A thin flexible wire of length L is connected to two adjacent fixed points and carries a current I in the clockwise direction, as shown in the figure. When the system is put in a uniform magnetic field of strength B going into the plane of the paper, the wire takes the shape of a circle. The tension in the wire is

$\mathrm{T}=\frac{\mathrm{Bil}}{2\mathrm{\pi }}$
Force on the loop will be zero Consider a small element dl of loop making 2$\mathrm{\theta }$ at centre. Let
‘T’ be the tension on loop.
The magnetic force on the element is dF = Bidl .The horizontal. components of tension get cancelled the vertical components of tension = 2Tsin$\mathrm{\theta }$
$=2\mathrm{T}\times \mathrm{\theta }=2\mathrm{T}\times \frac{\mathrm{dl}}{2\mathrm{R}}=\frac{\mathrm{T}\times \mathrm{dl}}{\mathrm{R}}$
In equilibrium of loop element
$2\mathrm{Tsin}\mathrm{\theta }=\mathrm{Bidl};\mathrm{T}\times \frac{\mathrm{dl}}{\mathrm{R}}=\mathrm{Bidl}$
$\therefore \mathrm{T}=\mathrm{BiR}( \mathrm{or} )\frac{\mathrm{Bil}}{2\mathrm{\pi }}$