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Inductance of hollow conductor (copper pipe)

  1. Feb 20, 2013 #1
    Hi Everyone,

    I'm working on some inductance calculations and was wondering if anyone knows of any formulas (or methods) to determine the inductance of a straight piece of copper pipe, if the ID, wall thickness, and length are known? The closest thing I have found so far was the inductance of a straight conductor but the formula assumes it is solid.

    Any help/pointers appreciated.

    Jason O
  2. jcsd
  3. Feb 20, 2013 #2

    Doing some snooping around, I found an equation which relates the inductance to the total Flux and current:

    [itex]L = \frac{\Phi}{i}[/itex]

    I also found an equation in my physics book, which gives the flux density at a point inside a cylindrical conductor, using this formula:

    [itex]B = \frac{\mu_0 i}{4r}[/itex]


    r is the radius inside the tube where the B field is to be calculated
    i is the current through the tube (assuming even current density distribution)

    I reasoned that I could calculate the total flux through the cross section by integrating B over the radius of the tube from a to b, assuming a is the inside radius, and b is the outside radius. So I got to this point here:

    [itex]\Phi = \frac {\mu_0 i}{4} \int ^{b}_{a}\frac{1}{r} dr = \frac {\mu_0 i}{4} [Ln(b) - Ln(a)][/itex]

    Combining the above result with the first equation, I get:

    [itex]L = \frac{\frac {\mu_0 i}{4} [Ln(b) - Ln(a)]}{i} = \frac {\mu_0}{4} [Ln(b) - Ln(a)][/itex]

    However, I'm pretty sure there is something wrong with this equation as it doesn't take into account the length of the hollow copper tube. Can anyone see what I did wrong here?

    Jason O
  4. Feb 21, 2013 #3
    Your equation yields "henry/meter" or inductance per unit length. Multiplying your answer by the length should give the actual inductance, assuming a mistake was not made in deriving it. The Ln(b) - ln(a) can be expressed as ln(b/a), which is dimensionless. The mu factor has units of henry/meter. So your answer is in henry/meter.

  5. Feb 21, 2013 #4
    Hi Claude,

    Thank you very much for the input. That does make sense now that I think about it.

    I did notice there was a booboo in one of the equations so I needed to redo it again.

    The equation for the magnetic field inside of a conductor is supposed to be

    [itex]B = \frac{\mu_0 i}{2\pi r}[/itex]

    Also, after asking around more about this problem, I was given an old Physics II exam where there was an expression derived for the B field inside any point in the hollow conducting tube. It looks different that the equation I came up with but here it is:

    [itex]B = \frac{\mu_0 i}{2\pi r}(\frac{r^2-a^2}{b^2-a^2})[/itex]


    r is the radius inside the tube where the B field is to be calculated
    i is the current through the tube (assuming even current density distribution)
    a = tube inner radius
    b = tube outer radius

    So, Integrating this from a to b to get the flux, I get:

    [itex]\Phi = \frac{-(2a^2Ln(a)-a^2(2Ln(b)+1)+b^2)i\mu_0}{4\pi(a^2-b^2)}[/itex]

    Dividing out the current, i to get the inductance per unit length, I get:

    [itex]L = \frac{-(2a^2Ln(a)-a^2(2Ln(b)+1)+b^2)\mu_0}{4\pi(a^2-b^2)}[/itex]

    Based on Claude's comment, I'll simple add a length term to specify the length of the tube. Also, rearranging terms to remove the minus sign and separate constants, I get this as my final equation:

    [itex]L = \frac{\mu_0l}{4\pi}\frac{2a^2Ln(a)-a^2(2Ln(b)+1)+b^2}{b^2-a^2}[/itex]

    So that is my final answer. If anyone sees any mistakes here, please let me know.

    Jason O
    Last edited: Feb 21, 2013
  6. Feb 23, 2013 #5
    Hi Jason

    You need to read up on "Skin effect"
    Try Google for starters.
    A pulse, or any high-frequency signal, tends to send current on the outer SURFACE of a conductor. Whether the "energy flow" is "inside" or "outside and guided by" the conductor is a question philosophers labour over!
    The upshot is the "inductance" (Henrys) is a function of how QUICK is the rate-of change!!

    The current is in the form of TRAVELLING WAVES surging back and forth along the conductor and (at hig hfrequency ) may reverse in DIRECTION with radius.

    You are right to use a hollow pipe as all the current (nearly all of it!) is on the
    SURFACE (good for lightning conductors too because of the ultra-fast pulses)

    Your pipe is not an inductor - it is a TRANSMISSION LINE with its own impedance, group and phase velocities.
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