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Phase of a magnetic field

  1. Apr 5, 2016 #1

    tech99

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    I should like to ask about the magnetic field of a wire carrying an alternating current (the dimensions of the experiment are too small for radiation to be significant). If I measure the phase of the magnetic field a distance from the surface of the wire, is it still in phase with the current? I think it will always be in phase but want to hear ideas.
     
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  3. Apr 5, 2016 #2

    mfb

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    How large is the distance, what is the frequency of the alternating current? Divide distance by the speed of light and compare to the period of your AC.
     
  4. Apr 5, 2016 #3

    berkeman

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    These two statements would seem to be at odds. The field propagates away from the wire at the speed of light c, so there will be a phase shift related to the frequency and the propagation distance, no?
     
  5. Apr 6, 2016 #4

    tech99

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    Thanks for the comments so far. This was my initial assumption. However, the energy in the magnetic field is not being radiated, and does not permanently leave the wire. It is just stored, and then returns to the wire. So my idea is that equal energy flows away from and towards the wire, so there is no net outflow. So I don't think it is possible to detect a propagation delay, and it appears that the whole field at whatever distance from the wire cycles simultaneously. It seems similar to the case of the phase of a standing wave, where energy also flows both ways.
    If there were a propagation delay, maybe the phase shift of the inductor would be more than 90 degrees.
     
  6. Apr 6, 2016 #5

    mfb

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    Well, it depends on the scales of your setup, where d is the distance (scale) and f is the AC frequency:
    ##df \ll c##? -> no notable phase shift, radiative terms (you can never avoid that completely) are negligible.
    ##df \gg c##? -> large phase shift, radiative terms are dominant.
    In between, it is complicated.
     
  7. Apr 6, 2016 #6

    tech99

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    Thank you. Yes, agree. If the wire is driven, for instance, by a step function, the magnetic field will expand outwards at the speed of light, so there is a time delay. But if the wire is driven by a continuous sinusoid, then energy is flowing both ways equally (away from and towards the wire) so my idea is that no propagation delay can then be discerned.
     
  8. Apr 6, 2016 #7

    berkeman

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    There is still a loss due to radiation, so that part of the signal does propagate away with the appropriate delay...
     
  9. Apr 6, 2016 #8

    marcusl

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    There is always a delay since information about the phase of the current travels from the wire to a distant point at the speed of light.
     
  10. Apr 7, 2016 #9

    tech99

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    But at that distant point, there are two waves. One going outward and one coming inward. Both contain equal energy. So the phasor diagram will show the two rotating in opposite directions, and give constant phase.
     
  11. Apr 7, 2016 #10

    berkeman

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    Where did the inward-moving wave come from?
     
  12. Apr 7, 2016 #11

    sophiecentaur

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    You should do the suggested sums yourself, based on c, and come to a conclusion. c is about 1m per 3 nanoseconds and the time period at 50Hz is 20ms. What sort of a fraction of a cycle does that represent for a distance of 10m?
     
  13. Apr 7, 2016 #12

    mfb

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    Who said 50 Hz?

    @berkeman: it is a solution to the Maxwell equations.
     
  14. Apr 7, 2016 #13

    tech99

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    For an alternating current, when the current increases, the field moves outwards as it is built, and when the current decreases, the field moves inwards as it collapses.
     
  15. Apr 7, 2016 #14

    berkeman

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    But EM is a transverse wave, not a longitudinal wave. I guess I'm not understanding what you and @mfb are saying...
     
  16. Apr 7, 2016 #15

    tech99

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    I understand your problem. What I am suggesting is that the energy flows outwards, then inwards, so there is no net flow, rather like a standing wave.
    So maybe if two coils of wire are separated by a distance, one driven and the other open circuit, we will not see a phase shift due to the distance. But if the second coil is terminated with a resistance, then we will see a phase shift because more energy is flowing outward than inward.
     
  17. Apr 7, 2016 #16

    berkeman

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    Your OP was about the field generated by a wire, but now you are talking about coils separated by a distance? Now I'm really confused...
     
  18. Apr 7, 2016 #17

    marcusl

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    A localized source generates only outgoing waves. An incoming wave can only be generated by an external source (think of a pair of communications antennas). The field of the source is computed by use of retarded potentials (see the wiki article) or the retarded Green function, both of which replace t by (t-r/c) as mfb pointed out.
     
  19. Apr 8, 2016 #18

    tech99

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    Does not that apply only to radiated energy? Reactive energy is just stored so cannot be outflow only; net energy is not lost from the wire unless radiation takes place.
     
  20. Apr 8, 2016 #19

    sophiecentaur

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    Would it be better to say that the Energy of the Magnetic Field moves out and in? If the wire is carrying AC, the energy that exists in the fields around it will 'return' to the circuit and be dissipated in a resistive load, in the source resistance or stored in the Capacitances (or equivalent)in the system.
    In all cases, the effect at any point is delayed by finite value of c.
     
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