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(B) The power of the electromagnetic radiation emitted by the lamp is $P = 100\,W 	imes 0.03 = 3\,W$.The intensity $I$ at a distance $r = 5\,m$ is gi

Vedclass · Accepted Answer

$2.68$

Vedclass · Answer

(B) The power of the electromagnetic radiation emitted by the lamp is $P = 100\,W 	imes 0.03 = 3\,W$.The intensity $I$ at a distance $r = 5\,m$ is given by $I = \frac{P}{4\pi r^2} = \frac{3}{4\pi (5)^2} = \frac{3}{100\pi}\,W/m^2$.The relationship between intensity and the amplitude of the electric field $E_0$ is $I = \frac{1}{2} c \varepsilon_0 E_0^2$.Rearranging for $E_0$,we get $E_0 = \sqrt{\frac{2I}{c \varepsilon_0}}$.Substituting the values $c = 3 	imes 10^8\,m/s$ and $\varepsilon_0 = \frac{1}{4\pi 	imes 9 	imes 10^9}$,we have $E_0 = \sqrt{\frac{2 	imes (3 / 100\pi)}{(3 	imes 10^8) 	imes (1 / (4\pi 	imes 9 	imes 10^9))}}$.Simplifying the expression: $E_0 = \sqrt{\frac{6}{100\pi} 	imes (4\pi 	imes 9 	imes 10^9) / (3 	imes 10^8)} = \sqrt{\frac{6 	imes 36 	imes 10^9}{100 	imes 3 	imes 10^8}} = \sqrt{\frac{216 	imes 10}{300}} = \sqrt{7.2} \approx 2.68\,V/m$.

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