A long straight wire carries a current $\mathcal{i}$. A particle having a positive charge $q$ and mass $m$, kept at a distance ${\mathcal{x}}_0$ from the wire is projected towards it with speed $\mathcal{v}$. Find the closest distance of approach of charged particle to the wire.

Magnetic field due to long wire is

$\overrightarrow{B}=-\frac{{\mu }_{0}i\hat{k}}{2\pi x}$

Let the velocity of the charged particle at any instant be

$\overrightarrow{v}=\left(v_x\hat{i}+v_y\hat{j}\right)$

Magnetic force on the charge,

$\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})=-\frac{{\mu }_{0}qi{v}_y}{2\pi x}\hat{i}+\frac{{\mu }_{0}qi{v}_x}{2\pi x}\hat{j}$                            ……..$\left(1\right)$

From eqn. $\left(1\right)$, y-component of acceleration is

Now, $a_y=\frac{d{v}_y}{dt}=\frac{{\mu }_{0}qi{v}_x}{2\pi mx}=\frac{{\mu }_{0}q}{2\pi mx}i\frac{dx}{dt}$

At the minimum separation, velocity of particle is $-\mathrm{v}\hat{\mathrm{j}}$

or  ${\int }_0^{v}d{v}_y=\frac{{\mu }_{0}qi}{2\pi m}{\int }_{x_0}^{x_{min}}\frac{dx}{x}$

or  $-v=\frac{{\mu }_{0}qi}{2\pi m}\ell n\left|\frac{x_{\min }}{x_0}\right|$

Thus  $x_{\min }=x_0{e}^{-2\pi mv/{\mu }_{0}qi}$