Lecture 14: Generating EM Waves, Energy, Scattering

OCW Scholar

« Previous | Next »

Lecture Topics

A hand holds three smoking cigarettes.
  • Poynting Vector
  • Solar Constant
  • Emission and Scattering of EM Waves

Learning Objectives

By the end of this lecture, you should:

  • relate E and B fields to the Poynting vector.
  • know the solar constant and why it does not correspond simply to an E field.
  • describe in general terms the derivation of the Larmor formula.
  • describe antenna emission and Rayleigh scattering (including polarization).

Lecture Activities

Check Yourself

  • Contrast the energy conversion that was demonstrated in the lectures, in which a light bulb was lit through EM wave energy being received in an antenna, with what has to be done to collect solar energy that has a Poynting vector in principle corresponding to an electric field of 1000 V/m.

View/hide answer

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.


  • Using the solar constant of 1400 W/m2 and assuming the Sun radiates EM energy isotropically, calculate the power of the Sun knowing Earth is 150 million km from it. Contrast the number with the electric power capacity of the US mentioned in the lecture.

View/hide answer

The area of a spherical surface is \(4\pi {r^2}\)and the Poynting vector falls off inversely with r2. 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 × 1014 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.


  • Make a very rough argument based on the Larmor formula for why the sky is blue (i.e. why scattered light power is proportional to the frequency to the fourth power).

View/hide answer

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.


  • Based on the Larmor formula, explain why radio tower antennas are held vertically even though it would be much easier just to lay them on the ground as wires.

View/hide answer

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).


  • In photographs of conditions in large dust storms in China, or even of prevalent pollution conditions, the sky often appears yellow. Explain this deviation from the normal color of the sky.

View/hide answer

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).


  • A cell phone radiates about 1 W of power at a frequency of 2.4 GHz. Calculate the wavelength, Poynting vector, and the peak E and B fields 0.1 and 100 m away, assuming the energy spreads isotropically over a sphere (whose area is \(4\pi {r^2}\)).

View/hide answer

\[λ = \frac{c}{f} = \frac{{3 \times {{10}^8}}}{{2.4 \times {{10}^9}}} = 0.125m.\] 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.


  • US electrical energy consumption was about 4,000 billion kWh in recent years (i.e. 4×1012 kWh), where a kWh (kilowatt-hour) is the amount of energy corresponding to a power of 1 kW used for an hour. If the overall efficiency of deriving solar energy is about 10%, what area in square km would need to be covered in solar cells to meet US electrical energy needs?

View/hide answer

Accounting for night, the atmosphere, and angle of incidence effects, likely about 500 W/m2 of the 1400 W/m2 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.


« Previous | Next »