The television station KAAH in Hawaii broadcasts at a power of 262kW. Assume tha

Question

The television station KAAH in Hawaii broadcasts at a power of 262kW. Assume that the wave spreads out uniformly into a hemisphere above the ground. Assume that your home is 10km away from the antenna. a) What average pressure does this wave exert on a totally reflecting surface? b) What are the amplitudes of the electric and magnetic fields of the wave? c) What is the average density of the energy this wave carries? What percentage is due to the electric field? What percentage is due to the magnetic field?

Explanation / Answer

Given that

power P=262 kW

distance d=10km

now we find the average power reflecting surface

avergae pressure =2*P/A*c=2*262*10^3/3.14*10^8*3*10^8=55.63*10^-13 pa

now we find the amplitude of electric field and magnetic field of waves

the intensity of wave I=power/Area =262 *10^3 /3.14*(10^4)^2=83.4*10^-5 W/m^2

amplitude of electric field Eo=[2I/ec]^1/2=[2*83.4*10^-5/3*10^8*8.85*10^-12]^1/2=[16.68/26.55]^1/2=0.793 N/c

amplitude of magnetic field B o=Eo/c=0.793/3*10^8=0.3*10^-8 T

now we find the average density of energy

the avergae density of energy =1/2*8.85*10^-12*0.793^2=2.8*10^-12 J.c/N

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