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A point source of electromagnetic radiation has an average power output of `1500W`. The maximum value of electric field at a distance `3m` from this source in `Vm^-1` isA. 500B. 100C. `(500)/(3)`D. `(250)/(3)` |
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Answer» Correct Answer - B Total energy density of electromagnetic wave is `u =(1) /(2) in _(0) R _(m) ^(2) & i=(I) /(C ) =(P) /( AC)` or `u= (p) /( 4pi r^(2) C) =(1) /(2) in _(0) E _(m)^(2)` ` E _(m) ^(2) =( 2P ) /( 4 pi in _(0) r^(2) C) = (2xx1500xx9xx10^(5) )/( 3xx3xx3xx10^(8))` `E_(m) = 100 Vm^(-1)` |
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