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(D) The intensity $I$ of an electromagnetic wave is given by the average power $P_{av}$ distributed over a spherical surface area $4\pi r^2$:$I = \fra

Vedclass · Accepted Answer

$62.6$

Vedclass · Answer

(D) The intensity $I$ of an electromagnetic wave is given by the average power $P_{av}$ distributed over a spherical surface area $4\pi r^2$:$I = \frac{P_{av}}{4\pi r^2}$Also,the intensity in terms of the maximum electric field $E_m$ is:$I = \frac{1}{2} c \epsilon_0 E_m^2 = \frac{E_m^2}{2 \mu_0 c}$Equating the two expressions:$\frac{P_{av}}{4\pi r^2} = \frac{E_m^2}{2 \mu_0 c}$$E_m = \sqrt{\frac{2 \mu_0 c P_{av}}{4\pi r^2}} = \sqrt{\frac{\mu_0 c P_{av}}{2\pi r^2}}$Substituting the given values ($P_{av} = 800 \, W$,$r = 3.5 \, m$,$\mu_0 = 4\pi 	imes 10^{-7} \, T \cdot m/A$,$c = 3 	imes 10^8 \, m/s$):$E_m = \sqrt{\frac{(4\pi 	imes 10^{-7}) 	imes (3 	imes 10^8) 	imes 800}{2\pi 	imes (3.5)^2}}$$E_m = \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 output power of $800 \, W$. What is the maximum value of the electric field at a distance of $3.5 \, m$ from the source in $V/m$?

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