Em Waves Question 98
Question: A point source of electromagnetic radiation has an average power output of 800 W. The maximum value of electric field at a distance $ 4.0m $ from the source is
Options:
A) $ 64.7V/m $
B) $ 57.8V/m $
C) $ 56.72V/m $
D) $ 54.77V/m $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Intensity of EM wave is given by $ I=\frac{P}{4\pi R^{2}}=v _{av}.c=\frac{1}{2}{\varepsilon _{0}}E _{0}^{2}\times c $
$ \Rightarrow E _{0}=\sqrt{\frac{P}{2\pi R^{2}{\varepsilon _{0}}c}} $
$ =\sqrt{\frac{800}{2\times 3.14\times {{(4)}^{2}}\times 8.85\times {{10}^{-12}}\times 3\times 10^{8}}} $
$ =54.77\frac{V}{m} $
BETA
