Answer:0.1759 v
Explanation:
Intensity of wave at receiver end is
I=[tex]\frac{P_{avg}}{A}[/tex]
I=[tex]\frac{3.80\times 10^3}{4\times \pi \times \left ( 4\times 1609.34\right )^2}[/tex]
I=[tex]7.296\times 10^{-6} W/m^2[/tex]
Amplitude of electric field at receiver end
[tex]E_{max}=\sqrt{2I\mu _0c}[/tex]
Amplitude of induced emf
=[tex]E_{max}d[/tex]
=[tex]\sqrt{2\times 7.29\times 10-6\times 4\pi \times 3\times 10^8}\times 0.75[/tex]
=[tex]17.591\times 10^{-2}=0.1759 v[/tex]
