
By the end of this lecture, you should:
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As one of the demonstrations showed, the EM radiation is highly polarized. If the antenna is in the plane of polarization, there is in principle direct acceleration of electrons in the wire of the antenna, and this could be fairly efficient at GHz or less frequencies. By contrast, the polarization in solar energy is random, so one could not align an antenna with it. Further, at the high frequency of light, antennas are not effective and absorption of energy involves molecules and atoms, so getting a current flowing is not as simple as in an antenna. If solar energy conversion was easy, the relatively large power density would have motivated us to have it long ago.
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The area of a spherical surface is \(4\pi {r^2}\)and the Poynting vector falls off inversely with r^{2}. As a result, it does not matter which radius of sphere we use to calculate the power output of the Sun as long as it is centered on the Sun. At the radius of Earth’s orbit, the total power is simply the area times the Poynting vector: \[4\pi {(150 \times {10^6} \times {10^3})^2}(1400) = 4 \times {10^{26}}W\] This is about 6 × 10^{14} times the power capacity of the US mentioned. Of course only a very tiny proportion of this even hits Earth, but even that is very large compared to our artificial power generating capacity.
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The power from the Larmor formula is \[P = \frac{{{q^2}{a^2}}}{{6\pi {\varepsilon _0}{c^3}}}\] and in harmonic motion the acceleration is proportional to \({\omega ^2}\)since it is the second time derivative of a function varying with \({e^{j\omega t}}\). In scattering, power has to go in and be absorbed, which will also lead to a similar frequency dependence. That the overall frequency dependence is on \({\omega ^4}\) is not surprising even without considering the argument about resonance made in the lectures.
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One reason is that the emitted wave has zero intensity along the axis of the acceleration, i.e. along the axis of the antenna. A radio station broadcasting from a horizontal antenna would have dead zones near the directions its antenna lined up with. Since this would be very annoying to people driving in cars, and that is a large part of the market for radio stations, it is better to have a vertical antenna which will give a signal that is uniform over large regions of the ground (although of course falling off in intensity far from the antenna).
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These dust particles are made up of large clusters of molecules, which although small by our macroscopic standards (perhaps hazardously so since they are small enough to enter the lungs), are large compared to the wavelength of light. They do not Rayleigh scatter effectively and thus reflect more the whitish yellow color of the Sun (possibly adding color by being more efficient at reflecting certain colors).
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\[λ = \frac{c}{f} = \frac{{3 \times {{10}^8}}}{{2.4 \times {{10}^9}}} = 0.125m.\] The power density is \[S = \frac{P}{{4\pi {r^2}}} = \frac{1}{{4\pi {{(0.1)}^2}}} = 7.95W/m^2\] 0.1 m away and 10^{6} times less 100 m away. But this is the averaged Poynting flux so \[\frac{1}{{4\pi {r^2}}} = \frac{{E_0^2}}{{2{\mu _0}c}} = \frac{{E_0^2}}{{2(4\pi \times {{10}^{  7}})c}}\]or \[{E_0} = \sqrt {\frac{{2 \times {{10}^{  7}}(3 \times {{10}^8})}}{{{r^2}}}} = \sqrt {\frac{{60}}{{{r^2}}}} = \frac{{7.75}}{r}\] At r=0.1 m this is 77.5 V/m and \[{B_0} = \frac{{{E_0}}}{c} = \frac{{77.5}}{{3 \times {{10}^8}}} = 2.6 \times {10^{  7}}T.\] Both fields go down as 1/r, so are 1000 times smaller at 100 m, i.e. 78 mV/m and .26 nT.
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Accounting for night, the atmosphere, and angle of incidence effects, likely about 500 W/m^{2} of the 1400 W/m^{2} solar constant is available on average in a clear desert area. There are 24 × 365 = 8760 hours in a year so the power needed is \[\frac{{4 \times {{10}^{12}} \times 1000}}{{8760}} = 4.5 \times {10^{11}}W.\] The area would be \[\frac{{4.6 \times {{10}^{11}}}}{{500 \times 0.1}} = 9 \times {10^9}m^2\] or 9000 sq. km. This is surprisingly small: only about a square 95 km on a side.