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)

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"