Question

A point source of electromagnetic radiation has an average power output of 1500 W.The maximum value of electric field at a distance of 3 m from this source $\left(inV{m}^{-1}\right)is$ in:

• A. 500
• B. 73
• C. $\frac{500}{3}$
• D. $\frac{250}{3}$

Answer 1

The correct option is B 73

Intensity of $EM$ wave is given by:-

$\begin{array}{c}I=\frac{P}{4\pi R^2}\\ =V_{av}.c=\frac{1}{2}{\epsilon }_0{E}_0^2\times c\\ \Rightarrow E_0=\sqrt{\frac{P}{2\pi R^2{\epsilon }_{0}c}}\\ =\sqrt{\frac{1500}{2\times 3.14\times {\left(3\right)}^2\times 8.85\times {10}^{-12}\times 3\times {10}^8}}\\ =73\frac{V}{m}\end{array}$