Question

A radio antenna broadcasts a 1.0 MHz radio wave with 25 kW power. Assume 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?

Answer 1

Concepts and reason

The concepts required to solve this problem are intensity of wave and the amplitude of electric field.

Initially, calculate the wave intensity by using the relation between Intensity and power. Finally, calculate the amplitude of electric field by using the relation between intensity and amplitude of the electric field.

Fundamentals

The expression for the intensity of wave is as follows:

$I = \frac{P}{{4\pi {r^2}}}$

Here, P is the power of the source and r is the distance from the antenna.

The expression for the intensity of the electromagnetic wave in terms of electric field is given as,

$I = \frac{{E_0^2}}{{2{\mu _0}c}}$

Here, ${\mu _0}$ is the permeability of free space, ${E_0}$ is the amplitude of the electric field, and c is the speed of the light.

(1)

The intensity of the wave is given as,

$I = \frac{P}{{4\pi {r^2}}}$

Here, P is the power of the source and r is the distance from the antenna.

Substitute 25 kW for P and 30 km for r in the above equation.

$\begin{array}{c}\\I = \frac{{25{\rm{ kW}}}}{{4\pi {{\left( {30{\rm{ km}}} \right)}^2}}}\\\\ = \frac{{25{\rm{ kW}}\left( {\frac{{{{10}^3}{\rm{ W}}}}{{1{\rm{ kW}}}}} \right)}}{{4\pi {{\left( {30{\rm{ km}}\left( {\frac{{{{10}^3}{\rm{ m}}}}{{1{\rm{ km}}}}} \right)} \right)}^2}}}\\\\ = 2.21 \times {10^{ - 6}}{\rm{ W/}}{{\rm{m}}^2}\\\\ \approx 2.2 \times {10^{ - 6}}{\rm{ W/}}{{\rm{m}}^2}\\\end{array}$

(2)

Intensity of the wave in terms of electric field is given as,

$I = \frac{{E_0^2}}{{2{\mu _0}c}}$

Rearrange the above equation for ${E_0}$.

$\begin{array}{c}\\I = \frac{{E_0^2}}{{2{\mu _0}c}}\\\\{E_0} = \sqrt {2{\mu _0}Ic} \\\end{array}$

Substitute $2.21 \times {10^{ - 6}}{\rm{ W/}}{{\rm{m}}^2}$ for I, $3 \times {10^8}{\rm{ m/s}}$ for c, and $4\pi \times {10^{ - 7}}{\rm{ H/m}}$ for ${\mu _0}$ in the above equation ${E_0} = \sqrt {2{\mu _0}Ic}$.

$\begin{array}{c}\\{E_0} = \sqrt {2\left( {4\pi \times {{10}^{ - 7}}{\rm{ H/m}}} \right)\left( {2.21 \times {{10}^{ - 6}}{\rm{ W/}}{{\rm{m}}^2}} \right)\left( {3 \times {{10}^8}{\rm{ m/s}}} \right)} \\\\ = 0.0408{\rm{ V/m}}\\\\ \approx {\rm{0}}{\rm{.041 V/m}}\\\end{array}$

Ans: Part 1

The magnitude of the intensity of wave is equal to$2.2 \times {10^{ - 6}}{\rm{ W/}}{{\rm{m}}^2}$.

Part 2

The amplitude of electric field is equal to 0.041 V/m.