1) An 800 kHz radio signal is detected at a point 2.4 km distant from a transmit

Question

1) An 800 kHz radio signal is detected at a point 2.4 km distant from a transmitter tower. The electric field amplitude of the signal at that point is 300 mV/m. Assume that the signal power is radiated uniformly in all directions and that radio waves incident upon the ground are completely absorbed. The average electromagnetic energy density at that point is closest to: A) 5.6x10^-13 B)4.0x10^-13 C)1.1x10^-12 D)1.6x10-12 E)8.0x10^-13

2)An 800 kHz radio signal is detected at a point 3.4 km distant from a transmitter tower. The electric field amplitude of the signal at that point is 910 mV/m. Assume that the signal power is radiated uniformly in all directions and that radio waves incident upon the ground are completely absorbed. The intensity of the radio signal at that point is closest to: A)2.2x10^-3 B)5.5x10^-4 C)1.6x10^-3 D)1.1x10^-3 E)7.8x10^-4

Explanation / Answer

1) the average electromagnetic energy density at that point is u = eo*E^2

eo is the permittivity of free space = 8.85*10^-12 C^2/(N-m^2)

then u = 0.5*eo*E^2 = 8.85*10^-12*(300*10^-3)^2 = 8*10^-13 J/m^3

so the answer is E) 8*10^-3 J/m^3

2) Intensity S and energy density u is related as S = c*u = c*eo*E^2 = (3*10^8*8.85*10^-12*0.91^2) = 2.2*10^-3 W/m^2

so the answer is A) 2.2*10^-3 W/m^2

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