Q157 / 160 · AP EAMCET 2021 (24 Aug Shift 2) · Physics

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

  • A
    $1.23 \mu \mathrm{V}$
  • Correct answer
    B
    $3.1 \mu \mathrm{V}$
  • C
    $31 \mu \mathrm{V}$
  • D
    $7.76 \mu \mathrm{V}$

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, $V_0=E_0 l=4 \times 0.776 \times 10^{-6} \mathrm{~V}=3.1 \mu \mathrm{V}$

Hence, the correct answer is option: 'B'.

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