A square frame projected towards an infinite current carrying wire.

The magnetic flux through the square frame when the left part is at a distance $x$ from wire is $\phi = \displaystyle \int B dA = \displaystyle \int_{0}^{l} \dfrac{\mu_{0} I}{2 \pi r} l dr = \dfrac{\mu_{0} I l}{2 \pi} \ln\bigg(1 + \frac{l}{x}\bigg)$

Now, the induced emf is $|\epsilon| = \bigg|\frac{d \phi}{dt}\bigg| = - \dfrac{\mu_{0} I}{2 \pi x(l+x)} v$ (Note that $v = \frac{dx}{dt}$ is -ve)

Now, the induced anticlockwise current is $i = \dfrac{|\epsilon|}{R} = -\dfrac{\mu_{0} I}{2 \pi R x(l+x)} v$

Net force experienced = $F = \dfrac{\mu_{0} I i}{2 \pi} \bigg(\dfrac{1}{x} - \dfrac{1}{l+x}\bigg)$

$= - \bigg(\dfrac{\mu_{0} I l}{2 \pi x(l+x) } \bigg)^2 \dfrac{v}{R}$

$\Rightarrow m \dfrac{v dv}{dx} = -\bigg(\dfrac{\mu_{0} I l}{2 \pi x(l+x) } \bigg)^2 \dfrac{v}{R}$

$\displaystyle \int_{\infty}^{a} \dfrac{dx}{(x(l+x))^2} = -\bigg(\dfrac{2 \pi}{\mu_{0} I l} \bigg)^2 m R \int_{-v_{0}}^{0} dv$

Since, $l=1m$ , we get $\displaystyle \int_{a}^{\infty} \dfrac{dx}{(x(1+x))^2} = 250000 m^{-3}$