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


by Fernsanz
Tags: current, direction, effect, electric, field, induces
Fernsanz
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#1
Jun30-10, 12:30 PM
P: 56
Hi,

Lately I've been concerned about the Ohm's law [tex]J=\sigma E[/tex] 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 [tex]E[/tex] over a ohmic conductor will cause a current [tex]J[/tex]. 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 [tex]J[/tex] injected somehow in a ohmic conductor will cause a electric field [tex]E[/tex], 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.
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cabraham
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#2
Jun30-10, 12:45 PM
P: 997
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
Fernsanz
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Jun30-10, 01:46 PM
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Quote Quote by cabraham View Post
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.
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 im asking about which is the electrical field that appears in Ohm's low, not any other mechanical forces)

cabraham
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Jun30-10, 05:34 PM
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Ohm's law, why current induces electric field and cause effect direction


Quote Quote by Fernsanz View Post
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 loss all his 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 im 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
Fernsanz
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Jul1-10, 01:01 AM
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Quote Quote by cabraham View Post
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 [tex]F=qE[/tex].
kmarinas86
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#6
Jul1-10, 12:10 PM
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Quote Quote by Fernsanz View Post
If mechanical and electrical are the same for you good for you, but there is no explanation in your words on how an ELECTRICAL field arise 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 ELECTRICAL field arise; 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 ELECTRICAL field, i.e., a field that would cause a test charge placed inside the conductor to move purely by electrical force [tex]F=qE[/tex].
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.
Fernsanz
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#7
Jul1-10, 12:36 PM
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Quote Quote by kmarinas86 View Post
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 thats 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
ZapperZ
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Jul1-10, 01:03 PM
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Quote Quote by Fernsanz View Post
Hi,

Lately I've been concerned about the Ohm's law [tex]J=\sigma E[/tex] 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 [tex]E[/tex] over a ohmic conductor will cause a current [tex]J[/tex]. 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 [tex]J[/tex] injected somehow in a ohmic conductor will cause a electric field [tex]E[/tex], 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.
Maybe I'm missing something here. Can you tell me where in Ohm's law does it state that moving charges in a conductor will can an electric field to be induced, and that this electric field is the "E" in Ohm's law?

Note that I can shoot electrons in vacuum (we do that all the time in a particle accelerator) and after the accelerating structure, it can simply coast without any applied field. So you have current, but no applied field that cause this current, nor does it induced the SAME field in reverse. It will have other fields, such as magnetic field or time varying electric field, but this electric field is NOT the same electric field as referred to in Ohm's Law, i.e. constant axial field in the direction of motion.

Zz.
Fernsanz
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#9
Jul1-10, 02:14 PM
P: 56
Quote Quote by ZapperZ View Post
Maybe I'm missing something here. Can you tell me where in Ohm's law does it state that moving charges in a conductor will can an electric field to be induced, and that this electric field is the "E" in Ohm's law?
In an Ohmic conductor the Ohm's law says that [tex]J=\sigma E[/tex]. So, according to this constitutive relation whenever a [tex]J[/tex] exists in the conductor you have a electric field [tex]E=\frac{1}{\sigma}J[/tex]. Pretty obvious, isn't it?

Are you suggesting that we can have the current with or without electric field depending on how we generate that current? i.e., are you suggesting that we can't read the Ohm's law in the direction [tex]J[/tex] implies [tex]E[/tex] but we can read it in the direction [tex]E[/tex] implies [tex]J[/tex]? If you are claiming this then we should warn electrical engineers around the world because the relation V=IR is not valid when the current is the first cause (for example when the current is generated in a photodiode); and obviously this is not the case
Fernsanz
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Jul1-10, 03:13 PM
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Quote Quote by Fernsanz View Post
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 thats 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?
Interestingly enough I have come across this post from the thread "Why current attain steady state": http://www.physicsforums.com/showpos...31&postcount=5
It is quite clarifying and supports my previous picture of sound wave to explain the propagation of the electric field.

This also explain that electric field induced when the current is the "first cause" is exactly the same as the one obtained from Ohm's law, because in any case the current has to be the dynamic solution of an existing electric field. So now it's clear for me: the Ohm's law is a true constitutive relation which relates two quantities no matter which one is the "first cause"

Thanks AJ Bentley
cabraham
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#11
Jul1-10, 10:44 PM
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The answer to the OP question "how is it that if the current is first that it can produce an E field?" is that the injected charge carrriers provide the said E field.

Remember that electrons & holes being charged, possess an E field of their own. Let's say we have a conductor, a good one but not a superconductor. Let's say it is connected across a current source suddenly. The holes & electrons enter the conductor. Upon arrival, they bang into the lattice resulting in heat dissipation known as resistance. Each charge carrier can approach a limited average velocity owing to collisions.

Hence a charge distribution will be effected. At the end where electrons enter, an accumulation of e- results in an E field since e- possess such a field. Likewise with holes. h+ at the opposite end. If you prefer, a hole entering is equivalent to an e- exiting. So one end has an accumulation of h+, the other has e-. There is an E field whose lines start on a hole & end on an electron, as well as a potential. The potential V, is just the line integral of E over the path.

In a superconductor, SC, there is no resistance, hence no E field. When e- enter a SC, they incur no collisions, & their tendency is to diffuse apart from one another. When they reach the surface, they cannot continue outward since energy is needed. So a charge accumulation exists on the surface of a SC. Hence, in the absence of interior charge unbalance, there can be no E field on the interior of the SC.

Did this help?

Claude
Fernsanz
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Jul2-10, 02:31 AM
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Quote Quote by cabraham View Post
The answer to the OP question "how is it that if the current is first that it can produce an E field?" is that the injected charge carrriers provide the said E field.
I'm not searching an answer anymore, I have found out by which mechanism an E-field is induced. As I said in my last post is by acelerating charges producing a e-field wave down the conductor.

Quote Quote by cabraham View Post
Remember that electrons & holes being charged, possess an E field of their own. Let's say we have a conductor, a good one but not a superconductor. Let's say it is connected across a current source suddenly. The holes & electrons enter the conductor. Upon arrival, they bang into the lattice resulting in heat dissipation known as resistance. Each charge carrier can approach a limited average velocity owing to collisions.

Hence a charge distribution will be effected. At the end where electrons enter, an accumulation of e- results in an E field since e- possess such a field. Likewise with holes. h+ at the opposite end. If you prefer, a hole entering is equivalent to an e- exiting. So one end has an accumulation of h+, the other has e-. There is an E field whose lines start on a hole & end on an electron, as well as a potential. The potential V, is just the line integral of E over the path.
First, in a metalic conductor like, for instance, copper, there are no holes. So the reasoning has nothing to do with holes.

Second, and once again, the E-field is not produced as a result of charge accumulation. An accumulation of electrons radiates an E-field in all directions, not just in the direction of current, so an accumulation would accelerate electrons which are behind it and would slow down electrons which are before it, i.e. would spread electrons out of the accumulation in all directions. Furthermore you are seeing the current inside the conductor as a succession of charged zones which is not true since the conductor carrying a current is homogeneous

Quote Quote by cabraham View Post
In a superconductor, SC, there is no resistance, hence no E field. When e- enter a SC, they incur no collisions, & their tendency is to diffuse apart from one another. When they reach the surface, they cannot continue outward since energy is needed. So a charge accumulation exists on the surface of a SC. Hence, in the absence of interior charge unbalance, there can be no E field on the interior of the SC.
Again touching the resistance subject which is not related which my question. And, I insist a last time, you can not explain a directed current on the basis of charge accumulation. Charge acummulation does not push neighbour electrons in a preferred direction, so no net current can be generated (as much electrons will be pushed forward as backwards).

As I said I have found the answer I wrote in my previous post. Thank you anyway.
Fernsanz
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Jul2-10, 02:42 AM
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Quote Quote by cabraham View Post
The answer to the OP question "how is it that if the current is first that it can produce an E field?" is that the injected charge carrriers provide the said E field.
I'm not searching an answer anymore, I have found out by which mechanism an E-field is induced. As I said in my last post is by acelerating charges producing a e-field wave down the conductor.

Quote Quote by cabraham View Post
Remember that electrons & holes being charged, possess an E field of their own. Let's say we have a conductor, a good one but not a superconductor. Let's say it is connected across a current source suddenly. The holes & electrons enter the conductor. Upon arrival, they bang into the lattice resulting in heat dissipation known as resistance. Each charge carrier can approach a limited average velocity owing to collisions.

Hence a charge distribution will be effected. At the end where electrons enter, an accumulation of e- results in an E field since e- possess such a field. Likewise with holes. h+ at the opposite end. If you prefer, a hole entering is equivalent to an e- exiting. So one end has an accumulation of h+, the other has e-. There is an E field whose lines start on a hole & end on an electron, as well as a potential. The potential V, is just the line integral of E over the path.
First, in a metalic conductor like, for instance, copper, there are no holes. So the reasoning has nothing to do with holes.

Second, and once again, the E-field is not produced as a result of charge accumulation. An accumulation of electrons radiates an E-field in all directions, not just in the direction of current, so an accumulation would accelerate electrons which are behind it and would slow down electrons which are before it, i.e. would spread electrons out of the accumulation in all directions. Furthermore you are seeing the current inside the conductor as a succession of charged zones which is not true since the conductor carrying a current is homogeneous

Quote Quote by cabraham View Post
In a superconductor, SC, there is no resistance, hence no E field. When e- enter a SC, they incur no collisions, & their tendency is to diffuse apart from one another. When they reach the surface, they cannot continue outward since energy is needed. So a charge accumulation exists on the surface of a SC. Hence, in the absence of interior charge unbalance, there can be no E field on the interior of the SC.
Again touching the resistance subject which is not related which my question. And, I insist a last time, you can not explain a directed current on the basis of charge accumulation. Charge acummulation does not push neighbour electrons in a preferred direction, so no net current can be generated (as much electrons will be pushed forward as backwards).

As I said I have found the answer I wrote in my previous post. Thank you anyway.
cabraham
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Jul2-10, 06:49 AM
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Quote Quote by Fernsanz View Post
I'm not searching an answer anymore, I have found out by which mechanism an E-field is induced. As I said in my last post is by acelerating charges producing a e-field wave down the conductor.

First, in a metalic conductor like, for instance, copper, there are no holes. So the reasoning has nothing to do with holes.

Second, and once again, the E-field is not produced as a result of charge accumulation. An accumulation of electrons radiates an E-field in all directions, not just in the direction of current, so an accumulation would accelerate electrons which are behind it and would slow down electrons which are before it, i.e. would spread electrons out of the accumulation in all directions. Furthermore you are seeing the current inside the conductor as a succession of charged zones which is not true since the conductor carrying a current is homogeneous

Again touching the resistance subject which is not related which my question. And, I insist a last time, you can not explain a directed current on the basis of charge accumulation. Charge acummulation does not push neighbour electrons in a preferred direction, so no net current can be generated (as much electrons will be pushed forward as backwards).

As I said I have found the answer I wrote in my previous post. Thank you anyway.
I used "holes" in a loose sense. Also, peer-reviewed e/m fields texts insist that the charge distribution inside the conductor is indeed a source of E field. Where did you read otherwise? Of course a single e- radiates an E field in all directions. But a group of e- at one end, & a void of e- at the other end of a conductor has an E field directed from +ve to the -ve end. There is an E field thast extends outward as well, & it does slow down e- "before it".

Why would peer reviewed uni texts say that if it were not so? What are your references? I never expalined a directed current based on charge accumulation. The charge accumulation accounts for the E field inside the good conductor (not SC). Please don't put words in my mouth.

I try to help you by giving free advice which takes years to acquire, & you rudely rebuke me with straw man arguments, correcting me on issues I never raised. Just out of curiosity, how far along did you get regarding education. Are you a practicing scientist/EE? Just curious. You have a lot of sass & moxie with little chops to back it up. The problem with these forums is that everybody thinks they know more than everyone else. You exemplify that attitude.

Claude
ZapperZ
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Jul2-10, 07:12 AM
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Quote Quote by Fernsanz View Post
In an Ohmic conductor the Ohm's law says that [tex]J=\sigma E[/tex]. So, according to this constitutive relation whenever a [tex]J[/tex] exists in the conductor you have a electric field [tex]E=\frac{1}{\sigma}J[/tex]. Pretty obvious, isn't it?

Are you suggesting that we can have the current with or without electric field depending on how we generate that current? i.e., are you suggesting that we can't read the Ohm's law in the direction [tex]J[/tex] implies [tex]E[/tex] but we can read it in the direction [tex]E[/tex] implies [tex]J[/tex]? If you are claiming this then we should warn electrical engineers around the world because the relation V=IR is not valid when the current is the first cause (for example when the current is generated in a photodiode); and obviously this is not the case
In physics, you cannot simply rearrange an equation and then reinterpret it and expect it to make sense. That's why physics isn't mathematics. There is such a thing as the dependent and independent variable! It is based on causality - A causes B, but B may not necessarily causes A.

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!

I still want to see this experiment where you shoot electrons into a conductor, and this process, in turn, generates the corresponding E field that is similar to the E-field that causes the current in the reverse process. I shoot electrons into metals all the time (it's part of my job) - we call it either a beam stop, or a Faraday Cup. In none of these do you generate the identical E-field in the conductor that is described by Ohm's Law.

Zz.
kmarinas86
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Jul2-10, 09:10 AM
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Quote Quote by ZapperZ View Post
Maybe I'm missing something here. Can you tell me where in Ohm's law does it state that moving charges in a conductor will can an electric field to be induced, and that this electric field is the "E" in Ohm's law?
He can't. It cannot be done. Ohm's law considers no such things. The "E" in Ohm's law is not the "E" of the real world. It is fictitious. It does not even necessarily equal the projection of the "E" over a differential length of the conductor. That only works if the electric field inside the wire is uniform and if the electric field outside the wire approaches each wire element at the same angle. Only in very a contrived circumstance with a bizarrely shaped electric field could you meet those conditions in a curved wire element. If the real "E" varies over the length of the conductor, speaking of the "E" in Ohm's law is like talking about the tooth fairy.

Quote Quote by ZapperZ View Post
Note that I can shoot electrons in vacuum (we do that all the time in a particle accelerator) and after the accelerating structure, it can simply coast without any applied field. So you have current, but no applied field that cause this current, nor does it induced the SAME field in reverse. It will have other fields, such as magnetic field or time varying electric field, but this electric field is NOT the same electric field as referred to in Ohm's Law, i.e. constant axial field in the direction of motion.
And because an electric field (changing or not) is needed to generate a current, the "E" of Ohm's law is not the electric field.

Quote Quote by ZapperZ View Post
In physics, you cannot simply rearrange an equation and then reinterpret it and expect it to make sense. That's why physics isn't mathematics. There is such a thing as the dependent and independent variable! It is based on causality - A causes B, but B may not necessarily causes A.
Some people allege that a changing electric field does not "generate" a magnetic field and that it's just a correlation. Many of those same people allege the opposite to be the case as well.

Ohm's "law" is very quaint. For example, [tex]\mathbf{J}=\sigma\mathbf{E}[/tex] applies only for a uniform field in the direction of the current. What kind of "law" is that?

There is no "right-hand law" in physics. Left-handed materials exist. We have something called "the right-hand rule". Ohm's "law" should be called "Ohm's rule". I did a Google search, and it turns out many concur.

Quote Quote by ZapperZ View Post
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 to give an example of a situation without an electric field that can generate a current? You could try gravity, but you have to assume it is not fundamentally due to electromagnetism. You could also hook up a battery to a remote control. One could argue that it is external to the conductor, but it is part of the same circuit, so it still works because its field is "internal" to that circuit just as the current it produces will be.

Quote Quote by ZapperZ View Post
Want another example? Look at the current from a photoemission process. Look ma! No electric field!
The photon has an electric field. It is changing, but so what? It is still an electric field.

Quote Quote by ZapperZ View Post
I still want to see this experiment where you shoot electrons into a conductor, and this process, in turn, generates the corresponding E field that is similar to the E-field that causes the current in the reverse process. I shoot electrons into metals all the time (it's part of my job) - we call it either a beam stop, or a Faraday Cup. In none of these do you generate the identical E-field in the conductor that is described by Ohm's Law.

Zz.
You're right, it's not identical.
jtbell
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Jul2-10, 10:40 AM
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Quote Quote by kmarinas86 View Post
[tex]\mathbf{J}=\sigma\mathbf{E}[/tex] applies only for a uniform field in the direction of the current.
Neither the field nor the current density need to be uniform. If the conductor is isotropic (conducts equally well in all directions), then at each point the current density and the electric field have the same direction, but possibly different directions at different points.

If the conductor is anisotropic, the current density and electric field can be in different directions at a given point. In this case you have to use a tensor for the conductivity.
cmos
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Jul2-10, 12:01 PM
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Quote Quote by ZapperZ View Post
In physics, you cannot simply rearrange an equation and then reinterpret it and expect it to make sense. That's why physics isn't mathematics. There is such a thing as the dependent and independent variable! It is based on causality - A causes B, but B may not necessarily causes A.
I think this is one of the best quotes I have ever seen on this forum. I encourage everybody reading this thread to really let ZapperZ's quote sink in. It is directly relevant to the topic at hand as well as to many of the misconceptions that commonly occur to students (both formal and informal) of physics.

Quite often, equations will be written in an "aesthetic" form. The left hand side (LHS) may sometimes be the cause, other times it may be the effect, yet other times the equation will simply be a mutual relation.

In the present case of Ohm's law,
[tex]J=\sigma E[/tex],
J is the effect and E is the cause. I suppose this makes perfect sense writing the equation this way because it places the dependent variable on the LHS and the independent variable on the RHS, i.e. it is in the from y(x)=mx.

Not to get off-topic, but I believe a few examples might be helpful to some....

Newton's second law: F=ma
F is the cause and a is the effect; forces cause bodies to accelerate, not vice versa. Why is it commonly written in the opposite form as Ohm's law? Easier to remember, looks nicer.... who cares?

The most famous equation in all of physics:
[tex]E=mc^2[/tex]
This is simply, what I called above, a mutual relation. Rest energy is mass, mass is rest energy.


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