The energy flow to the earth from sunlight is about $\mathbf{1}\mathbf{.}\mathbf{4}\mathbf{}\mathbf{kW}\mathbf{/}{\mathbf{m}}^{\mathbf{2}}$. (a) Find the maximum values of the electric and magnetic fields for a sinusoidal wave of this intensity. (b) The distance from the earth to the sun is about$\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{11}}\mathbf{}\mathbf{m}$ . Find the total power radiated by the sun.

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

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

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

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

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

$$\begin{gathered} E_{max}=\sqrt{\frac{2l}{{\epsilon }_{0}c}} \\ {B}_{max}=\frac{E_{max}}{c} \end{gathered}$$

Where, ${\epsilon }_0=8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}.{\mathrm{m}}^2$and c is the speed of light that is equal to $3.0\times {10}^8\mathrm{m}/\mathrm{s}$.

Given that$I=1.4\mathrm{kW}/{\mathrm{m}}^2$

The maximum value of electric field is:

$E_{max}=\sqrt{\frac{2l}{{\epsilon }_{0}c}}$

Substitute the values

$$\begin{gathered} E_{max}=\sqrt{\frac{2\times \left(1400\right)}{\left(8.85\times {10}^{-12}\right)\left(3\times {10}^8\right)}} \\ =1026\mathrm{V}/\mathrm{m} \end{gathered}$$

The amplitude of magnetic field is:

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

Substitute the values

$$\begin{gathered} B_{max}=\frac{1026}{3\times {10}^8} \\ =3.42\times {10}^{-6}\mathrm{T} \end{gathered}$$

Hence, the maximum values of the electric and magnetic fields for a sinusoidal wave with intensity 1.4 kW/m² are 1026 V/m and $3.42\times {10}^{-6}\mathrm{T}$respectively.

Given that,

$$\begin{gathered} I=1.4\mathrm{kW}/{\mathrm{m}}^2 \\ r=1.5\times {10}^{11}\mathrm{m} \end{gathered}$$

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

$$\begin{gathered} I=\frac{P}{A} \\ \Rightarrow P=IA \end{gathered}$$

Substitute the values

$$\begin{gathered} P=\left(1400\right)\left[4\pi \left(1.5\times {10}^{11}\right)\right] \\ =1400\times 2.82\times {10}^{23} \\ =3.95\times {10}^{26}W \end{gathered}$$

Hence, the total power radiated by the sun is $3.95\times {10}^{26}W$.