1. Magnetic force on a small element

For a small element of the wire of length $d\ell$ carrying current $I$ in a uniform magnetic field $\vec B$ into the plane, the magnetic force is

$dF = I B\, d\ell$

because the current element is perpendicular to $\vec B$.

This force acts everywhere normal to the wire. When the wire becomes circular, this magnetic force acts radially outward at every point.

2. Force due to tension on a small arc

Let the wire form a circle of radius $R$. Consider a small arc subtending angle $d\theta$ at the center.

Its length is

$d\ell = R\, d\theta$

If the tension in the wire is $T$, then the two tensions at the ends of this small arc have a resultant inward force

$dF_T = 2T \sin\left(\frac{d\theta}{2}\right) \approx T\, d\theta$

for small $d\theta$.

This inward force balances the outward magnetic force on the same arc.

So,

$T\, d\theta = I B\, d\ell$

Using $d\ell = R d\theta$,

$T\, d\theta = I B R\, d\theta$

Hence,

$T = IBR$

3. Use the total length of the wire

Since the wire forms a complete circle,

$L = 2\pi R$

So,

$R = \frac{L}{2\pi}$

Substitute into $T = IBR$:

$T = IB\left(\frac{L}{2\pi}\right)$

Therefore,

$\boxed{T = \frac{IBL}{2\pi}}$

4. Match with options

The correct option is

$\boxed{\text{C: } \frac{IBL}{2\pi}}$