An electron revolves around an infinite cylindrical wire having uniform linear charge density $$2 \times 10^{-8}$$ C m$$^{-1}$$ in circular path under the influence of attractive electrostatic field as shown in the figure. The velocity of electron with which it is revolving is _______ $$\times 10^6$$ m s$$^{-1}$$. Given mass of electron $$= 9 \times 10^{-31}$$ kg

Electric field at distance r from infinite line charge: $$E = \frac{\lambda}{2\pi\varepsilon_0 r}$$

For circular motion: $$eE = \frac{mv^2}{r}$$, so $$\frac{e\lambda}{2\pi\varepsilon_0 r} = \frac{mv^2}{r}$$

$$v^2 = \frac{e\lambda}{2\pi\varepsilon_0 m} = \frac{1.6 \times 10^{-19} \times 2 \times 10^{-8}}{2\pi \times 8.85 \times 10^{-12} \times 9 \times 10^{-31}}$$

$$= \frac{3.2 \times 10^{-27}}{5.0 \times 10^{-41}} = 6.4 \times 10^{13}$$

$$v = 8 \times 10^6$$ m/s. The answer is 8.