Public television station KQED in San Francisco broadcasts a sinusoidal radio si
ID: 1508371 • Letter: P
Question
Public television station KQED in San Francisco broadcasts a sinusoidal radio signal at a power of 316 kW. Assume that the wave spreads out uniformly into a hemisphere above the ground.
A) At a home 5.50km away from the antenna, what average pressure does this wave exert on a totally reflecting surface? answer in PA units
B)At a home 5.50km away from the antenna, what are the amplitudes of the electric and magnetic fields of the wave? answer in N/C, T
C) At a home 5.50km away from the antenna, what is the average density of the energy this wave carries? answer in J/m^3
D) For the energy density in part (c), what percentage is due to the electric field?
E)For the energy density in part (c), what percentage is due to the magnetic field?
Explanation / Answer
intesity = Pav/A
A = for hemisphere = 2pi*r^2
I = 316 kW
r = 5.5km = 5.5 x 10^3 m
I = 1.66 x 10^-3
radiation pressure = 2I/c
c = speed of light
pa = 1.108 x 10^-11 Pa
part b )
I = 1/2 * eo*c/ * Emax^2
Emax = sqrt(2I/(eo*c))
Emax = 1.12 N/C
Bmax = Emax/c = 3.73 x 10^-9 T
part c )
u = eo*E^2
uav = eo(Erms)^2 = Erms = Emax/sqrt(2)
uav = eo*Emax^2/2
uav = 5.54 x 10^-12 J/m^3
part d and e )
the energy density is the same for the electric and magnetic fields, so each one has 50% of the energy density
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