Question

A metallic wire is folded to form a square loop of side $a$ . It carries a current $i$ and is kept perpendicular to a uniform magnetic field. If the shape of the loop is changed from square to a circle without changing the length of the wire and current, the amount of work done in doing so is

• A. $\left(\textit{iBa}\right)^{2}\left(\pi + 2\right)$
• B. $\left(\textit{iBa}\right)^{2}\left(\pi - 2\right)$
• C. $\left(\textit{iBa}\right)^{2}\left(4 / \pi + 1\right)$
• D. $\left(\textit{iBa}\right)^{2}\left(1 - \frac{4}{\pi }\right)$

Answer 1

Answer: (D)

$W \, = \, U_{f} \, - \, U_{i}$
$=\left(- M_{\text{f}} B \left(\text{ cos 0}\right)^{\text{o}}\right)-\left(- M_{\text{i}} B \text{ cos } 0^{\text{o}}\right)$
$=iB\left(A_{\text{i}} - A_{\text{f}}\right)$
where $A_{\text{i}}=a^{2}\text{,A}_{\text{f}}=\pi R^{2}=\frac{4 a^{2}}{\pi }$
as $2\pi R=4a\text{ or }\textit{R}=\frac{2 a}{\pi }$
$∴ \, \, W=iBa^{2}\left(1 - \frac{4}{\pi }\right)$