Question

A circle coil has two turns and it is made from a wire of length $L$, carrying a current $I$. Find its magnetic moment ;

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

Answer 1

Answer: (A)

$\frac{I{L}^2}{8\pi }$

Given

A coil of two turns, N=2

Length of wire = L

current=I

Solution

Magnetic moment of a coil is

M=NIA

A= area of coil

L=N* circumference of one coil.

Therefore $L=4\pi R$

$A=\pi R^2$

$A=\pi \ast \frac{L^2}{16\ast {\pi }^2}$

$A=\frac{L^2}{16\pi }$

M=NIA

M=2*I*$\frac{L^2}{16\pi }$

$M=\frac{I{L}^2}{8\pi }$

The correct option is A