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(A) The power radiated as electromagnetic waves is $P = 3\% 	ext{ of } 100\,W = 3\,W$.The intensity $S_{av}$ at a distance $R = 10\,m$ is given by $S

Vedclass · Accepted Answer

$1.34$

Vedclass · Answer

(A) The power radiated as electromagnetic waves is $P = 3\% 	ext{ of } 100\,W = 3\,W$.The intensity $S_{av}$ at a distance $R = 10\,m$ is given by $S_{av} = \frac{P}{4\pi R^2}$.Also,the intensity of an electromagnetic wave is related to the electric field amplitude $E_0$ by $S_{av} = \frac{1}{2} \varepsilon_0 c E_0^2$.Equating the two expressions: $\frac{1}{2} \varepsilon_0 c E_0^2 = \frac{P}{4\pi R^2}$.Solving for $E_0$: $E_0 = \sqrt{\frac{P}{2\pi R^2 \varepsilon_0 c}}$.Substituting the values: $E_0 = \sqrt{\frac{3}{2 	imes 3.14 	imes (10)^2 	imes 8.85 	imes 10^{-12} 	imes 3 	imes 10^8}}$.$E_0 = \sqrt{\frac{3}{1665.72 	imes 10^{-4}}} \approx \sqrt{1.8} \approx 1.34\,V/m$.

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