TS EAMCET · Physics · Electromagnetic Waves
The radiation energy emitted per second by a point source is \(100 \mathrm{~W}\). If the efficiency of the source is \(4 \%\), then the rms value of the electric field at distance of \(2 \mathrm{~m}\) is [use \(\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9\) in SI unit]
- A \(\sqrt{60} \mathrm{~V} / \mathrm{m}\)
- B \(\sqrt{30} \mathrm{~V} / \mathrm{m}\)
- C \(\sqrt{50} \mathrm{~V} / \mathrm{m}\)
- D \(\sqrt{40} \mathrm{~V} / \mathrm{m}\)
Answer & Solution
Correct Answer
(B) \(\sqrt{30} \mathrm{~V} / \mathrm{m}\)
Step-by-step Solution
Detailed explanation
Average intensity, \(I_{\text {avg }}=\frac{P_{\text {avg }}}{A}\)....(i) where, \(P_{\text {avg }}=\) average power generated by source \(=4 \% \text { of } 100 \mathrm{~W}=\frac{4}{100} \times 100 \mathrm{~W}=4 \mathrm{~W}\) \(A=\) area to which energy is transmitted…
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