Consider an small element AB of length dl of the circle of radius R subtending an angle $$\theta$$ at the centre O.

If T is the tension in the wire, then force towards the centre will be equal to $$2T\sin \left( {{\theta \over 2}} \right)$$ which is balanced by outward magnetic force on the current carrying element $$( = IdlB)$$

$$2T\sin \left( {{\theta \over 2}} \right) = IdlB$$

For small angle $$\theta$$, $$\sin {\theta \over 2} \approx {\theta \over 2}$$

or, $$T = {{IBdl} \over \theta } = IBR$$ ($$\because$$ $$\theta = {{dl} \over R}$$)

$$ = {{IBL} \over {2\pi }}$$ ($$\because$$ $$R = {L \over {2\pi }}$$)