A radio receiver antenna that is 2 , m long i | vedclass

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(A) The intensity $I$ of an electromagnetic wave is given by $I = \frac{1}{2} \epsilon_0 c E_0^2$,where $E_0$ is the amplitude of the electric field.R

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$1.23 \, \mu V$

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(A) The intensity $I$ of an electromagnetic wave is given by $I = \frac{1}{2} \epsilon_0 c E_0^2$,where $E_0$ is the amplitude of the electric field.Rearranging for $E_0$: $E_0 = \sqrt{\frac{2I}{\epsilon_0 c}}$.Given $I = 5 	imes 10^{-16} \, W/m^2$,$\epsilon_0 = 8.854 	imes 10^{-12} \, F/m$,and $c = 3 	imes 10^8 \, m/s$:$E_0 = \sqrt{\frac{2 	imes 5 	imes 10^{-16}}{8.854 	imes 10^{-12} 	imes 3 	imes 10^8}} = \sqrt{\frac{10 	imes 10^{-16}}{26.562 	imes 10^{-4}}} = \sqrt{0.376 	imes 10^{-12}} \approx 0.613 	imes 10^{-6} \, V/m$.The maximum potential difference $V_0$ across an antenna of length $L = 2 \, m$ is $V_0 = E_0 	imes L$.$V_0 = 0.613 	imes 10^{-6} 	imes 2 = 1.226 	imes 10^{-6} \, V \approx 1.23 \, \mu V$.

$A$ radio receiver antenna that is $2 \, m$ long is oriented along the direction of the electromagnetic wave and receives a signal of intensity $5 \times 10^{-16} \, W/m^2$. The maximum instantaneous potential difference across the two ends of the antenna is

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