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JEE Main 2025 · Physics · Electromagnetic Induction and Alternating Current · easy

A point source radiates 1 kW of electromagnetic power isotropically in vacuum. Take ε0 = 8.85×10^-12 F m^-1 and c = 3×10^8 m s^-1. The amplitude of…

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A point source radiates 1 kW1\ \mathrm{kW} of electromagnetic power isotropically in vacuum. Take ε0=8.85×10−12 F m−1\varepsilon_0 = 8.85\times10^{-12}\ \mathrm{F\,m^{-1}} and c=3×108 m s−1c = 3\times10^{8}\ \mathrm{m\,s^{-1}}. The amplitude of the electric field of the wave at a distance of 5 m5\ \mathrm{m} from the source is nearly
  1. A. 35 V m−135\ \mathrm{V\,m^{-1}}
  2. B. 49 V m−149\ \mathrm{V\,m^{-1}}
  3. C. 69 V m−169\ \mathrm{V\,m^{-1}}
  4. D. 98 V m−198\ \mathrm{V\,m^{-1}}
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Answer: (B) 49 V m−149\ \mathrm{V\,m^{-1}}

The intensity at distance rr is I=P4πr2=10004π(25)=3.18 W m−2I = \dfrac{P}{4\pi r^2} = \dfrac{1000}{4\pi(25)} = 3.18\ \mathrm{W\,m^{-2}}. For a plane wave, I=12ε0c E02I = \tfrac12\varepsilon_0 c\,E_0^2, so E0=2Iε0c=2×3.188.85×10−12×3×108≈49 V m−1E_0 = \sqrt{\dfrac{2I}{\varepsilon_0 c}} = \sqrt{\dfrac{2\times3.18}{8.85\times10^{-12}\times3\times10^{8}}} \approx 49\ \mathrm{V\,m^{-1}}. The value 35 V m−135\ \mathrm{V\,m^{-1}} is the rms field, obtained by omitting the factor 22 under the root.
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