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Electromagnetic Waves question

2024 · 6 Apr · Shift 2 · Q79
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  5. /2024 · 6 Apr · Shift 2 · Q79

Electromagnetic Waves question

2024 · 6 Apr · Shift 2 · Q79

JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1
In the given electromagnetic wave Ey=600sin⁡(ωt−kx)Vm−1\mathrm{E}_{\mathrm{y}}=600 \sin (\omega t-\mathrm{kx}) \mathrm{Vm}^{-1}Ey​=600sin(ωt−kx)Vm−1, intensity of the associated light beam is (in W/m2\mathrm{W} / \mathrm{m}^2W/m2: (Given ϵ0=9×10−12C2 N−1 m−2\epsilon_0=9 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}ϵ0​=9×10−12C2 N−1 m−2 )
  1. A
    486
  2. B
    729
  3. C
    243
  4. D
    972
View written solutionFree

Correct answer: A

  1. For a plane electromagnetic wave, E=E0sin⁡(ωt−kx)E = E_0 \sin(\omega t-kx)E=E0​sin(ωt−kx) so the amplitude of electric field is E0=600 V m−1.E_0 = 600\ \text{V m}^{-1}.E0​=600 V m−1.

  2. The average intensity of an electromagnetic wave is I=12cϵ0E02.I = \frac{1}{2}c\epsilon_0 E_0^2.I=21​cϵ0​E02​.

  3. Substitute the given values: c=3×108 m s−1,ϵ0=9×10−12,E0=600.c = 3\times 10^8\ \text{m s}^{-1}, \qquad \epsilon_0 = 9\times 10^{-12}, \qquad E_0 = 600.c=3×108 m s−1,ϵ0​=9×10−12,E0​=600.

    I=12(3×108)(9×10−12)(600)2.I = \frac{1}{2}(3\times 10^8)(9\times 10^{-12})(600)^2.I=21​(3×108)(9×10−12)(600)2.

  4. Now compute step-by-step: 12×3×9=13.5\frac{1}{2}\times 3 \times 9 = 13.521​×3×9=13.5 and 108×10−12=10−4.10^8\times 10^{-12} = 10^{-4}.108×10−12=10−4. So, 12cϵ0=13.5×10−4=1.35×10−3.\frac{1}{2}c\epsilon_0 = 13.5\times 10^{-4} = 1.35\times 10^{-3}.21​cϵ0​=13.5×10−4=1.35×10−3.

    Also, (600)2=360000=3.6×105.(600)^2 = 360000 = 3.6\times 10^5.(600)2=360000=3.6×105.

  5. Therefore, I=(1.35×10−3)(3.6×105)I = (1.35\times 10^{-3})(3.6\times 10^5)I=(1.35×10−3)(3.6×105) =4.86×102= 4.86\times 10^2=4.86×102 =486 W m−2.= 486\ \text{W m}^{-2}.=486 W m−2.

  6. Hence the correct option is A  (486).\boxed{A\; (486)}.A(486)​.

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