JEE Main 2019 — Electromagnetic Waves Question with Solution
From: JEE Main 2019 (Online) 9th April Morning Slot
Question
The magnetic field of a plane electromagnetic
wave is given by :
$\overline B = {B_0}\widehat i\left[ {\cos (kz - \omega t)} \right] + {B_i}\widehat j\cos (kz + \omega t)$ B0 = 3 × 10–5 T and B1 = 2 × 10–6 T.
The rms value of the force experienced by a stationary charge Q = 10–4 C at z = 0 is closest to :
$\overline B = {B_0}\widehat i\left[ {\cos (kz - \omega t)} \right] + {B_i}\widehat j\cos (kz + \omega t)$ B0 = 3 × 10–5 T and B1 = 2 × 10–6 T.
The rms value of the force experienced by a stationary charge Q = 10–4 C at z = 0 is closest to :
Choose an option
Show full solutionCorrect option: A
Correct answer
A0.6 N
Step-by-step explanation
Maximum electric field E = (B) (C)
Maximum force
(approx)
Maximum force
(approx)
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This is a previous-year question from JEE Main 2019, covering the Electromagnetic Waves chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.