# The energy flow to the earth from sunlight is about

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.
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Vivian Soares
(a) The given value represents the intensity $I=1.4k\frac{W}{{m}^{2}}$ of the light. The intenity of a sinusoidal electromagnetic wave in vacuum is related to the electric-field amplitude ${E}_{max}$ 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
${E}_{max}=\sqrt{\frac{2I}{{ϵ}_{0}c}}$
$=\sqrt{\frac{2\left(1400\frac{W}{{m}^{2}}\right)}{\left(8.85×{10}^{-12}\frac{{C}^{2}}{N}×{m}^{2}\right)\left(3×{10}^{8}\frac{m}{s}\right)}}$
The maximum electric field is related to the maximum magnetic field in form:
${B}_{max}=\frac{{E}_{max}}{c}$
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 distance between the Earth and the Sun represents the radius of the path of the Earth around the Sun and it is considered to be sphere. So, the area is calculated:
$A=4\pi {r}^{2}=4\pi {\left(1.5×{10}^{11}m\right)}^{2}=2.82×{10}^{23}{m}^{2}$
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) $3.95×{10}^{26}W$