A point source of electromagnetic radiation h | vedclass

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Question

Intensity of electromagnetic wave given is by

${\mathrm{I}=\frac{\mathrm{P}_{\mathrm{av}}}{4 \pi \mathrm{r}^{2}}=\frac{\mathrm{E}_{0}^{2}}{2 \mu_{0

Vedclass · Accepted Answer

$62.6$

Vedclass · Answer

Intensity of electromagnetic wave given is by

${\mathrm{I}=\frac{\mathrm{P}_{\mathrm{av}}}{4 \pi \mathrm{r}^{2}}=\frac{\mathrm{E}_{0}^{2}}{2 \mu_{0} \mathrm{c}}} $

${\mathrm{E}=\sqrt{\frac{\mu_{0} \mathrm{cP}_{\mathrm{av}}}{2 \pi \mathrm{r}^{2}}}} $

${=\sqrt{\frac{\left(4 \pi 	imes 10^{-7}ight) 	imes\left(3 	imes 10^{8}ight) 	imes 800}{2 \pi 	imes(3.5)^{2}}}=62.6 \mathrm{\,V} / \mathrm{m}}$

A point source of electromagnetic radiation has an average power output of $800\,W$ . The maximum value of electric field at a distance $3.5\,m$ from the source will be.....$V/m$

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