The radiation energy emitted per second by a | vedclass

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(B) The average power emitted by the source is $P_{	ext{avg}} = 4 \% 	ext{ of } 100 \,W = \frac{4}{100} 	imes 100 \,W = 4 \,W$.At a distance $r = 2

Vedclass · Accepted Answer

$\sqrt{30} \,V / m$

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(B) The average power emitted by the source is $P_{	ext{avg}} = 4 \% 	ext{ of } 100 \,W = \frac{4}{100} 	imes 100 \,W = 4 \,W$.At a distance $r = 2 \,m$, the radiation spreads over a spherical surface area $A = 4 \pi r^2 = 4 \pi (2)^2 = 16 \pi \,m^2$.The average intensity $I_{	ext{avg}}$ is given by $I_{	ext{avg}} = \frac{P_{	ext{avg}}}{A} = \frac{4}{16 \pi} = \frac{1}{4 \pi} \,W/m^2$.Also, the average intensity of an electromagnetic wave is related to the rms electric field $E_{	ext{rms}}$ by $I_{	ext{avg}} = \varepsilon_0 c E_{	ext{rms}}^2$, where $c = 3 	imes 10^8 \,m/s$ is the speed of light.Equating the two expressions for intensity: $\frac{1}{4 \pi} = \varepsilon_0 c E_{	ext{rms}}^2$.Rearranging for $E_{	ext{rms}}^2$: $E_{	ext{rms}}^2 = \frac{1}{4 \pi \varepsilon_0 c} = \left( \frac{1}{4 \pi \varepsilon_0} ight) 	imes \frac{1}{c}$.Substituting the given values: $E_{	ext{rms}}^2 = (9 	imes 10^9) 	imes \frac{1}{3 	imes 10^8} = 3 	imes 10 = 30$.Therefore, $E_{	ext{rms}} = \sqrt{30} \,V/m$.

The radiation energy emitted per second by a point source is $100 \,W$. If the efficiency of the source is $4 \%$, then the rms value of the electric field at a distance of $2 \,m$ is [use $\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9$ in $SI$ units].

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