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(C) Consider a small element of the wire of length $d\ell$ subtending an angle $d	heta$ at the center of the circle formed by the wire.From the geome

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$\frac{IBL}{2\pi}$

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(C) Consider a small element of the wire of length $d\ell$ subtending an angle $d	heta$ at the center of the circle formed by the wire.From the geometry,$d\ell = R d	heta$,where $R$ is the radius of the circle.The magnetic force on this small element is $dF = B I d\ell = B I R d	heta$,directed radially outwards.This force is balanced by the radial component of the tension $T$ at the two ends of the element.The radial component of tension is $2T \sin(\frac{d	heta}{2})$.For small $d	heta$,$\sin(\frac{d	heta}{2}) \approx \frac{d	heta}{2}$.Equating the forces: $2T (\frac{d	heta}{2}) = B I R d	heta$.$T d	heta = B I R d	heta \Rightarrow T = B I R$.Since the total length of the wire is $L = 2\pi R$,we have $R = \frac{L}{2\pi}$.Substituting $R$ in the expression for tension: $T = B I (\frac{L}{2\pi}) = \frac{IBL}{2\pi}$.

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