A point source of electromagnetic radiation has an average power output of 800W. The maximum value of electric field at a distance 3.5m from the source will be:
Answer & explanation
Correct answer: option 2
Intensity of electromagnetic wave given is by
I = $\frac{P_{a v}}{4 \pi r^2}=\frac{E^2 m}{2 \mu_0 c}$
$E_m=\sqrt{\frac{\mu_0 c P_{a v}}{2 \pi r^2}}$
$=\sqrt{\frac{\left(4 \pi \times 10^{-7}\right) \times\left(3 \times 10^8\right) \times 800}{2 \pi \times 3.5^2}}$ = 62.6 V/m