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Electromagnetic Waves question

2022 · 25 Jun · Shift 1 · Q55
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  5. /2022 · 25 Jun · Shift 1 · Q55

Electromagnetic Waves question

2022 · 25 Jun · Shift 1 · Q55

JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1
The electric field in an electromagnetic wave is given by E = 56.5 sin ω\omegaω(t −-− x/c) NC −-− 1. Find the intensity of the wave if it is propagating along x-axis in the free space. (Given : ε\varepsilonε 0 = 8.85 ×\times× 10 −-− 12C2N −-− 1m −-− 2)
  1. A
    5.65 Wm −-− 2
  2. B
    4.24 Wm −-− 2
  3. C
    1.9 ×\times× 10 −-− 7 Wm −-− 2
  4. D
    56.5 Wm −-− 2
View written solutionFree

Correct answer: B

  1. Given electric field

The electromagnetic wave is E=56.5sin⁡ω(t−xc) N C−1E = 56.5\sin\omega\left(t-\frac{x}{c}\right)\ \text{N C}^{-1}E=56.5sinω(t−cx​) N C−1

So the amplitude of electric field is E0=56.5 N C−1E_0 = 56.5\ \text{N C}^{-1}E0​=56.5 N C−1

  1. Formula for intensity of an electromagnetic wave

For a plane electromagnetic wave in free space, the average intensity is I=12cε0E02I = \frac{1}{2}c\varepsilon_0 E_0^2I=21​cε0​E02​

where

  • c=3×108 m s−1c = 3\times 10^8\ \text{m s}^{-1}c=3×108 m s−1
  • ε0=8.85×10−12 C2N−1m−2\varepsilon_0 = 8.85\times 10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}ε0​=8.85×10−12 C2N−1m−2
  • E0=56.5 N C−1E_0 = 56.5\ \text{N C}^{-1}E0​=56.5 N C−1
  1. Substitute values

I=12(3×108)(8.85×10−12)(56.5)2I = \frac{1}{2}(3\times 10^8)(8.85\times 10^{-12})(56.5)^2I=21​(3×108)(8.85×10−12)(56.5)2

First, 12×3×8.85=13.275\frac{1}{2}\times 3\times 8.85 = 13.27521​×3×8.85=13.275

and 108×10−12=10−410^8\times 10^{-12} = 10^{-4}108×10−12=10−4

So, I=13.275×10−4×(56.5)2I = 13.275\times 10^{-4}\times (56.5)^2I=13.275×10−4×(56.5)2

Now, (56.5)2=3192.25(56.5)^2 = 3192.25(56.5)2=3192.25

Hence, I=13.275×10−4×3192.25I = 13.275\times 10^{-4}\times 3192.25I=13.275×10−4×3192.25

I≈4.24 W m−2I \approx 4.24\ \text{W m}^{-2}I≈4.24 W m−2

  1. Match with options

The correct option is B: 4.24 W m−2\boxed{\text{B: } 4.24\ \text{W m}^{-2}}B: 4.24 W m−2​

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