Television Broadcasting. Public television station KQED in San Francisco broadcasts a sinusoidal radio signal at a power of 777 KW. Assume that the wave spreads out uniformly into a hemisphere above the ground. At a home 5.00 km away from the antenna, (a) what average pressure does this wave exert on a totally reflecting surface, (b) what are the amplitudes of the electric and magnetic fields of the wave, and (c) what is the average density of the energy this wave carries? (d) For the energy density in part (c), what percentage is due to the electric field and what percentage is due to the magnetic field?

The power transported per unit area is known as the intensity (l).

The formula used to calculate the intensity (l) is:

$I=\frac{P_{avg}}{A}$

Where, A is area measured in the direction perpendicular to the energy and P is the power in watts.

The average force per unit area due to the wave is known as radiation pressurerole="math" localid="1664346816206" ${\mathit{p}}_{\mathbf{rad}}$. Radiation pressure is the average dp/dtdivided by the absorbing areaA.

The formula for radiation pressure of the wave totally reflected light is:

$p_{rad}=\frac{2I}{c}$

The formula used to determine the amplitude of electric and magnetic fields of the wave are:

$$\begin{gathered} E{\frac{2I}{{\epsilon }_{0}c}}_{max} \\ B{\frac{E_{max}}{c}}_{max} \end{gathered}$$

The average density of wave$u_{avg}={\epsilon }_0(E_{rms}{)}^2$where,

$$\begin{gathered} E_{ms}=\frac{E_{max}}{\sqrt{2}} \\ {\epsilon }_0=8.85\times {10}^{-12}{C}^2/N·m^2 \end{gathered}$$

Given that,

$$\begin{gathered} P_{avg}=777KW \\ r=5.00km \end{gathered}$$

The radiation pressure of the wave totally reflected light is:

$p_{rad}=\frac{2I}{c}$

Where,$I=\frac{P_{avg}}{A}$

$$\begin{gathered} p_{rad}=\frac{2\left({\displaystyle \frac{P_{avg}}{A}}\right)}{c} \\ =\frac{2\times \left({\displaystyle \frac{777000}{2\mathrm{\pi }{\left(5000\right)}^2}}\right)}{3\times {10}^8} \\ =\frac{2\times \left(0.005\right)}{3\times {10}^8} \\ =3.3\times {10}^{-11}Pa \end{gathered}$$

Hence, the average radiation pressure exerted by the wave on a totally reflecting surface is $3.3\times {10}^{-11}Pa$.

The amplitude of electric field is:

$E{\sqrt{\frac{2I}{{\epsilon }_{0}c}}}_{max}$

Substitute the values

$$\begin{gathered} E_{max}=\sqrt{\frac{2\times \left({\displaystyle \frac{777000}{2\mathrm{\pi }{\left(5000\right)}^2}}\right)}{\left(8.85\times {10}^{-12}\right)\left(3\times {10}^8\right)}} \\ =1.94N/c \end{gathered}$$

The amplitude of magnetic field is:

$B{\frac{E_{max}}{c}}_{max}$

Substitute the values

$$\begin{gathered} B_{max}=\frac{1.94}{3\times {10}^8} \\ =6.5\times {10}^{-9}T \end{gathered}$$

Hence, the amplitudes of the electric and magnetic fields of wave are 1.94 N/c and $6.5\times {10}^{-9}T$respectively.

The average density of wave$u_{avg}={\epsilon }_0(E_{rms}{)}^2$

Substitute$E_{rms}=\frac{E_{max}}{\sqrt{2}}$in formula od average density

$u_{avg}={\epsilon }_0\left(\frac{E_{max}}{\sqrt{2}}({)}^2\right)$

Again, substitute the value

$$\begin{gathered} u_{avg}=(8.85\times {10}^{-12}){\left(\frac{1.94}{\sqrt{2}}\right)}^2 \\ =1.67\times {10}^{-11}J/m^3 \end{gathered}$$

Hence, average density of the energy wave carries$1.67\times {10}^{-11}J/m^3$.

As the energy density of electric and magnetic fields are same therefore, both has 50% of energy density.

Hence, 50% is due to the both electric and magnetic field.