Question

Asteady 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 ${\overline{\mu }}_1$ and ${\overline{\mu }}_2$ respectively denote the magetic moments of the current loop before and after folding. Then:

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

Answer 1

Answer: (C)

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

The Initial magnetic moment will be $=\mu =iL$

After folding the loop,

$M=$ magnetic moment due to each part $=i\left(\frac{L}{2}\right)\times =\frac{i{L}^2}{2}=\frac{{\mu }_1}{2}$

$\Rightarrow {\mu }_0=M\sqrt{2}=\frac{{\mu }_1}{2}\times \sqrt{2}=\frac{{\mu }_1}{\sqrt{2}}$

So, option $C$ is correct.