Q. 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
A) $64.7\ Vm^{- 1}$
B) $57.8\ Vm^{- 1}$
C) $56.72\ Vm^{- 1}$
D) $54.77\ Vm^{- 1}$
Explanation
(d) Intensity of electromagnetic wave is $I = \frac{P_{av}}{4\pi \times r^{2}} = \frac{E_{0}^{2}}{2\mu_{0}c}$
or $E_{0} = \sqrt{\frac{\mu_{0}cP_{av}}{2\pi r^{2}}}$
$= \sqrt{\frac{\left( 4\pi \times 10^{- 7} \right) \times \left( 3 \times 10^{8} \right) \times 800}{2\pi \times (4)^{2}}}$
$= 54.77\ Vm^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.