A point source of electromagnetic radiation h | vedclass

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Question

(A) Given,average power output,$P = 960 \,W$. Distance,$r = 400 \,cm = 4 \,m$. The intensity $I$ of electromagnetic waves at a distance $r$ from a poi

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(A) Given,average power output,$P = 960 \,W$. Distance,$r = 400 \,cm = 4 \,m$. The intensity $I$ of electromagnetic waves at a distance $r$ from a point source is given by $I = \frac{P}{4 \pi r^2}$. Also,the intensity in terms of the peak electric field $E_0$ is $I = \frac{1}{2} \varepsilon_0 E_0^2 c$. Equating the two expressions: $\frac{P}{4 \pi r^2} = \frac{1}{2} \varepsilon_0 E_0^2 c$. Solving for $E_0$: $E_0 = \sqrt{\frac{P}{2 \pi r^2 \varepsilon_0 c}}$. Substituting the values: $E_0 = \sqrt{\frac{960}{2 	imes 3.14 	imes 4^2 	imes 8.85 	imes 10^{-12} 	imes 3 	imes 10^8}}$. $E_0 = \sqrt{\frac{960}{2 	imes 3.14 	imes 16 	imes 8.85 	imes 3 	imes 10^{-4}}}$. $E_0 = \sqrt{\frac{960}{2662.656 	imes 10^{-4}}} \approx \sqrt{3600} = 60 \,Vm^{-1}$.

$A$ point source of electromagnetic radiation has an average power output of $960 \,W$. The peak value of the electric field at a distance $400 \,cm$ from the source is (in $\,Vm^{-1}$)

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