Question

A wire of a length 2 m carrying a current of 1 A is bend to form a circle. The magnetic moment of the coil is $(inA-m^2)$:
(a) $1/\pi$
(b) $\pi /2$

Answer 1

length of the wire $=2m,$ when bent becomes circumference $\left(c\right)$ of the loop, whose radius is given by :

$r=\frac{c}{2\pi }=\frac{1}{\pi }m.$

Thus, area of loop :-

$A=\pi r^2=\pi {\left(\frac{1}{\pi }\right)}^2=\frac{1}{\pi }{m}^2$

So, the magnetic moment of the loop in,

$M=IA=\frac{1}{\pi }A{m}^2$.