JEE Main 2015PhysicsElectromagnetic WavesMediumMCQ

JEE Main 2015 — Electromagnetic Waves Question with Solution

JEE Main 2015 (10 Apr Online)

Question

An electromagnetic wave travelling in the x- direction has frequency of  2×1014 H z  and electric field amplitude of  27 V m–1  oscillates in Y-direction. From the options given below, which one describes the magnetic field for this wave?

Choose an option

Show full solutionCorrect option: D
Correct answer
D B → x,t=9×10-8Tk^sin2πx1.5×10-6-2×1014t

Step-by-step explanation

When undefined

Then  B = B0sin⁡ ( kx−ωt )

Of light in travelling along i^ then B→ in either along j ^ or k ^ .

∴ speed of light  C= E0B0 ⇒B0= E0C

⇒     B0= 273×108=9 ×10-8 T

also, ω=2π f=2π ×2 ×1014=4 π ×1014

k= ω c = 4π× 10 14 3× 10 8

=( 2π 1.5× 10 −6 )

Looking into the option the correct

Answer is B→=9×10-8sin⁡2πx1.5×10-6-2×1014tk^

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About this question

This is a previous-year question from JEE Main 2015, 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.