Motion of Charges in an Electric Field question from JEE Main 2024 (4 Apr, Shift 1)

Answer: (d)

1. Field of the line: $E = \frac{\lambda}{2\pi\varepsilon_0 r}$; attraction gives the centripetal force.
2. $\frac{mv^2}{r} = \frac{e\lambda}{2\pi\varepsilon_0 r} \Rightarrow mv^2 = \frac{e\lambda}{2\pi\varepsilon_0}$
3. $KE = \frac{1}{2}mv^2 = \frac{e\lambda}{4\pi\varepsilon_0}$, independent of $r$.
4. So the answer is (d), a horizontal line.