A satellite 575 km above the earth’s surface transmits sinusoidal electromagnetic waves of frequency 92.4 MHz uniformly in all directions, with a power of 25.0 kW. (a) What is the intensity of these waves as they reach a receiver at the surface of the earth directly below the satellite? (b) What are the amplitudes of the electric and magnetic fields at the receiver? (c) If the receiver has a totally absorbing panel measuring 15.0 cm by 40.0 cm oriented with its plane perpendicular to the direction the waves travel, what average force do these waves exert on the panel? Is this force large enough to cause significant effects?

The intensity I is proportional to ${\mathrm{E}}_{\max }^2$and it represents the incident power P per area A is given as $\mathrm{l}=\frac{\mathrm{P}}{\mathrm{A}}$The intensity of a sinusoidal electromagnetic wave in vacuum is related to the electric-field amplitude E the amplitude of magnetic field B and it is given by equation in the form $\mathrm{l}=\frac{1}{2}{\mathrm{\epsilon }}_0{\mathrm{cE}}_{\max }^2$Where ${\mathrm{\epsilon }}_0$is the electric constant, c is the speed of light

The radius represents the distance from the satellite to the Earth r=$575\times {10}^3\mathrm{m}$. So, the area of the is calculated by$\mathrm{A}=4{\mathrm{\pi r}}^2=4\mathrm{\pi }{\left(575\times {10}^3\mathrm{m}\right)}^2=4.15\times {10}^{12}{\mathrm{m}}^2$

Substitute the values in equation $\mathrm{I}=\frac{\mathrm{P}}{\mathrm{A}}$we have,

$\mathrm{I}=\frac{\mathrm{P}}{\mathrm{A}}=\frac{25\times {10}^3\mathrm{W}}{4.15\times {10}^{12}{\mathrm{m}}^2}=6.02\times {10}^{-9}\mathrm{W}/{\mathrm{m}}^2$

The intensity of a sinusoidal electromagnetic wave in vacuum is given as$\mathrm{I}=\frac{1}{2}{\mathrm{\epsilon }}_0{\mathrm{cE}}_{\max }^2$which can be written as ${\mathrm{E}}_{\max }=\sqrt{\frac{2\mathrm{l}}{{\mathrm{\epsilon }}_0\mathrm{c}}}$Substitute the values in equation we have

$$\begin{gathered} {\mathrm{E}}_{\max }=\sqrt{\frac{2\left(6.02\times {10}^{-9}\mathrm{W}/{\mathrm{m}}^2\right)}{8.85\times {10}^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2\times 3\times {10}^8\mathrm{m}/\mathrm{s}}} \\ =2.12\times {10}^{-3}\mathrm{V}/\mathrm{m} \end{gathered}$$

The maximum electric field is related to the maximum magnetic field and the relationship between both of them is given by equation ${\mathrm{B}}_{\max }=\frac{{\mathrm{E}}_{\max }}{\mathrm{c}}$Substitute the values in equation we have

${\mathrm{B}}_{\max }=\frac{2.12\times {10}^{-3}\mathrm{V}/\mathrm{m}}{3\times {10}^8\mathrm{m}/\mathrm{s}}=7.07\times {10}^{-12}\mathrm{T}$

The wave exerts an average force F per unit area and this is the radiation pressure Prad and it is the average value of dp/dt divided by the area. So, the radiation pressure of the wave that totally absorbed is given by equation ${\mathrm{p}}_{\mathrm{rad}}=\frac{\mathrm{I}}{\mathrm{c}}$Substitute the values in equation we have

$$\begin{gathered} {\mathrm{p}}_{\mathrm{rad}}=\frac{\left(6.02\times {10}^{-9}\mathrm{W}/{\mathrm{m}}^2\right)}{3\times {10}^8\mathrm{m}/\mathrm{s}}=2\times {10}^{-17}\mathrm{Pa} \\ \mathrm{F}={\mathrm{p}}_{\mathrm{rad}}\mathrm{A}=\left(2\times {10}^{-17}\mathrm{Pa}\right)\left(0.15\mathrm{m}\times 0.40\mathrm{m}\right)=12\times {10}^{-10}\mathrm{N} \end{gathered}$$