A straight wire carrying a current (I) is turned into a circular...

Let the length of the wire be $L$.

When the wire is bent into a circular loop of radius $R$, its circumference is equal to the length of the wire:

$2\pi R = L \Rightarrow R = \dfrac{L}{2\pi}$

The area of the circular loop is:

$A = \pi R^2 = \pi \left(\dfrac{L}{2\pi}\right)^2 = \dfrac{L^2}{4\pi}$

The magnetic moment $M$ of the current-carrying loop is given by:

$M = IA = I \left(\dfrac{L^2}{4\pi}\right)<br><br>Rearranging the above equation to solve for$L$:<br><br>$L^2 = $\dfrac{4\pi M}{I}$

$L = \sqrt{\dfrac{4\pi M}{I}}$

Answer: $\sqrt{\dfrac{4\pi M}{I}}$