Skip to main content

The electric field of a plane electromagnetic wave is given by E=E_0 i+ j √2 cos⁡(kz+ω t) . At t=0 , a positively charged particle is at the point…

JEE Main 2020 — Physics Electromagnetism

2020mcqmedium

The electric field of a plane electromagnetic wave is given by E⃗=E0i^+j^2cos⁡(kz+ωt)\vec{E}={E}_{0}\frac{\hat{i}+\hat{j}}{\sqrt{2}}\mathrm{cos}⁡(kz+\omega t)E=E0​2​i^+j^​​cos⁡(kz+ωt) . At t=0t=0t=0 , a positively charged particle is at the point (x,y,z)=(0,0,πk)(x,y,z)=(0,0,\frac{\pi }{k})(x,y,z)=(0,0,kπ​) . If its instantaneous velocity at (t=0)(t=0)(t=0) is v0k^{v}_{0}\hat{k}v0​k^ , the force acting on it due to the wave is:

Official previous-year question

Held on 7 Jan 2020 · Verified 6 Jul 2026.

Options

  1. A

    parallel to i^+j^2\frac{\hat{i}+\hat{j}}{\sqrt{2}}2​i^+j^​​

  2. B

    zero

  3. C

    antiparallel to i^+j^2\frac{\hat{i}+\hat{j}}{\sqrt{2}}2​i^+j^​​

  4. D

    parallel to k^\hat{k}k^

Did you get this right?

Sign in to track your attempts and accuracy.

Your note

Sign in to keep a private note on this question. Nothing you write is ever public.

JEE Main Electromagnetism in other years

More JEE Main Electromagnetism questions

Other Physics topics for JEE Main

Keep practicing Electromagnetism

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

Browse all Physics topics