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Answer :
Answer:
0.17074 V
Explanation:
[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi \times 10^{-7}\ H/m[/tex]
c = Speed of light = [tex]3\times 10^8\ m/s[/tex]
[tex]\mu_0 c=377\ \Omega[/tex]
r = Distance = 3.46 km
P = Power = 6.02 kW
Intensity is given by
[tex]I=\dfrac{P}{4\pi r^2}\\\Rightarrow I=\dfrac{6020}{4\pi\times 3460^2}\\\Rightarrow I=4.00161\times 10^{-5}\ W/m^2[/tex]
Also
[tex]I=\dfrac{E_m^2}{2\mu_0c}\\\Rightarrow E_m=\sqrt{2I\mu_0c}\\\Rightarrow E_m=\sqrt{2\times 4.00161\times 10^{-5}\times 377}[/tex]
Amplitude induced is given by
[tex]\epsilon=dE_m\\\Rightarrow \epsilon=0.983\times \sqrt{2\times 4.00161\times 10^{-5}\times 377}\\\Rightarrow \epsilon=0.17074\ V[/tex]
The amplitude of the induced emf by this signal is 0.17075 V
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