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AP EAMCET · PHYSICS · Electromagnetic Waves

The rms value of the electric field of an electromagnetic wave emitted by a source is \(660 \mathrm{NC}^{-1}\). The average energy density of the electromagnetic wave is

  1. A \(1.75 \times 10^{-6} \mathrm{~J} \mathrm{~m}^{-3}\)
  2. B \(2.75 \times 10^{-6} \mathrm{~J} \mathrm{~m}^{-3}\)
  3. C \(4.85 \times 10^{-6} \mathrm{~J} \mathrm{~m}^{-3}\)
  4. D \(3.85 \times 10^{-6} \mathrm{~J} \mathrm{~m}^{-3}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(3.85 \times 10^{-6} \mathrm{~J} \mathrm{~m}^{-3}\)

Step-by-step Solution

Detailed explanation

\begin{aligned} & \mathrm{E}_{\mathrm{rms}}=660 \mathrm{NC}^{-1} \\ & \therefore \mathrm{E}_0=\sqrt{2} \mathrm{E}_{\mathrm{rms}}=\sqrt{2} \times 660 \mathrm{NC}^{-1} \\ & \therefore \text { Average density, } \mathrm{U}_{\mathrm{av}}=\frac{1}{2} \varepsilon_0 \mathrm{E}_0^2 \\ &…

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