Question
A point source of electromagnetic radiation has an average power output of 800 W. The maximum value of electric field at a distance 4.0 m from the source is
Answer: Option D
:
D
Intensity of EM wave is given by
I=P4πR2=vav.C=12ϵ0E20×c
⇒E0=√P2πR2ϵ0c
=√8002×3.14×(4)2×8.85×10−12×3×108
=54.77Vm
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:
D
Intensity of EM wave is given by
I=P4πR2=vav.C=12ϵ0E20×c
⇒E0=√P2πR2ϵ0c
=√8002×3.14×(4)2×8.85×10−12×3×108
=54.77Vm
Was this answer helpful ?
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