An electron revolves around an infinite cylindrical wire having uniform linear charge density \(2×10^{–8}\ Cm^{-1}\) 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 = \(9\times10^{-31}\) kg.
To determine the velocity \(v\) of the electron, we follow these steps: 1. Electric Field due to an Infinite Cylindrical Wire: The electric field \(E\) at a distance \(r\) from an infinite wire with linear charge density \(\lambda\) is given by: \[ E = \frac{\lambda}{2\pi\epsilon_0 r} \] where \(\epsilon_0\) is the permittivity of free space (\(\epsilon_0 \approx 8.85 \times 10^{-12}\) C\(^2\)/N·m\(^2\)). 2. Force on the Electron: The electrostatic force \(F\) acting on the electron is: \[ F = eE = \frac{e\lambda}{2\pi\epsilon_0 r} \] where \(e\) is the charge of the electron (\(e \approx 1.6 \times 10^{-19}\) C). 3. Centripetal Force: For the electron to move in a circular path, the electrostatic force must provide the necessary centripetal force: \[ F = \frac{mv^2}{r} \] Equating the two expressions for \(F\): \[ \frac{e\lambda}{2\pi\epsilon_0 r} = \frac{mv^2}{r} \] Simplifying, we get: \[ v^2 = \frac{e\lambda}{2\pi\epsilon_0 m} \] \[ v = \sqrt{\frac{e\lambda}{2\pi\epsilon_0 m}} \] 4. Substitute the Given Values: Plugging in the values: \[ v = \sqrt{\frac{(1.6 \times 10^{-19})(2 \times 10^{-8})}{2\pi(8.85 \times 10^{-12})(9 \times 10^{-31})}} \] \[ v = \sqrt{\frac{3.2 \times 10^{-27}}{5.0 \times 10^{-41}}} \] \[ v = \sqrt{6.4 \times 10^{13}} \] \[ v \approx 8 \times 10^6 \text{ m/s} \] Final Answer The velocity of the electron is \(\boxed{8 \times 10^{6}}\) m/s.
