An electron revolves around an… - JEE Main 2023 Physics PYQ

JEE Main 2023 — Physics Electromagnetism

2023integerhard

An electron revolves around an infinite cylindrical wire having uniform linear charge density $2\times {10}^{-8}$ $C{m}^{-1}$ 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 ______________ $\times {10}^{6}m{s}^{-1}$ . Given mass of electron $=9\times {10}^{-31}\mathrm{kg}$

JEE Main 2023 Physics, Electrostatics — question figure

Official previous-year question

Held on 10 Apr 2023 · Verified 6 Jul 2026.

Did you get this right?

Solution

The electric field is given by

$E=(\frac{2k\lambda }{r})$

The required centripetal force is

$F=(\frac{m{v}^{2}}{r})$

The force due to the field is equal to the required centripetal force,

$\Rightarrow (\frac{2k\lambda }{r})e=\frac{m{v}^{2}}{r}$

$\Rightarrow v=\sqrt{\frac{2k\lambda e}{m}} =\sqrt{\frac{2\times 9\times {10}^{9}\times 2\times {10}^{-8}\times 1.6\times {10}^{-19}}{9\times {10}^{-31}}}$

$=\sqrt{1.6\times 4\times {10}^{13}} =4\times 2\times {10}^{6}m{s}^{-1}$

$=8\times {10}^{6}m{s}^{-1}$

Your note

JEE Main Electromagnetism in other years

Electromagnetism 2026 Electromagnetism 2025 Electromagnetism 2024 Electromagnetism 2022 Electromagnetism 2021 Electromagnetism 2020 Electromagnetism 2019 Electromagnetism 2018 Electromagnetism 2017 Electromagnetism 2016 Electromagnetism 2015 Electromagnetism 2014

Keep practicing Electromagnetism

Work through every JEE Main Electromagnetism PYQ, year by year.

Browse all Physics topics

Solution

The electric field is given by

$E=(\frac{2k\lambda }{r})$

The required centripetal force is

$F=(\frac{m{v}^{2}}{r})$

The force due to the field is equal to the required centripetal force,

$\Rightarrow (\frac{2k\lambda }{r})e=\frac{m{v}^{2}}{r}$

$\Rightarrow v=\sqrt{\frac{2k\lambda e}{m}} =\sqrt{\frac{2\times 9\times {10}^{9}\times 2\times {10}^{-8}\times 1.6\times {10}^{-19}}{9\times {10}^{-31}}}$

$=\sqrt{1.6\times 4\times {10}^{13}} =4\times 2\times {10}^{6}m{s}^{-1}$

$=8\times {10}^{6}m{s}^{-1}$

Did you get this right?

Solution

Your note

JEE Main Electromagnetism in other years

More JEE Main Electromagnetism questions

Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Electromagnetism

Did you get this right?

Solution

Your note

JEE Main Electromagnetism in other years

More JEE Main Electromagnetism questions

Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Electromagnetism