Magnetic field due to long wire is

$\overrightarrow{\mathrm{B}}=\frac{-{\mathrm{\mu }}_0\mathrm{i}\hat{\mathrm{k}}}{2\mathrm{\pi x}}$

let the charged particle's velocity 

at any instant be $({\mathrm{v}}_{\mathrm{x}}\hat{\mathrm{i}}+{\mathrm{v}}_{\mathrm{y}}\hat{\mathrm{j}}).$

magnetic force on the charge

$\overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{v}}\times \overrightarrow{\mathrm{B}})$

$=-\frac{{\mathrm{\mu }}_0{\mathrm{qiv}}_{\mathrm{y}}}{2\mathrm{\pi x}}\hat{\mathrm{i}}+\frac{{\mathrm{\mu }}_0\mathrm{q}.{\mathrm{iv}}_{\mathrm{x}}}{2\mathrm{\pi x}}\hat{\mathrm{j}}$  .....(I)

${\text{a}}_{\text{y}} =\frac{{\mathrm{dv}}_{\mathrm{y}}}{\mathrm{dt}}=\frac{{\mathrm{\mu }}_0\mathrm{q}.{\mathrm{iv}}_{\mathrm{x}}}{2\mathrm{\pi mx}}=\frac{{\mathrm{\mu }}_0\mathrm{q}}{2\mathrm{\pi mx}}\mathrm{i}.\frac{\mathrm{dx}}{\mathrm{dt}}$

at minimum separation velocity of particle is $-\mathrm{v}\hat{\mathrm{j}}$

$\underset{0}{\overset{-\mathrm{v}}{\int }}{\mathrm{dv}}_{\mathrm{y}}=\frac{{\mathrm{\mu }}_0\mathrm{qi}}{2\mathrm{\pi m}}\underset{{\mathrm{x}}_0}{\overset{{\mathrm{x}}_{\min }}{\int }}\frac{\mathrm{dx}}{\mathrm{x}}$

on integrating

${\mathrm{x}}_{\min }={\mathrm{x}}_0{\mathrm{e}}^{-\frac{2\mathrm{\pi mv}}{{\mathrm{\mu }}_0\mathrm{qi}}}$