An AM radio station broadcasts with an average power of 6.02 kW. A dipole receiving antenna, 98.3 cm long, is located 3.46 km from the transmitter. Assume the transmitter is a point source, the waves travel perpendicular to the axis of the receiving antenna, and the source is far enough away that the wave amplitude is constant along the receiving antenna.

Compute the amplitude of the induced emf by this signal between the ends of the receiving antenna. Use $\mu_0 c = 377 \, \Omega$ and $1.609 \, \text{km} = 1 \, \text{mi}$.

Answer in units of V.

Question

High School

Answer :

boffeemadrid boffeemadrid · Answer 1 · 2024-04-18T14:38:44-04:00

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