# Inductance of hollow conductor (copper pipe)

1. Feb 20, 2013

### Jdo300

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.

Thanks,
Jason O

2. Feb 20, 2013

### Jdo300

Update:

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

$L = \frac{\Phi}{i}$

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

$B = \frac{\mu_0 i}{4r}$

Where:

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:

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

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

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

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?

Thanks,
Jason O

3. Feb 21, 2013

### cabraham

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.

Claude

4. Feb 21, 2013

### Jdo300

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

$B = \frac{\mu_0 i}{2\pi r}$

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:

$B = \frac{\mu_0 i}{2\pi r}(\frac{r^2-a^2}{b^2-a^2})$

Where:

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:

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

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

$L = \frac{-(2a^2Ln(a)-a^2(2Ln(b)+1)+b^2)\mu_0}{4\pi(a^2-b^2)}$

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:

$L = \frac{\mu_0l}{4\pi}\frac{2a^2Ln(a)-a^2(2Ln(b)+1)+b^2}{b^2-a^2}$

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

Thanks!
Jason O

Last edited: Feb 21, 2013
5. Feb 23, 2013

### Drmarshall

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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