RES.8-005 | Fall 2012 | Undergraduate

Vibrations and Waves Problem Solving

Electromagnetic Waves in a Vacuum

Problems - Electromagnetic Waves in a Vacuum

Electromagnetic Waves in a Vacuum

Problem 1

Two parallel strips of metal of negligible resistance have width w and length D. They are separated by a distance d with dw and wD. At time t=0 a constant potential difference, V0, is applied across the two strips at the left end of the strips and kept at V0 for all t>0. The system is shown in the figure below.

figure8_1

Choose some instant of time, t=T>0, such that the voltage across the two strips at the far end is still zero. You may ignore the fringe fields. At time t=T:

  1. Plot the magnitude of the electric field, E, and magnetic field, B (between the strips) as a function of the longitudinal position along the strips.
  2. What are the directions of the vectors E and B?
  3. Compute the power supplied by the voltage source.

View/Hide Hint

You can do this problem either by transmission line analysis (inductance, capacitance per unit length, etc.) or by physical reasoning on how electromagnetic fields propagate between the plates (free space).

View/Hide Answers

  1. E0=V0d,B0=E0c,   plots shown below.  

     

figure8_2
  1. E points from upper to lower strip, B is parallel to the strips (pointing into the paper).
  2. Power supplied by voltage source = V02wμ0cd

Problem 2

Sunlight with B-field B=106(Bx,2,0)cos(ωtkxx+kyy) webers/m2, where kx=8π/a, ky=6π/a and a = 2500 nanometers, shines on a mirror-like polar ice lying in the x-z plane. The area of the ice is 100,000 square kilometers.

  1. Find Bx, the direction, the wavelength and the frequency of the incident sunshine;
  2. Find the E-field of the incident sunshine;
  3. Find the Poynting vector of the incident sunshine;
  4. Find the reflected E-field;

    View/Hide Hint

    What are the boundary conditions at the surface of a perfect conductor?

  5. Find the total E-field;
  6. Due to global warming, this piece of ice has just melted, exposing the totally absorbing black rocks underneath. How much additional energy per second is the Earth absorbing due to this melted ice? How does it compare with the average electrical power consumption of the world (about four times the US power consumption, which is the average power consumption per person times the US population)?

View/Hide Answers

  1. The direction n^=(4/5,3/5,0); the wavelength is λ=a/5=500 nm; the frequency of the incident sunshine f=c/λ=61014 Hz; and Bx=3/2, thus B=2.5106(3/5,4/5,0)cos(ωtkxx+kyy) webers/m2.
  2. The incident E-field = ck^×B=7.5×102(0,0,1)cos(ωtkxx+kyy) Newtons/Coulomb
  3. The Poynting vector of the incident light: S=1.5×103cos2(ωtkxx+kyy)n^ Watts/m2<cos2(ωtkxx+kyy)>=0.5and so<S>=7.5×102 Watts/m2
  4. The reflected E-field = 7.5×102(0,0,1)cos(ωtkxxkyy) Newtons/Coulomb
  5. The total E-field = 1.5×103(0,0,1)sin(ωtkxx)sin(kyy) Newtons/Coulomb, i.e. a propagating wave in the x direction and a standing wave in the y direction;
  6. The additional energy per second absorbed by the Earth is SA=4.5×1013 Watts which is much bigger than the 2×1012 Watts of the average electrical power consumption of the world (10 billion people world-wide times 100 Watts/person, or see e.g. Orders of magnitude (power)), so global warming will accelerate faster and faster, as more and more ice melts.

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