Question

A regular hexagon has side $a$. $A$ wire of length $24a$ is coiled on that hexagon. If current in hexagon is $I$ then find the magnetic moment

• A. $3\sqrt{3}I{a}^2$
• B. $6\sqrt{3}I{a}^2$
• C. $\frac{3\sqrt{3}}{2}I{a}^2$
• D. $6I{a}^2$

Answer 1

Answer: (B)

Let number of turns $=n$
$n\times 6a=24a$
$n=4$
Now magnetic moment $M=nIA=4.I.A$
Now area of hexagon $\frac{1}{2}{a}^2\sin (120)+a2a\sin 60+\frac{1}{2}{a}^2\sin (120)$
$=\frac{\sqrt{3}{a}^2}{4}+\sqrt{3}{a}^2+\frac{\sqrt{3}{a}^2}{4}=\frac{3\sqrt{3}{a}^2}{2}$
So magnetic moment $=4$.I. $\left(\frac{3\sqrt{3}{a}^2}{2}\right)=6\sqrt{3}I{a}^2$