To solve the problem we will apply the concepts related to the Intensity as a function of the power and the area, as well as the electric field as a function of the current, the speed of light and the permeability in free space, as shown below.
The intensity of the wave at the receiver is
[tex]I = \frac{P_{avg}}{A}[/tex]
[tex]I = \frac{P_{avg}}{4\pi r^2}[/tex]
[tex]I = \frac{3.4*10^3}{4\pi(4*1609.34)^2} \rightarrow 1mile = 1609.3m[/tex]
[tex]I = 6.529*10^{-6}W/m^2[/tex]
The amplitude of electric field at the receiver is
[tex]I = \frac{E_{max}^2}{2\mu_0 c}[/tex]
[tex]E_{max}= \sqrt{2I\mu_0 c}[/tex]
The amplitude of induced emf by this signal between the ends of the receiving antenna is
[tex]\epsilon_{max} = E_{max} d[/tex]
[tex]\epsilon_{max} = \sqrt{2I \mu_0 cd}[/tex]
Here,
I = Current
[tex]\mu_0[/tex] = Permeability at free space
c = Light speed
d = Distance
Replacing,
[tex]\epsilon_{max} = \sqrt{2(6.529*10^{-6})(4\pi*10^{-7})(3*10^{8})(60.0*10^{-2})}[/tex]
[tex]\epsilon_{max} = 0.05434V[/tex]
Thus, the amplitude of induced emf by this signal between the ends of the receiving antenna is 0.0543V
