The energy flow to the earth from sunlight is about 1.4kW/m^{2} . (a) Find

kenziebabyyy4e 2021-11-24 Answered
The energy flow to the earth from sunlight is about \(\displaystyle{1.4}{k}\frac{{W}}{{m}^{{{2}}}}\) . (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 \(\displaystyle{1.5}\times{10}^{{{11}}}{m}\). Find the total power radiated by the sun.

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soniarus7x
Answered 2021-11-25 Author has 9353 answers
Step 1 (a) The given value represents the intensity \(\displaystyle{I}={1.4}{k}\frac{{W}}{{m}^{{{2}}}}\) of the light. The intensity of a sinusoidal electromagnetic wave in vacuum is related to the electric-field amplitude \(\displaystyle{E}_{{\max}}\) and the amplitude of magnetic field \(\displaystyle{B}_{{\max}}\) and it is given by equation (32.29) in the form
\(\displaystyle{I}={\frac{{{1}}}{{{2}}}}\epsilon_{{{o}}}{c}{{E}_{{\max}}^{{{2}}}}\) (1)
Where \(\displaystyle\epsilon_{{{o}}}\) is the electric constant, cc is the speed of light. Solve equation (1) for \(\displaystyle{E}_{{\max}}\)
\(\displaystyle{E}_{{\max}}=\sqrt{{{\frac{{{2}{I}}}{{\epsilon_{{{o}}}{c}}}}}}\) (2)
Plug the values for I, c, and \(\displaystyle\epsilon_{{{o}}}\) into equation (2) to get \(\displaystyle{E}_{{\max}}\)
\(\displaystyle{E}_{{\max}}=\sqrt{{{\frac{{{2}{I}}}{{\epsilon_{{{o}}}{c}}}}}}\)
\(\displaystyle=\sqrt{{{\frac{{{2}{\left({1400}\frac{{W}}{{m}^{{{2}}}}\right)}}}{{{\left({8.85}\times{10}^{{-{12}}}\frac{{C}^{{{2}}}}{{N}}\cdot{m}^{{{2}}}\right)}{\left({3}\times{10}^{{{8}}}\frac{{m}}{{s}}\right)}}}}}}\)
\(\displaystyle={1026}\frac{{V}}{{m}}\)
The maximum electric field is related to the maximum magnetic field and the relationship between both of them is given by equation (32.4) in the form
\(\displaystyle{B}_{{\max}}={\frac{{{E}_{{\max}}}}{{{c}}}}\)
Where c is the speed of the light. Now, plug the values for \(\displaystyle{E}_{{\max}}\) and c to get \(\displaystyle{B}_{{\max}}\)
\(\displaystyle{B}_{{\max}}={\frac{{{E}_{{\max}}}}{{{c}}}}={\frac{{{1026}\frac{{V}}{{m}}}}{{{3}\times{10}^{{{8}}}\frac{{m}}{{s}}}}}={3.42}\times{10}^{{-{6}}}{T}\)
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Lounctirough
Answered 2021-11-26 Author has 8545 answers
Step 2
(b) The intensity I is proportional to \(\displaystyle{{E}_{{\max}}^{{{2}}}}\) and it represents the incident power P per area A.
\(\displaystyle{P}={I}{A}\) (3)
The distance between Earth and the Sun represents the radius of the path of the Earth around the Sun and the path is considered to be a sphere. So, the area of this path is calculated by
\(\displaystyle{A}={4}\pi{r}^{{{2}}}={4}\pi{\left({1.5}\times{10}^{{{11}}}{m}\right)}^{{{2}}}={2.82}\times{10}^{{{23}}}{m}^{{{2}}}\)
Now, plug the values for P and A into equation (2) to get I
\(\displaystyle{P}={I}{A}={\left({1400}\frac{{W}}{{m}^{{{2}}}}\right)}{\left({2.82}\times{10}^{{{23}}}{m}^{{{2}}}\right)}={3.95}\times{10}^{{{26}}}{W}\)
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