Question

A given length $L$ of wire carries a current $I$. The wire can be formed into a circular coil of any number of turns. The maximum value of the torque $\tau$ that can be developed on the coil when placed in a given magnetic field $B$ is $\tau =\frac{IB{L}^2}{x\pi }$. The value of $x$ is ______.

Answer 1

Let us suppose $n$ be the total number of turns and $r$ is the radius of each turn.

Then,

$n\left(2\pi r\right)=L$

$\Rightarrow nr=\frac{L}{2\pi }...\left(1\right)$

Now, torque on the coil,

$\overrightarrow{\tau }=nI(\overrightarrow{A}\times \overrightarrow{B})=nI\left(\pi r^2\right)B\sin \theta \hat{\eta }$

$\Rightarrow \tau =\frac{IBLr\sin \theta }{2}$ [From (1)]

For maximum torque, $\sin \theta =1$ and r should be maximum.

$\Rightarrow$ n should be minimum. [From (1)]

$\therefore n=1$ [Minimum value]

Hence,

${\tau }_{max}=\frac{IBLr}{2}=\frac{IBL}{2}\times \frac{L}{2\pi }$ [From(1)]

$\Rightarrow {\tau }_{max}=\frac{IB{L}^2}{4\pi }=\frac{IB{L}^2}{x\pi }$

$\therefore x=4$