An 800-kHz radio signal is detected at a point 8.5 km distant from a transmitter tower. The electric field amplitude of the signal at that p

Question

An 800-kHz radio signal is detected at a point 8.5 km distant from a transmitter tower. The electric field amplitude of the signal at that point is 0.90 V/m. Assume that the signal power is radiated uniformly in all directions and that radio waves incident upon the ground are completely absorbed. What is the average electromagnetic energy density at that point

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Thiên Hương 5 years 2021-07-17T17:01:50+00:00 2 Answers 133 views 0

Answers ( )

  1. mocmien2
    0
    2021-07-17T17:03:00+00:00
    Reply

    Answer:

    4.96*10^-20 J

    Explanation:

    To find the average electromagnetic energy of the wave you use the fact that the wave can be taken as a spherical electromagnetic wave. IN this case, the amplitude of the wave change in space according to:

    |E|=\frac{E_o}{r}

    for a distance of 8.5km = 5.5*10^3m you have:

    |E|=\frac{0.90\ V/m}{8.5*10^3m}=1.058*10^{-4}\ V/m

    Next, to find the average energy at this point you use the following formula:

    u_E=\frac{1}{2}\epsilon_0|E|^2

    εo : dielectric permittivity of vacuum = 8.85*10^-12C^2/Nm^2

    u_E=\frac{1}{2}(8.85*10^{-12})|1.058*10^{-4}|^2\ J=4.96*10^{-20}J

    hence, the average energy of the wave is 4.96*10^-20 J

  2. Eirian
    0
    2021-07-17T17:03:14+00:00
    Reply

    Answer:

    Explanation:

    Given that,

    Frequency of radio signal is

    f = 800kHz = 800,000 Hz.

    Distance from transmitter

    d = 8.5km = 8500m

    Electric field amplitude

    E = 0.9 V/m

    The average energy density can be calculated using

    U_E = ½•ϵo•E²

    Where ϵo = 8.85 × 10^-12 F/m

    Then,

    U_E = ½ × 8.85 × 10^-12 × 0.9²

    U_E = 3.58 × 10^-12 J/m²

    The average electromagnetic energy density is 3.58 × 10^-12 J/m²

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