Question

Public television station KQED in San Francisco broadcasts a sinusoidal radio signal at a power of 316 kW. Assume that the wave spreads out uniformly into a hemisphere above the ground.

A) At a home 5.50km away from the antenna, what average pressure does this wave exert on a totally reflecting surface? answer in PA units

B)At a home 5.50km away from the antenna, what are the amplitudes of the electric and magnetic fields of the wave? answer in N/C, T

C) At a home 5.50km away from the antenna, what is the average density of the energy this wave carries? answer in J/m^3

D) For the energy density in part (c), what percentage is due to the electric field?

E)For the energy density in part (c), what percentage is due to the magnetic field?

Answer 1

Concepts and reason

The concepts required to solve this problem is properties of electromagnetic waves.

Initially, calculate the average pressure by using the intensity of the wave. Later, use the amplitude equation to calculate the electric and magnetic field amplitude. Then, calculate the average density of the energy of the electromagnetic wave. Finally, calculate the energy density due to electric and magnetic field and their percentage in the total energy density of the wave.

Fundamentals

The intensity of electromagnetic wave is given as,

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

Here, P is the power, and $A$ is the surface area over which wave spreads.

The amplitude of electric field is given as,

${E_{\max }} = \sqrt {\frac{{2I}}{{{\varepsilon _0}c}}}$

Here, ${\varepsilon _0}$ is the vacuum permittivity, $I$ is the intensity of the incident light, and $c$ is the speed of light.

The average energy density of the electromagnetic wave is,

$u = \frac{1}{2}{\varepsilon _0}E_{{\rm{avg}}}^2 + \frac{1}{{2{\mu _0}}}B_{{\rm{avg}}}^2$

Here, ${\varepsilon _0}$ is the vacuum permittivity, ${\mu _0}$ is the vacuum permeability, ${E_{{\rm{avg}}}}$ is the average value of the electric field, and ${B_{{\rm{avg}}}}$ is the average value of the magnetic field.

The energy density due to electric field is,

${u_{\rm{E}}} = \frac{1}{2}{\varepsilon _0}E_{{\rm{avg}}}^2$

Here, ${\varepsilon _0}$ is the vacuum permittivity and ${E_{{\rm{avg}}}}$ is the average value of the electric field.

The energy density due to magnetic field is,

${u_{\rm{B}}} = \frac{1}{{2{\mu _0}}}B_{{\rm{avg}}}^2$

Here, ${\mu _0}$ is the vacuum permeability and ${B_{{\rm{avg}}}}$ is the average value of the magnetic field.

(A)

Substitute $2\pi {r^2}$ for $A$ in the equation $I = \frac{P}{A}$ .

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

Substitute $316{\rm{ kW}}$ for $P$ and $5.50{\rm{ km}}$ for $r$ in the equation $I = \frac{P}{{2\pi {r^2}}}$ and calculate the intensity of the electromagnetic wave.

$\begin{array}{c}\\I = \frac{{\left( {316{\rm{ kW}}} \right)}}{{2\pi {{\left( {5.50{\rm{ km}}} \right)}^2}}}\\\\ = \frac{{\left( {316{\rm{ kW}}\left( {\frac{{{{10}^3}{\rm{ W}}}}{{1{\rm{ kW}}}}} \right)} \right)}}{{2\pi {{\left( {5.50{\rm{ km}}\left( {\frac{{{{10}^3}{\rm{ m}}}}{{1{\rm{ km}}}}} \right)} \right)}^2}}}\\\\ = 1.66 \times {10^{ - 3}}{\rm{ W/}}{{\rm{m}}^2}\\\end{array}$

Use the pressure equation.

Substitute $1.66 \times {10^{ - 3}}{\rm{ W/}}{{\rm{m}}^2}$ for $I$ and $3.0 \times {10^8}{\rm{ m/s}}$ for $c$ in the equation $p = \frac{{2I}}{c}$ and calculate the average pressure due to electromagnetic wave for a totally reflecting surface.

$\begin{array}{c}\\p = \frac{{2\left( {1.66 \times {{10}^{ - 3}}{\rm{ W/}}{{\rm{m}}^2}} \right)}}{{3.0 \times {{10}^8}{\rm{ m/s}}}}\\\\ = 1.1 \times {10^{ - 11}}{\rm{ pa}}\\\end{array}$

(B)

Use the electric field amplitude equation.

Substitute $1.66 \times {10^{ - 3}}{\rm{ W/}}{{\rm{m}}^2}$ for $I$ , $8.854 \times {10^{ - 12}}{\rm{ }}{{\rm{C}}^2}{\rm{/N}} \cdot {{\rm{m}}^2}$ for ${\varepsilon _0}$ , and $3.0 \times {10^8}{\rm{ m/s}}$ for $c$ in the equation ${E_{\max }} = \sqrt {\frac{{2I}}{{{\varepsilon _0}c}}}$ .

$\begin{array}{c}\\{E_{\max }} = \sqrt {\frac{{2\left( {1.66 \times {{10}^{ - 3}}{\rm{ W/}}{{\rm{m}}^2}} \right)}}{{\left( {8.854 \times {{10}^{ - 12}}{\rm{ }}{{\rm{C}}^2}{\rm{/N}} \cdot {{\rm{m}}^2}} \right)\left( {3.0 \times {{10}^8}{\rm{ m/s}}} \right)}}} \\\\ = 1.12{\rm{ N/C}}\\\end{array}$

The amplitude of the magnetic field is,

${B_{\max }} = \frac{{{E_{\max }}}}{c}$

Substitute $1.12{\rm{ N/C}}$ for ${E_{\max }}$ and $3.0 \times {10^8}{\rm{ m/s}}$ for $c$ in the equation ${B_{\max }} = \frac{{{E_{\max }}}}{c}$ .

$\begin{array}{c}\\{B_{\max }} = \frac{{1.12{\rm{ N/C}}}}{{3.0 \times {{10}^8}{\rm{ m/s}}}}\\\\ = 3.73 \times {10^{ - 9}}{\rm{ T}}\\\end{array}$

(C)

Use the average energy density equation

Substitute $\frac{{{E_{\max }}}}{{\sqrt 2 }}$ for ${E_{{\rm{avg}}}}$ and $\frac{{{B_{\max }}}}{{\sqrt 2 }}$ for ${B_{{\rm{avg}}}}$ in the equation $u = \frac{1}{2}{\varepsilon _0}E_{{\rm{avg}}}^2 + \frac{1}{{2{\mu _0}}}B_{{\rm{avg}}}^2$ .

$u = \frac{1}{4}{\varepsilon _0}E_{\max }^2 + \frac{1}{{4{\mu _0}}}B_{\max }^2$

Substitute $8.854 \times {10^{ - 12}}{\rm{ }}{{\rm{C}}^2}{\rm{/N}} \cdot {{\rm{m}}^2}$ for ${\varepsilon _0}$ , $1.12{\rm{ N/C}}$ for ${E_{\max }}$ , $4\pi \times {10^{ - 7}}{\rm{ T}} \cdot {\rm{m/A}}$ for ${\mu _0}$ , and $3.73 \times {10^{ - 9}}{\rm{ T}}$ for ${B_{\max }}$ in the equation $u = \frac{1}{4}{\varepsilon _0}E_{\max }^2 + \frac{1}{{4{\mu _0}}}B_{\max }^2$ .

$\begin{array}{c}\\u = \frac{1}{4}\left( {8.854 \times {{10}^{ - 12}}{\rm{ }}{{\rm{C}}^2}{\rm{/N}} \cdot {{\rm{m}}^2}} \right){\left( {1.12{\rm{ N/C}}} \right)^2} + \frac{1}{{4\left( {4\pi \times {{10}^{ - 7}}{\rm{ T}} \cdot {\rm{m/A}}} \right)}}{\left( {3.73 \times {{10}^{ - 9}}{\rm{ T}}} \right)^2}\\\\ = 5.54 \times {10^{ - 12}}{\rm{ J/}}{{\rm{m}}^3}\\\end{array}$

(D)

Use the average energy density equation

Substitute $\frac{{{E_{\max }}}}{{\sqrt 2 }}$ for ${E_{{\rm{avg}}}}$ in the equation ${u_{\rm{E}}} = \frac{1}{2}{\varepsilon _0}E_{{\rm{avg}}}^2$ .

${u_{\rm{E}}} = \frac{1}{4}{\varepsilon _0}E_{\max }^2$

Substitute $8.854 \times {10^{ - 12}}{\rm{ }}{{\rm{C}}^2}{\rm{/N}} \cdot {{\rm{m}}^2}$ for ${\varepsilon _0}$ , and $1.12{\rm{ N/C}}$ for ${E_{\max }}$ in the equation ${u_{\rm{E}}} = \frac{1}{4}{\varepsilon _0}E_{\max }^2$ .

$\begin{array}{c}\\{u_{\rm{E}}} = \frac{1}{4}\left( {8.854 \times {{10}^{ - 12}}{\rm{ }}{{\rm{C}}^2}{\rm{/N}} \cdot {{\rm{m}}^2}} \right){\left( {1.12{\rm{ N/C}}} \right)^2}\\\\ = 2.77 \times {10^{ - 12}}{\rm{ J/}}{{\rm{m}}^3}\\\end{array}$

The percentage of the energy due to electric field in the total energy density is,

$percentage = \frac{{{u_{\rm{E}}}}}{u} \times 100\%$

Substitute $2.77 \times {10^{ - 12}}{\rm{ J/}}{{\rm{m}}^3}$ for ${u_{\rm{E}}}$ and $5.54 \times {10^{ - 12}}{\rm{ J/}}{{\rm{m}}^3}$ for $u$ in the equation $percentage = \frac{{{u_{\rm{E}}}}}{u} \times 100\%$ .

$\begin{array}{c}\\percentage = \frac{{2.77 \times {{10}^{ - 12}}{\rm{ J/}}{{\rm{m}}^3}}}{{5.54 \times {{10}^{ - 12}}{\rm{ J/}}{{\rm{m}}^3}}} \times 100\% \\\\ = 50\% \\\end{array}$

(E)

Use the average energy density equation

Substitute $\frac{{{B_{\max }}}}{{\sqrt 2 }}$ for ${B_{{\rm{avg}}}}$ in the equation ${u_{\rm{B}}} = \frac{1}{{2{\mu _0}}}B_{{\rm{avg}}}^2$ .

${u_{\rm{B}}} = \frac{1}{{4{\mu _0}}}B_{\max }^2$

Substitute $4\pi \times {10^{ - 7}}{\rm{ T}} \cdot {\rm{m/A}}$ for ${\mu _0}$ , and $3.73 \times {10^{ - 9}}{\rm{ T}}$ for ${B_{\max }}$ in the equation ${u_{\rm{B}}} = \frac{1}{{4{\mu _0}}}B_{\max }^2$ .

$\begin{array}{c}\\{u_{\rm{B}}} = \frac{1}{{4\left( {4\pi \times {{10}^{ - 7}}{\rm{ T}} \cdot {\rm{m/A}}} \right)}}{\left( {3.73 \times {{10}^{ - 9}}{\rm{ T}}} \right)^2}\\\\ = 2.77 \times {10^{ - 12}}{\rm{ J/}}{{\rm{m}}^3}\\\end{array}$

The percentage of the energy due to magnetic field in the total energy density is,

$percentage = \frac{{{u_{\rm{B}}}}}{u} \times 100\%$

Substitute $2.77 \times {10^{ - 12}}{\rm{ J/}}{{\rm{m}}^3}$ for ${u_{\rm{B}}}$ and $5.54 \times {10^{ - 12}}{\rm{ J/}}{{\rm{m}}^3}$ for $u$ in the above equation.

$\begin{array}{c}\\percentage = \frac{{2.77 \times {{10}^{ - 12}}{\rm{ J/}}{{\rm{m}}^3}}}{{5.54 \times {{10}^{ - 12}}{\rm{ J/}}{{\rm{m}}^3}}} \times 100\% \\\\ = 50\% \\\end{array}$

Ans: Part A

The average pressure due to incident intensity of electromagnetic wave for a totally reflecting surface is $1.1 \times {10^{ - 11}}{\rm{ pa}}$ .

Part B

The amplitude of electric field is $1.12{\rm{ N/C}}$ and magnetic field is $3.73 \times {10^{ - 9}}{\rm{ T}}$ .

Part C

The average density of the energy this wave carries is $5.54 \times {10^{ - 12}}{\rm{ J/}}{{\rm{m}}^3}$ .

Part D

The percentage of energy density due to the electric field is 50%.

Part E

The percentage of energy density due to the magnetic field is 50%.