Ohm's law, why current induces electric field and cause effect direction

[cabraham]

It has become clear to me that you are unable to understand the essence, neither the statement of the problem.

FACTS:

1. A current can be generated without electric field
2. A current crossing a resistor (ohmic resistor) will induce a voltage, i.e., an E-field

EXPERIMENT:

1. Take a current generated without electric field (possible by fact 1) and connect it to a resistor (ohmic resistor)
2. Measure the voltage across the resistor (it will happen by fact 2)

CONCLUSIONS:

1. A current generated without electric field will induce an electric field i.e. a voltage acrross the resistor
2. The value of that voltage is V=IR

QUESTION:
Where does that elecric field arise from? By which mechanism?

The question is clear and direct, so answer it. When I told you "when don't you apply Ohm's law" your answer was "in superconductivity". What's wrong with you? didn't I say enough times that I'm talking about ohmic conductors?

Provide an answer for that question which is the only question in this thread. Any other subjects are only brought about by you and has no relevance to the question.

That's all, clear and simple.

Very good explanation. We agree on these points, but you still don't like my collision explanation of resistance. Any references that you can recommend?

Claude

 

[Fernsanz]

I don't believe what I'm reading. The inductor terminated in a perfect short has 0 volts across it. Not a voltage source, but a dead short. Please research this. It is elementary. Ask a professor. Inductors terminated in a SC short retain their current indefinitely. When the path is switched so that said inductor is terminated in a resistance, the current at the instant of transition remains as it was, then decays. The initial voltage across said resistor is I*R, then it decays. In other words I determines V.

Exactly. In fact this is the reason to use what is called "freewheeling" or "flyback" diodes: they protect against the peak voltage observed at the very moment of switching.

 

[Fernsanz]

Very good explanation. We agree on these points, but you still don't like my collision explanation of resistance. Any references that you can recommend?
Claude

I'm still trying to get that references. I'm in talks with AJ Bentley, which is the one who provide the explanation in term of "E shockwaves" in the thread I referenced. I will let you know when I have a good reference.

What is clear now is that we are the only one who, up to now, are aware of the problem: to explain why a current generates a E-field. So I'm afraid you and me are the only one to work out the problem.

Thank you for explanation (even if I don't like). A pleasure discussing with you.

 

[kmarinas86]

Up to now cabraham has been the only person who has taken for granted what every electrical engineer knows (even by pure experience): the Ohm's law is always appliable no matter which is the cause. From there he has tried to provide an explanation on why a current implies an E-field. He and me can agree or disagree in the explanation but he is the only one trying to answer the OP question. The rest of you are more worried trying to pretend that you have answer for everything and clear ideas and, obviously, is not the case. Anyway, as I have said, I found the explanation in another thread in this forum, and that thread support my ideas.

Ohm's law is valid under the majority of cases. This is because the material that we use for most circuits are metals that obey the Drude model under the majority of the situation that they are used, i.e. the APPROXIMATION that the conduction electrons can be modeled as free electron gas. This explicitly described in Chapter 1 of Ashcroft and Mermin. However, go a bit further and look at Chapter 3, and you'll see a whole chapter titled "Failure of the Free Electron Model"! There are materials and situations in which the Drude Model, i.e. the foundation for Ohm's Law, no longer works! This is not my opinion, it is a FACT!

I still don't understand why we are debating if Ohm's Law works or not. I don't ever recall in my original post of contesting the validity of Ohm's Law. I did contest the validity of your interpretation, i.e. that any kind of interpretation is valid based on mathematics. Since you claim that you can reverse the "order" of the cause-and-effect in this, I asked for you to go back to the http://people.seas.harvard.edu/~jones/es154/lectures/lecture_2/drude_model/drude_model_cc/drude_model_cc.html" , and show where, at the microscopic level, that you can get the electric field (the identical field in Ohm's Law) as a consequence of the drift velocity of the electrons. To me, this is the only way to show a convincing argument of what you are claiming.

Zz.

Ohm's law is simply wrong when the current in an inductor has yet to reach the full value implied by (V/R). This is most likely in systems with a high ratio of inductance/resistance (which translates to very large coils) whose circuit is being switched on and off at very high frequency.

 

[Fernsanz]

Ohm's law is simply wrong when the current in an inductor has yet to reach the full value implied by (V/R). This is most likely in systems with a high ratio of inductance/resistance (which translates to very large coils) whose circuit is being switched on and off at very high frequency.

Lol, lol. Please inform of this to universities where nobody teach that the laws to analyzing circuits depend on the values of inductances and resistances, lol.

Also inform to all engineers around the world (including me) who always use Ohm's law. I will inform my mates that we have wrongly used the Ohm's law in the magnetotorquers of the satellite to rate the freewheeling diodes.

Lol, after this claim of yours (strongly supported by a lot of reasoning in your post by the way) it is pointless to keep arguing.

 

[cabraham]

Ohm's law is simply wrong when the current in an inductor has yet to reach the full value implied by (V/R). This is most likely in systems with a high ratio of inductance/resistance (which translates to very large coils) whose circuit is being switched on and off at very high frequency.

Don't be offended when I say that you have much to learn about circuit theory. There is no shame in not knowing it, the shame is claiming to know more than the experts & not backing it up. I know absolutely nothing about veterinary medicine. I'm not a problem as long as I let the world know that. But if I were to claim to be a vet, & people bring their pets to me in full trust, that is a problem for which I am liable.

Ohm's law is always valid in an R/L network. When the switch is closed, inductor current starts at zero for example, & the source & resistor are in series with L. As the current increases in L & R, the voltage across R, Vr, increases likewise. At every point in time, Vr = R*I. Ohm's law is valid from the start until the steady statevalue of I = V/R. No exceptions have ever been found.

Who says otherwise. Again, when I ask how much formal circuit theory you've had, I'm just asking what your sources are, not trying to be confrontational. What credible reference claims that Ohm's law is valid only some of the time? Please share.

Claude

 

[ZapperZ]

It has become clear to me that you are unable to understand the essence, neither the statement of the problem.

FACTS:

1. A current can be generated without electric field
2. A current crossing a resistor (ohmic resistor) will induce a voltage, i.e., an E-field

EXPERIMENT:

1. Take a current I generated without electric field (possible by fact 1) and connect it to a resistor (ohmic resistor)
2. Measure the voltage V across the resistor (it will happen by fact 2)

Sorry, but what experiment is this? I've looked at all the various "experiments" that you've described in this thread, and none of them have anything to do current being generated in an ohmic conductor that isn't due to an electric field.

CONCLUSIONS:

1. A current generated without electric field will induce an electric field i.e. a voltage acrross the resistor
2. The value of that voltage satisfies the Ohm's law: V=IR

QUESTION:
Where does that elecric field arise from? By which mechanism?

The question is clear and direct, so answer it. When I told you "when don't you apply Ohm's law" your answer was "in superconductivity or semiconductors". What's wrong with you? didn't I say enough times that I'm talking about ohmic conductors? Are you avoiding by all means facing the problem?

Provide an answer for that question which is the only question in this thread. Any other subjects are only brought about by you and has no relevance to the question.

That's all, clear and simple.

So was my assertion that you cannot have such a thing via the Drude model. You DO know that, at the most fundamental level, these "current" are charge carriers (i.e. electrons in a conductor), don't you? So now, if we AGREE on that, and we are sticking by the idealized view of a conductor and resistor per the Drude model that PRODUCES Ohm's law, I had asked you, at least twice, to show me where it is possible to produce such drift velocity without any applied field. In other words, in this model that produces Ohm's law as a result, there are no other mechanism for charge transport other than the E-field!

One could confirm this another using a more generalized approach by applying the Boltzmann transport equation, which contains more than one mechanism for charge transport. Now, look at what needs to be non-zero for you to get back the Drude model and thus, Ohm's law?

Zz.

 

[cabraham]

Sorry, but what experiment is this? I've looked at all the various "experiments" that you've described in this thread, and none of them have anything to do current being generated in an ohmic conductor that isn't due to an electric field.



So was my assertion that you cannot have such a thing via the Drude model. You DO know that, at the most fundamental level, these "current" are charge carriers (i.e. electrons in a conductor), don't you? So now, if we AGREE on that, and we are sticking by the idealized view of a conductor and resistor per the Drude model that PRODUCES Ohm's law, I had asked you, at least twice, to show me where it is possible to produce such drift velocity without any applied field. In other words, in this model that produces Ohm's law as a result, there are no other mechanism for charge transport other than the E-field!

One could confirm this another using a more generalized approach by applying the Boltzmann transport equation, which contains more than one mechanism for charge transport. Now, look at what needs to be non-zero for you to get back the Drude model and thus, Ohm's law?

Zz.

I've answered this question in spades. The question was "can a current give rise to an E field?" The answer is yes, & I gave an example plus an explanation from peer reviewed uni e/m fields texts. However, the current which gives rise to this conductor's local E field is driven by an external energy conversion source with an associated E field. In other words, the E field produced by the current is a different E field from that of the power source producing the current. Of course the E field produced by the current cannot also drive that same current. Nobody ever implied that.

Also, photoelectric effect describes currents produced due to photon energy transfer. The current in a photodiode is photon induced, not field induced.

Does this help?

Claude

 

[Fernsanz]

Sorry, but what experiment is this? I've looked at all the various "experiments" that you've described in this thread, and none of them have anything to do current being generated in an ohmic conductor that isn't due to an electric field.

? You are fantastic: instead of answering, you ask (pretending to be surprised) about this generic experiment, and the whole thread deal with this experiment from the very beginning!, lol. By the way, who said that the generation of the current has to take place in the ohmic conductor? Read the experiment again please, is clear enough.

[...]I had asked you, at least twice, to show me where it is possible to produce such drift velocity without any applied field. In other words, in this model that produces Ohm's law as a result, there are no other mechanism for charge transport other than the E-field!

Answered by yourself:

I've just shown you an example where I could have a current without the use of an external electric field to generate that current. Want another example? Look at the current from a photoemission process. Look ma! No electric field!

Are you reaching contradictions with yourself?

 

[kmarinas86]

Ohm's law is simply wrong when the current in an inductor has yet to reach the full value implied by (V/R). This is most likely in systems with a high ratio of inductance/resistance (which translates to very large coils) whose circuit is being switched on and off at very high frequency.

Lol, lol. Please inform of this to universities where nobody teach that the laws to analyzing circuits depend on the values of inductances and resistances, lol.

Also inform to all engineers around the world (including me) who always use Ohm's law. I will inform my mates that we have wrongly used the Ohm's law in the magnetotorquers of the satellite to rate the freewheeling diodes.

Lol, after this claim of yours (strongly supported by a lot of reasoning in your post by the way) it is pointless to keep arguing.

Don't be offended when I say that you have much to learn about circuit theory. There is no shame in not knowing it, the shame is claiming to know more than the experts & not backing it up. I know absolutely nothing about veterinary medicine. I'm not a problem as long as I let the world know that. But if I were to claim to be a vet, & people bring their pets to me in full trust, that is a problem for which I am liable.

Ohm's law is always valid in an R/L network. When the switch is closed, inductor current starts at zero for example, & the source & resistor are in series with L. As the current increases in L & R, the voltage across R, Vr, increases likewise. At every point in time, Vr = R*I. Ohm's law is valid from the start until the steady statevalue of I = V/R. No exceptions have ever been found.

Who says otherwise. Again, when I ask how much formal circuit theory you've had, I'm just asking what your sources are, not trying to be confrontational. What credible reference claims that Ohm's law is valid only some of the time? Please share.

Claude

You both assume that "V" stands for the resistive voltage drop.

If V is the resistive voltage drop, then what letter is used your inductive voltage drop? Do you even care about the inductive voltage drop?

What letter do you use for applied voltage?

What about the capacitive voltage drop?

Haven't you ever heard of the analogy between RLC circuits and mechanical harmonic oscillators?

Voltage is not limited to I*R. I*R is a voltage drop (V_R). It is not voltage related to stored energy. It is voltage related to dissipated energy. Voltage related to stored energy can use symbols (V_C) or (V_L) depending on what is discussed.

There is such a thing as static charge inside a non-shorted capacitor. Where is the current in that? What R in a capacitor has any meaning? A capacitor stores voltage does it not?

Why are other meanings of V ignored in this discussion? It may be fine in the field to assume that V means resistive voltage drop, but this is utterly misleading to outsiders. How can I believe even 50% of what I read here when so much goes around being equivocated?

The topic of the thread is "Ohm's law, why current induces electric field and cause effect direction".

The truth is that current induces a magnetic field. The magnetic field is a first-order effect of the electric fields of the charges which the current is comprised of. If this current induces an electric field, it must be a secondary-order effect of those electric fields, meaning that the magnetic fields they produce must have a changing magnitude. A macroscopically viewable example of combined second-order effects can be easily demonstrated with a coil electromagnet. The inductive voltage drop which exists in the inductor is in fact a second-order effect of those electric fields. That voltage drop behaves like an applied voltage (in this case, an induced electrical field) in the opposite direction, and this creates a delay of the current in reaching its final value determined by V_(source)/R. A charge undergoing acceleration produces such a secondary-order effect. The same is true for deceleration of charge.

The truth is that current inducing electric fields has absolutely nothing to do with a constant DC current, so any analysis in this thread that assumes a constant DC current is utterly beside the point and totally irrelevant. The so-called "constant" AC current may cause an oscillating electric field, but what a surprise that is! NOT. "Why current induces electric field and cause effect direction" is about circuit laws[/url] because to quantify from scratch the electric fields due to changes of currents at a local level requires the use of differential equations.

When reactive elements such as capacitors, inductors, or transmission lines are involved in a circuit to which AC or time-varying voltage or current is applied, the relationship between voltage and current becomes the solution to a differential equation, so Ohm's law (as defined above) does not directly apply since that form contains only resistances having value R, not complex impedances which may contain capacitance ("C") or inductance ("L").

http://en.wikipedia.org/wiki/Ohm%27s_law

It becomes even harder to define Ohm's law when your waveforms are neither flat nor sinusoidal. I am not an AC dude. AC circuits are not what I care about. I care about circuits that do not have uniform currents. I care about second-order, third-order, and higher-order electromagnetic effects. I care about circuits that have unusually high inductance values for their resistance which are switched at high frequency. If you have a coil that has 10 henries of inductance and 2 ohms of resistance, how can it reach the full value of current if you switch the circuit on and off every other millisecond?

The above image is from http://www.intmath.com/Differential-equations/5_RL-circuits.php

This is really basic stuff, and I always try to keep it on the top of my head. Every model I question on these forums I connect with fundamental basics which must go on in any situation, regardless of its magnitude, and even if the "scenario" denies, contradicts, or fails to mention that such effects must exist!

 

[kmarinas86]

If moving charges collide into a capacitor, it induces an electrical field due to the collapse of the magnetic field. The collision of like-charged particles which stores potential energy in the electric field is what induces the electrical field in a wire. The potential energy results from the increase of the volume-averaged RMS value of the electric field. This is because the volume density of the energy in the electric field varies by (1/2)*(electric constant of permittivity)*E^2. The RMS value of E increases as like charges increase their proximity to one another. So the volume-averaged value of E^2 is greater in that case.

Thanks thanks thanks! That's the type of reasoning I was looking for. I don't see it clear though: if you get electrons closer to each other than they would be in equilibrium

Equilibrium requires that forces must be in equilibrium. If you attain it exactly, you have no leak. Otherwise, you can have a leak of charge.

You have electrons and things that contain electrons. Without something to contain electrons, there is no equilibrium for them because then the electrons just repel each other away.

any of those electron will feel a stronger field from its neighbours, but that's true in all direction! so the sum of stronger fields at any point would cancel out, even though it is true that the density of potential energy will be higher.

Well that is what happens when charge is stored in a bottle yo.

Or do you mean that it should be viewed more like a compresion-rarefaction process like a sound wave? That is, the first avalanche of electrons moving fast get really close to the front-end of standing-still electrons; then, because of proximity this standing still electrons will move forward getting closer to the next ones and so on. Is this a correct visualization?

Thanks again

This part really does answer the question, "Why do currents induce electric fields and cause/effect their direction?" In fact, one could care less whether you're using an ohmic conductor or not because either way the strength of the electric fields can be produced by the interaction of electric charges. What should worry you more are the differential equations for the electric and magnetic fields.

Talking about the basics is obviously sufficient to show that the question posed does have a valid approach to an answer. In this case, it is far too simple to arrive at this answer. You don't even need decades of experience to do it. You just need knowledge of the laws of physics.

Imposing an "ohmic" view on this scenario "currents causing electric fields" is ridiculous.

 

[ZapperZ]

? You are fantastic: instead of answering, you ask (pretending to be surprised) about this generic experiment, and the whole thread deal with this experiment from the very beginning!, lol. By the way, who said that the generation of the current has to take place in the ohmic conductor? Read the experiment again please, is clear enough.

So now we are going out of Ohmic conductor after you insist that we HAVE to deal with one?

Answered by yourself:


Are you reaching contradictions with yourself?

No, because the examples I gave were NOT in a conductor. There were in vacuum as my initial attempt to show you that one can have current without an applied field (i.e. not via Ohm's law).

In a conductor, the Ohm's Law is a direct consequence of the Drude Model. When an electron enters such a system, it has something that is called the mean-free path. This is the characteristic length it travels before it collides into something, be it other electrons, other ions, other impurities, etc. This defines the resistivity of the material. For a typical metal, the mean-free path isn't any longer than an order of 10's or 100's of nanometers. You cannot have a "current" at the macroscopic length with such a length without any applied field. That is why I kept telling you to tell me how you would get the drift velocity without any applied field. You can't simply force an electron to enter a conductor and then to travel such distances. We already have trouble getting them to do that even when we have an applied field. Without one, you have no current.

Ohm's law is a phenomenological model. When you are told that something is the microscopic description of Ohm's Law, it means that it is the underlying explanation for something that is based on observation, and so, it is the fundamental description of it. I take it that, since you have continued to not address this model, that you either have never heard of it, or not aware of it. Therefore, before you continue to insist that you can simply rearrange Ohm's Law and make a different interpretation of it simply on your ability to do algebraic manipulation, I strongly suggest you look up the link I've given you on the Drude model and the free electron gas.

If you don't wish to do that, then there's nothing I can do to correct what I know to be a mistake in your view. If you are OK with that, then there's nothing I need to care about.

Zz.

 

[kmarinas86]

I've answered this question in spades. The question was "can a current give rise to an E field?" The answer is yes, & I gave an example plus an explanation from peer reviewed uni e/m fields texts. However, the current which gives rise to this conductor's local E field is driven by an external energy conversion source with an associated E field. In other words, the E field produced by the current is a different E field from that of the power source producing the current. Of course the E field produced by the current cannot also drive that same current. Nobody ever implied that.

You are right up to this point.

It turns out that zeroth, second, and fourth order derivatives of the E-field lead back to E-fields. Any even-order derivative does. These derivatives will not all point the same way (for the obvious reason that changes of acceleration are not always aligned with velocity), so these even-order effects are not always "parallel" or "anti-parallel" to each other. Guess what B-fields are generated by? B-fields are caused by odd-order effects of E-fields. ...Duhsville!

 

[kmarinas86]

In a conductor, the Ohm's Law is a direct consequence of the Drude Model. When an electron enters such a system, it has something that is called the mean-free path. This is the characteristic length it travels before it collides into something, be it other electrons, other ions, other impurities, etc. This defines the resistivity of the material. For a typical metal, the mean-free path isn't any longer than an order of 10's or 100's of nanometers. You cannot have a "current" at the macroscopic length with such a length without any applied field. That is why I kept telling you to tell me how you would get the drift velocity without any applied field. You can't simply force an electron to enter a conductor and then to travel such distances. We already have trouble getting them to do that even when we have an applied field. Without one, you have no current.

Ohm's law is a phenomenological model. When you are told that something is the microscopic description of Ohm's Law, it means that it is the underlying explanation for something that is based on observation, and so, it is the fundamental description of it.

A coherent summary of a truth.

 

[Born2bwire]

So now we are going out of Ohmic conductor after you insist that we HAVE to deal with one?



No, because the examples I gave were NOT in a conductor. There were in vacuum as my initial attempt to show you that one can have current without an applied field (i.e. not via Ohm's law).

In a conductor, the Ohm's Law is a direct consequence of the Drude Model. When an electron enters such a system, it has something that is called the mean-free path. This is the characteristic length it travels before it collides into something, be it other electrons, other ions, other impurities, etc. This defines the resistivity of the material. For a typical metal, the mean-free path isn't any longer than an order of 10's or 100's of nanometers. You cannot have a "current" at the macroscopic length with such a length without any applied field. That is why I kept telling you to tell me how you would get the drift velocity without any applied field. You can't simply force an electron to enter a conductor and then to travel such distances. We already have trouble getting them to do that even when we have an applied field. Without one, you have no current.

Ohm's law is a phenomenological model. When you are told that something is the microscopic description of Ohm's Law, it means that it is the underlying explanation for something that is based on observation, and so, it is the fundamental description of it. I take it that, since you have continued to not address this model, that you either have never heard of it, or not aware of it. Therefore, before you continue to insist that you can simply rearrange Ohm's Law and make a different interpretation of it simply on your ability to do algebraic manipulation, I strongly suggest you look up the link I've given you on the Drude model and the free electron gas.

If you don't wish to do that, then there's nothing I can do to correct what I know to be a mistake in your view. If you are OK with that, then there's nothing I need to care about.

Zz.

I won't be weighing in on some of the other discussions but in terms of the OP's question, ZapperZ has hit upon it several posts back and this is just another good summarization of the answer.

Physics is not the same as mathematics in that we can not simply rearrange an equation at our leisure and find that all rearrangements, while mathematically viable, are still physically valid. Ohm's Law is a macroscopic description of the movement of carriers through a medium as given by the Drude Model. The idea being that as the carriers move through the material, they collide with the constituent particles of the material (atoms, molecules, lattice). In doing so, they give up some of their kinetic energy in the form of vibrational energy in the material bulk (phonons = heat) and also undergo a change in momentum.

Now we could of course have this happen without any applied electric field. We have an electron gun that shoots a stream of electrons into a bulk. But this is not going to give rise to the same physical phenomenon because the electrons will continually bleed off their energy (thus allowing us to simply make a bulk thick enough such that no appreciable amount of carriers can penetrate through). In addition, the constant changes in directions from the collisions will hamper any real net movement of the carriers. Yes, the charge carriers themselves will give rise to an electric field, but it will not be the same electric field as an applied field because it will obviously be one that is not conducive to the propagation of the charges through the material (by virtue of the fact that the individual fields will try and separate the carriers).

Only if we have an applied field will we see a continuous injection of energy back into the carriers so that they can recover from the energy lost to the collisions and progress through the bulk. In addition, the applied field also reasserts the drift direction which would eventually be lost if we allowed the carriers to continually collide randomly.

So Ohm's Law does not make physical sense without the applied electric field because it is an assumption that is required by the underlying physics that gave rise to the relation. Of course, the above is nothing more than another restatement of ZapperZ's post.

 

[Fernsanz]

It has become clear to me that you are unable to understand the essence, neither the statement of the problem. Now I add: this thread is futile except for cabraham contributions. I will make a last attempt though, in case brains wake up.

FACTS:

1. A current can be generated without electric field
2. A current crossing a resistor (ohmic resistor) will induce a voltage, i.e., an E-field

EXPERIMENT:

1. Take a current I generated without electric field (possible by fact 1) and connect it to a resistor (ohmic resistor)
2. Measure the voltage V across the resistor (it will happen by fact 2)

CLARIFICATIONS:

1. Step 1 of the experiment involves generating a current withouth electric field. It can be photoemission process, SC, or God moving electrons one by one without electric field.
2. Once again for the slowest ones: step 1 involves taking an active component/device which provides current without electrical field. Obviosuly enough it can't be an ohmic conductor, it might be a semiconductor, a SC or God force.
3. Step 2 of the experiment is to take an ohmic resistor, an ohmic passive component, a resistor which is ohmic, a piece of conductor which is ohmic, always and only ohmic, and feed it with the current provided by the component in step 1.
4. Once again for the slowest ones: step 2 involves taking a passive ohmic resistor and pump into it the current generated by the "whatever-it-is-active-current-source-without-electric-field". It is here and only here where we stick to ohmic conductor.
5. To sum up one more time in case someone need: the only condition for the active current source is to generate current without electrical field. The only condition for the passive element is to be an ohmic resistor. Take the current source on one hand, take the resistor on the other hand, bless them both, and then connect them.
   

CONCLUSIONS:

1. A current generated without electric field will induce an electric field i.e. a voltage acrross the resistor
2. The value of that voltage satisfies the Ohm's law: V=IR

QUESTION:
Where does that elecric field arise from? By which mechanism?

The question is clear and direct, yet still unanswered (except for an explanation, right or not, provided by cabraham).

It would be useless to "answer" with another question like "could you tell me in the Drude model bla bla bla". Is me who is raising a question; is me who expect an answer which starts something like "the electric field is induced because electrons will accumulate bla bla bla" or "the electric field is induced because a shockwave of electric nature will arise bla bla bla" or "the electric field arise cause God comes down to Earth bla bla bla". If someone need explanations or comments on Drude model, harmonic oscillator or turbo engines in cars, please open another thread and feel free to talk about it for ages.

The Drude model is of no relevance in this discussion because it is a model to explain the dynamics of electrons once an electric field has been established. My question goes previous to that: how was that electrical field established when you connect the ohmic conductor to a current with no electric field? And you keep answering that the Drude model does not allow a current in an ohmic conductor without electric field. That is not the fact under question. The question is, once again and again, how was that electric field created from a situation when a current started traveling along the ohmic resistor with no previous electric field. When you connect the ohmic resistor to the "no-electric-field-current-source" a electric field arise which was not there before (and which is the E-field the Drude model takes as a starting point to explain electrons dynamics, the model does not address how it was generated, where does it come from or nature of the field). And lastly, is there any people on Earth who think that V=IR is not valid for the ohmic resistor? Apparently only kmarinas86; for the rest of the world the Ohm's law is always true even in this situation where the feed to the resistor is a current which conveyed not electric field in its origin (but which subsequently generated an electric field across the resistor by a mechanism that is the subject under question)

Here is an example of a suitable and plausible explanation and the type of answer anyone would expect: at the very moment when the current star entering the ohmic conductor, the leading electrons will get closer to the first electrons in the ohmic conductor; thus, this leading group of electrons in the current will slow down by repulsive force and the electrons in the omhic resistor will start moving. So, the electric field could be the repulsive force between electrons. This can be viewed as a sound wave in which the "slowed down" part and the "repulsed" part are called compression and rarefaction respectively. So, the electric field can be viewed as a lot of pulses, a lot of "electric shockwaves" that all together create a macroscopic E-field that one could measure placing a test charge inside the conductor. It would be also the explanation that the Ohm's law is applicable even if the feeding to the resistor is a non-electric-field-current.

You see? that would be something addressing the questions in this thread. An explanation of this type is what Cabraham provided and not a disgression with no constructive attitude showing a total lack of understanding..

 

[ZapperZ]

It has become clear to me that you are unable to understand the essence, neither the statement of the problem. Now I add: this thread is futile except for cabraham contributions. I will make a last attempt though, in case brains wake up.

FACTS:

1. A current can be generated without electric field
2. A current crossing a resistor (ohmic resistor) will induce a voltage, i.e., an E-field

EXPERIMENT:

1. Take a current I generated without electric field (possible by fact 1) and connect it to a resistor (ohmic resistor)
2. Measure the voltage V across the resistor (it will happen by fact 2)

CLARIFICATIONS:

1. Step 1 of the experiment involves generating a current withouth electric field. It can be photoemission process, SC, or God moving electrons one by one without electric field.

1. Perfect.
   
   1. I have a "current source", i.e. free electron beams from my particle accelerator. After the last accelerating structure, it is now simply coasting, i.e. no electric field, but the electrons are still moving, so it has a current.
   
   2. I shoot them into a piece of copper (it is what you would qualify as an "ohmic conductor", no?).
   
   3. I then measure the field across it. What do I get?
   
   4. I actually do this! I mentioned WAY early in my responses of something called the "Faraday Cup". In fact, 2 summers ago, I assigned an undergraduate summer intern to actually build a super fast Faraday Cup to measure very short electron bunches. Therefore, I strongly suggest you look up what a "Faraday Cup" is, and come back to me and tell me that you actually can get the same E-field as described in Ohm's Law across the conductor. You do NOT! You don't have to believe what I said, you are welcome to look it up.
   
   5. In Ohm's law, you have a field E, that creates a current I, in the SAME CONDUCTOR. In your "experiment", there is NO "I" in the conductor (refer to my earlier post on the mean free path). The "I" or current in your scenario is external to the conductor and only comes in because it hits the conductors. Any kind of E-field that is generated in the conductors is NOT in correspondence with the external current. This is NOT Ohm's Law!
   
   6. I study the process of secondary emission, which involves electrons with energy ranging from 200 eV all the way to hundreds of keV, hitting materials ranging from conductors to insulators. So the mechanism of electron transport once it hits a material is something I don't just read on, but it is something I experiment on! I truly would like you to do that exact experiment that you just described before you make such claims, because some of us actually HAVE done such experiments!
   
   7. You posted this in the physics part of PF, not in the engineering part. Presumably, you wanted some fundamental physics issues being resolved here. I've tried, several times, to refer you to the Drude model with the hope that you might learn that there's a more fundamental description of the physics that produces Ohm's law, and discover for yourself the difficulty of having a current in a conductor without any preexisting E-field.
   
   8. I can't make you learn and I think I've tried several times. It is no longer any of my concern to try and correct the errors in your understanding. You originated this thread and asked a question, but it appears that you already have made up your mind of the answer. All you appear to want to do is carry out an argument. I'll end my participation here, because I've already wasted too much time on a futile effort.
   
   Zz.

 

[Fernsanz]

Perfect.

1. I have a "current source", i.e. free electron beams from my particle accelerator. After the last accelerating structure, it is now simply coasting, i.e. no electric field, but the electrons are still moving, so it has a current.

2. I shoot them into a piece of copper (it is what you would qualify as an "ohmic conductor", no?).

3. I then measure the field across it. What do I get?

4. I actually do this! I mentioned WAY early in my responses of something called the "Faraday Cup". In fact, 2 summers ago, I assigned an undergraduate summer intern to actually build a super fast Faraday Cup to measure very short electron bunches. Therefore, I strongly suggest you look up what a "Faraday Cup" is, and come back to me and tell me that you actually can get the same E-field as described in Ohm's Law across the conductor. You do NOT! You don't have to believe what I said, you are welcome to look it up.

5. In Ohm's law, you have a field E, that creates a current I, in the SAME CONDUCTOR. In your "experiment", there is NO "I" in the conductor (refer to my earlier post on the mean free path). The "I" or current in your scenario is external to the conductor and only comes in because it hits the conductors. Any kind of E-field that is generated in the conductors is NOT in correspondence with the external current. This is NOT Ohm's Law!

Are you the one who decides what is Ohm's law and what isn't? The current in my scenarios is external? What the hell is a current that travels across the resistor? If i have a current in a SC loop and in t=0 I insert a resistor in the loop, what the hell is external? The E-field that will arise across the resistor is not the one Ohm's law? Which part of the Ohm's law says if an electrical field along a conductor is the Ohm's law or not? Will the resistor violate the relation V=IR while the current in the loop semicondcutor-resistor is dacaying? No, it will not violate, so V=IR, so that's Ohm Law.

Do you even know my job? It will be interesting when I inform my mates that when we anlyze the effetc of space radiation in the circuits of the satellite (for example electrons being pumped into satellite ohmic conductors) we are worngly using the Ohm's law V=IR because the electrons come from outside. Ridiculous

You will never admit you are wrong becasue, as has become clear, any E-field that it is generated you will say it is not the Ohm's law. Stick yourself to explain how the hell that "kind of" electrical field is generated; I will decide if it is Ohm's law or not at my own.

7. You posted this in the physics part of PF, not in the engineering part. Presumably, you wanted some fundamental physics issues being resolved here. I've tried, several times, to refer you to the Drude model with the hope that you might learn that there's a more fundamental description of the physics that produces Ohm's law, and discover for yourself the difficulty of having a current in a conductor without any preexisting E-field.

8. I can't make you learn and I think I've tried several times. It is no longer any of my concern to try and correct the errors in your understanding. You originated this thread and asked a question, but it appears that you already have made up your mind of the answer. All you appear to want to do is carry out an argument. I'll end my participation here, because I've already wasted too much time on a futile effort.

It seems you can't stop typing "Drude model"! Read what I said about it in my previous post because you haven't even read it. Drude model has nothing to say here!

You can not teach what you ignore. Yes please, end you participation here and let others who have answers participate.

 

[kmarinas86]

The question is clear and direct, yet still unanswered (except for an explanation, right or not, provided by cabraham).

It would be useless to "answer" with another question like "could you tell me in the Drude model bla bla bla". Is me who is raising a question; is me who expect an answer which starts something like "the electric field is induced because electrons will accumulate bla bla bla" or "the electric field is induced because a shockwave of electric nature will arise bla bla bla" or "the electric field arise cause God comes down to Earth bla bla bla". If someone need explanations or comments on Drude model, harmonic oscillator or turbo engines in cars, please open another thread and feel free to talk about it for ages.

The Drude model is of no relevance in this discussion because it is a model to explain the dynamics of electrons once an electric field has been established. My question goes previous to that: how was that electrical field established when you connect the ohmic conductor to a current with no electric field? And you keep answering that the Drude model does not allow a current in an ohmic conductor without electric field. That is not the fact under question. The question is, once again and again, how was that electric field created from a situation when a current started traveling along the ohmic resistor with no previous electric field. When you connect the ohmic resistor to the "no-electric-field-current-source" a electric field arise which was not there before (and which is the E-field the Drude model takes as a starting point to explain electrons dynamics, the model does not address how it was generated, where does it come from or nature of the field). And lastly, is there any people on Earth who think that V=IR is not valid for the ohmic resistor? Apparently only kmarinas86; for the rest of the world the Ohm's law is always true even in this situation where the feed to the resistor is a current which conveyed not electric field in its origin (but which subsequently generated an electric field across the resistor by a mechanism that is the subject under question)

I emphasize the following:

When reactive elements such as capacitors, inductors, or transmission lines are involved in a circuit to which AC or time-varying voltage or current is applied, the relationship between voltage and current becomes the solution to a differential equation, so Ohm's law (as defined above) does not directly apply since that form contains only resistances having value R, not complex impedances which may contain capacitance ("C") or inductance ("L").

http://en.wikipedia.org/wiki/Ohm%27s_law

It becomes even harder to define Ohm's law when your waveforms are neither flat nor sinusoidal. I am not an AC dude. AC circuits are not what I care about. I care about circuits that do not have uniform currents. I care about second-order, third-order, and higher-order electromagnetic effects. I care about circuits that have unusually high inductance values for their resistance which are switched at high frequency. If you have a coil that has 10 henries of inductance and 2 ohms of resistance, how can it reach the full value of current if you switch the circuit on and off every other millisecond?

The above image is from http://www.intmath.com/Differential-equations/5_RL-circuits.php

The problem with the topic of this thread is that because of its constraints you stubbornly do not consider other physically valid scenarios, and thereby you limit yourself to explanations for which there are better alternatives. The sound analogy is the best one you got, and both cabraham and I both mentioned something of that nature much earlier in this thread. The first seven responses of this thread go exactly this:

Hi,

Lately I've been concerned about the Ohm's law $$J=\sigma E$$ and the physical interpretation of this law depending on what is considered the cause and what the effect.

More concretely, it is quite natural and intuitive for me the interpretation of this law in one direction: a electric field $$E$$ over a ohmic conductor will cause a current $$J$$. No one would have problems working out that electric force acting upon electrons will cause them to move, i.e. setting up a current. We can certainly say that the electrical field is the cause in this case and the current is the effect, the result.

However, in the reverse direction, namely that a current $$J$$ injected somehow in a ohmic conductor will cause a electric field $$E$$, is by far more counterintuitive. Anyhow the current is generated previously by other means (for instance photoelectric or thermoionic effects) and injected into the conductor an electric field will arise. What is the explanation that acounts for that electric field caused just by movement of electrons in a conductor (which is nothing but a cristaline structure "covered" by a cloud of electrons)?

I would gladly read your explanations.

Thanks.

It is counter-intuitive indeed, but they are mutually inclusive, neither being the cause nor the effect.

I view it in the following way to see it better. By definition a constant voltage source is one that maintains a constant voltage despite varying load resistance. Let's say a current enters a conductor, the charge carriers in conductors are electrons. They have energy, but lose a portion of that energy when they collide with the lattice structure.

When an electron in the conduction band collides with an electron in the valence band, energy is transferred. So a current is a transfer of energy among electrons, where the actual motion/displacement of an electron is small, but the propagation of energy is near light speed.

The energy loss due to lattice collisions constitutes resistive loss. Thus the conduction electrons carry energy, a portion of which they lose due to resistance i.e. Lattice collisions. By definition, the energy lost per unit charge is the voltage drop. Hence current entering a conductor incurs energy loss due to resistance/lattice collisions, resulting in a drop in voltage.

The converse also holds. A conductor with no current is instantly connected across a small voltage. The e field results in charge motion i.e. Current.

Ohm's law is bi-lateral, where j & e are inclusive. Which one comes first is a moot question. You are thinking the right way.

Claude

The energy lost per unit charge is not the voltage drop because that loss is caused by "mechanical" collision among electrons. Reasoning that way one could take an electron, throw it against a wall and conclude that, since the electron has lost all its energy after hitting the wall, there must be an electric field out of the wall and hence a voltage drop.

So the question is still in the air: Why an electrical field appear just because electrons are moving? (don't lose of sight the "electrical" nature of the force I am asking about which is the electrical field that appears in ohm's low, not any other mechanical forces)

Mechanical - electrical?! What is the difference? How do you think resistance takes place? If the electron can travel all the way through the conductor w/o colliding w/ the lattice, there is zero resistance, hence zero voltage drop. Superconductors achieve this property.

The "thermal energy" associated w/ an electron is given by 0.5*m*v^2 = k*t, where m is mass, v is velocity, k is boltzmann's constant, & t is temperature (absolute).

If "resistance" is not a "mechanical" entity, then what is it? I'd like to know. Thanks in advance.

Claude

If mechanical and electrical are the same for you good for you, but there is no explanation in your words on how an electric field arises as the result of moving charges in that lattice. You have started talking about resistance, which i did not mention. Resistance is what steal energy from electrons. I'm not asking about where the electrons lose their energy, I'm asking about why an electric field arises; an electrical field that would be measurable by measuring the force exherted on a test charge placed inside the conductor even if such test charge is not moving!

Please, I'm not expecting simple or naive answer; i expect a physical explanation on why moving charges in a ohmic conductor induce an electric field, i.e., a field that would cause a test charge placed inside the conductor to move purely by electric force $$f=qe$$.

If moving charges collide into a capacitor, it induces an electrical field due to the collapse of the magnetic field. The collision of like-charged particles which stores potential energy in the electric field is what induces the electrical field in a wire. The potential energy results from the increase of the volume-averaged rms value of the electric field. This is because the volume density of the energy in the electric field varies by (1/2)*(electric constant of permittivity)*e^2. The rms value of e increases as like charges increase their proximity to one another. So the volume-averaged value of e^2 is greater in that case.

Thanks thanks thanks! That's the type of reasoning i was looking for. I don't see it clear though: If you get electrons closer to each other than they would be in equilibrium, any of those electron will feel a stronger field from its neighbours, but that's true in all direction! So the sum of stronger fields at any point would cancel out, even though it is true that the density of potential energy will be higher.

Or do you mean that it should be viewed more like a compresion-rarefaction process like a sound wave? That is, the first avalanche of electrons moving fast get really close to the front-end of standing-still electrons; then, because of proximity this standing still electrons will move forward getting closer to the next ones and so on. Is this a correct visualization?

Thanks again

What do you mean by "Thanks again"? Did you think I was cabraham?

Here is an example of a suitable and plausible explanation and the type of answer anyone would expect: at the very moment when the current star entering the ohmic conductor, the leading electrons will get closer to the first electrons in the ohmic conductor; thus, this leading group of electrons in the current will slow down by repulsive force and the electrons in the omhic resistor will start moving. So, the electric field could be the repulsive force between electrons. This can be viewed as a sound wave in which the "slowed down" part and the "repulsed" part are called compression and rarefaction respectively. So, the electric field can be viewed as a lot of pulses, a lot of "electric shockwaves" that all together create a macroscopic E-field that one could measure placing a test charge inside the conductor. It would be also the explanation that the Ohm's law is applicable even if the feeding to the resistor is a non-electric-field-current.

You see? that would be something addressing the questions in this thread. An explanation of this type is what Cabraham provided and not a disgression with no constructive attitude showing a total lack of understanding.

Both cabraham and I contributed to this explanation, per above. Yet you only give credit to him. You even responded more thankfully to my explanation, but you thought I was someone else. Nevermind about this then. We've got a plausible explanation that doesn't go into many details as to why electrons behave this way. Do you want another explanation or deepen this one? Do you want to define the differential equations that are required to describe the second-order and higher-order effects of the E-field? Do you want to discussion in this thread to advance? Or do you want some kind of "totally exclusive and equally plausible explanation" for this? Seems like work doesn't it?

The 8th post was from ZapperZ. At this point, all hell in this thread broke loose. After a recent set of questions by ZapperZ, this following conservation was spawned:

Sorry, but what experiment is this?

You DO know that, at the most fundamental level, these "current" are charge carriers (i.e. electrons in a conductor), don't you?

I had asked you, at least twice, to show me where it is possible to produce such drift velocity without any applied field.

Now, look at what needs to be non-zero for you to get back the Drude model and thus, Ohm's law?

I've answered this question in spades. The question was "can a current give rise to an E field?" The answer is yes, & I gave an example plus an explanation from peer reviewed uni e/m fields texts. However, the current which gives rise to this conductor's local E field is driven by an external energy conversion source with an associated E field. In other words, the E field produced by the current is a different E field from that of the power source producing the current. Of course the E field produced by the current cannot also drive that same current. Nobody ever implied that.

You are right up to this point.

It turns out that zeroth, second, and fourth order derivatives of the E-field lead back to E-fields. Any even-order derivative does. These derivatives will not all point the same way (for the obvious reason that changes of acceleration are not always aligned with velocity), so these even-order effects are not always "parallel" or "anti-parallel" to each other. Guess what B-fields are generated by? B-fields are caused by odd-order effects of E-fields. ...Duhsville!

ZapperZ thinks an applied field is not restricted to an "internal source", yet cabraham thinks otherwise. If an E-field is external, does that meant it is not "applied field"? It seems really arbitrary to assume either way. This thread is littered with disagreements that stem from mere terminological differences. The above is just one example.

ZapperZ is about to leave this thread for good, if he has not already:

Perfect.

1. I have a "current source", i.e. free electron beams from my particle accelerator. After the last accelerating structure, it is now simply coasting, i.e. no electric field, but the electrons are still moving, so it has a current.

2. I shoot them into a piece of copper (it is what you would qualify as an "ohmic conductor", no?).

3. I then measure the field across it. What do I get?

4. I actually do this! I mentioned WAY early in my responses of something called the "Faraday Cup". In fact, 2 summers ago, I assigned an undergraduate summer intern to actually build a super fast Faraday Cup to measure very short electron bunches. Therefore, I strongly suggest you look up what a "Faraday Cup" is, and come back to me and tell me that you actually can get the same E-field as described in Ohm's Law across the conductor. You do NOT! You don't have to believe what I said, you are welcome to look it up.

5. In Ohm's law, you have a field E, that creates a current I, in the SAME CONDUCTOR. In your "experiment", there is NO "I" in the conductor (refer to my earlier post on the mean free path). The "I" or current in your scenario is external to the conductor and only comes in because it hits the conductors. Any kind of E-field that is generated in the conductors is NOT in correspondence with the external current. This is NOT Ohm's Law!

Are you the one who decides what is Ohm's law and what isn't? The current in my scenarios is external? What the hell is a current that travels across the resistor? If i have a current in a SC loop and in t=0 I insert a resistor in the loop, what the hell is external? The E-field that will arise across the resistor is not the one Ohm's law? Which part of the Ohm's law says if an electrical field along a conductor is the Ohm's law or not? Will the resistor violate the relation V=IR while the current in the loop semicondcutor-resistor is dacaying? No, it will not violate, so V=IR, so that's Ohm Law.

Do you even know my job? It will be interesting when I inform my mates that when we anlyze the effetc of space radiation in the circuits of the satellite (for example electrons being pumped into satellite ohmic conductors) we are worngly using the Ohm's law V=IR because the electrons come from outside. Ridiculous

You will never admit you are wrong becasue, as has become clear, any E-field that it is generated you will say it is not the Ohm's law. Stick yourself to explain how the hell that "kind of" electrical field is generated; I will decide if it is Ohm's law or not at my own.

6. I study the process of secondary emission, which involves electrons with energy ranging from 200 eV all the way to hundreds of keV, hitting materials ranging from conductors to insulators. So the mechanism of electron transport once it hits a material is something I don't just read on, but it is something I experiment on! I truly would like you to do that exact experiment that you just described before you make such claims, because some of us actually HAVE done such experiments!

7. You posted this in the physics part of PF, not in the engineering part. Presumably, you wanted some fundamental physics issues being resolved here. I've tried, several times, to refer you to the Drude model with the hope that you might learn that there's a more fundamental description of the physics that produces Ohm's law, and discover for yourself the difficulty of having a current in a conductor without any preexisting E-field.

8. I can't make you learn and I think I've tried several times. It is no longer any of my concern to try and correct the errors in your understanding. You originated this thread and asked a question, but it appears that you already have made up your mind of the answer. All you appear to want to do is carry out an argument. I'll end my participation here, because I've already wasted too much time on a futile effort.

Zz.

It seems you can't stop typing "Drude model"! Read what I said about it in my previous post because you haven't even read it. Drude model has nothing to say here!

You can not teach what you ignore. Yes please, end you participation here and let others who have answers participate.

Let's face it. You are more an applied physicist than a theoretical physicist. Why would someone like you care to put priority on the opinions of those who lean more toward the science of theoretical physics over those who lean more toward the practice of applied physics? You insist in using the conventions of engineers. Physics has more advanced explanations than are typically used in practice. Aren't you asking about something that your years of experience has not told you yet?

 

[AJ Bentley]

My two-pennyworth.

IMO
The classical ideas of fields and the movement of charges (not specifically electrons) don't sit well in the study of electrodynamics.

I rather like the approach taken by Prof Mead of 'explaining' electricity as a quantum effect from the outset. What's more, he points out (not in so many words) that when we study physical systems, we inevitably begin with a simple situation.
Typically we would avoid friction effects (often by setting ourselves up in a separate universe for the experiment!). For that reason, he elects to only consider superconductors.

It sounds crazy but it works. Within the first chapter of his book 'collective electrodynamics' he disposes of the E field and the B field as totally unnecessary artificial constructs and just uses the scalar and vector potentials along with simple QM concepts to completely reconstruct the subject.
He doesn't destroy Maxwell's work, he simply redirects it along the lines Maxwell would have gone if he'd been aware of facts we now know.

I can't recommend it strongly enough.

Just to give you the Flavour:-
He points out that the voltage across the ends of a loop (scalar potential) and the Vector potential A around the loop together satisfy the de Broglie relationship of frequency to wavenumber. By considering charge as a wave, it's therefore possible to specify the potentials as a four-vector throughout space - they depend only on J (also a four vector with the charge density.)
E and B - you don't need - you can easily calculate a value for them at any point if you want - but it turns out that most of the time you don't need to.

 

[cabraham]

Above there is a waveform of an R-L circuit whose current builds up to V/R over many time constants. May I re-emphasize that Ohm's law is valid in the resistor from time 0 out to steady state.

At t=0, Vr = 0, I = 0.

When I = 0.1*V/R, Vr = 0.1*V.

When I = 0.5*V/R, Vr = 0.5*V.

As the current builds up in the resistor the voltage builds up in unison. At every point in time, Vr = I*R. At t=0, there is no current, I=0. The entire voltage, V, is acoss the inductor, while Vr (voltage across resistor) is zero. Then the current builds up. During this time the voltage across the inductor is decreasing & the voltage across the resistor is increasing.

Ohm's law is in effect the entire time, transient as well as steady state.

Again, the resistor voltage increases simultaneously w/ the current, such that the ratio of Vr/I is always equal to the resistance R.

Regarding E fields & currents, the fact that charges drift in the presence of an E field is indisputable. But whenever an E field imparts kinetic energy to a charge, the E field energy is reduced, as conservation of energy is always happening. The E field energy lost is replaced by an energy conversion process, i.e. chemical reaction in battery (redox), mechanical power input to generator, etc. So E fields when exerting force on charges give up energy, & if not replenished cannot sustain a cuuent.

Likewise a current injected into a conductor can produce an E field but again, this E field does not impart force to the network current.

Neither J nor E can drive the other. The power source gives rise to both. Without energy conversion, neither J nor E will be sustained. When a battery is connected across a lamp, what drives E & J? It is the redox chemical process. If a cap is charged, then disconnected from the source, the cap can be switched across the lamp & it will momentarily light up, then fade.

If a current source like the inductor is switched into a lamp, it will glow momentarily, then fade. Just as J produces E, E can also produce J. But to sustain J & E, energy conversion must take place. The solid state physics aspect is intriguing, but sophomore level EE circuits & junior level EE fields is all that is needed. I think some make it way harder than needed. Any comments/feedback welcome.

Claude

 

[kmarinas86]

Above there is a waveform of an R-L circuit whose current builds up to V/R over many time constants. May I re-emphasize that Ohm's law is valid in the resistor from time 0 out to steady state.

At t=0, Vr = 0, I = 0.

When I = 0.1*V/R, Vr = 0.1*V.

When I = 0.5*V/R, Vr = 0.5*V.

As the current builds up in the resistor the voltage builds up in unison. At every point in time, Vr = I*R. At t=0, there is no current, I=0. The entire voltage, V, is acoss the inductor, while Vr (voltage across resistor) is zero. Then the current builds up. During this time the voltage across the inductor is decreasing & the voltage across the resistor is increasing.

Ohm's law is in effect the entire time, transient as well as steady state.

Okay, now I see that "Vr" is your resistive voltage drop. OK.

I just discovered that you can use subscripts.

V_R.

It's a lot better that show it that way.

Again, the resistor voltage increases simultaneously w/ the current, such that the ratio of Vr/I is always equal to the resistance R.

Regarding E fields & currents, the fact that charges drift in the presence of an E field is indisputable. But whenever an E field imparts kinetic energy to a charge, the E field energy is reduced, as conservation of energy is always happening. The E field energy lost is replaced by an energy conversion process, i.e. chemical reaction in battery (redox), mechanical power input to generator, etc. So E fields when exerting force on charges give up energy, & if not replenished cannot sustain a cuuent.

I totally agree with this.

Likewise a current injected into a conductor can produce an E field but again, this E field does not impart force to the network current.

This does not sound right at all. Why should current injected into a conductor be any different than a lightning strike electrocution causing a current to be forced through one's body?

Neither J nor E can drive the other. The power source gives rise to both. Without energy conversion, neither J nor E will be sustained.

What if the power source consists of changing J and changing E?

When a battery is connected across a lamp, what drives E & J? It is the redox chemical process. If a cap is charged, then disconnected from the source, the cap can be switched across the lamp & it will momentarily light up, then fade.

This could be due to changing J and changing E at quantum level.

If a current source like the inductor is switched into a lamp, it will glow momentarily, then fade. Just as J produces E, E can also produce J. But to sustain J & E, energy conversion must take place. The solid state physics aspect is intriguing,

Totally agreed.

but sophomore level EE circuits & junior level EE fields is all that is needed. I think some make it way harder than needed. Any comments/feedback welcome.

Claude

Needed for what? People have different plans, so maybe what they need depends on what they need to do.

 

[Gordianus]

Ohm´s law used to be simple and the Drude model explained very well the electric behaviour of metals. Now, having read many posts I feel totally confused. Do we really need QM to predict how much current flows in a wire?
My simple mind always believed the electric field was the stimulus and the current the result.
I´ve been teching elementary electricity and magnetism and I´ve never seen so much confusion around Ohm´s law.

 

[cabraham]

Ohm´s law used to be simple and the Drude model explained very well the electric behaviour of metals. Now, having read many posts I feel totally confused. Do we really need QM to predict how much current flows in a wire?
My simple mind always believed the electric field was the stimulus and the current the result.
I´ve been teching elementary electricity and magnetism and I´ve never seen so much confusion around Ohm´s law.

This type of prejudice is quite common. But consider what it takes to generate an E field. Charges must be displaced, i.e. separated. First, the mere separation of charges involves moving them , which constitutes current. Second, work must be done to achieve this. Third any E field which imparts force & energy to a charge carrier must lose the same amopunt of energy it imparted. CEL is immutable (conservation of energy).

An ac generator is a little more involved than a battery, so we'll look at the latter. If a battery powers a circuit consisting of conductors & a lamp, one can say that the current in the conductors & lamp is related to the E field per Ohm, i.e. J = sigma*E. But the current in the battery is oriented in the opposite direction. Let's use positive current convention.

In a passive element, current enters the terminal that is the more positive of the two, & exits the more negative terminal. The charges drift in the presence of E field per F = q*E.

But in the battery, positive current exits the positive terminal & enters at the negative terminal. This completely opposes the notion that "the E field drives the current". In the battery, the current is oriented in a direction against the E field. Electrons are moving from the pos to neg terminals inside the battery. This is the exact opposite of what happens if the E field "drives" the current.

So what "actually drives the current?" It is the chemical redox reaction, lead acid, nickel cadmium, or whatever. Energy is spent creating the E field. This E field does indeed impart energy to the conductors & lamp resulting in charge motion, i.e. current. But the E field is diminishing with every electron it moves. The chemical redox reaction keeps replenishing the E field, & current keeps going.

Brilliant minds have analyzed this Ohm's law question since the mid 19th century. "Does J drive E, or vice-versa"? The answer arrived at for more than a century & a half is "Why does one of them have to be the cause of the other? What if both J & E are caused by another entity?" This is the only logical answer. The cause of anything mechanical, electrical, chemical, nuclear, etc. is the transfer of energy. An inductor energized & shorted has energy per 0.5*L*I^2. If the current is diverted from the short to a resistor, this energy is dissipated as heat. First there was J w/o E, then E appeared, then both decayed to zero. The energy associated with each was converted to heat.

Likewise a charged capacitor open has energy per 0.5*C*V^2. When placed across a resistor, we get a J where we formerly had only E. But they soon both vanish. Again, think energy, not J or E being "first". Either can be first, but the cause is always the delivery of energy, while the effect is the receiving of energy.

Nothing else makes any sense at all. Cheers.

Claude

 

[kmarinas86]

Likewise a charged capacitor open has energy per 0.5*C*V^2. When placed across a resistor, we get a J where we formerly had only E. But they soon both vanish. Again, think energy, not J or E being "first". Either can be first, but the cause is always the delivery of energy, while the effect is the receiving of energy.

Nothing else makes any sense at all. Cheers.

Claude

So what caused the delivery of the energy...

Here's a better idea. Our understanding of cause and effect (i.e. as a before and after thing) is an old and outdated understanding. New understanding of cause and effect is that everything results from change. A cause=A change. Everything that is conserved has always existed and will continue to exist in the same quantities. You must explain the transfer of energy as the result of changes without violating conservation. So this change is actually the change of something or more than one something. So the magnitudes of changes are what determine an effect. Cause is therefore more about the rate of change, the change of that rate of change, and so forth. Cause is therefore an ensemble of change, and therefore it cannot ascribed to an unchanging property or object. All that unchanging properties or objects do (assuming they exist) is put permanent limits on the boundaries of change.

 

[kmarinas86]

Brilliant minds have analyzed this Ohm's law question since the mid 19th century. "Does J drive E, or vice-versa"? The answer arrived at for more than a century & a half is "Why does one of them have to be the cause of the other? What if both J & E are caused by another entity?" This is the only logical answer.

J/E is equal to conductivity. Since resistivity is equal to inverse conductivity, the verdict is quite clear. A current is produced by shorting an electric field. This creates a low-resistance path through which charges may travel. If E does not exist, then what is J? On the other hand, E may exist without J in a system with zero conductivity, but without J it cannot increase or decrease. The truth is that they all cause and effect each other simultaneously. What do you think causes the pathway to short? Perhaps its the build up of an electric field at a cathode. Perhaps it is infinite regress.

 

[kmarinas86]

So what "actually drives the current?" It is the chemical redox reaction, lead acid, nickel cadmium, or whatever. Energy is spent creating the E field. This E field does indeed impart energy to the conductors & lamp resulting in charge motion, i.e. current. But the E field is diminishing with every electron it moves. The chemical redox reaction keeps replenishing the E field, & current keeps going.

So what makes you think that this by itself explains how much amps are going to come out of the battery? What is the with the obsession with a "singular" cause? The philosophy of "singular" cause is utter baloney nonsense!

Everything that science is capable of explaining is "caused". What do you think triggers the chemical redox reaction? Something must change outside the cathode to trigger the current yo.

 

[ZapperZ]

My two-pennyworth.

IMO
The classical ideas of fields and the movement of charges (not specifically electrons) don't sit well in the study of electrodynamics.

I rather like the approach taken by Prof Mead of 'explaining' electricity as a quantum effect from the outset. What's more, he points out (not in so many words) that when we study physical systems, we inevitably begin with a simple situation.
Typically we would avoid friction effects (often by setting ourselves up in a separate universe for the experiment!). For that reason, he elects to only consider superconductors.

It sounds crazy but it works. Within the first chapter of his book 'collective electrodynamics' he disposes of the E field and the B field as totally unnecessary artificial constructs and just uses the scalar and vector potentials along with simple QM concepts to completely reconstruct the subject.
He doesn't destroy Maxwell's work, he simply redirects it along the lines Maxwell would have gone if he'd been aware of facts we now know.

I can't recommend it strongly enough.

Just to give you the Flavour:-
He points out that the voltage across the ends of a loop (scalar potential) and the Vector potential A around the loop together satisfy the de Broglie relationship of frequency to wavenumber. By considering charge as a wave, it's therefore possible to specify the potentials as a four-vector throughout space - they depend only on J (also a four vector with the charge density.)
E and B - you don't need - you can easily calculate a value for them at any point if you want - but it turns out that most of the time you don't need to.

I know of Carver Mead's work quite well, In fact, if you do a search on PF, I've cited his PNAS paper on his Collective Electrodynamics several times. This is because he cited superconductivity as being the clearest manifestation of QM at the macroscopic scale. His reformulation of E&M fields is actually quite interesting and refreshing.

However, his work cannot derive, for example, Ohm's law, nor can he arrive at an explanation for "resistivity" in metals. Maxwell equations can't either. This is because this is not an issue of electromagnetic field, but rather a materials property (it is why such derivations are very seldom done in E&M classes, but rather, in solid state classes). Mead was able to describe, using his picture, various phenomena of superconductivity because the charge carrier responsible for that phenomenon has long-range coherence. The supercurrent does not interact with the microscopic details of the bulk material, what David Pines termed as a state of "quantum protectorate". So in essence, dealing with just the supercurrent made it "easier".

This is not the case with ordinary metals in the normal state. The question is, how are charges transported from one location to another, resulting in what we call a "current". We know that the charge carriers undergo several interactions: (i) electron-electron interactions, (ii) electron-ion (or phonon) interactions (iii) electron-impurities interactions), etc. (refer to, say, Valla et al., PRL 83, 2085(1999)). At room temperature, in a typical metal, the electron-ion/phonon interactions dominates, resulting in the fact that a free electron trying to move, will not make it way past the mean-free path. That's it. Without any external field, any electrons moving through such a material will thermalize. This is the origin of resistivity. One can derive from First Principles the resistivity of various materials using such a model.

This area is such a well-studied subject in condensed matter physics, because transport phenomenon is one of the most important aspects of that field. I could easily point out to the QFT approach to such a phenomenon (to answer the question from another member who wanted to know if such a thing has to be handled quantum mechanically) using what we call as the propagator, and arrive as the single-particle spectral function. Here it is even clearer that, analogous to the random walk problem, each charge carrier WILL experience multiple interactions impeding its motion. It is why this is a many-body physics problem. It is why the "electron" that we detect in a material does not have the same "mass" as the bare material. In heavy fermionic system, the charge carrier can have a mass more than 200 times the bare mass!

In all of these, there is a clear cause-and-effect, especially in how one gets Ohm's Law, how charges move through a material, etc.

Zz.

 

[Fernsanz]

This type of prejudice is quite common. But consider what it takes to generate an E field. Charges must be displaced, i.e. separated. First, the mere separation of charges involves moving them , which constitutes current. Second, work must be done to achieve this. Third any E field which imparts force & energy to a charge carrier must lose the same amopunt of energy it imparted. CEL is immutable (conservation of energy).

An ac generator is a little more involved than a battery, so we'll look at the latter. If a battery powers a circuit consisting of conductors & a lamp, one can say that the current in the conductors & lamp is related to the E field per Ohm, i.e. J = sigma*E. But the current in the battery is oriented in the opposite direction. Let's use positive current convention.

In a passive element, current enters the terminal that is the more positive of the two, & exits the more negative terminal. The charges drift in the presence of E field per F = q*E.

But in the battery, positive current exits the positive terminal & enters at the negative terminal. This completely opposes the notion that "the E field drives the current". In the battery, the current is oriented in a direction against the E field. Electrons are moving from the pos to neg terminals inside the battery. This is the exact opposite of what happens if the E field "drives" the current.

So what "actually drives the current?" It is the chemical redox reaction, lead acid, nickel cadmium, or whatever. Energy is spent creating the E field. This E field does indeed impart energy to the conductors & lamp resulting in charge motion, i.e. current. But the E field is diminishing with every electron it moves. The chemical redox reaction keeps replenishing the E field, & current keeps going.

Brilliant minds have analyzed this Ohm's law question since the mid 19th century. "Does J drive E, or vice-versa"? The answer arrived at for more than a century & a half is "Why does one of them have to be the cause of the other? What if both J & E are caused by another entity?" This is the only logical answer. The cause of anything mechanical, electrical, chemical, nuclear, etc. is the transfer of energy. An inductor energized & shorted has energy per 0.5*L*I^2. If the current is diverted from the short to a resistor, this energy is dissipated as heat. First there was J w/o E, then E appeared, then both decayed to zero. The energy associated with each was converted to heat.

Likewise a charged capacitor open has energy per 0.5*C*V^2. When placed across a resistor, we get a J where we formerly had only E. But they soon both vanish. Again, think energy, not J or E being "first". Either can be first, but the cause is always the delivery of energy, while the effect is the receiving of energy.

Nothing else makes any sense at all. Cheers.

Claude

A really good post.

But, for the sake of simplicity and to focus on the essence of the problem, let's get rid off bateries and sources. Just imagine a current flowing in a superconductor loop with no resistance. Then, at some time, let's label it $$t=t_0$$, I insert an ohmic resistor in the loop. At that very instant I will see a voltage i.e. an electric field across the resistor. Of course, both the current and the voltage will quickly decay to 0 since there is no source to keep it flowing but that is irrelevant for our question: how did that E-field arise?

 

[cabraham]

So what makes you think that this by itself explains how much amps are going to come out of the battery? What is the with the obsession with a "singular" cause? The philosophy of "singular" cause is utter baloney nonsense!

Everything that science is capable of explaining is "caused". What do you think triggers the chemical redox reaction? Something must change outside the cathode to trigger the current yo.

I never believed in a singular cause. I was just emphasizing that the energy has to be conserved, & must originate somewhere, i.e. redox. The OP question was related to J & E, as to one causing the other. I was merely explaining that they are interactive, mutual, & require energy conversion/source to sustain.

As far was what determines how many amps are supplied by the battery, Sir Oliver Heaviside laid that to rest in the 1870's.

In a previous post you stated

"The truth is that they all cause and effect each other simultaneously"

That has been my position since forever. Reread my posts on this forum or any other & that is what I/ve been saying for years. I've never bought into the belief that 1 specific variable is the cause of another under all conditions.

Claude

 

[Fernsanz]

I attach this fantastic paper I've found with title: "Electric Field and Plasma Flow: What Drives What?"

It is no exactly the same question that the OP one but very very similar. This is exactly the type of argument I was expecting. As clearly stated by the author, definitions of cause and effect, who goes first and who goes after, can be rigurously obtained with no philosophical burden.

I suspect that, as the subject of the paper is magnetohydrodynamics, something similar can be done for our subject, if it has not been done already. In fact his paper contain the words "Ohm's law" and touch the subject very close.

I would love an analogous paper to exist for Ohm's relation.

PS: The paper concludes, among other things, that flow also drives electrical field; hence, the equation can be interpreted symetrically depending on what is considered the stimulus and what the response. This is just the case for Ohm's law also.

 

[Fernsanz]

Ohm´s law used to be simple and the Drude model explained very well the electric behaviour of metals. Now, having read many posts I feel totally confused. Do we really need QM to predict how much current flows in a wire?
My simple mind always believed the electric field was the stimulus and the current the result.
I´ve been teching elementary electricity and magnetism and I´ve never seen so much confusion around Ohm´s law.

That is because most people take statements as they are given with little critical attitude towards it, if any. If you dig just a for a moment into many things we think we know well you will end up with more questions than certainties. And that, in turn, will lead you to gain deep insight on subjects most people haven't even realized. Ohm's law is an example as it is the question "why water evaporates below 100ºC when everybody knows that below 100ºC it is liquid". Just an example to comment on your appreciated post.

I hope that, in the end of this thread your view of Ohm's law be really deeper and different. Less simple it seems at first glance. I don't think your mind is simple, by the way.

 

[kmarinas86]

That is because most people take statements as they are given with little critical attitude towards it, if any. If you dig just a for a moment into many things we think we know well you will end up with more questions than certainties. And that, in turn, will lead you to gain deep insight on subjects most people haven't even realized. Ohm's law is an example as it is the question "why water evaporates below 100ºC when everybody knows that below 100ºC it is liquid". Just an example to comment on your appreciated post.

I hope that, in the end of this thread your view of Ohm's law be really deeper and different. Less simple it seems at first glance. I don't think your mind is simple, by the way.

Word.