InterviewSolution
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InterviewSolution
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A point source of electromagnetic radiation has an average power output of 800W. (a) Find the maximum value of electric field at a distance 3.5 m from the source. (b) What will be the maximum value of magnetic field? (c) What will be the energy density at a distance at a distance 3.5m from the source? |
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Answer» (a) `I=P/(4pir^2)=u_(av)c=1/2 in_0E_0^2c` `:. E_0=sqrt(P/(2pir^2in_0c))=sqrt(800/(2pixx(3.5)^2xx8.854xx10^-12xx3xx10^8))=62.6 V//m` (b) Maximum value of magnetic field, `B_0=(E_0)/c=62.6/(3xx10^8)=2.09xx10^-7T`. (c) Total energy density at 3.5 m is, `U=1/2 in_0E_0^2=1/2xx(8.85xx10^-12)xx(62.6)^2=1.73xx10^-8Jm^-3`. |
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