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The rms value of the electric field of the light coming from the sun is \mathrm{720 \mathrm{NC}^{-1}.} The average total energy density of the electromagnetic wave is:
 

Option: 1

\mathrm{3.3 \times 10^{-3} \mathrm{~J} \mathrm{~m}^{-3}}


 


Option: 2

\mathrm{4.58 \times 10^{-6} \mathrm{Jm}^{-3}}


Option: 3

\mathrm{6.37 \times 10^{-9} \mathrm{Jm}^{-3}}


Option: 4

\mathrm{81.35 \times 10^{-12} \mathrm{Jm}^{-3}}


Answers (1)

best_answer

Total average energy density of electromagnetic wave is

\mathrm{ <\mathrm{u}>=\frac{1}{2} \varepsilon_0 \mathrm{E}_{\mathrm{rms}}^2+\frac{1}{2 \mu_0} \mathrm{~B}_{\mathrm{rms}}^2 }

\mathrm{ =\frac{1}{2} \varepsilon_0 \mathrm{E}_{\mathrm{rms}}^2+\frac{1}{2 \mu_0}\left(\frac{\mathrm{E}_{\mathrm{rms}}^2}{\mathrm{c}^2}\right) \quad\left(\because \mathrm{B}_{\mathrm{rms}}=\frac{\mathrm{E}_{\mathrm{rms}}}{\mathrm{c}}\right) }

\mathrm{ =\frac{1}{2} \varepsilon_0 \mathrm{E}_{\mathrm{rms}}^2+\frac{1}{2 \mu_0} \mathrm{E}_{\mathrm{rms}}^2 \varepsilon_0 \mu_0 }

\mathrm{ =\frac{1}{2} \varepsilon_0 \mathrm{E}_{\mathrm{rms}}^2+\frac{1}{2} \varepsilon_0 \mathrm{E}_{\mathrm{rms}}^2=\varepsilon_0 \mathrm{E}_{\mathrm{rms}}^2 }

\mathrm{ =\left(8.85 \times 10^{-12}\right) \times(720)^2=4.58 \times 10^{-6} \mathrm{~J} \mathrm{~m}^{-3} }

Hence option 2 is correct.





 

Posted by

jitender.kumar

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