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

Consider a small element of the circular wire-loop subtending an angle $d\theta$ at the center. The length of this element is $dl = a d\theta$.

The magnetic force acting on this small element is given by:
$dF = i dl B = i (a d\theta) B$

According to the right-hand rule (or Fleming's left-hand rule), since the current is counter-clockwise and the magnetic field is directed into the page, the magnetic force on each element is directed radially inward. This inward force compresses the loop.

To maintain equilibrium, a compressive force (tension) $T$ is developed in the wire. The outward radial component of this compressive force from the two ends of the element balances the inward magnetic force:
$2T \sin\left(\dfrac{d\theta}{2}\right) = dF$

Since $d\theta$ is very small, $\sin\left(\dfrac{d\theta}{2}\right) \approx \dfrac{d\theta}{2}$.
$2T \left(\dfrac{d\theta}{2}\right) = i a B d\theta$
$T = i a B$

The compressive stress $\sigma$ in the wire is the force per unit cross-sectional area:
$\sigma = \dfrac{T}{A} = \dfrac{i a B}{\pi r^2}$

The longitudinal strain $\epsilon$ in the wire is given by Hooke's law:
$\epsilon = \dfrac{\sigma}{Y} = \dfrac{i a B}{\pi r^2 Y}$

The strain is also equal to the fractional change in the circumference, which is the same as the fractional change in the radius:
$\epsilon = \dfrac{\Delta L}{L} = \dfrac{\Delta (2\pi a)}{2\pi a} = \dfrac{\Delta a}{a}$

Equating the two expressions for strain:
$\dfrac{\Delta a}{a} = \dfrac{i a B}{\pi r^2 Y}$

$\Delta a = \dfrac{i a^2 B}{\pi r^2 Y}$

When the magnetic field is switched off, the compressive force vanishes, and the loop expands back to its natural length. Therefore, the increase in the radius of the loop is $\dfrac{i a^2 B}{\pi r^2 Y}$.