Question

A radio receiver can detect signals with electric field amplitudes as small as $300\mathrm{\mu V}/\mathrm{m}$. What is the intensity of the smallest detectable signal$?$

Short answer

Intensity of the smallest detectable signal is$1.19532\times {10}^{-10}\mathrm{W}/{\mathrm{m}}^2$.

Step-by-step solution

Formula for intensity

Intensity is,

$\mathrm{I}=\frac{1}{2}{\mathrm{E}}_{\mathrm{o}}{\left|{\mathrm{E}}_{\mathrm{o}}\right|}^2\mathrm{c}$

${\mathrm{E}}_0$is Electric field amplitude.

localid="1648922508504" $\mathrm{c}$-Speed of the light in vacuum,

localid="1648922516996" $\mathrm{c}=3\times {10}^8\mathrm{m}/\mathrm{s}$

Calculation for intensity of the signal

As it stands,

$\left|{\mathrm{E}}_{{\mathrm{o}}_{\min }}\right|=300\mathrm{\mu V}/\mathrm{m}$

$\mathrm{c}=3\times {10}^8\mathrm{m}/\mathrm{s}$

$\mathrm{I}=\frac{1}{2}{\mathrm{E}}_{\mathrm{o}}{\left|{\mathrm{E}}_{\mathrm{o}}\right|}^2\mathrm{c}$

localid="1648786141912">I=12300|300|2×3×108

$\mathrm{I}=1.19532\times {10}^{-10}\mathrm{W}/{\mathrm{m}}^2$