Can varying magnetic field cause eddy current in aluminum?

[yungman]

As the tittle, can varying magnetic field ( well EM) cause eddy current in non magnetic conductors like aluminum? If so, why?

Thanks

Alan

 

[Emilyjoint]

Yes ! the important thing is that aluminium is a conductor so changing magnetic fields will induce an emf and induce a current.
We have a brilliant demo of an aluminium plate that swings between the poles of a very strong horseshoe magnet...it stops very quickly because of the eddy currents.
An identical aluminium plate with slots cut in it swings for much longer because the slots reduce the effect of eddy currents.
This demo was used to explain to us why transformer cores are laminated steel sheets

 

[OmCheeto]

As the tittle, can varying magnetic field ( well EM) cause eddy current in non magnetic conductors like aluminum? If so, why?

Thanks

Alan

We had a foray into this topic in the "You Tube Classics" thread. It went on for two pages. Ivan's MRI videos were the most impressive, IMHO.

 

[Emilyjoint]

great videos...he should also try it with an aluminium block with slots cut in it.
proof beyond doubt

 

[yungman]

Thanks all of you.

 

[yungman]

Is eddy current the same current of the surface current density of the magnetic boundary condition according to Maxwell's equation:

[tex] \nabla \times \vec B = \mu \vec J +\mu\frac{\partial \vec D}{\partial t}[/tex]

 

[Bob S]

Eddy currents arise from [itex]\nabla \mathrm{x} \overrightarrow{E}= -\mu\mu_o \frac{\partial \overrightarrow{H} }{\partial t}[/itex]
A varying magnetic field can induce currents in any electrical conductor, both magnetic and non-magnetic.

 

[Creator]

Eddy currents arise from [itex]\nabla \mathrm{x} \overrightarrow{E}= -\mu\mu_o \frac{\partial \overrightarrow{H} }{\partial t}[/itex]
A varying magnetic field can induce currents in any electrical conductor, both magnetic and non-magnetic.

Is there an equation that gives me the depth of the current induced ?
...

..

 

[Bob S]

Is there an equation that gives me the depth of the current induced ?

For flat surfaces, it is straight exponential
[tex] i\left(z \right)=i\left(0 \right)e^{-\left(\frac{\mu\omega}{2\rho} \right)^{1/2}z} [/tex]
For curved surfaces, it is given by the ratio of two modified Bessel functions of order 0 with complex arguments. See chapter 10 in Smythe
Static and Dynamic Electricity (3rd Edition)

 

[yungman]

For flat surfaces, it is straight exponential
[tex] i\left(z \right)=i\left(0 \right)e^{-\left(\frac{\mu\omega}{2\rho} \right)^{1/2}z} [/tex]
For curved surfaces, it is given by the ratio of two modified Bessel functions of order 0 with complex arguments. See chapter 10 in Smythe
Static and Dynamic Electricity (3rd Edition)

Is ρ supposed to be σ ?

 

[Creator]

For flat surfaces, it is straight exponential
i\left(z \right)=i\left(0 \right)e^{-\left(\frac{\mu\omega}{2\rho} \right)^{1/2}z}

Thanks Bob, but I can't read the tiny print ...looks like inverse exp. sqrt power of: {(magnetic permiability of material x frequency) / 2 x current density} x z ??
Correct?

I want a depth, say, where I get 90 or 95% of the total current...
OK; let's see if I can do this: (without latex of course, forgive me...)
Thus ...rearranging and taking the nat. log...

ln[I(z)/I(0)] = ln[e^-(...)] = -sqrt(uw/2p)(z)...ln = nat log; I = current.

Thus...solving for z...
z = ln[I(z)/I(0)] / -sqrt(uw/2p)...where u = mu; w = omega, and p = rho.

Correct?
Now are we talking current density at z depth?
so how can I estimate rho at z depth, independently? The other variables are determinable.

Thanks for the help.

Creator

 

[Bob S]

Is ρ supposed to be σ ?

The skin depth is [itex] \delta=\left(\frac{2\rho}{\omega\mu} \right)^{\frac{1}{2}} [/itex] where ρ is resistivity.

 

[Bob S]

Thanks Bob, but I can't read it ...looks like inverse exp. sqrt power of: {(magnetic perm
Thus ...
ln[I(z)/I(0)] = ln[e^-(...)] = -sqrt(uw/2p)(z)...ln = nat log; I = current.

Thus...
z = ln[I(z)/I(0)] / -sqrt(uw/2p)...where u = mu; w = omega, and p = rho.

Correct?

[tex] Ln\left(\frac{i\left(z \right)}{i\left(0 \right)} \right)=-\left(\frac{\mu\omega}{2\rho} \right)^{\frac{1}{2}}z [/tex]

 

[yungman]

The skin depth is [itex] \delta=\left(\frac{2\rho}{\omega\mu} \right)^{\frac{1}{2}} [/itex] where ρ is resistivity.

I see, that makes the whole world of sense! I was going in circle as I use ρ as charge! And even when I use σ instead it still don't work.

So Eddy current is a totally different mechanism from free surface current according to the magnetic boundary condition?

Thanks

 

[Bob S]

So Eddy current is a totally different mechanism from free surface current according to the magnetic boundary condition?

See http://www.rfcafe.com/references/electrical/skin-depth.htm

 

[yungman]

See http://www.rfcafe.com/references/electrical/skin-depth.htm

I know about the skin effect, all the formulas and how it works.

I just want to verify that eddy current is from a totally different mechanism from boundary condition.

I know J in the boundary condition is free current density. Eddy current is also free current where electrons are actually moving?

I am still digging up notes on the current in high frequency. From talking to you long time ago, my understanding AC signals are actually EM wave propagation, as electrons don't move in any speed like the signal. All sort of connections like pcb trace, coax or even single wires, are actually guided structures that EM wave propagates in. Current and voltage are just the consequence of the boundary condition of the EM wave.

I want to understand the different types of current and how it relate to EM wave propagation. This is the whole thing I am confuse as you posted the Ultra Cad article that derive skin effect totally by Lentz Law using current where books I have using EM wave with attenuation constant [itex]\alpha\;[/itex].

 

[Creator]

[tex] Ln\left(\frac{i\left(z \right)}{i\left(0 \right)} \right)=-\left(\frac{\mu\omega}{2\rho} \right)^{\frac{1}{2}}z [/tex]

OK; good, thanks, Bob; that's what I wrote in non-latex; except that I carried it further and solved for z.

The reason for my question was that I noticed (a few years ago) that dropping a magnet through a 'thick' walled Aluminum pipe had a longer transit time (greater 'drag') than that of a thin walled Al pipe, and assumed the field penetration (& current depth) must be significant,...

I can assume then that for a permanent magnet a single drop should count as a very low frequency pass of the field and that low freq. accounts for the high depth...requiring a thicker wall to develop a fuller Lenz response to the applied field. ??

My further pragmatic query ...
was thinking that instead of using a thick block of aluminum (in the video of the slowly dropping aluminum at the entrance of a MRI machine) could I wrap foil around a brick for the same delayed response ? http://www.youtube.com/watch?v=liDjr439-fY&feature=player_embedded
Accordingly, since MRI is high frequency, there should still be enough surface current to develop a full Lenz response to the field.

Just speculating.

Creator

 

[Creator]

I see, that makes the whole world of sense! I was going in circle as I use ρ as charge!
Thanks

Yes, thanks, that makes sense; I assumed it was current density.

Creator