Tension in Current Carrying Wire in Uniform Magnetic Field

To find the tension in the current-carrying wire, we analyze the mechanical equilibrium of a small element of the wire.

Let $R$ be the radius of the circular loop formed by the wire of length $L$. The relation between the total length $L$ and the radius $R$ is given by:
$L = 2\pi R \implies R = \frac{L}{2\pi}$

Now, consider a small arc element of the wire subtending a small angle $d\theta$ at the center of the circular loop. The length of this small element is:
$dl = R \, d\theta$

The magnetic force acting on a current element $I d\vec{l}$ in a uniform magnetic field $\vec{B}$ is given by:
$d\vec{F}_m = I (d\vec{l} \times \vec{B})$

Given that the current $I$ flows in the clockwise direction and the magnetic field $\vec{B}$ is directed into the plane of the paper, the direction of $d\vec{F}_m$ is radially outward by the right-hand rule. The magnitude of this magnetic force on the element is:
$dF_m = I (dl) B = I B R \, d\theta$

To oppose this outward radial force and keep the element in equilibrium, the tension $T$ acts tangentially at both ends of the arc element.

Resolving the tension forces along the radial direction toward the center, the net inward force due to tension is:
$F_{\text{inward}} = 2 T \sin\left(\frac{d\theta}{2}\right)$

For a very small angle $d\theta$, using the approximation $\sin\left(\frac{d\theta}{2}\right) \approx \frac{d\theta}{2}$, we get:
$F_{\text{inward}} \approx 2 T \left(\frac{d\theta}{2}\right) = T \, d\theta$

For radial equilibrium of the element, the net inward tension force must balance the net outward magnetic force:
$F_{\text{inward}} = dF_m$
$T \, d\theta = I B R \, d\theta$
$T = I B R$

Substituting $R = \frac{L}{2\pi}$ into the expression for tension $T$:
$T = I B \left(\frac{L}{2\pi}\right) = \frac{IBL}{2\pi}$

Therefore, the tension in the wire is $\frac{IBL}{2\pi}$, corresponding to Option C.