Q. 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 ${\widehat{M}}_{1}\ \ and\ \ {\widehat{M}}_{2}$ respectively denote the magnetic moments due to current loop before and after folding. Then
A) ${\widehat{M}}_{2} = 0$
B) ${\widehat{M}}_{1}\ \ and\ \ {\widehat{M}}_{2}$ are in the same direction
C) $M_{1}/M_{2} = \ \sqrt{2}$
D) $M_{1}/M_{2} = \ \ 1/\sqrt{2}$
Explanation
(c) $\ M_{1} = l^{2}\ i,\ $ and
$M_{2} = \sqrt{\left( \frac{l}{2} \times l \times i \right)^{2} + \left( \frac{l}{2} \times l \times i \right)^{2}} = \frac{l^{2}i}{\sqrt{2}}$
$\ \therefore\ \ \ \frac{M_{1}}{M_{2}} = \ \sqrt{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.