Question

The wire of length $l$ is bent into a circular loop of a single turn and is suspended in a magnetic field of induction $B$. When a current $I$ is passed through the loop, the maximum torque experienced by it is:

• A. $\left(\frac{1}{4\pi }\right)BI{l}^2$
• B. $\frac{1}{4}B{I}^{2}l$
• C. $\left(\frac{1}{4\pi }\right)BIl$
• D. $\left(\frac{1}{4\pi }\right)B^{2}Il$
• E. $\left(\frac{1}{4\pi }\right)B^2{I}^2{l}^2$

Answer 1

The correct option is A $\left(\frac{1}{4\pi }\right)BI{l}^2$

When the wire of length $l$ is bent into circular loop of radius $r$,
$l=2\pi r\Rightarrow r=\frac{l}{2\pi }$
When current $I$ is passed through the loop, the maximum torque acting on the loop is given by, $\tau =IBA$
$\tau =I\times B\times \pi r^2$
$\tau =IB\pi {\left(\frac{l}{2\pi }\right)}^2=IB\frac{l^2}{4\pi }=\left(\frac{1}{4\pi }\right)BI{l}^2$.