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A radio antenna broadcasts a 1.0 MHz radio wave with 20.0 kW of power. Assume that the radiation is emitted uniformly in all directions.
1) What is the wave's intensity 30 km from the antenna
2) What is the electric field amplitude at this distance?


Sagot :

Answer:

1) [tex]I=1.8*10^{-6}\: W/m^{2}[/tex]  

2) [tex]E=0.037 \: V/m[/tex]

Explanation:

1)

The intensity equation is given by:

[tex]I=\frac{P}{4\pi r^{2}}[/tex]

Where:

  • P is the power of the radio wave
  • r is the distance from the source

[tex]I=\frac{20000}{4\pi (30000)^{2}}[/tex]

[tex]I=1.8*10^{-6}\: W/m^{2}[/tex]  

2)

Now, the intensity and the electric field are related as:

[tex] I=0.5c\epsilon_{0}E^{2}[/tex]

Here:

  • c is the speed of light
  • ε₀ is the electric permittivity
  • E is the electric field

We need to solve it for E.

[tex]E= \sqrt{\frac{2I}{c\epsilon_{0}}}[/tex]

[tex]E= \sqrt{\frac{2(1.8*10^{-6})}{(3*10^{8})(8.85*10^{-12})}}[/tex]

[tex]E=0.037 \: V/m[/tex]

I hope it helps you!