A radio receiver antenna that is $4 \mathrm{~m}$ long is oriented along the direction of the electromagnetic…
A radio receiver antenna that is $4 \mathrm{~m}$ long is oriented along the direction of the
electromagnetic wave and it receives a signal of intensity $8 \times 10^{-16} \mathrm{Wm}^{-2}$. The maximum instantaneous potential difference across the two ends of the antenna is
$1.23 \mu \mathrm{V}$
$3.1 \mu \mathrm{V}$
$31 \mu \mathrm{V}$
$7.76 \mu \mathrm{V}$
Solution
Given that, length of antenna, $l=4 \mathrm{~m}$
Intensity of signal, $I=8 \times 10^{-16} \mathrm{~W} / \mathrm{m}^2$
Using, $I=\frac{1}{2} \varepsilon_0 c E_0^2 \Rightarrow E_0=\sqrt{\frac{2 I}{\varepsilon_0 c}}$
By substituting the values, we get
$E_0=\sqrt{\frac{2 \times 8 \times 10^{-16}}{8.85 \times 10^{-12} \times 3 \times 10^8}}$
$=0.776 \times 10^{-6} \mathrm{~V} / \mathrm{m}$
Now, maximum potential difference,
$V_0=E_0 l=4 \times 0.776 \times 10^{-6} \mathrm{~V}=3.1 \mu \mathrm{V}$