Let number of turns in length $l$ is $n$ so $l = n\,(2\pi r)$ or $\alpha = \frac{l}{{2\pi n}}$
$ \Rightarrow {\tau _{\max }} = \frac{{ni\pi B{l^2}}}{{4{\pi ^2}{n^2}}} = \frac{{{l^2}iB}}{{4\pi \,{n_{\min }}}} $
$\Rightarrow {\tau _{\max }} \propto \frac{1}{{{n_{\min }}}} \Rightarrow {n_{\min }} = 1$

