JEE Main 2014 — Electromagnetic Waves Question with Solution
From: JEE Main 2014 (Online) 11th April Morning Slot
Question
An electromagnetic wave of frequency 1 1014 hertz is propagating along z - axis. The amplitude of electric field is 4 V/m. If 0 = 8.8 1012 C2/N-m2, then average energy density of electric field will be :
Choose an option
Show full solutionCorrect option: C
Correct answer
C35.2 1012 J/m3
Step-by-step explanation
We have
Since, energy density is
Since, energy density is
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This is a previous-year question from JEE Main 2014, 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.