A light bulb emits power of 800 , W. What is | vedclass

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

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$62.6$

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(B) The average intensity $I$ of an electromagnetic wave at a distance $r$ 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 $E_m$ is $I = \frac{E_m^2}{2\mu_0 c}$.Equating the two expressions: $\frac{P_{av}}{4\pi r^2} = \frac{E_m^2}{2\mu_0 c}$.Solving for $E_m$: $E_m = \sqrt{\frac{\mu_0 c P_{av}}{2\pi r^2}}$.Substituting the given values: $P_{av} = 800 \, W$, $r = 3.5 \, m$, $\mu_0 = 4\pi 	imes 10^{-7} \, T \cdot m/A$, and $c = 3 	imes 10^8 \, m/s$.$E_m = \sqrt{\frac{(4\pi 	imes 10^{-7}) 	imes (3 	imes 10^8) 	imes 800}{2\pi 	imes (3.5)^2}} = \sqrt{\frac{2 	imes 3 	imes 10 	imes 800}{12.25}} = \sqrt{\frac{48000}{12.25}} \approx 62.6 \, V/m$.

$A$ light bulb emits power of $800 \, W$. What is the maximum value of the electric field at a distance of $3.5 \, m$ from the bulb in $V/m$?

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