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  • A point source of electromagnetic radiation has an average power output of . The maximum

A point source of electromagnetic radiation has an average power output of 800 \mathrm{~W}. The maximum value of electric field at a distance 3.5 \mathrm{~m} from the source will be:
 

Option: 1

56.7 \mathrm{~V} / \mathrm{m}



 


Option: 2

62.6 \mathrm{~V} / \mathrm{m}
 


Option: 3

39.3 \mathrm{~V} / \mathrm{m}
 


Option: 4

47.5\mathrm{V/m}


Answers (1)

best_answer

Intensity of electromagnetic wave given is by

\mathrm{ I=\frac{P_{a v}}{4 \pi r^2}=\frac{E^2 m}{2 \mu_0 c} }

\mathrm{ E_m=\sqrt{\frac{\mu_0 c P_{a v}}{2 \pi r^2}} }

\mathrm{ =\sqrt{\frac{\left(4 \pi \times 10^{-7}\right) \times\left(3 \times 10^8\right) \times 800}{2 \pi \times 3.5^2}}=62.6 \mathrm{~V} / \mathrm{m} }

Hence option 2 is correct.



 

Posted by

jitender.kumar

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