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8.Electromagnetic waves
medium
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
A
$1.23 \mu \,V$
B
$1.23 \mu V$
C
$1.23 \,V$
D
$12.3 \,mV$
Solution
(a)$I = \frac{1}{2}{\varepsilon _0}CE_0^2$
==> ${E_0} = \sqrt {\frac{{2I}}{{{\varepsilon _0}C}}} = \sqrt {\frac{{2 \times 5 \times {{10}^{ – 16}}}}{{8.85}}} = 0.61 \times {10^{ – 6}}\frac{V}{m}$
Also ${E_0} = \frac{{{V_0}}}{d}$ ==> ${V_0} = {E_0}d = 0.61 \times {10^{ – 6}} \times 2 = 1.23\mu V$
Standard 12
Physics
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