }_2}$ are in the same direction
• C. $\frac{|\overrightarrow{{\mu }_1}|}{|\overrightarrow{{\mu }_2}|}=\sqrt{2}$
• D. $\frac{|\overrightarrow{{\mu }_1}|}{|\overrightarrow{{\mu }_2}|}=\left(\frac{1}{\sqrt{2}}\right)$

Answer 1

Answer: (C)

Initial magnetic moment $={\mu }_1=i{L}^2$
After folding the loop, $M=$ magnetic moment due to each
part $=i\left(\frac{L}{2}\right)\times L=\frac{i{L}^2}{2}=\frac{{\mu }_1}{2}$
$\Rightarrow {\mu }_2=M\sqrt{2}=\frac{{\mu }_1}{2}\times \sqrt{2}=\frac{{\mu }_1}{\sqrt{2}}$
or $\frac{{\mu }_1}{{\mu }_2}=\sqrt{2}$