Blah, Inductance question

[TecHNooB]

I'm trying to solve for inductance.

If I have two parallel plates with two different materials such that each plate sandwiches half of both materials, how do the materials affect the current distribution? I know the B field normal to the interface is supposed to be the same, but Ampere's law seems to tell me the H fields normal to the interface are the same as well (unless the current distributions are non-uniform).

Or, see #3 http://cobweb.ecn.purdue.edu/~djiao/ee311/handout/EE311%20Homework 10.pdf

I'm guessing if I solve for H in freespace using Ampere's Law, I could get to a solution, but I don't even understand what's happening.

I know the H field between the plates is equal to the current density divided by the width. But if the current density is uniform along the plates, H would be the same along the interface, which is incorrect. The only thing that would make them different would be having a different current density. An explanation of what physically is happening would help.

 

[FelixDeCat]

An explanation of what physically is happening would help.

No.

 

[TecHNooB]

No.

About FelixDeKat

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[FelixDeCat]

About FelixDeKat

Biography
Helping Hand to those in need.

....and who offer money....duh!


edit: I couldnt help you anyway as I havent studied EE.

 

[Leros]

I could have given you a really good explanation a year ago when I was taking this class.

 

[Ken g6]

It's been a long time since I did any EE, but that looks like two parallel-plate capacitors, with differing dielectrics, wired in parallel, to me.

 

[Leros]

It's been a long time since I did any EE, but that looks like two parallel-plate capacitors, with differing dielectrics, wired in parallel, to me.

It is pretty simple to figure out what the induced current would be. You just find the magnetic field produced by the first plate. Then figure out what kind of current that would induce on the second plate.

The material interface makes it a little more difficult. I don't remember exactly how that works.

I'm not sure about this part:
Doesn't the material in between essentially bend the magnetic field lines? Would this not effectively make the plates further apart so you would get lower mutual inductance? Assuming the plates are infinitely long, it would be the same as if there was no material in between but with the two plates further apart.

 

[Minjin]

If you can figure out the dielectric constant, can't you find the unit capacitance and then convert it to unit inductance?

edit: actually, I think you need to use telegrapher's equations
http://farside.ph.utexas.edu/teaching/em/lectures/node86.html
Scroll down to 1003 to see how it is done with a cylindrical, non filled transmission line.

edit2: Here you go. Plug and chug:
http://www.ivorcatt.com/6_2.htm
http://hibp.ecse.rpi.edu/~connor/education/Fields/TwoWire-ParallelPlateLines.PDF
Do that for each side and add them, remembering that inductors in parallel are like resistors in parallel.

 

[TecHNooB]

Someone knows the answer to this!

 

[Born2bwire]

You want the current distribution? Good luck. I would recommend using a numerical method like method of moments of finite element method.

If you want the total current or say capacitance, that's different. All it is is two capacitors in parallel. The upper plates are shorted together, with a simple capacitor model that is no different than just having a wire shot the pins of two capacitors together. Now just figure out what their capacitances are from the simple capacitor equations and add them up according to parallel capacitor rules and you're done.

 

[TecHNooB]

You want the current distribution? Good luck. I would recommend using a numerical method like method of moments of finite element method.

If you want the total current or say capacitance, that's different. All it is is two capacitors in parallel. The upper plates are shorted together, with a simple capacitor model that is no different than just having a wire shot the pins of two capacitors together. Now just figure out what their capacitances are from the simple capacitor equations and add them up according to parallel capacitor rules and you're done.

I'm looking for inductance :O

Btw, I fixed that code I was bitching about. All I did was recode the whole thing and it magically worked. Now I feel twice as dumb.

 

[Matthiasa]

I have a hard time imagining that you don't have a book for that class...
I suggest opening it.

There's probably even an example of a question similar in it.

That said I would have probably been able to tell you last semester, and will be able to again next semester. Or I could right now but since your to lazy to use google so am I.

 

[esun]

I had written up a post with references and such but the forums died when I clicked post and I don't feel like writing it all up again.

Anyway, using Google (try "parallel plate" inductance) you should be able to find results indicating that in general for a parallel plate structure with a uniform material you have:

L/l = mu_0 * d / w

So in your case you have

L1/l = 2 * mu_0 * d / w
L2/l = 200 * mu_0 * d / w

L/l = L1/l || L2/l = (200/101) * mu_0 * d / w

 

[TecHNooB]

meh.

 

[SSSnail]

ATOT is not your personal, eh... tutor.

Do your own homework.

 

[Born2bwire]

I'm looking for inductance :O

Meh, same diff. As long as the parallel plate thing is homogeneous in the direction between the plates, then it can be split up into multiple parallel plates that are circuit elements placed in parallel since most simplifications of their behavior ignore fringing effects.

Btw, I fixed that code I was bitching about. All I did was recode the whole thing and it magically worked. Now I feel twice as dumb.

Righto then. Probably just a coding but then. Learning to use a debugger to step through the code and follow the process for debugging purposes is a good skill to learn for stuff like this.