A long cylindrical wire carries a positive charge of linear density l. An electron (- e,m) revolves around it in a circular path under the influence of the attractive electrostatic force.The speed of electron is

Electric field at perpendicular distance r,

$\mathrm{E}=\frac{\mathrm{\lambda }}{2{\mathrm{\pi \epsilon }}_0\mathrm{r}}$

Electric force on electron, $\mathrm{F}=\mathrm{eE}$

To move in circular path, necessary centripetal force is provided by electric force,

$$\begin{gathered} {\mathrm{F}}_{\mathrm{c}}=\frac{{\mathrm{m\nu }}^2}{\mathrm{r}}=\mathrm{eE}=\frac{\mathrm{e\lambda }}{2{\mathrm{\pi \epsilon }}_0\mathrm{r}} \\ \mathrm{Hence}, \mathrm{speed} \mathrm{of} \mathrm{the} \mathrm{electron}, \\ \mathrm{\nu }=\sqrt{\frac{\mathrm{e\lambda }}{2{\mathrm{\pi \epsilon }}_0\mathrm{m}}}\Rightarrow \mathrm{\nu } \mathrm{\alpha } {\mathrm{r}}^0 \end{gathered}$$