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Electromagnetic Waves question

2021 · 27 Aug · Shift 1 · Q56
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  5. /2021 · 27 Aug · Shift 1 · Q56

Electromagnetic Waves question

2021 · 27 Aug · Shift 1 · Q56

JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1
Electric field in a plane electromagnetic wave is given by E = 50 sin(500x −-− 10 ×\times× 1010 t) V/m The velocity of electromagnetic wave in this medium is : (Given C = speed of light in vacuum)
  1. A
    32{3 \over 2}23​ C
  2. B
    C
  3. C
    23{2 \over 3}32​ C
  4. D
    C2{C \over 2}2C​
View written solutionFree

Correct answer: C

  1. Write the standard form of a plane wave

A plane electromagnetic wave can be written as

E=E0sin⁡(kx−ωt)E = E_0 \sin(kx - \omega t)E=E0​sin(kx−ωt)

where:

  • kkk = wave number
  • ω\omegaω = angular frequency
  • wave speed v=ωkv = \dfrac{\omega}{k}v=kω​
  1. Compare with the given equation

Given:

E=50sin⁡(500x−10×1010t) V/mE = 50\sin(500x - 10 \times 10^{10} t)\ \text{V/m}E=50sin(500x−10×1010t) V/m

Comparing with E=E0sin⁡(kx−ωt)E = E_0 \sin(kx - \omega t)E=E0​sin(kx−ωt),

k=500 rad/mk = 500\ \text{rad/m}k=500 rad/m

and

ω=10×1010=1011 rad/s\omega = 10 \times 10^{10} = 10^{11}\ \text{rad/s}ω=10×1010=1011 rad/s

  1. Calculate the wave velocity

v=ωk=1011500v = \frac{\omega}{k} = \frac{10^{11}}{500}v=kω​=5001011​

v=2×108 m/sv = 2 \times 10^8\ \text{m/s}v=2×108 m/s

  1. Express in terms of CCC

Given speed of light in vacuum is C=3×108 m/sC = 3 \times 10^8\ \text{m/s}C=3×108 m/s,

v=2×108=23×3×108v = 2 \times 10^8 = \frac{2}{3} \times 3 \times 10^8v=2×108=32​×3×108

So,

v=23Cv = \frac{2}{3}Cv=32​C

  1. Check options
  • A: 32C\dfrac{3}{2}C23​C ❌
  • B: CCC ❌
  • C: 23C\dfrac{2}{3}C32​C ✅
  • D: C2\dfrac{C}{2}2C​ ❌

Hence, the correct option is C.

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