An electron moves in a circular path of radius 0.1 m around an infinitely long straight line charge of linear charge density \( +0.5\,\mu\mathrm{C\,m^{-1}} \). The speed of the electron is (Take \( \frac{1}{4\pi\varepsilon_0}=9\times10^{9}\,\mathrm{N\,m^2\,C^{-2}} \), \( e=1.6\times10^{-19}\,\mathrm{C} \), \( m_e=9\times10^{-31}\,\mathrm{kg} \).)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 4\times10^{7}\,\mathrm{m\,s^{-1}} \)
The field of an infinitely long line charge at distance \( r \) is \( E=\lambda/2\pi\varepsilon_0r \), directed radially outward for positive \( \lambda \). The electron is pulled towards the line by the force \( eE \), and this force provides the centripetal force \( m_ev^{2}/r \) for the circular motion.
Equating the two forces:
\( \displaystyle \frac{e\lambda}{2\pi\varepsilon_0r}=\frac{m_ev^{2}}{r} \)
\( \displaystyle \Rightarrow\; v^{2}=\frac{e\lambda}{2\pi\varepsilon_0m_e} \)
From the above, the speed does not depend on the radius of the orbit. In SI units the line charge density is:
\( \lambda=0.5\,\mu\mathrm{C\,m^{-1}}=0.5\times10^{-6}\,\mathrm{C\,m^{-1}} \)
\( \begin{aligned} v^{2} &= 2\times\frac{1}{4\pi\varepsilon_0}\times\frac{e\lambda}{m_e} \\ &= \frac{2\times9\times10^{9}\times1.6\times10^{-19}\times0.5\times10^{-6}}{9\times10^{-31}}\,\mathrm{m^{2}\,s^{-2}} \\ &= \frac{1.44\times10^{-15}}{9\times10^{-31}}\,\mathrm{m^{2}\,s^{-2}} \\ &= 1.6\times10^{15}\,\mathrm{m^{2}\,s^{-2}} \end{aligned} \)
\( \begin{aligned} v &= \sqrt{1.6\times10^{15}}\,\mathrm{m\,s^{-1}} \\ &= 4\times10^{7}\,\mathrm{m\,s^{-1}} \end{aligned} \)