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 ……. $\times 10^{6} {ms}^{- 1}$ . Given mass of electron = $9 \times 10^{- 31} kg$ .

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answer is 8.

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Detailed Solution

For Circular motion
$\begin{array}{l} eE = \frac{{mv}^{2}}{r} \\ \Rightarrow e \cdot \frac{λ}{2 {πϵ}_{0} r} = \frac{{mv}^{2}}{r} \\ \Rightarrow v = \sqrt{\frac{eλ}{2 {πϵ}_{0} m}} = \sqrt{\frac{1.6 \times 10^{- 19} \times 2 \times 10^{- 8} \times 2 \times 9 \times 10^{9}}{9 \times 10^{- 31}}} \\ v = 8 \times 10^{6} m / \sec \end{array}$

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