I am asking for the electric field E inside the conductor - to be more precise, the tangential component that contribute to the integral of E.dl .
I can tell you that in my case it would be in the mV/m range.
What value do you get in your case?
What are you afraid of? If your claims are sound, you would instantly know the answer for the questions we've posed. Even for the one Sredni found amusing. If you think that that question is out of proportion, think again. That could very well be traces on a PCB or wires on any real installation. If your Kirchhoff only works for perfectly rectangular or round loops, with known resistors within tolerances, it is a useless theory.
I am asking for the electric field E inside the conductor - to be more precise, the tangential component that contribute to the integral of E.dl .
I can tell you that in my case it would be in the mV/m range.
What value do you get in your case?
OK I think I understand what part you do not understand so I will try to find a bit of a different example
Can you or can you not compute this electric field?
Our circuit is stationary, nothing is moving.
What is the electric field inside your conductor?
You are not telling because you do not know how to compute it, or because you cannot justify a 0.25 V (yeah, minus 0.002V) voltage difference at its ends?
Because that's the point.
Yes that generator example is spot on.
The last two pages of arguing in this thread seam to be all because of different definitions of voltage.
The correct formal definition used by Dr Lewin: Voltage is the integral of all forces on charges along a given path
The common definition used in circuit analysis: Voltage is the difference in charge density between two points
[...]
So I suggest that you make it clear in further discussion what kind of voltage you are talking about.
I sort of feel like I deal with trolls
Me too. What a coincidence
You agreed with ogden, who expressed electric field in volts. You should have read better, before saying that his reply was more to the point.
What is the field inside the copper conductor and how do you justify - with formulas - that there is a (0.25-0.002) volts difference across it?
Assume standard conductivity for copper, say 5.8 10^7 mhos per meter
QuoteWhat is the field inside the copper conductor and how do you justify - with formulas - that there is a (0.25-0.002) volts difference across it?BTW wire fragment resistance is 0.001 Ohms so it is (0.25-0.001) Volts, not (0.25-0.002).
Inside our conductor there are two E fields: E.induced and E.coloumb. Total electric field E = E.coloumb + E.induced. Coulomb electric field in the wire is opposite the direction of the induced electric field
- that's the justification of voltage difference.
Potential difference (integral of E.dl) at the ends of that copper conductor you calculate using same formula as for 0.25V chemical battery that has 0.001 Ohm internal resistance and 1A current load. Answer is mentioned already here in this thread.
QuoteAssume standard conductivity for copper, say 5.8 10^7 mhos per meterDo not introduce new unnecessary conditions.
Yes that generator example is spot on.
The last two pages of arguing in this thread seam to be all because of different definitions of voltage.
The correct formal definition used by Dr Lewin: Voltage is the integral of all forces on charges along a given path
The common definition used in circuit analysis: Voltage is the difference in charge density between two points
[...]
So I suggest that you make it clear in further discussion what kind of voltage you are talking about.
Just out of curiosity...
Have you ever heard the term "dimensional analysis"?
(Also, I guess you won't tell us what the field inside the conductor is... Kirchhoffians are allergic to electric fields, it appears)
I am not sure I can sort this mess.
I'd like to kindly ask the Kirchhoff experts what tools do Kirchhoff rules give me to calculate the "right" voltage of the loop below. Except for the irregular perimeter, all conditions are the same as for my rectangular loop above.
I need to know the exact points where to connect the voltmeter and its precise location. Any reply will be appreciated.
The field inside the copper conductor is the sum of E.coloumb with E.induced, you said (and I agree). How do you think the copper can tell which is which?
The copper sees the net, resulting, field. (and THIS is the point)
What is this sum?
QuotePotential difference (integral of E.dl) at the ends of that copper conductor you calculate using same formula as for 0.25V chemical battery that has 0.001 Ohm internal resistance and 1A current load. Answer is mentioned already here in this thread.Please, indulge me. Give me the number in V/m (volts per meter).
Oh and also i solved your curvy wire example.
Here is where you have to put the voltmeter for it to read 0V
Solved using Solidworks Fusion 360 due to it having a convenient area measurement tool.
I sort of feel like I deal with trolls
Me too. What a coincidence
More like my probing technique is bigger in 'theory' than yours
Far to much theoretical waffle and going around in circles or in some cases non circular circuits. Given this is a forum for EE's not theoretical Physicists with very average probing technique, get on design the experiment, test it and prove or disprove it and then get it repeated by others.
Oh and also i solved your curvy wire example.
Here is where you have to put the voltmeter for it to read 0V
Solved using Solidworks Fusion 360 due to it having a convenient area measurement tool.
Wonderful. Much appreciated. Now we need to measure the voltage indicated by the calculations so as to confirm that they are right. But, alas, in the real circuit there is a physical obstruction, that in no way affects the magnetic field. This obstruction goes all the way with the field while it is perpendicular to the surface.
How can we measure measure that voltage? Thanks in advance for your kind reply.
Wonderful. Much appreciated. Now we need to measure the voltage indicated by the calculations so as to confirm that they are right. But, alas, in the real circuit there is a physical obstruction, that in no way affects the magnetic field. This obstruction goes all the way with the field while it is perpendicular to the surface.
How can we measure measure that voltage? Thanks in advance for your kind reply.
Does Ohm's Law still work? I've got this LED I have to turn on and I need to know which side of the LED to put the resistor...
The field inside the copper conductor is the sum of E.coloumb with E.induced, you said (and I agree). How do you think the copper can tell which is which?What are you smoking?
Integral of E.dl where E = E.coloumb + E.induced. EMF of wire segment is EMF.total/4 (because segment is 1/4 of loop) = 1/4V and voltage drop due to current is 0.001Ohm*1A = 0.001V. So, this sum is 0.25+(-0.001) Volts. What's the point to ask question so many times?
Please, indulge me. Give me the number in V/m (volts per meter).Quote from: ogdenWith same success you can ask me weight of the wire used in experiment. Before asking V/m, make sure you give enough data to calculate such
So, let me ask you again, because this is important:
What is the field inside the copper conductor and how do you justify - with formulas - that there is a (0.25-0.002) volts difference across it?
Assume standard conductivity for copper, say 5.8 10^7 mhos per meter and a copper section of 1 mm in diameter (or any real world value you can attribute to a circuit similar to those shown by Lewin, Mehdi or Mabilde - it's about 10 cm diameter loop, suppose half of it is allocated by the big resistors, but it's not important).
I am asking for the electric field E inside the conductor - to be more precise, the tangential component that contribute to the integral of E.dl .
I can tell you that in my case it would be in the mV/m range.
What value do you get in your case?
Quote from: ogdenIntegral of E.dl where E = E.coloumb + E.induced. EMF of wire segment is EMF.total/4 (because segment is 1/4 of loop) = 1/4V and voltage drop due to current is 0.001Ohm*1A = 0.001V. So, this sum is 0.25+(-0.001) Volts. What's the point to ask question so many times?Because you guys keep telling me the voltage. I want to know the electric field.
Quote from: SredniAssume standard conductivity for copper, say 5.8 10^7 mhos per meter and a copper section of 1 mm in diameter (or any real world value you can attribute to a circuit similar to those shown by Lewin, Mehdi or Mabilde - it's about 10 cm diameter loop, suppose half of it is allocated by the big resistors, but it's not important).
I am asking for the electric field E inside the conductor - to be more precise, the tangential component that contribute to the integral of E.dl .
I can tell you that in my case it would be in the mV/m range.
What value do you get in your case?
And still, no answer.
Because you guys keep telling me the voltage. I want to know the electric field.I said (E = E.coloumb + E.induced). Are you satisfied now?
You can either provide solution yourself and tell what you want to say with it or stick that tangential component where it hurts. I do not see the point of solving your tasks. "Trail of the troll" was at least funny.
Does Ohm's Law still work? I've got this LED I have to turn on and I need to know which side of the LED to put the resistor...
Does Ohm's Law still work? I've got this LED I have to turn on and I need to know which side of the LED to put the resistor...
This question is actually a lot more relevant than it appears here (due to its ironic nature),
Ok, you have no idea on how to compute the electric field inside a conductor. It's not a crime. Maybe all that facepalming has interfered with your mental processes but, fine.
Any other Kirchhoffian who believes that the 'real' voltage across the 0.9 ohm resistor is 0.65 V and the real voltage across one of the two arcs of copper is 0.25-0.001 V care to tell us what the electric field is inside said copper?