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Maxwell's Fourth equation

  1. Jan 16, 2015 #1
    maxwell.PNG
    Can someone please explain to me under what circumstance the second part of the RHS is not equal to zero.

    I only remember the video:


    deriving it 44:30 minutes in, I don't really understand it.

    Cheers
     
  2. jcsd
  3. Jan 16, 2015 #2

    sophiecentaur

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    If the capacitor is charging up (i.e. not DC conditions) then that second term will not be zero. I don't think he mentions that.
    Nice video, btw and his presentation is good without being 'flashy', in front of an audience.
     
  4. Jan 16, 2015 #3
    I understand that, but in the real world can you give me some examples in industry (transformers or power dist or something) or in physics (no idea, maybe like the ionosphere off the top of my head) where the second term is important.

    Cheers
     
  5. Jan 16, 2015 #4

    sophiecentaur

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    The first example that comes to mind, for me, is a radio transmitting antenna. There are changing Electric and Magnetic fields around the radiating wire (in the near field) that do not radiate, but fall to zero after less than a wavelength. In that situation, there will be a displacement current which is there because of the changing Electric Field and an Electric Field component that is due to the changing current.
     
  6. Jan 16, 2015 #5

    jasonRF

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    It is important for all electromagnetic wave phenomena - visible light, wifi signals, cell phone signals, radiowaves your AM/FM radio receives, cable tv signals in a coax cable, cosmic microwave background radiation, etc.
     
  7. Jan 17, 2015 #6
    Thanks for the replies.

    Interesting.
    Ok, so say I've got a closed loop of wire, I put a current J, say DC, through it, it gives me this closed path of B around the wire (on the LHS) and this 'displacement current' component on the RHS. Correct? And is this Electric field parallel or perpendicular to the wire?

    Thanks a lot!
     
  8. Jan 17, 2015 #7

    sophiecentaur

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    If it's DC then dE/dt is zero. Also, the E field will be zero for a loop of zero resistance.
     
  9. Jan 17, 2015 #8

    sophiecentaur

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    This link and this link give an idea about the fields near a dipole antenna. It is worth while getting this sorted in your head first, I think, before you look at loop antennae.
     
  10. Jan 17, 2015 #9
    Ok if it was AC would my statment be true?

    I've started looking at your links, so you were implying previously that the displacement current around a wire will just be in the 'reactive area'?
    Just as a side question I read "In the reactive area, the E and H fields are the strongest and can be measured separately. One field or the other will likely dominate, depending on antenna type" this is what always confuses me, so one dominates the other, yet in the far field of an EM wave both E and H have the same energy (equipartitioning) as they propagate, does this mean that even although one is dominated in the near field that they still both have the same energy?

    Thanks a lot
     
  11. Jan 18, 2015 #10

    sophiecentaur

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    I wasn't. I was trying to give an example that was as near as possible to the simple example with a Capacitor that was quoted in the video.

    Where are you 'at' with your EM theory? If you have got this far from basics and can understand and reproduce the sums then surely you can appreciate that displacement current is just a consequence of the model and that it will always be there, when E changes with time. If you are just cherry picking the ideas from what you have found out from the massive edifice of EM theory then you will be constantly finding such surprising results.

    The fact that the ratio of E to H varies from antenna to antenna can be regarded as a difference in Impedance. Nothing particularly special and it would surely be a surprise if the situation around a 'short circuited' loop of wire was the same as around an 'open circuited ' pair of dipole wires. Calling it the "reactive" region is fair enough and the term just implies a build up of stored energy in the very near field.
     
  12. Jan 18, 2015 #11
    To be sure you're asking when will displacement current be zero?

    Displacement current is associated with the accumulation of charge: the more charge accumulating at a point, the greater the electric field preventing further accumukation of charge. We assume that perfect conductors have no displacement current and this leads us to say that current is not a function in position due to the displacement current, which is the basic postulate for KCL.

    I recommend that you read Introduction to Electrodynamic, David J. Griffiths

    Edit: the displacement current is not zero wherever charge deposits. As in transmission lines where this accumulation of charge is modelled by a capactiance per unit length. Also in uniform plane waves travelling in lossy media, the displacement current will in general be non zero.
     
  13. Jan 20, 2015 #12
    Embarrassingly I have actually done all this from first principles, but that was years ago and when I was doing some revision for a magnetics simulation project I'm going to be working on I've realised I've forgotten almost EVERYTHING. I can understand why It would look like I'm just cherry picking, but I know first hand how illadivsed that would be on this topic.

    I have so many questions I could ask about that sentence, but for the sake of embarrssing myself further I won't. I vaguely remember stuff like the E field goes to zero at the face of a surface or something, and stuff about 'pill boxes' and rectangular loops. I thought they did have closed loop antennas though...
    I will pry slightly about, could you please remind me what varies the E and H, and how can they vary but when the propagate they have the same energy again?

    Wow, excellent succinct way to put it. So KCL as a law breaks down under circumstances or degress of precision? So would KCL need to include charge deposits on long transmission lines, or would this only be a transient thing?


    I have that book in PDF so I will read that section, anything else of interest in I should revise?

    How are the "uniform plane waves travelling in lossy media" related to displacement current?


    Thanks for the replies!
     
  14. Jan 20, 2015 #13
    KCL is only applicable for the lumped element model (check that on wikipedia) we apply KCL to transmission lines by modelling it as a lumped element, and that is how we derive the telegraphist equations. Notice how we model the displacement current by a capacitor in the derivation. I am not very well informed on transmission line theory so make sure to keep looking for answers :)
     
  15. Jan 20, 2015 #14
    Its a long story "how displacement current is related to uniform plane waves" so you should read into that, it took my book around 20 pages to explain this concept, so i am not sure how i can fit it in a comment on a thread :) but to get you prepped the four maxwell's equations are in decoupled format and once you couple them (by some vector algebra) you arrive at the uniform plane wave equation. I should have been more careful with my wording ALL uniform plane waves (in general) depend on displacement current. In fact you can say that displacement current is the most essential piece of information in electromagnetic theory. It is very beautiful how it seld corrects charge accumulation and how it also leads to waves of nothing but fluctuating electric fields and magnetic fields. Without displacement current there would be no electromagnetic waves. Like if you consider the electomagnetic wave in vacuum (which is lossless and here again i apologise for my previous mistake of saying that its only relevant for lossy mediums) faraday's law and ampere's law comspire beautifully to self-propel the electric and magnetic fields (one curls and causes the other to time vary and the time variation of that one causes the curling of the first) at the speed of light which came up to maxwell in an exhilarating form, namely 1/(muo_naught*epsilon_naught)^1/2 which appears in the decoupled (B-E format as opposed to the E-H format) ampere's law (the same one you have up there with 1/c^2 attached to the displacement current!). Apparently, i couldnt restrain my excitement, so be sure to revise these concepts, they are too beautiful to be forgotten ;)
     
  16. Jan 20, 2015 #15

    sophiecentaur

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    The fact that they are equal in a free space wave is 'just one of those things' - like the speed of light relating to permittivity and permeability. Why wouldn't this be the case everywhere? Well, why should it? You can produce regions (inside Capacitors and Transformers, for instance) where there is a vast amount of energy in the form of E field and B field, respectively so that has established the principle that they aren't always the same. The energy in the fields around a radiating structure doesn't all propagate outwards - because there are components of the fields that are in quadrature. In a highly resonant antenna, you would expect a lot of energy stored around it and only a small amount radiated.
     
  17. Jan 21, 2015 #16
    Very interesting stuff, what is the name of your book? Is it published?

    Oh yeah, I remember doing a lab on that, polarisation or something, I need to go and dig out the report on it: All I really remember is that you want the impedance of the antenna to be the same as free space so nothing is reflected and all is transmitted and you don't think about impedance being like Ohms or Ohms/m it's like intrinsic impedance is a ratio of something, so distance is irrelevant.
    This is probably a stupid question, but if some transmitting antenna is a ratio of E and H where, say, E is smaller, will the propagating EM wave be of the magnitude that E was, or H was, on the antenna? When they equal out in freespace.


    Thanks!
     
  18. Jan 21, 2015 #17
    Introduction to Electrodynamics, David J. Griffiths
    But i also refer
    Microwave Engineering, David M. Pozar
    Engineering Electromagnetics, William H. Hayt
     
    Last edited: Jan 21, 2015
  19. Jan 21, 2015 #18

    sophiecentaur

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    Excellent question (dammit) - not totally sure of the full answer but. . . I think the answer could be that the total power radiated will be equal to the transmitter drive volts / the resistive part of the Input Impedance. (A bit of a cop-out, maybe). No antenna is isotropic, so you can't really assign a particular E or H because they vary all over the region of the antenna. Once the wave is far enough away then the Impedance of Free space determines the E/H ratio, as you know, so the extremes of E and H near the antenna could be said to be 'directing the power' to form the Radiation Pattern. It isn't just the wire that's responsible for the way an antenna behaves. If it were, then a thin wire dipole wouldn't have the effective area of greater than 1λ2.

    You could address your question to a wave launched into any transmission line of Z0 impedance from a source of another Impedance.
     
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