A radiation is emitted by a 1000 , W bulb and | vedclass

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Question

(A) The power radiated by the bulb is $P_{rad} = 	ext{Efficiency} 	imes P_{total} = \frac{1.25}{100} 	imes 1000 \, W = 12.5 \, W$.The intensity $I$

Vedclass · Accepted Answer

$137$

Vedclass · Answer

(A) The power radiated by the bulb is $P_{rad} = 	ext{Efficiency} 	imes P_{total} = \frac{1.25}{100} 	imes 1000 \, W = 12.5 \, W$.The intensity $I$ at a distance $r = 2 \, m$ is given by $I = \frac{P_{rad}}{4 \pi r^2} = \frac{12.5}{4 \pi (2)^2} = \frac{12.5}{16 \pi} \, W/m^2$.The relationship between intensity and peak electric field $E_0$ is $I = \frac{1}{2} \varepsilon_0 E_0^2 c$.Equating the two expressions for intensity:$\frac{12.5}{16 \pi} = \frac{1}{2} 	imes 8.85 	imes 10^{-12} 	imes 3 	imes 10^8 	imes E_0^2$.$E_0^2 = \frac{12.5 	imes 2}{16 	imes 3.14159 	imes 8.85 	imes 10^{-12} 	imes 3 	imes 10^8} \approx \frac{25}{132.73} 	imes 10^4 \approx 188.35$.$E_0 = \sqrt{188.35} \approx 13.724 \, V/m$.Expressing this as $x 	imes 10^{-1} \, V/m$, we get $E_0 = 137.24 	imes 10^{-1} \, V/m$.Rounding to the nearest integer, $x = 137$.

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