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Several questions in electromagnetics

by CheyenneXia
Tags: electromagnetics
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yungman
#19
Dec18-12, 05:17 PM
P: 3,898
Ha ha, you guys!!! You ready to spend your whole Christmas on this? Go post your disagreement in the Classical Physics section. You are going to have a very Merry Christmas. I can't even finish watch this video, it is very similar to the one I spent my Christmas, it's the same professor Levin.

I serve my time, read the thread I show first. I went through similar thoughts, hell, I did experiments, I drew equivalent circuits, I explained voltage source from magnetic induction. I even show how you can make the reading change on the scope just by moving the ground lead of the scope probe. Go to post #11 here, watch the video, go to post #224 of that thread where I showed the result of the experiment and my argument.

Read my work to save you a lot of time first and then continue the argument. Don't argue here, go to the classical physics forum where they have people with strong electrodynamics knowledge. I am out, I hope you win.

Late editing: I have every intention to have a Merry Christmas.......Not on the professor Levin. You guys can carry the torch!!!
yungman
#20
Dec22-12, 02:40 PM
P: 3,898
Quote Quote by marcusl View Post
This post contains numerous errors. Electrons do not jump from atom to atom, they are in the metallic crystal's conduction band where they act as an "electron gas". The electron drift velocity is usually directly proportional to electric field. Better conductors do not have "slower" electrons, I don't even know what this means. Better conductors have fewer scattering defects, hence a longer mean free path between collisions, which results in higher net drift velocity.

There are also misconceptions in other posts in this thread. Caveat emptor, beware...
Please reply with more specific error that I made. It's been a week and I want to hear the specifics. Since you pretty much said "buyers beware", it's important to know where did I get it wrong so I can learn, or if I don't agree, I can present the counter argument. I already response to the specific point that you mentioned.

You are the Science Adviser, I value your input.
sophiecentaur
#21
Dec23-12, 06:05 PM
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Ohm's Law is not a 'Law', as such. It is really only a description of how metals behave at a constant temperature - and that isn't something you can 'challenge', exactly. It may be that people misinterpret it but that's another matter. Resistivity of a metal hardly changes with frequency but skin effect will limit the effective CSA of the metal that is conducting.
yungman
#22
Dec23-12, 07:05 PM
P: 3,898
Actually what I presented in the thread I posted is not about resistivity, it is about the EM nature of signal making the current follow a path different than the conventional Ohm's Law prediction that most current takes on the path of least resistance at the given voltage.

Quote Quote by sophiecentaur View Post
Ohm's Law is not a 'Law', as such. It is really only a description of how metals behave at a constant temperature - and that isn't something you can 'challenge', exactly. It may be that people misinterpret it but that's another matter. Resistivity of a metal hardly changes with frequency but skin effect will limit the effective CSA of the metal that is conducting.
That's the whole point, people regard it as law of the land. Ohm's Law work in most case, people just have to understand there are limitation of using. Just like professor Levin's demo and argument. The one month of debating make me realize that a lot of the theorems that EE hold dearly do have some holes and it's important to realize it. They work in most case, it is mostly safe to use them, just have to keep in mind that there is limitation. It is by no means I claimed those laws and theorems are wrong. I think that's the reason people use Ohm's Law in the point form where you use V=IR at a given point.

Regarding to the KVL, If you look at the long thread of Levin, there is a collision between KVL and Electromagnetics. It cannot no be both true. If you read my experiment, there is no way that I can think of to measure the voltage without being affected by the magnetic induction.

If any of the EE here can definitively proof Levin is wrong, we might be able to poke a hole in Maxwell's equation.............But I won't hold my breath on this!!! But this will be big if we can proof Levin wrong. I still have everything for the original experiment, if someone comes up with an idea, I can still play with it. But do read my posts on the thread, I did really cover a lot on the arguments.
yungman
#23
Dec23-12, 08:00 PM
P: 3,898
Hey guys, if you really think Levin is wrong, we can go in again. I was alone, out gunned, out theory. If there are a few of us, it might be different. To the best of my understanding, Levin's argument is E is no longer conservative under varying B field, whereby KVL don't hold. And this is according to Maxwell's equation:

[tex]\nabla X \vec E=-\frac {\partial \vec B}{\partial t}[/tex]

I tried so hard to say because of [itex]EMF=-\frac {\partial \Phi}{\partial t}[/itex], you have to include this as a distributed voltage source. If you take into consideration of the distributed voltage source, KVL holds.

That will be something to win this argument......That Maxwell's equation need to consider the induced voltage source and the definition of conservative field has to be modified.
jim hardy
#24
Dec23-12, 08:17 PM
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yungman - i don't do well at higher math.

But i did think Levin was mis-applying KVL when he ignored some induction.

Where's that thread ? - i'll see if i can get my alleged brain around it

but be advised - in math skills i dont even come up to your knees.

it is about the EM nature of signal making the current follow a path different than the conventional Ohm's Law prediction that most current takes on the path of least resistance at the given voltage.
isn't that Lenz's law - it'll try to make the current oppose the changing flux?
And what do you mean "at the given voltage"?


old jim
yungman
#25
Dec23-12, 08:30 PM
P: 3,898
Ha ha. I know curiosity will eventually get one of you.....hopefully more of you!!!

http://www.physicsforums.com/showthr...3575&highlight

And at post #224 on page 14, I did my experiment and draw the equivalent circuit.

http://www.physicsforums.com/showthr...453575&page=14

Then in post #226 and #242, I explained the path where the ground lead of the scope probe got voltage induced and thereby have different reading as I swing the ground lead of the probe.

http://www.physicsforums.com/showthr...453575&page=16
yungman
#26
Dec23-12, 08:36 PM
P: 3,898
Quote Quote by jim hardy View Post
yungman - i don't do well at higher math.

But i did think Levin was mis-applying KVL when he ignored some induction.

Where's that thread ? - i'll see if i can get my alleged brain around it

but be advised - in math skills i dont even come up to your knees.


isn't that Lenz's law - it'll try to make the current oppose the changing flux?
And what do you mean "at the given voltage"?


old jim
This is related to this post:

http://www.physicsforums.com/showthread.php?t=659307

The return current path follow right under the trace no matter how you snake the trace around.
marcusl
#27
Dec24-12, 06:39 PM
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I will expand on the technical issues that I alluded to before, then address some additional issues.
1) The reason that the integral of [tex] \oint_c \vec E\cdot d\vec l [/tex] is zero is not that the area of the path goes to zero. It is zero because the line integral of E parallel to and on the outside of the boundary separating material 1 from 2 becomes equal and opposite to that along the inside of the boundary, as the separation between the two paths approaches zero. Thus the two contributions cancel.

2) In the last integral in post #4, J is current density, not surface current as you state; it is allowed to cross the boundary, and it does not have the dimensions of surface current density. Your discussion about EM waves seems to have little direct relevance to the question, furthermore.
Evaluating the integral gives [tex] \vec{J}\cdot\vec{m} \Delta l = \vec{K}\cdot\vec{m}, [/tex] where K is surface current density and m is the normal to the surface that is bounded by the integration curve. This equation gives rise to the boundary condition on H_tangential that the original poster asked about, namely [tex]\vec{n}\times(\vec{H_2} - \vec{H_1})=\vec{K} .[/tex]
3) Conduction does not occur by electrons jumping from one atom to the next.

4) Current (or conduction) is always spoken of as being driven ultimately by potential energy differences. Hence current in a charge-neutral wire is the gradient of a potential, leading to Ohm’s law being written [tex]\vec{J}=\sigma \vec{E}[/tex] rather than the other way around. Furthermore, we always measure "I-V" curves where V is the independent variable. As a result, drift velocity increases in copper samples of greater purity and decreases in dirtier samples.

5) Post 6 gives a rather roundabout argument that a time-invariant form of Kirchoff’s laws cannot be applied to a time varying circuit. This, of course, merely restates the condition that gave rise to the original question. The answer to the question is to form Kirchoff’s laws to include the time-varying terms in Maxwell’s equations. Including dB/dt introduces into the circuit some emf’s due to self and mutual induction, as the_emi_guy has already pointed out. Including dD/dt allows the treatment of displacement currents in capacitors.

6) “Electromagnetic is the most difficult subject in EE by a mile.” There are topics (stability of a system, information capacity of a multipath scattering communication channel, maximum entropy spectral analysis, error correction coding, etc.) that others could claim are much more difficult than the problems we are dealing with here.

Now some non-technical comments.

yungman, I have asked you in previous threads not to post erroneous answers in areas where you are not an expert. You state here, instead, that you post what you like and want other people to correct you, and that you like to argue with the corrections. This is counterproductive from many standpoints. I'll give three:
1. It is tedious for those (myself in this case, others in previous threads of yours) whom you expect to a) wade through your voluminous posts, b) try to correct your errors and misconceptions, and c) then argue with you.
2. You spread confusion and misinformation. In this case, you confused the original poster CheyenneXia, at the very least.
3. You undermine your own credibility, and also make it less likely that experts will be interested in engaging with you.

I believe restraint is the better approach.
berkeman
#28
Dec24-12, 06:47 PM
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Your concerns are valid marcusl. Sorry yungman, please take his feedback to heart.
marcusl
#29
Dec24-12, 06:48 PM
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Also I'll point out that you have completely hijacked this thread so it is focused on a topic taken from your own, different thread.
yungman
#30
Dec25-12, 12:47 AM
P: 3,898
Quote Quote by marcusl View Post
I will expand on the technical issues that I alluded to before, then address some additional issues.
1) The reason that the integral of [tex] \oint_c \vec E\cdot d\vec l [/tex] is zero is not that the area of the path goes to zero. It is zero because the line integral of E parallel to and on the outside of the boundary separating material 1 from 2 becomes equal and opposite to that along the inside of the boundary, as the separation between the two paths approaches zero. Thus the two contributions cancel.

It is the integration around a closed path is ZERO in a conservative field.

http://en.wikipedia.org/wiki/Conservative_field.

Read the definition

And under magnetic induction, the E field is no longer conservative. Watch the Professor Lavine's discussion on this. This is written in page 309 of Field and Wave Electromagnetics by David K Cheng.

http://www.amazon.com/Field-Wave-Ele...ectromagnetics


2) In the last integral in post #4, J is current density, not surface current as you state; it is allowed to cross the boundary, and it does not have the dimensions of surface current density. Your discussion about EM waves seems to have little direct relevance to the question, furthermore.
Evaluating the integral gives [tex] \vec{J}\cdot\vec{m} \Delta l = \vec{K}\cdot\vec{m}, [/tex] where K is surface current density and m is the normal to the surface that is bounded by the integration curve. This equation gives rise to the boundary condition on H_tangential that the original poster asked about, namely [tex]\vec{n}\times(\vec{H_2} - \vec{H_1})=\vec{K} .[/tex]

Which part you don't get that the OP was asking about the tangential H field and current in question 3? That EM wave produce current by boundary condition?
Read Field and Wave Electromagnetics by David K Cheng page 262 that clearly state the tangential H field create surface current. AND page 331 to 332 CLEARLY explains FREE current density created at the boundary between dielectric and good conductor in equation 7-70.

[tex] \hat A_{n2}\;\times\;(\vec H_1-\vec H_2)= \vec J_s [/tex]
Then in page 430 and 431. It explain the boundary condition creates SURFACE CURRENT on the conductor plates of the parallel plate tx line.
[tex]\hat a_y \;\times\; \vec H=\vec J_{su}\;[/tex]

See the attached image of p430 below. If you don't think the boundary condition produce surface current, why are you saying K in your equation is surface current? Your equation is almost exactly the same as I quoted from the book. Are you confused?

AND in Engineering Electromagnetic page 182:

http://www.amazon.com/Electromagneti...agnetics+ulaby

[tex]\int_s(\nabla \times \vec H)\cdot d\vec s =\int_s \vec J\cdot d\vec s \;+\; \int_s\frac{\partial \vec D}{\partial t}\cdot d\vec s = I_c+I_D[/tex]
Where Ic is CONDUCTIVE current and Id is DISPLACEMENT current.

Then refer to Introduction to electrodynamics by David Griffiths P332. It explains discontinued tangential H across the boundary produces FREE surface current density.

http://www.amazon.com/Introduction-E...lectrodynamics

This one, you can actually read the content of the book. Go to page 332.

This is CLEARLY STATED in the book. In the boundary condition, it is the SURFACE CONDUCTION CURRENT.


3) Conduction does not occur by electrons jumping from one atom to the next.
As I said, you can look at it as it's a cloud of conductive band or the atom share the valency electrons, they move around, and they can drop back to the atom depend on the energy. That's is electron move from one to the other in my book.

4) Current (or conduction) is always spoken of as being driven ultimately by potential energy differences. Hence current in a charge-neutral wire is the gradient of a potential, leading to Ohm’s law being written [tex]\vec{J}=\sigma \vec{E}[/tex] rather than the other way around. Furthermore, we always measure "I-V" curves where V is the independent variable. As a result, drift velocity increases in copper samples of greater purity and decreases in dirtier samples.
I response to this already in post #12.
5) Post 6 gives a rather roundabout argument that a time-invariant form of Kirchoff’s laws cannot be applied to a time varying circuit. This, of course, merely restates the condition that gave rise to the original question. The answer to the question is to form Kirchoff’s laws to include the time-varying terms in Maxwell’s equations. Including dB/dt introduces into the circuit some emf’s due to self and mutual induction, as the_emi_guy has already pointed out. Including dD/dt allows the treatment of displacement currents in capacitors.
You better argue with Professor Lavine of MIT. I can see his arguement that the electric field under varying magnetic field is not conservative.
6) “Electromagnetic is the most difficult subject in EE by a mile.” There are topics (stability of a system, information capacity of a multipath scattering communication channel, maximum entropy spectral analysis, error correction coding, etc.) that others could claim are much more difficult than the problems we are dealing with here.
That's is an opinion, You have yours and I have mine.
Now some non-technical comments.

yungman, I have asked you in previous threads not to post erroneous answers in areas where you are not an expert. You state here, instead, that you post what you like and want other people to correct you, and that you like to argue with the corrections. This is counterproductive from many standpoints. I'll give three:
1. It is tedious for those (myself in this case, others in previous threads of yours) whom you expect to a) wade through your voluminous posts, b) try to correct your errors and misconceptions, and c) then argue with you.
2. You spread confusion and misinformation. In this case, you confused the original poster CheyenneXia, at the very least.
3. You undermine your own credibility, and also make it less likely that experts will be interested in engaging with you.

I believe restraint is the better approach.
You make the accusation, it's up to you to make the correction. You are the adviser and you make the statement. So it is your responsibility to correct any inaccuracy here.

Condescending comments have no place in this forum. If you don't agree, state the reason.

Attached is an image of page 430. It is blur as I have to shrink the size to 300K. The equation in question is 9-7b for the upper plate. Ignore all my hand written notes, just read the text of the book. The second image is my drawing according to the figure from the book, I added color and more detail. The figure at the bottom of the text page is Fig.9-3 if you care to read the text.

[tex] -\hat a_y\times \vec H=\vec J_{su}[/tex]
Attached Thumbnails
TX_L.png   EM propagation L.jpg  
yungman
#31
Dec25-12, 05:13 AM
P: 3,898
Quote Quote by marcusl View Post
Also I'll point out that you have completely hijacked this thread so it is focused on a topic taken from your own, different thread.
You read the original question? Question 1 is on KVL under magnetic field condition. Question 3 about tangential boundary condition which talk about free charge and free current. Where is it off the topics when I talk about MIT professor Levine. He made a video demonstrated KVL don't hold under varying magnetic field. AND current in transmission line are consequence of EXACTLY the tangential boundary condition between good conductor and dielectric. You watch the video I posted before you speak?
micromass
#32
Dec25-12, 11:17 AM
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Locked pending moderation.
berkeman
#33
Dec26-12, 04:48 PM
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This thread will remain closed. I've asked yungman to take the debate to PMs if he wants to continue it.


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