A point source of electromagnetic radiation has an average power output of \(960 \mathrm{~W}\). The peak…
A point source of electromagnetic radiation has an average power output of \(960 \mathrm{~W}\). The peak value of the electric field at a distance \(400 \mathrm{~cm}\) from the source is
\(60 \mathrm{Vm}^{-1}\)
\(120 \mathrm{Vm}^{-1}\)
\(30 \mathrm{Vm}^{-1}\)
\(180 \mathrm{Vm}^{-1}\)
Solution
Given, average power output,
\(P=960 \mathrm{~W}\)
Distance, \(r=400 \mathrm{~cm}=4 \mathrm{~m}\)
Intensity of \(E M\) waves is given by
\(\begin{aligned}
I & =\frac{P}{4 \pi r^2}=\frac{1}{2} \varepsilon_0 E_0^2 c \Rightarrow E_0^2=\frac{P}{2 \pi r^2 \varepsilon_0 c} \\
\therefore \quad E & =\sqrt{\frac{P}{2 \pi r^2 \varepsilon_0 c}} \\
& =\sqrt{\frac{960}{2 \times 3.14 \times 4^2 \times 8.85 \times 10^{-12} \times 3 \times 10^8}} \\
& =\sqrt{0.36 \times 10^4}=0.6 \times 10^2=60 \mathrm{Vm}^{-1}
\end{aligned}\)