Question

The current carrying wire of length $l$ and current $I$ is coiled into a circular coil of $n$ number of turns. The magnetic dipole moment of the coil is

• A. $\frac{I{l}^2}{\pi n}$
• B. $\frac{I{l}^2}{2\pi n}$
• C. $\frac{nI{l}^2}{4\pi }$
• D. $\frac{I{l}^2}{4\pi n}$

Answer 1

The correct option is D $\frac{I{l}^2}{4\pi n}$

When turned into coil, its radius becomes;

$n\left(2\pi r\right)=l$

$\Rightarrow r=\frac{l}{2\pi n}$

$\mu =niA$

$\mu =\frac{ni\times \pi l^2}{4\times {\pi }^2\times n^2}$

$\mu =\frac{i{l}^2}{4\pi n}$