Question
An infinitely long positively charged straight wire has a linear charge density $\lambda$. An electron is revolving around the wire as its centre with a constant speed in a circular plane perpendicular to the wire. Deduce the expression for its kinetic energy.

Answer

Let $r$ be the radius of the uniform circular motion of the electron. The electric field intensity $\vec{E}$ at every point on the circular path is radially outward and has the same magnitude $E=$ $\frac{\lambda}{2 \pi \varepsilon r}$.
$\therefore$ The centripetal force on the electron,
$
F _{ c }=\frac{m_{ e } v^2}{r}= eE =\frac{e \lambda}{2 \pi \varepsilon r}
$
where $m_e$ and $v$ are the mass and linear speed of the electron.
$
\therefore m _{ e } v ^2=\frac{e \lambda}{2 \pi \varepsilon}
$
$\therefore$ The kinetic energy of the electron,
$
\frac{1}{2} m _{ e } v ^2=\frac{e \lambda}{4 \pi \varepsilon}
$

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