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Poynting Vector and Mobile phones in ovens

  1. Jun 20, 2014 #1
    Hi, sorry if this is in the wrong forum but 2 Quick hopefully simple questions about Poynting vector and EM waves.

    Q1)
    Say I have two antenna (antenni?)...dipoles maybe, both with say, 50W going into them, one oscillating on the kHz range, one on the MHz range (GHz, whatever). But for this example same number of oscilating charges in both antenna.

    so ave S = Eo X Bo / 2μ (watts/square meter)

    because the average power of sin^2 is a half, irrespective of frequency? (EoSin(kx-ωt) = E)

    To maintain equal output power...So either Eo is reduced with increased frequency ω, or there is just less EM per square meter (but since higher frequency the same power) ? Which and how? (It is my understanding that Eo is propertional to intensity, as inensity is proportional to power; and from intensity I imagine that is determined by the amount of oscilating charges and the voltage that induces their movement. Please correct me if I'm wrong. I'd love for someone to explain how U = (epsilon*E)/2 + (B/(2*mu)) relates to U = h.f)

    P.S: Ignore Q2:

    Q2)
    What I've read about TEM waveguides tells me that my mobile phone (GHz λ) shouldn't work inside my bird cage or microwave oven, yet it does. How can this be? (since given that when the oscilating electric field contacts a conductor it creates surface charge)
     
    Last edited: Jun 20, 2014
  2. jcsd
  3. Jun 20, 2014 #2
    Not sure about Q1(I think the antenna with the higher frequency will radiate more power and you ll have to reduce the input power on that antenna in order to have equal radiated power from both antennas, Poynting vector give us the radiated power and not the input power)

    but for Q2 though the mobile phone might work inside a microwave oven, i am sure the signal level drops dramatically once you put it inside it and close the window. The metal grid on the window reduces dramatically the EM power that enters (for Ghz wavelengths) but cannnot eliminate it completely.
     
  4. Jun 20, 2014 #3
    Thanks anyway but I'm really after at least a partial mathematical explanation, that's my point that a higher frequency SHOULD radiate more power, however that is not suppored by the maths of the poynting vector (at least not obviously). And no, I have to disagree that, if both have a source of the same magnitude then concervation of energy would say they both would radiate the same power. So something is different between signals for it to be: same power in each, same poynting vector out, (yet different ω)
    For Q2, I really echo the first part of this response that the mathematics SHOULD reduce it to zero theoretically, however given that there is no conductive gap as big as the wavelength, the fact that the signal wasn't reduced even dramatically enough to disturb the functionality, this is puzzeling. But P.S it really is the first question that is of interest.
     
    Last edited: Jun 20, 2014
  5. Jun 20, 2014 #4
    Infact I'm only really interested in the first question and opinions on "either Eo is reduced with increased frequency ω, or there is just less EM per square meter (but since higher frequency the same power) ? Which and how? (It is my understanding that Eo is propertional to intensity, as inensity is proportional to power; and from intensity I imagine that is determined by the amount of oscilating charges and the voltage that induces their movement. Please correct me if I'm wrong. I'd love for someone to explain how U = (epsilon*E)/2 + (B/(2*mu)) relates to U = h.f)"
     
  6. Jun 20, 2014 #5
    Well, i dont know the exact math about specific types of antennas (like half-wave dipole antenna for example) but you may find this page in wikipedia helpfull

    http://en.wikipedia.org/wiki/Electric_Dipole_Radiation#Dipole_radiation.
    Eo (and Bo) is proportional to the square of the angular frequency w, so Eo is increasing as w increases for the same dipole moment Po. In real antennas one doesnt have a dipole moment but a current density (which you refer to as the "amount of oscilating charges") which depends on the input power. I suppose for real antennas similar relation would hold , thus for the "same" current density Eo and Bo would be proportional to the angular frequency. My point anyway was that Poynting vector give us only the radiated power which is a fraction of the input power. The input power in the antenna equals radiated power+near field power+power lost as heat in the ohmic resistance of the antenna.

    About the mobile phone, how do you know its working properly inside the oven? I suppose all you can do is to put it inside and then call your number and hear it ringing. Even though its ringing, if you could pick it up and try to talk you would see that you would have problem in the communication(interrupts, losing signal e.t.c).
     
    Last edited: Jun 20, 2014
  7. Jun 21, 2014 #6
    Thanks for the reply Delta, and for the link. I'm still a little confused about interpretation though.
    So same current density Eo and Bo would be proportional to the angular frequency, indicating that Eo is larger for the MHz or GHz antenna, so same amount of waves, different amplitudes and the difference between the two antennas is the higher frequency one might have better radiated power efficiency. But is Eo only proportional to ω? Because if you were to have a big satelite dish broad casting the same frequency as a smaller one, you would expect Eo to be bigger from the source emmiting more photons.


    P.S
    Because if it rings, or if it is put in the microwave after it has been answered it doesn't drop out. I suppose you could put a tape playing in with it and walk far away to listen to the quality of the reception, or a camera to film the bars of signal.
     
  8. Jun 21, 2014 #7
    Well the formula's at wikipedia link are for the elementary hertzian dipole antenna and probably apply more ore less for all the dipole antennas. For antennas with dish the formulas would be different, because the dish reflects to the front part of the electromagnetic field power that would otherwise go to the back of the dish, the larger the dish is the more reflection of energy to the front is done.
     
  9. Jun 21, 2014 #8
    Perhaps mentioning 'dish' was an unnecessary red herring; what I'm getting at is, we've established that Eo is proportional to frequency with a dipole, so keeping in mind the peak electric field Eo, is this also proportional to the amount of waves (intensity) too? so like:

    Number of photons * frequency*h = Number of photons *1/2 * [epsilon*Eo^2 + mu*Ho^2]


    Thanks
     
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