# Eddy current in an infinitely long cylindrical conductor in an solenoid!

1. Mar 6, 2012

### Hassan2

Problem:

Calculating the eddy current distribution inside an infinitely long cylindrical conductor inside a straight and infinitely long solenoid energized by an alternating sinusoidal current.

The problem has a perfect summery and we know that magnetic field and the eddy current depend on radial dimension only ( and of course on time). The problem looks simple and I'm sure this problem has been solved perhaps in 19th century, but I can't find the solution in the net. In fact when I attempted to solve it by myself, It was required to solve Bessel's differential equation and i made doubt about my approach. Is the problem this much complicated really?

Last edited: Mar 6, 2012
2. Mar 6, 2012

### marcusl

Yes, Bessel functions are the correct orthogonal functions for problems with cylindrical symmetry. As for references, you might try Smythe, Static and Dynamic Electricity. He has a good chapter on eddy currents, and might work out this problem.

3. Mar 6, 2012

### Philip Wood

Divide up rod (length L, radius a) into co-axial 'pipes'.
Flux through pipe of radius r and wall thickness dr is $\Phi$ = $\pi$ r2 B.
emf is $\epsilon$ = $\pi$ r2 $\dot{B}$
Current in annulus is Iann =$\frac{\epsilon}{R}$ = $\frac{\pi r^{2}\dot{B}}{\rho 2\pi r/L dr}$ = $\frac{Lr dr \dot{B}}{2 \rho}$

Integrating between r = 0 and r = a, to find total current circulating in rod:

Iann = $\frac{L a^{2}\dot{B}}{4 \rho}$

in which $\dot{B} = \mu_{0}\frac{N}{L} I_{0} \omega cos (\omega t)$
N = number of turns in solenoid, I = peak solenoid current.

This neglects flux due to the induced currents in the rod.

Last edited: Mar 6, 2012
4. Mar 6, 2012

### marcusl

The question was about eddy currents. The correct equation will be the diffusion equation. It is separable, and since the problem has circular symmetry, only the radial part is non-trivial.

5. Mar 6, 2012

### Philip Wood

You're saying there isn't an eddy-current solution of the type I've given?

6. Mar 6, 2012

### Hassan2

Thanks a lot. In fact i have heard of the solution with neglecting flux due to the eddy current but i think in a solid cylinder even for 50 Hz frequency, the field is not negligible. I guess the results have a big difference and I would like to know the difference.

Last edited: Mar 6, 2012
7. Mar 6, 2012

### Philip Wood

It might not be too bad to work out flux due to eddy currents, as the rod is a system of nested long solenoids (assuming my original part-solution was valid). I'll have a go.

8. Mar 6, 2012

### Philip Wood

What I've done is to consider a hollow rod, i.e. a pipe, inside the solenoid, parallel to its axis. I'm fairly confident that by performing one or two integrations w.r.t. r, we could modify the treatment to give the eddy currents in the solid rod (assuming that they are circles around the rod axis).

Anyway, the case of the pipe is in the attachment. My conclusion is that for a copper pipe of diameter 1 cm and wall thickness 1 mm, the flux density inside it due to the induced current in its walls is small compared with the field due to the solenoid, assuming a frequency of 50 Hz. The frequency, though, is crucial. Around 1000 Hz, the eddy current contribution overtakes the solenoid contribution (if my calculations are correct).

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Last edited: Mar 6, 2012
9. Mar 6, 2012

### Hassan2

Thanks for the calculations and the attachment.

Your calculation is based on a uniform field in the cylinder. Perhaps when the cylinder is made of a non-ferromagnetic material, the flux due to the induced current is small. However for an iron cylinder, which is usually the case for cores, the flux due to the induced current is quite large, so that it makes the flux in the cylinder non-uniform and causes the skin effect of the flux ( as well as the skin effect of the induced current).

In general, the equation is :

$\nabla^{2} H=σμ\frac{\partial H}{\partial t}H$

which becomes a Bessel's differential equation when H depends on the radial distance (r) only. It's clear than H can't be uniform unless it is time-invariant.
the current is also given by

Je=-σ($\frac{\partial A}{\partial t}+\nabla V$)

where A is the magnetic vector potential and V is the electric potential.

Last edited: Mar 7, 2012
10. Mar 7, 2012

### Philip Wood

Yes. I think I've pushed elementary methods to their limit (or beyond it!) with the case of the copper pipe! I'll be interested in what you find for the ferromagnetic cylinder.

11. Mar 7, 2012

### Hassan2

Well, It is required to solve this equation. It doesn't seem easy!

$\frac{\partial^{2}H}{\partial r^{2}}$+$\frac{1}{r}\frac{\partial H}{\partial r}$=$\mu\sigma \frac{\partial H}{\partial t}$

and in frequency domain
$\frac{\partial^{2}H(ω)}{\partial r^{2}}$+$\frac{1}{r}\frac{\partial H(ω)}{\partial r}$=jωμσH(ω)

Now I doubt this is Bessel's equation.

12. Mar 7, 2012

### Philip Wood

Yes, I reached your equations (albeit clumsily) when I extended my treatment of the copper pipe a little, to deal with the solid rod. I see why you doubt if it's Bessel's equation. It's the imaginary multiplied constant on the right which is the problem? I'm afraid DEs whose solutions are special functions aren't my line of country. Good luck!

13. Mar 7, 2012

### Hassan2

Same here. Let's leave it to mathematicians!

I have solved eddy current problems numerically using a finite element software. The result is so that the flux is more concentrated near the surface and it is minimum on the axis of the cylinder. Increase in ω , σ or μ all cause the flux to become less in the center. The flux magnitude decrease " apparently exponentially" as we move from the surface toward the axis. The same goes with the current. This is the skin effect we have heard of. In relatively high frequencies. the flux and the current in the interior is nearly zero. This can be explained physically too.

14. Mar 7, 2012

### Philip Wood

These are interesting conclusions. Thank you for sharing them – and the problem itself.

15. Mar 7, 2012

### jim hardy

This effect is described in Sylvanus P Thompson's 1901 textbook "Dynamo Electric Machinery" but without your mathematical analysis.
He placed coils at various depths inside an iron bar of substantial diameter to measure progression of flux from periphery toward center. His interest was railway dynamo poles.

It was a problem in control rod position indication for some early nuclear power reactors.
Eddy currents changed with temperature and affected indication, which was based on flux.
There's still some investigation to be done on that system.

Would your formula from post #11 lend itself to calculating flux in a ferritic stainless steel cylinder placed inside a solenoid with AC excitation?
Perhaps you know of a reference book.

thanks,

old jim

16. Mar 10, 2012

### marcusl

That's right. You show no understanding of eddy currents here.
Hi Hassan, it may not be easy but don't give up! Your equations are correct (this is the diffusion equation), but you should write B instead of H since you want the induction within the material.

The standard trick found in advanced books on E&M is to change variables
$$\rho=r\sqrt{jr},$$ transforming the equation into the standard modified Bessel equation. You can find the modified Bessel equation and its solutions in books on E&M or mathematical methods. (Here is a link to an online math handbook, but the presentation is very terse http://dlmf.nist.gov/10.25. In the present case, the order nu turns out to be zero.) The ultimate answer for the radial dependence of the induction is a ratio of modified Bessel functions I_0 with complex arguments. Each is usually broken into a second set of functions called ber and bei that give the real and imaginary parts, so that you can compute and plot the results. True that it's not easy, but not impossible either...

You will find that the longitudinal current follows the same equation and behavior.

EDIT: BTW, as frequency increases the current concentrates into a thin shell near the surface of the cylinder (the skin effect). At sufficiently high frequencies this skin becomes so thin that the curvature of the cylinder may be ignored. The usual skin depth equation
$$\delta=\sqrt{\frac{2}{\mu\sigma\omega}}$$
which derives from an infinite half-plane geometry may then be used.

Last edited: Mar 10, 2012
17. Mar 10, 2012

### marcusl

I think ferritic stainless is non-magnetic, correct? If so, then the diffusion equation will, indeed, be appropriate. See the book by Smythe mentioned in my earlier post. I once owned a copy of Eddy Currents in Linear Conducting Media by Tegopoulos, which was a good reference as well.

18. Mar 11, 2012

### Hassan2

Thanks you very much marcusl. With your hint, the problem becomes solvable!
In fact one of my motivations to solve this equation was the approximate expression given for the skin depth. The formula is derived for an infinite half-plane as you mentioned, but it used in cylinders too.

Thanks again,

Hassan