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Eddy Current Calculations 
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#1
Aug2011, 06:36 PM

P: 4

I searched a lot of google for calculations on eddy currents, and got a lot of things that describe how eddy currents work, but almost nothing about how to quantify the induced currents and the resulting magnetic field. Does anyone here know much about where to start?
In particular, I'm looking to quantify the induced eddy currents in a flat aluminum plate with a changing, perpindicular magnetic field. 


#2
Aug2111, 04:10 AM

PF Gold
P: 963

Maybe it would help to look at a simple special case: a disc. By symmetry, the eddy currents will follow circular paths centred on the centre, O, of the disc. The emf acting in a circular path of radius r will, from Faraday’s law be given by
[tex]\varepsilon = \frac{d\Phi}{dt} = \pi r^2 \dot B [/tex] But if the current density is J around a path of radius r, we have [tex]\varepsilon = \rho 2 \pi r J [/tex], in which [itex]\rho[/itex] is the resistivity. So [tex]J = \frac{r \dot B}{2\rho} [/tex]. Say if this isn't clear. 


#3
Aug2111, 11:01 AM

P: 4

This Helps a Ton and it's very clear! thank you.
I'm still a little confused about how to derive the induced magnetic field. The current density that you've given is now a function of radius. Does this mean we can use Ampere's law to model the Magnetic Field as a function of radius? [tex]J=\frac{r \dot B}{2\rho}[/tex] [tex]\oint B \cdot \partial L = \mu_0 I_{enc}[/tex] [tex]I_{enc} = \int J \cdot \partial A = J \pi r^2 [/tex] [tex]\int \int_0^{2\pi} Br \cdot \partial \theta \cdot \partial r = \pi \mu_0 \int Jr^2 \cdot \partial r[/tex] Or is this just bad calculus? 


#4
Aug2111, 11:20 AM

PF Gold
P: 963

Eddy Current Calculations
Ampère's law



#5
Aug2111, 12:20 PM

PF Gold
P: 963

There are only a few cases where Ampère's law can be used to find B. These are cases in which there's enough symmetry for B to be effectively the same all along a particular integration path. Looking at your post, it seems that you don't have a particular path in mind. And the bad news is that for these circular eddy currents, as for a single circular loop of wire, there is no path along which B is constant. Ampère's law, beautiful though it is, can't help.
In fact the general problem of finding B at points in the vicinity of the disc, as for a circular loop, is very difficult. The only easy cases are for points on the axis of the disc, and, simplest of all, at its centre. To find the field at the centre of the disc (of thickness b, say), think if it as made up of annuli of crosssectional area b dr. Then the current in an annulus is Jbdr. But from the BiotSavart law we know that the field at the centre of a ring carrying current I is [itex]\mu[/itex]_{0}I/2r. Using the J from my previous post, and integrating for a disc of radius a, I find [tex]B_{ind} = \frac{\mu_0 ba \dot B}{4\rho}[/tex] A neat result, I thought. But I'm prone to slips... 


#6
Aug2111, 11:12 PM

P: 4

this makes a good amount of sense. As for the overall behavior of the areas outside the center, though, is it safe to say that the induced B field is greatest at the center, falls to zero where the Magnetic Field Source ends, and then goes a bit negative before returning to zero?
(Positive being the direction dictated by Lenz's Law) 


#7
Aug2211, 02:49 AM

PF Gold
P: 963

Hadn't thought of the field ending; was thinking of the whole disc being subjected to a uniform normal field. But if the field 'covered' only an inner part, D_{B}, of the disc, I wouldn't expect the induced field to drop to zero at exactly the edge of D_{B}. Disc annuli outside D_{B} will still have emfs induced in them, because changing flux will still be linked with them. The emf will be
[tex]\varepsilon = \pi r_B^2 \dot B[/tex] in which r_{B} is the radius of D_{B}. But beyond D_{B}, the current density will fall because of the increasing value of 2[itex]\pi[/itex]r, and at some point, I think that B will indeed drop to zero and then reverse, as would happen for an ordinary currentcarrying loop. Thanks for such an interesting question. 


#8
Aug2211, 08:39 AM

P: 4

You've been a tremendous help. Thank you!



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