Questions
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
a
64.7 V/m
b
57.8 V/m
c
56.72 V/m
d
54.77 V/m
detailed solution
Correct option is D
Intensity of EM wave is given byI=P4πR2=uav.c=12ε0E02×c ⇒E0=P2πR2ε0c =8002×3.14×(4)2×8.85×10−12×3×108 = 54.77 VmTalk to our academic expert!
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