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(B) Given that, length of antenna, $l = 4 \, m$.Intensity of signal, $I = 8 	imes 10^{-16} \, W/m^2$.The intensity of an electromagnetic wave is give

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(B) Given that, length of antenna, $l = 4 \, m$.Intensity of signal, $I = 8 	imes 10^{-16} \, W/m^2$.The intensity of an electromagnetic wave is given by $I = \frac{1}{2} \varepsilon_0 c E_0^2$, where $E_0$ is the maximum electric field amplitude.Rearranging for $E_0$, we get $E_0 = \sqrt{\frac{2I}{\varepsilon_0 c}}$.Substituting the values ($\varepsilon_0 = 8.85 	imes 10^{-12} \, F/m$, $c = 3 	imes 10^8 \, m/s$):$E_0 = \sqrt{\frac{2 	imes 8 	imes 10^{-16}}{8.85 	imes 10^{-12} 	imes 3 	imes 10^8}} = \sqrt{\frac{16 	imes 10^{-16}}{26.55 	imes 10^{-4}}} = \sqrt{0.6026 	imes 10^{-12}} \approx 0.776 	imes 10^{-6} \, V/m$.The maximum potential difference $V_0$ across the antenna is $V_0 = E_0 	imes l$.$V_0 = 0.776 	imes 10^{-6} \, V/m 	imes 4 \, m = 3.104 	imes 10^{-6} \, V \approx 3.1 \, \mu V$.

$A$ radio receiver antenna that is $4 \, m$ long is oriented along the direction of the electromagnetic wave and it receives a signal of intensity $8 \times 10^{-16} \, W/m^2$. The maximum instantaneous potential difference across the two ends of the antenna is (in $ \, \mu V$)

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