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Impedance of a Cylinder

  1. Oct 31, 2011 #1
    Friends:

    I was recently introduced to a [published] method of calculating the impedance of a metallic cylinder [applicable to both DC and AC] which I thought was interesting but counter-intuitive. This method is:

    Lets say you have a metallic cylinder that is inhomogeneous. In other words, this cylinder is made up of more than one material. Lets say you have two materials: the inner part of the cylinder is steel and the outer part of the cylinder is copper (so that the steel forms a solid cylinder and the copper forms a tubular cylinder that surrounds the steel). I have been told that if you wished to find only the impedance of the steel part of this inhomogeneous cylinder you would simply do the following:

    Zsteel = (Esteel/Hsteel) / Circumference,steel

    Finding the impedance of only the copper part of the cylinder is more involved, so I will not show it here, but suffice to say that the impedance of the copper part is a function of both the copper portion AND the steel portion.

    The point is this: The impedance of the steel portion is unaffected by the copper portion, which is outside of the steel. However, the impedance of the copper portion IS affected by the steel portion, which is inside of it. Is this true??

    I always thought that because of mutual inductance, every part of a metallic cylinder affected every other part of a cylinder. So the properties of the steel portion would affect the impedance of the copper portion, and the properties of the copper portion would affect the impedance of the steel portion.

    Any help is appreciated and sorry for being long-winded!!!
     
  2. jcsd
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