Get Answers to all your Questions

header-bg qa

A point source of electromagnetic radiation has an average power output of 800 W. The maximum value of electric filed at a distance 4.0 m from the source is

Option: 1

64.7 \mathrm{Vm}^{-1}


Option: 2

57.8 \mathrm{Vm}^{-1}


Option: 3

56.72 \mathrm{Vm}^{-1}


Option: 4

54.77 \mathrm{Vm}^{-1}


Answers (1)

best_answer

Intensity of electromagnetic wave is \mathrm{I=\frac{P_{\mathrm{av}}}{4 \pi \times r^2}= }
\mathrm{\begin{aligned} & \frac{E_0^2}{2 \mu_0 c} \\ & \text { or } E_0=\sqrt{\frac{\mu_0 c P_{\mathrm{av}}}{2 \pi r^2}} \\ & =\sqrt{\frac{\left(4 \pi \times 10^{-7}\right) \times\left(3 \times 10^8\right) \times 800}{2 \pi \times(4)^2}} \\ & =54.77 \mathrm{Vm}^{-1} \end{aligned} }

Posted by

vinayak

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE