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JEE Main 2019 · Physics · Electromagnetic Induction and Alternating Current · easy

A point source radiates 60 W of electromagnetic power uniformly in all directions. Assuming the medium to be vacuum, the amplitude of the electric…

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A point source radiates 60 W60\ \mathrm{W} of electromagnetic power uniformly in all directions. Assuming the medium to be vacuum, the amplitude of the electric field at a distance of 3 m3\ \mathrm{m} from the source is nearly (ε0=8.85×10−12 F m−1\varepsilon_0=8.85\times10^{-12}\ \mathrm{F\,m^{-1}}, c=3×108 m s−1c=3\times10^{8}\ \mathrm{m\,s^{-1}})
  1. A. 14 V m−114\ \mathrm{V\,m^{-1}}
  2. B. 28 V m−128\ \mathrm{V\,m^{-1}}
  3. C. 20 V m−120\ \mathrm{V\,m^{-1}}
  4. D. 40 V m−140\ \mathrm{V\,m^{-1}}
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Answer: (C) 20 V m−120\ \mathrm{V\,m^{-1}}

The intensity at r=3 mr=3\ \mathrm{m} is I=P4πr2=604π(9)≈0.53 W m−2I=\dfrac{P}{4\pi r^2}=\dfrac{60}{4\pi(9)}\approx0.53\ \mathrm{W\,m^{-2}}. For an electromagnetic wave, I=12ε0cE02I=\tfrac12\varepsilon_0cE_0^2, so E0=2Iε0c=2(0.53)(8.85×10−12)(3×108)≈400=20 V m−1E_0=\sqrt{\dfrac{2I}{\varepsilon_0c}}=\sqrt{\dfrac{2(0.53)}{(8.85\times10^{-12})(3\times10^{8})}}\approx\sqrt{400}=20\ \mathrm{V\,m^{-1}}. The rms value E0/2≈14 V m−1E_0/\sqrt2\approx14\ \mathrm{V\,m^{-1}} is the distractor obtained from I=ε0cErms2I=\varepsilon_0cE_{rms}^2 if one forgets to convert to amplitude.
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