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(B) The intensity $I$ of an electromagnetic wave from a point source is given by $I = \frac{P_{av}}{4 \pi r^2}$.Also,the intensity in terms of the max

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$62.6$

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(B) The intensity $I$ of an electromagnetic wave from a point source is given by $I = \frac{P_{av}}{4 \pi r^2}$.Also,the intensity in terms of the maximum electric field $E_0$ is $I = \frac{E_0^2}{2 \mu_0 c}$.Equating the two expressions: $\frac{P_{av}}{4 \pi r^2} = \frac{E_0^2}{2 \mu_0 c}$.Solving for $E_0$: $E_0 = \sqrt{\frac{2 \mu_0 c P_{av}}{4 \pi r^2}} = \sqrt{\frac{\mu_0 c P_{av}}{2 \pi r^2}}$.Substituting the given values: $P_{av} = 800\,W$,$r = 3.5\,m$,$\mu_0 = 4 \pi 	imes 10^{-7}\,T\cdot m/A$,and $c = 3 	imes 10^8\,m/s$.$E_0 = \sqrt{\frac{(4 \pi 	imes 10^{-7}) 	imes (3 	imes 10^8) 	imes 800}{2 \pi 	imes (3.5)^2}}$.$E_0 = \sqrt{\frac{2 	imes 10^{-7} 	imes 3 	imes 10^8 	imes 800}{12.25}} = \sqrt{\frac{48000}{12.25}} \approx \sqrt{3918.36} \approx 62.6\,V/m$.

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

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