Chapter 22, Problem 48GP

To determine

The average power output of the radio station.

Answer to Problem 48GP

  $54kW$

Explanation of Solution

Given info:

Amplitude of the electric field = $0.12V/m$

Permittivity of free space, ${\epsilon }_0=8.85\times {10}^{-12}{C}^2/N\cdot m^2$

Speed of light, $c=3.00\times {10}^{8}m/s$

Distance is $15km=15\times {10}^{3}m$

Formula used:

The average intensity over an extended period of time if $E$ and $B$ are sinusoidal $\overline{E^2}=\frac{E_0^2}{2}$ , is

  $\overline{I}=\frac{1}{2}{\epsilon }_{0}c{E}_0^2$

Where,

  $c$ is speed of light

  $E_0$ is the maximum value of $E$

  ${\epsilon }_0$ is permittivity of free space

Formula to calculate power, $P=\overline{I}A$

Where,

  $\overline{I}$ is average intensity

  $P$ is power

  $A$ is area

Calculation:

First calculate the average intensity. Substitute the values in formula: $\overline{I}=\frac{1}{2}{\epsilon }_{0}c{E}_0^2$ .

  $\begin{array}{l}\overline{I}=\frac{1}{2}\left(8.85\times {10}^{-12}{C}^2/N\cdot m^2\right)\left(3\times {10}^{8}m/s\right){\left(0.12V/m\right)}^2\\ \overline{I}=1.0116\times {10}^{-5}W/m^2\end{array}$

Now calculate the power. Substitute the values in formula: $P=\overline{I}A$

  $\begin{array}{l}P=\left(1.0116\times {10}^{-5}W/m^2\right)4\pi {\left(15\times {10}^{3}m\right)}^2\\ P=54kW\end{array}$

Conclusion:

The required power is $54kW$ .