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
- A \(1.75 \times 10^{-6} \mathrm{~J} \mathrm{~m}^{-3}\)
- B \(2.75 \times 10^{-6} \mathrm{~J} \mathrm{~m}^{-3}\)
- C \(4.85 \times 10^{-6} \mathrm{~J} \mathrm{~m}^{-3}\)
- D \(3.85 \times 10^{-6} \mathrm{~J} \mathrm{~m}^{-3}\)
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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\begin{gathered}
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