JEE Main 2019PhysicsElectromagnetic WavesHardMCQ

JEE Main 2019 — Electromagnetic Waves Question with Solution

JEE Main 2019 (10 Jan Shift 2)

Question

The electric field of a plane polarized electromagnetic wave in free space at time t=0 is given by the expression E→x, y=10j^cos6x+8z. The magnetic field B→x, z, t is given by (c is the velocity of light.)

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Show full solutionCorrect option: D
Correct answer
D1c6k^-8i^cos6x+8z-10 ct

Step-by-step explanation

E=E0cosωt-K  →· r →

∴ At any time t,

K→=6i^+8k^

B0=E0c=10c

λ=2πk=2π10=π5

Also, E→×B→=c→

∴ω=2πf=2πcλ

i^j^k^010BxByBz=6i^+8k^10

ω=2π×cπ5=10c

∴i^Bz-0+j^0+k^-Bx=0.6i^-0.8k^

Comparing the coefficients on both sides,

∴Bz=0.6; Bx=-0.8

B0→=10c-0.8i^+0.6k^=-8i^+6k^c

So the magnetic field,B→=1c-8i^+6k^cos⁡6x+8z-10ct tesla

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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.