The rms value of the electric field of the li | vedclass

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Question

(B) The average total energy density $\mu$ of an electromagnetic wave is the sum of the electric energy density $\mu_E$ and the magnetic energy densit

Vedclass · Accepted Answer

$4.58 	imes 10^{-6} \ J/m^3$

Vedclass · Answer

(B) The average total energy density $\mu$ of an electromagnetic wave is the sum of the electric energy density $\mu_E$ and the magnetic energy density $\mu_B$.$\mu = \mu_E + \mu_B = \frac{1}{2} \varepsilon_0 E_{rms}^2 + \frac{1}{2\mu_0} B_{rms}^2$Since $B_{rms} = \frac{E_{rms}}{c}$ and $c^2 = \frac{1}{\mu_0 \varepsilon_0}$,we have $\mu_B = \frac{1}{2\mu_0} \left(\frac{E_{rms}^2}{c^2}ight) = \frac{1}{2\mu_0} (E_{rms}^2 \mu_0 \varepsilon_0) = \frac{1}{2} \varepsilon_0 E_{rms}^2$.Therefore,$\mu = \frac{1}{2} \varepsilon_0 E_{rms}^2 + \frac{1}{2} \varepsilon_0 E_{rms}^2 = \varepsilon_0 E_{rms}^2$.Given $E_{rms} = 720 \ N/C$ and $\varepsilon_0 = 8.85 	imes 10^{-12} \ F/m$.$\mu = (8.85 	imes 10^{-12}) 	imes (720)^2$$\mu = 8.85 	imes 10^{-12} 	imes 518400$$\mu \approx 4.587 	imes 10^{-6} \ J/m^3$.

The rms value of the electric field of the light coming from the sun is $720 \ N/C$. The average total energy density of the electromagnetic wave is $:-$

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