 zgribestika

2021-12-15

The energy flow to the earth from sunlight is about $1.4{\left\{k\frac{W}{m}\right\}}^{2}$.
(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 $1.5×{10}^{11}m$ Find the total power radiated by the sun. Vivian Soares

(a) The indicated value indicates the intensity $I=1.4k\frac{W}{{m}^{2}}$ of the light. An electromagnetic wave's intensity in a vacuum is inversely proportional to the amplitude of the electric field. ${E}_{max}$ and the amplitude of magnetic field ${B}_{max}$ and is given by the equation:
$I=\frac{1}{2}{ϵ}_{0}c{E}_{max}^{2}$, where ${ϵ}_{0}$ is a electric constant, $c$ is speed of light.
Now, solve the equation for ${E}_{max}$
${E}_{max}=\sqrt{\frac{2I}{{ϵ}_{0}c}}$
Plug the values for $I,ϵ,c$ into equation
${E}_{max}=\sqrt{\frac{2I}{{ϵ}_{0}c}}$
$=\sqrt{\frac{2\left(1400\frac{W}{{m}^{2}}\right)}{\left(8.85×{10}^{-12}\frac{{C}^{2}}{N}×{m}^{2}\right)\left(3×{10}^{8}\frac{m}{s}\right)}}$
Maximum magnetic field and maximum electric field have a relationship in the following way:
${B}_{max}=\frac{{E}_{max}}{c}$
Now, plug the values into the equation
${B}_{max}=\frac{{E}_{max}}{c}=\frac{1026\frac{V}{m}}{3×{10}^{8}\frac{m}{s}}=3.42×{10}^{-6}T$
(b) The intensity is propotional to ${E}_{max}^{2}$ and represents the incident power $P$ per area $A$
$P=IA$
The radius of the Earth's path around the Sun is represented by the distance between it and the Sun, and this distance is thought to be a sphere. The area is computed as follows:
$A=4\pi {r}^{2}=4\pi {\left(1.5×{10}^{11}m\right)}^{2}=2.82×{10}^{23}{m}^{2}$
Now, plug in the values:
$P=IA=\left(1400\frac{W}{{m}^{2}}\right)\left(2.82×{10}^{23}{m}^{2}\right)=3.95×{10}^{26}W$
The answer is: (a) $3.42×{10}^{-6}T$, (b) $3.95×{10}^{26}W$

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