The ratio of contributions made by the electr | vedclass

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Question

Intensity of electromagnetic wave, $\mathrm{I}=\mathrm{U}_{\mathrm{av}} \mathrm{c}$

In terms of electric field, $\mathrm{U}_{\mathrm{av}}=\frac{1}{

Vedclass · Accepted Answer

$1:1$

Vedclass · Answer

Intensity of electromagnetic wave, $\mathrm{I}=\mathrm{U}_{\mathrm{av}} \mathrm{c}$

In terms of electric field, $\mathrm{U}_{\mathrm{av}}=\frac{1}{2} \varepsilon_{0} \mathrm{E}_{0}^{2}$

In terms of magnetic field $\mathrm{U}_{\mathrm{av}}=\frac{1}{2} \frac{\mathrm{B}_{0}^{2}}{\mu_{0}}$

Now, $U_{a v}$ (due to electric field) $=\frac{1}{2} \varepsilon_{0} E_{0}^{2}$

$=\frac{1}{2} \varepsilon_{0}\left(\mathrm{cB}_{0}ight)^{2}=\frac{1}{2} \varepsilon_{0} 	imes \frac{1}{\mu_{0} \varepsilon_{0}} \mathrm{B}_{0}^{2} \quad\left(\because \frac{\mathrm{E}_{0}}{\mathrm{B}_{0}}=\mathrm{c}ight)$

$=\frac{1 \mathrm{B}_{0}^{2}}{2 \mu_{0}}=\mathrm{U}_{\mathrm{av}}$ (due to magnetic field)

Therefore, the ratio of contributions by the electric field and magnetic field components to the intensity of electromagnetic wave is $1: 1 .$

The ratio of contributions made by the electric field and magnetic filed components to the intensity of an electromagnetic wave is

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