Question

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

Answer 1

The initial magnetic moment will be

${\mu }_1=i{L}^2$

After folding the loop,

$M=$ magnetic moment due to each part

$M=i\left(\frac{L}{2}\right)\times \left(L\right)=\frac{i{L}^2}{2}=\frac{{\mu }_1}{2}$

$\therefore$ the resultant magnetic moment after folding will be

$\Rightarrow {\mu }_2=M\sqrt{2}=\frac{{\mu }_1}{2}\times \sqrt{2}=\frac{{\mu }_1}{\sqrt{2}}$

$\Rightarrow \frac{|{\mu }_1|}{|{\mu }_2|}=\sqrt{2}$

Thus, the correct choice is option $\text{(C)}$.