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

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Question

$u =\frac{1}{2} \varepsilon_{0} E _{ rms }^{2}+\frac{1}{2 \mu_{0}} B _{ rms }^{2}$

$= \frac{1}{2} \varepsilon_{0} E _{ rms }^{2}+\frac{1}{2 \mu_{0}

Vedclass · Accepted Answer

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

Vedclass · Answer

$u =\frac{1}{2} \varepsilon_{0} E _{ rms }^{2}+\frac{1}{2 \mu_{0}} B _{ rms }^{2}$

$= \frac{1}{2} \varepsilon_{0} E _{ rms }^{2}+\frac{1}{2 \mu_{0}} \frac{ E _{ rms }^{2}}{ c ^{2}} \quad\left(\because B _{ rms }=\frac{ E _{ rms }}{ c }ight)$

$= \frac{1}{2} \varepsilon_{0} E _{ rms }^{2}+\frac{\varepsilon_{0} \mu_{0} E _{ rms }^{2}}{2 \mu_{0}} \left(\because c ^{2}=\frac{1}{\varepsilon_{0} \mu_{0}}ight)$

$= \frac{1}{2} \varepsilon_{0} E _{ rms }^{2}+\frac{1}{2} \varepsilon_{0} E _{ rms }^{2}$$=\varepsilon_{0} E _{ rms }^{2}$

$= 8.85 	imes 10^{-12} 	imes(720)^{2}$

$= 4.58 	imes 10^{-6} J / m ^{3}$

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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