Question

A uniform copper wire of length $L$ is bent into a circular coil of two turns and a current $i$ is passed through it. The coil now behaves like a magnetic dipole of moment

• A. $\frac{i{L}^2}{16\pi }$
• B. $\frac{i{L}^2}{8\pi }$
• C. $\frac{i{L}^2}{4\pi }$
• D. $\frac{i{L}^2}{2\pi }$

Answer 1

The correct option is B $\frac{i{L}^2}{8\pi }$

Wire is bent into a circular coils of two turns
$l=2\times 2\pi r$
$r=\frac{l}{r\pi }$
magnetic dipole moment$=n\times i\times A=n\times i\times \pi r^2=2i\left(\frac{l^2}{16{\pi }^2}\right)\pi =\frac{i{L}^2}{8\pi }$