Questions
The peak value of electric field produced by the radiation coming from a bulb at a distance of is ( assume that the efficiency of the bulb is 2.5% and it is a point source)
detailed solution
Correct option is B
Useful intensity, I=powerarea=100×2.51004π(3)2 ⇒I=2.536πWm−2Half of this intensity(I) belongs to electric field and half of that to magnetic field I2=14∈0E02C ⇒E0=2I∈0C ⇒E0=2×(2.5/36π)(14π×9×109)×(3×108)=4.08Vm−1Talk to our academic expert!
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