Question

An AM radio transmitter broadcasts 50.0 kw of power uniformly in all directions. (a) Assuming all of the radio waves that strike the ground are completely absorbed, and that there is no absorption by the atmosphere or other objects, what is the intensity 30.0 km away? (Hint: Half the power will be spread over the area of a hemisphere.) (b) What is the maximum electric field strength at this distance?

Short answer

(a) Intensity is $4.42\times {10}^{-6}\mathrm{W}/{\mathrm{m}}^2$.

(b) The electric field strength is $5.77\times {10}^{-2}\mathrm{V}/\mathrm{m}$.

Step-by-step solution

Define electric field strength

The intensity of an electric field at a certain area is quantified as electric field strength. The volt per meter is the standard unit.

Evaluating the intensity

The formula for calculating intensity in terms of power and area is well-known.

$\mathrm{I}=\frac{\mathrm{P}}{\mathrm{A}}$

Where is the intensity of the wave,P is power transmitted, and A is the area.

$$\begin{gathered} A=4\tau \tau r^2 \\ =4\times 3.14\times \left(3000{\mathrm{m}}^2\right) \\ =1.13\times {10}^{10}{\mathrm{m}}^2 \end{gathered}$$

Substitute the values in the above equation,

$$\begin{gathered} \mathrm{I}=\frac{50\mathrm{kW}}{1.13\times {10}^{10}{\mathrm{m}}^2} \\ =\frac{50000\mathrm{W}}{1.13\times {10}^{10}{\mathrm{m}}^2} \\ =4.42\times {10}^{-6}\mathrm{W}/{\mathrm{m}}^2. \\ \mathrm{Therefore},\mathrm{intensity}\mathrm{is}4.42\times {10}^{-6}\mathrm{W}/{\mathrm{m}}^2. \end{gathered}$$

Evaluating the electric field strength

(b)

In terms of maximum electric field and light speed, intensity is given by

$$\begin{gathered} \mathrm{I}=\frac{{\mathrm{c\epsilon }}_0{\mathrm{E}}_0^2}{} \\ {\mathrm{E}}_0=\sqrt{\frac{2\mathrm{I}}{{\mathrm{c\epsilon }}_0}} \\ \mathrm{Where}\mathrm{I}\mathrm{is}\mathrm{intensity},\mathrm{C}\mathrm{is}\mathrm{the}\mathrm{speed}\mathrm{of}\mathrm{light},{\mathrm{\epsilon }}_0\mathrm{is}\mathrm{the}\mathrm{permittivity}\mathrm{of}\mathrm{free}\mathrm{space},{\mathrm{E}}_0\mathrm{is} \\ \mathrm{the}\mathrm{maximum}\mathrm{electric}\mathrm{field}. \\ \mathrm{Substitute}\mathrm{the}\mathrm{values}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{equation}, \\ {\mathrm{E}}_0=\sqrt{\frac{2\left(4.42\times {10}^{-6}\mathrm{W}/{\mathrm{m}}^2\right)}{\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)\left(8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}-{\mathrm{m}}^2\right)}} \\ =5.77\times {10}^{-2}\mathrm{V}/\mathrm{m} \\ \mathrm{Therefore},\mathrm{the}\mathrm{electric}\mathrm{field}\mathrm{strength}\mathrm{is}5.77\times {10}^{-2}\mathrm{V}/\mathrm{m}. \end{gathered}$$