A wire of length $L$ carries a current $i$. If the wire is turned into a circular coil and kept in a uniform…
A wire of length $L$ carries a current $i$. If the wire is turned into a circular coil and kept in a uniform magnetic field $B$, the maximum magnitude of torque in the given magnetic field will be
$\frac{B L^2}{4 \pi}$
$\frac{B i L^2}{4 \pi}$
$\frac{B^2 L^2}{2}$
$\frac{B i L^2}{2}$
Solution
The torque on the current loop is $\tau=N I A B \sin \alpha---(1)$ If there are $N$ turns of the circular coil, each of radius $r$, then $L=2 \pi r N$
$\Rightarrow r=\frac{L}{2 \pi N}$
Area of the coil, $A=\pi r^2=\pi\left(\frac{L}{2 \pi N}\right)^2=\frac{L^2}{4 \pi N^2}$ Putting this value in (1), we get,
$\tau=N I\left(\frac{L^2}{4 \pi N^2}\right) B \sin \alpha=\frac{L^2 I B \sin \alpha}{4 \pi N}--(2)$
From (2), it is clear that $\tau$ is maximum if $N$ is minimum. Torque is maximum if $\sin \alpha=1$ and $N=1$
$\therefore \tau_{\max }=L^2 I B / 4 \pi$