A traveling sinusoidal electromagnetic wave in vacuum has an electric field amplitude of 83.7 V/m. Find the intensity of this wave and calcu

Question

A traveling sinusoidal electromagnetic wave in vacuum has an electric field amplitude of 83.7 V/m. Find the intensity of this wave and calculate the energy flowing during 15.5 s through an area of 0.0225 m2 that is perpendicular to the wave\’s direction of propagation.

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Yến Oanh 4 years 2021-07-29T05:58:27+00:00 1 Answers 11 views 0

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  1. ngochoa
    0
    2021-07-29T05:59:58+00:00
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    Answer with Explanation:

    We are given that

    Electric field,E=83.7V/m

    Time,t=15.5 s

    Area,A=0.0225m^2

    We have to find the intensity of the wave and energy .

    Intensity,I=\frac{1}{2}c\epsilon_0E^2

    Where c=3\times 10^8m/s

    \epsilon_0=8.85\times 10^{-12}

    Substitute the values

    I=\frac{1}{2}(3\times 10^8\times 8.85\times 10^{-12}\times (83.7)^2)=9.3W/m^2

    Energy,E=IAt

    Substitute the values

    E=9.3\times 0.0225\times 15.5=3.24 J

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