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I Displacement current in coaxial cables

  1. Jun 20, 2017 #1
    To calculate the displacement current in a coaxial cable (with equal and opposite currents on the inner and outer conductors), most standard texts use the magnetoquasistatic approximation, which ignores the time-varying electric field term in Ampere’s Law.

    Using this approximation, the time-varying magnetic field is calculated by drawing a circular Amperian loop concentric with the inner cylinder (perpendicular to the axis of the cable). The magnetic field is phi-directed and azimuthally symmetric.

    The induced electric field is then calculated by applying Faraday’s law to a rectangular loop with two sides parallel to the axis of the cable and the remaining two perpendicular. Of the two parallel sides, one lies in the region between the two conductors, and one outside. (This is the same geometrical arrangement that is used to calculate the magnetostatic field of a solenoid carrying a steady current). This gives a longitudinal electric field (ie parallel to the axis of the cable).

    In the light of this, I have the following questions:

    1. Does it make sense to calculate the induced electric field from the result obtained by ignoring it in the first place?
    2. If the induced electric field is in the longitudinal direction, how does a coaxial cable support the TEM mode?
     
  2. jcsd
  3. Jun 21, 2017 #2

    vanhees71

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    You can solve the full Maxwell equations by first evaluating the mode functions for the boundary-value problem of the coax cable. You'll get TEM, TE, and TM modes. Perhaps the following helps:

    http://www.ece.rutgers.edu/~orfanidi/ewa/ch09.pdf
     
  4. Jun 22, 2017 #3

    tech99

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    My understanding is that if the coax is small diameter compared with the wavelength, then the cable contains both the ordinary TEM mode plus a "single wire waveguide" mode. The latter is small if the coax is small diameter, and both modes have the same group velocity and characteristic impedance. For any wire or transmission line carrying high frequencies, I would expect to find the single wire mode, where a component of the E field is directed along the axis of the wire. For a wire well away from other conductors I would expect the single wire mode to be predominant. Sorry I can't help much with your Q1.
     
  5. Jun 23, 2017 #4

    jasonRF

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    The displacement current is, by definition, $$ \frac{\partial \mathbf{D}}{\partial t}. $$
    Is that what you are trying to compute? Or are you trying to compute the surface currents on the walls?
     
  6. Jun 23, 2017 #5

    jasonRF

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    If you are trying to compute the surface currents you just use the boundary condition for the magnetic field parallel to a perfect conductor,
    $$ \mathbf{\hat{n}\times H =J_s}$$

    This holds both for static and dynamic fields.
     
  7. Jul 4, 2017 #6
    Yes, I am trying to calculate the rate of change of the electric field (displacement current) rather than the surface currents on the walls (conduction current).
     
  8. Jul 4, 2017 #7

    jasonRF

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    I am wondering what texts you are looking at; I just looked at some of mine and none of them do it this way. Instead they start with the differential versions of Maxwell's equations and solve the partial differential equations to find the fields.

    Anyway, they are not making the magnetoquasistatic approximation. Ampere's law is,
    $$
    \int \mathbf{B \cdot} d\mathbf{r} = \mu_0 \int \mathbf{J \cdot} d\mathbf{a} + \mu_0 \epsilon \frac{\partial}{\partial t}\int \mathbf{E \cdot} d\mathbf{a} $$
    For your case, if we let the axis of the coax be in the z direction, the line integral of the magnetic field is in the azimuthal direction and ##d\mathbf{a} = \hat{\mathbf{z}}\, da##. If we are looking for TEM waves then there is no electric field in the z direction, so the last term is zero because the electric field is only in the radial direction. This is not a quasi-static approximation for the TEM wave, it is exact.

    If you are looking for TEM waves there is no longitudinal component. Any contributions to the line integral are from the radial sides of the path, not from the parallel sides.

    Jason
     
  9. Jul 5, 2017 #8
    The textbook I had in mind is "Introduction to Electrodynamics", by David J. Griffiths, 4th edition, PHI . More specifically, Example 7.9, Problem 7.16, and Problem 7.36.
     
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