91Ó°ÊÓ

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:

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

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

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

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

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

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

The amplitude of electric field is:

$E{\sqrt{\frac{2I}{{\mathrm{Î\mu }}_{0}c}}}_{max}$

Substitute the values

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

The amplitude of magnetic field is:

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

Substitute the values

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

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

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

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

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

Again, substitute the value

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

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

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.