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Intensity of an EM wave from its electric field amplitude

This type of question, asked 4 times: 25 Jun 2022, 27 Jan 2024, 6 Apr 2024, 23 Jan 2026 S1. All 4 questions below, oldest first.

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  1. Question 1

    25 Jun 2022

    12
    The electric field in an electromagnetic wave is given by E=56.5sin⁡ω(t−x/c)NC−1E = 56.5\sin\omega(t - x/c)NC^{- 1}. Find the intensity of the wave if it is propagating along xx-axis in the free space. (Given ε0=8.85×10−12C2 N−1 m−2\varepsilon_{0}= 8.85 \times10^{- 12}C^{2}{\text{ }N}^{- 1}{\text{ }m}^{- 2} )

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  2. Question 2

    27 Jan 2024

    16
    A plane electromagnetic wave propagating in xx-direction is described by Ey=(200Vm−1)sin⁡[1.5×107t−0.05x]E_{y}=\left( 200Vm^{- 1} \right)\sin\left\lbrack 1.5 \times10^{7}t - 0.05x \right\rbrack; The intensity of the wave is: (Use ∈0=8.85×10−12C2 N−1 m−2\in_{0}= 8.85 \times10^{- 12}C^{2}{\text{ }N}^{- 1}{\text{ }m}^{- 2} )

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  3. Question 3

    6 Apr 2024

    94
    In the given electromagnetic wave Ey=600sin⁡(ωt−kx)Vm−1E_{y}= 600\sin(\omega t - kx)Vm^{- 1}, intensity of the associated light beam is (in W/m2W/m^{2} ); (Given ϵ0=\epsilon_{0}= 9×10−12C2 N−1 m−29 \times10^{- 12}C^{2}{\text{ }N}^{- 1}{\text{ }m}^{- 2} )

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  4. Question 4

    23 Jan 2026 S1

    22
    The equation of the electric field of an electromagnetic wave propagating through free space is given by: E=377sin⁡(6.27×103t−2.09×10−5x) N/CE = \sqrt{377}\sin\left( 6.27 \times 10^{3}t - 2.09 \times 10^{- 5}x \right)\ N/C. The average power of the electromagnetic wave is (1α)W/m2\left( \frac{1}{\alpha} \right)W/m^{2}. The value of α\alpha is ____\_\_\_\_ (Take μ0ε0=377\sqrt{\frac{\mu_{0}}{\varepsilon_{0}}} = 377 in SI units)