Physics · Electrostatics

JEE Main 2024 — 4 April, Shift 1 — Question 45

An infinitely long positively charged straight thread has a linear charge density $\lambda \mathrm{Cm}^{-1}$ . An electron revolves along a circular path having axis along the length of the wire. The graph that correctly represents the variation of the kinetic energy of electron as a function of radius of circular path from the wire is :

Option A:
Option B:
Correct
Option C:
Option D:

Answer: B

Step-by-step solution

Electric field E at a distance r due to infinite long wire is

$\mathrm{E}=\frac{2 \mathrm{k} \lambda}{\mathrm{r}}$

\begin{aligned} & \mathrm{F}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\frac{2 \mathrm{k} \lambda \mathrm{e}}{\mathrm{r}} \begin{aligned} \mathrm{v} & =\sqrt{\frac{2 \mathrm{k} \lambda \mathrm{e}}{\mathrm{~m}}} \mathrm{KE} & =\frac{1}{2} \mathrm{mv}^{2}=\frac{1}{2} \mathrm{~m}\left(\frac{2 \mathrm{k} \lambda \mathrm{e}}{\mathrm{~m}}\right) & =\mathrm{k} \lambda \mathrm{e} \end{aligned} \end{aligned}

\begin{aligned} & \mathrm{F}=\mathrm{e}\left(\frac{2 \mathrm{k} \lambda}{\mathrm{r}}\right) & \mathrm{F}=\frac{2 \mathrm{k} \lambda \mathrm{e}}{\mathrm{r}} \end{aligned}

Solution figure

Answer key and solution verified before publishing.

Practise Electrostatics

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2024
Paper: 4 April, Shift 1
Subject: Physics
Chapter: Electrostatics
Topic: Electrostatic Potential Energy and Electric Potential

An infinite plane sheet of charge having uniform surface charge density +sigma s C / m^2 is placed on x- y plane. Another infinitely long…