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. $\frac{\left|\overrightarrow{{\mu }_1}\right|}{\left|\overrightarrow{{\mu }_2}\right|}=\sqrt{2}$
  
• D. $\frac{\left|\overrightarrow{{\mu }_1}\right|}{\left|\overrightarrow{{\mu }_2}\right|}=\left(\frac{1}{\sqrt{2}}\right)$

Answer 1

The correct option is C $\frac{\left|\overrightarrow{{\mu }_1}\right|}{\left|\overrightarrow{{\mu }_2}\right|}=\sqrt{2}$




M = magnetic moment due to each part = $i\left(\frac{L}{2}\right)\times L=\frac{i{L}^2}{2}=\frac{{\mu }_1}{2}$
$\therefore$ ${\mu }_2=M\sqrt{2}=\frac{{\mu }_1}{2}\times \sqrt{2}=\frac{{\mu }_1}{\sqrt{2}}$