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AP EAMCET · PHYSICS · Alternating Current

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. A \(20 \mathrm{Vm}^{-1}\)
  2. B \(40 \mathrm{Vm}^{-1}\)
  3. C \(60 \mathrm{Vm}^{-1}\)
  4. D \(90 \mathrm{Vm}^{-1}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(60 \mathrm{Vm}^{-1}\)

Step-by-step Solution

Detailed explanation

\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…