The average power output of a point source of an electromagnetic radiation is $1080 \mathrm{~W}$. The…

The average power output of a point source of an electromagnetic radiation is $1080 \mathrm{~W}$. The maximum value of the rms value of the electric field at a distance of $3 \mathrm{~m}$ from the source is
  1. $20 \mathrm{Vm}^{-1}$
  2. $40 \mathrm{Vm}^{-1}$
  3. $60 \mathrm{Vm}^{-1}$
  4. $90 \mathrm{Vm}^{-1}$

Solution

$\begin{aligned} & \text { Maximum emf, } \varepsilon_0=\sqrt{\frac{\mathrm{P}}{2 \pi \mathrm{R}^2 \mathrm{E}_0 \mathrm{c}}} \\ & =\sqrt{\frac{1080}{2 \times 3.14 \times 3^2 \times 8.85 \times 10^{-12} \times 3 \times 10^8}} \\ & =\sqrt{\frac{1080 \times 10^8}{54 \times 314 \times 885}=84.85} \\ & \therefore \varepsilon_{\mathrm{rms}}=\frac{\varepsilon_0}{\sqrt{2}}=\frac{84.85}{1.414} \simeq 60 \mathrm{Vm}^{-1}\end{aligned}$

Asked in: AP EAMCET 2023 (15 May Shift 1)

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