A length L of a wire carries a current I. Show that if the wire is formed into a circular coil, the maximum torque in a given magnetic fiel B, is developed when the coil has one turn only and the maximum torque has the value, `tau_(max)=L^2IB//4pi`.

Explanation

Torque on the current loop is `tau=NIAB sin alpha` …(i)
If there are N turns of circular coil, each of radius r, then
`L=2pirN` or `r=L//2piN`
Area of the coil, `A=pir^2=piL^2//(4pi^2N^2)=L^2//(4piN^2)`
Putting this value in (i),we get
`tau=NI[(L^2)/(4piN^2)]B sin alpha=(L^2IBsinalpha)/(4piN)`...(ii)
From (ii), it is clear that `tau` is maximum if N is minimum, i.e., the number of turns `N=1`. Torque is maximum if `sin alpha=1` and `N=1`
`:. tau_(max)=L^2IB//4pi`.