JEE Main 2019 — Electromagnetic Waves Question with Solution
From: JEE Main 2019 (Online) 12th April Evening Slot
Question
A plane electromagnetic wave having a frequency v = 23.9 GHz propagates along the positive z-direction in
free space. The peak value of the Electric Field is 60 V/m. Which among the following is the acceptable
magnetic field component in the electromagnetic wave ?
Choose an option
Show full solutionCorrect option: D
Correct answer
D = 2 × 10–7
sin(0.5 × 103
z - 1.5 × 1011t)
Step-by-step explanation
Since the wave is propagating in positive z-direction
So acceptable magnetic field component will be
C =
= = 2 10-7
and = 223.9109 = 1.5 × 1011 Hz
and k = = = 0.5 × 103
Note : When wave is propagating in positive z-direction then sign of kz and should be opposite.
From option you can see only option D can be correct.
So acceptable magnetic field component will be
C =
= = 2 10-7
and = 223.9109 = 1.5 × 1011 Hz
and k = = = 0.5 × 103
Note : When wave is propagating in positive z-direction then sign of kz and should be opposite.
From option you can see only option D can be correct.
Practice this on the real CBT interface
Solve this JEE Main question (and the rest of the Electromagnetic Waves chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.
Solve interactively →About this question
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.