A circular wire-loop of radius a , carrying a current i , is placed...

Consider a small element of the circular wire subtending an angle $d\theta$ at the center.

The length of this element is $dl = a d\theta$.

The magnetic force on this element is $dF = i(dl)B = i(a d\theta)B$.

By the right-hand rule for $d\vec{l} \times \vec{B}$, this force is directed radially inwards, causing a compressive force $T$ in the wire.

The compressive forces act tangentially at the ends of the small element. The radial component of these forces balances the inward magnetic force.

The net inward radial force due to compression is $2T \sin\left(\dfrac{d\theta}{2}\right)$.

For a small angle $d\theta$, $\sin\left(\dfrac{d\theta}{2}\right) \approx \dfrac{d\theta}{2}$.

Equating the forces for equilibrium:

$2T \left(\dfrac{d\theta}{2}\right) = i a B d\theta$

$T d\theta = i a B d\theta$

$T = i a B$

Alternatively, considering a semicircular portion of the wire, the total magnetic force is $F = i(2a)B = 2iaB$ directed towards the center. This is balanced by the compressive forces at the two ends, so $2T = 2iaB \Rightarrow T = iaB$.