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

 

Option: 1

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


Option: 2

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


Option: 3

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


Option: 4

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


Answers (1)

best_answer

Total average energy density of electromagnetic wave is

\begin{aligned} & \mathrm{U}_{\mathrm{av}}=\frac{1}{2} \varepsilon_0 \mathrm{E}_{\mathrm{mms}}^2+\frac{1}{2^\mu{ }_0} \mathrm{~B}_{\mathrm{rms}}^2=\frac{1}{2} \varepsilon_0 \mathrm{E}_{\mathrm{rms}}^2+\frac{1}{2^\mu{ }_0}\left(\frac{\mathrm{E}^2}{\mathrm{c}^2 \mathrm{~m}^{\mathrm{s}}}\right) \mid \quad\left(\because \mathrm{B}_{\mathrm{rms}}=\frac{\mathrm{E}_{\mathrm{rms}}}{\mathrm{c}}\right) \\ & =\frac{1}{2} \varepsilon_0 \mathrm{E}_{\mathrm{ms}}^2+\frac{1}{2 \mu_0} \mathrm{E}_{\mathrm{ms}}^2 \varepsilon_0 \mu_0 \quad\left(\because \mathrm{c}=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}\right) \\ & =\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{ms}}^2 \\ & =8 \\ & .85 \times 10^{-12} \times(720)^2 \quad 4.58 \times 10^{-6} \mathrm{Jm}_{-3} \\ & \end{aligned}

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