Electromagnetism question about moving a current-carrying wire instead of the test charge

  • #1
PhDnotForMe
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OK so a wire with a current creates a b field in accordance with the right hand rule. Now moving charge will feel a force (if its moving in the correct direction) while stationary charges will not feel a magnetic force.

What I am curious about is if we move the wire itself, will a stationary charge possibly feel a magnetic force? From the perspective of the wire, the charge is moving, so I assume there would be a magnetic force even though the charge is stationary.
 
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  • #2
A.T.
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...if we move the wire itself,...From the perspective of the wire, ...
You have to decide what reference frame you use.
 
  • #3
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What I am curious about is if we move the wire itself, will a stationary charge possibly feel a magnetic force? From the perspective of the wire, the charge is moving, so I assume there would be a magnetic force even though the charge is stationary.
The stationary charge cannot experience a magnetic force since its velocity is zero and the magnetic force is equal to ##q(\mathbf v \times \mathbf B)##.

However, for the symmetry reason that you mention it must experience a force. And since it is not moving this force must be an electric force. This implies that a purely magnetic field in one frame will transform to have at least some electric field in another frame.
 
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  • #4
PhDnotForMe
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The stationary charge cannot experience a magnetic force since its velocity is zero and the magnetic force is equal to ##q(\mathbf v \times \mathbf B)##.

However, for the symmetry reason that you mention it must experience a force. And since it is not moving this force must be an electric force. This implies that a purely magnetic field in one frame will transform to have at least some electric field in another frame.

So if I have a wire with AC current and then I physically move the wire back and forth along its length at the same frequency of the AC current, I would perceive my wire emitting an electric field?
 
  • #5
tech99
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The stationary charge cannot experience a magnetic force since its velocity is zero and the magnetic force is equal to ##q(\mathbf v \times \mathbf B)##.

However, for the symmetry reason that you mention it must experience a force. And since it is not moving this force must be an electric force. This implies that a purely magnetic field in one frame will transform to have at least some electric field in another frame.
I think Maxwell was aware of this symmetry problem and was on the edge of understanding Relativity.
 
  • #6
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So if I have a wire with AC current and then I physically move the wire back and forth along its length at the same frequency of the AC current, I would perceive my wire emitting an electric field?
Yes. Also yes if the current were DC or if the movement were at a different frequency or if the direction were different or if the motion were not sinusoidal.
 
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  • #7
vanhees71
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OK so a wire with a current creates a b field in accordance with the right hand rule. Now moving charge will feel a force (if its moving in the correct direction) while stationary charges will not feel a magnetic force.

What I am curious about is if we move the wire itself, will a stationary charge possibly feel a magnetic force? From the perspective of the wire, the charge is moving, so I assume there would be a magnetic force even though the charge is stationary.
There is only an electromagnetic field. The split into electric and magnetic components is frame dependent.

In this connection it is important to note that the wire is uncharged in the rest frame of the conduction electrons rather than the restframe of the wire. This is overlooked in almost all treatments of this problem, because a non-relativistic approximation for Ohm's Law is used and thus the Hall effect is neglected. For a nice treatment, see

P. C. Peters, In what frame is a current-conducting wire
neutral, Am. J. Phys. 53 (1985) 1156.
https://doi.org/10.1119/1.14075
 
  • #8
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P. C. Peters, In what frame is a current-conducting wire
neutral, Am. J. Phys. 53 (1985) 1156.
https://doi.org/10.1119/1.14075
Do you have a non-paywalled version of this paper? I am particularly interested in what assumptions they make regarding the grounding of the current-conducting wire. I have thought about this issue before but never pursued it in detail. When I thought about it I suspected that the bulk of the wire would be uncharged in the lab frame only if the center of the wire is grounded, which is indeed not the usual arrangement for a lab.
 
  • #9
vanhees71
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Unfortunately, I guess there's no (legal) non-paywalled version of the paper. The argument is simple: In the frame with the wire at rest the (quasi-)free conduction electrons are deflected by the magnetic field, which builds up a negative charge density within the wire and an electric field such that the corresponding force compensates the magnetic force, such that in the steady state there's no radial current (Hall effect). In the frame where the conduction electrons are at rest there's also a magnetic field due to the moving positive wire, but no force acts on the freely moving conduction electrons since these are at rest. The bulk of the wire (i.e., the wire without the freely moving conduction electrons) doesn't contain freely moving charges and thus no electric field is built up. In this frame thus the charge density must vanish.

I once derived this result without knowing about this paper in a different way, simply using the correct relativistic version of Ohm's Law,
$$\vec{j}=\sigma (\vec{E}+\vec{v} \times \vec{B}/c),$$
assuming that ##\vec{j}## is homogenous over the cross section of the wire and directed along the wire (as in the usual textbook ansatz, using the incomplete Ohm's law neglecting the magnetic force). The resulting charge density is ##\mathcal{O}(v^2/c^2)##, where ##v## is the speed of the conduction electrons, i.e., totally academic in view of the fact that house-hold currents imply speeds in the range of mm/s.

Here's a writeup of my treatment of the DC-carrying coax cable (though it's in German). The relativistic part is Sect. 6 starting at p. 11:

https://itp.uni-frankfurt.de/~hees/faq-pdf/coax.pdf
 
  • #10
PhDnotForMe
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If there is a wire pointed up with current moving upward, the wire creates a respective magnetic field due to the RHR. Say there is a stationary proton floating 1cm away from the wire. The proton does not move since particles only feel magnetic field when they move.
What if you considered this scenario under another frame of reference. Say there is a free floating neutron moving upward very slowly. From the neutron's frame it sees the proton moving. Does this mean the magnetic field of the wire does not exist in the neutrons frame of reference? If it did, wouldn't the neutron observe a magnetic force on the proton?
 
  • #11
PhDnotForMe
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In the frame where the conduction electrons are at rest there's also a magnetic field due to the moving positive wire, but no force acts on the freely moving conduction electrons since these are at rest.
I'm confused. How can electrons move outwards in one frame but not move outwards in another frame? In the frame with stationary electrons, wouldn't symmetry tell us there must be an electric force that pushes the electrons outward?
 
  • #12
vanhees71
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There's no force of the electrons pushing them outward in any frame. In the frame, where the wire is at rest (i.e., the positive "background") there's no force, because there's both a radial electric and a magnetic field from the other moving electrons such that the radial force is precisely 0 (Hall effect).

In the frame, where the electrons are at rest there's only a magnetic field, which doesn't lead to a force on the resting electron.

There's no force acting on the electrons at all in the stationary state, because the electrons move with constant speeds along the wire. In longitudinal direction you have the driving force of the applied electric field along the wire, which is (in the stationary state) compensated by the friction force, described by the "transport coefficient" electric conductivity.
 
  • #13
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If there is a wire pointed up with current moving upward, the wire creates a respective magnetic field due to the RHR. Say there is a stationary proton floating 1cm away from the wire. The proton does not move since particles only feel magnetic field when they move.
What if you considered this scenario under another frame of reference. Say there is a free floating neutron moving upward very slowly. From the neutron's frame it sees the proton moving. Does this mean the magnetic field of the wire does not exist in the neutrons frame of reference? If it did, wouldn't the neutron observe a magnetic force on the proton?
The electric and magnetic fields are frame dependent. In a frame moving with the charges, there is no magnetic field.

In this way there are different explanations in different frames for the same physical behaviour.

It was trying to explain this that led Einstein to SR.
 
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  • #14
PhDnotForMe
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In the frame with the wire at rest the (quasi-)free conduction electrons are deflected by the magnetic field, which builds up a negative charge density within the wire and an electric field such that the corresponding force compensates the magnetic force, such that in the steady state there's no radial current (Hall effect).
I think this is the part I'm stuck on. If there is no force acting on the electrons, what causes the deflection?
 
  • #15
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I'm confused. How can electrons move outwards in one frame but not move outwards in another frame?
They move outward in all frames, but the explanation is frame dependent. In some frames it is a Hall effect and in others it is Coulomb repulsion and in others it is a combination.
 
  • #16
PhDnotForMe
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They move outward in all frames, but the explanation is frame dependent. In some frames it is a Hall effect and in others it is Coulomb repulsion and in others it is a combination.
So they DO move outwards? I believe @vanhees71 is saying otherwise. "There's no force of the electrons pushing them outward in any frame. "
 
  • #17
vanhees71
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I think this is the part I'm stuck on. If there is no force acting on the electrons, what causes the deflection?
I should have said, there's no NET force acting on the electrons.
 
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  • #18
PhDnotForMe
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So if I have a moving proton, that proton creates a magnetic field via the RHR. Say this proton, proton A, moves past a stationary proton, proton S. The protons experience electric force, but because proton S is not moving, there is no magnetic force.

Now imagine a frame where a neutron is moving in the same direction as proton A but half as fast. In the neutrons frame, this neutron would see the moving proton A moving at a slower speed, and it would see proton S moving as well. This would mean that the neutron would observe a magnetic force due to F=qvB where B is the magnetic field created by the proton A and v is the observed velocity of proton S.

So from the frame of proton S there is no magnetic force, but from the frame of the neutron there is a magnetic force. How is this possible? I don't see how the electric force would be different in either scenario, so how would the magnetic force change?
 
  • #19
Vanadium 50
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This is the third thread you've started on moving charges. Perhaps you should start with what you don't understand the first two times this was explained to you.

<Moderator's note: all posts are now in this thread.>
 
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  • #20
PhDnotForMe
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This is the third thread you've started on moving charges. Perhaps you should start with what you don't understand the first two times this was explained to you.
The previous ones states that electric field and magnetic field change with respect to the frame. In this example I do not see how the electric field can be any different regardless of the chosen frame.
 
  • #23
A.T.
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Okay, so I'm looking at the line that reads
E⊥′ = γ (E⊥ + vB)
In "vB" what is v?
The symbols are explained in the paragraph above the equations.
 
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  • #25
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In "vB" what is v?
##v## is the velocity of the frame that you are transforming to relative to the frame that you are transforming from. That is described in both the picture and the text immediately preceding the equations.

B field created from moving point charge is proportionate to the v velocity of said point charge.
The velocity of the sources is not needed here. This is a transformation of the fields themselves and applies regardless of the motion of the sources in the first frame. To transform the sources see the section on ##\textbf{The } \mathbf{\rho} \textbf{ and J fields}##

Does that mean vB proportionate to v² ? Where v is the relative velocity of the particle?
Not in general, no.
 
  • #26
PhDnotForMe
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##v## is the velocity of the frame that you are transforming to relative to the frame that you are transforming from. That is described in both the picture and the text immediately preceding the equations.

The velocity of the sources is not needed here. This is a transformation of the fields themselves and applies regardless of the motion of the sources in the first frame. To transform the sources see the section on ##\textbf{The } \mathbf{\rho} \textbf{ and J fields}##

Not in general, no.
I see that v is the velocity of the frame that you are transforming to relative to the frame that you are transforming from. My issue is with B. B is dependent on v as well. And I can't seem to see how these two v's would be anything but equal.
 
  • #27
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My issue is with B. B is dependent on v as well.
Look at Maxwell’s equations. B depends on J, not v. There is no v anywhere in Maxwell’s equations.

The transformation for J is given in that other section.
 
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  • #28
PhDnotForMe
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Look at Maxwell’s equations. B depends on J, not v. There is no v anywhere in Maxwell’s equations.

The transformation for J is given in that other section.
It does depend of J. That is correct. From what I'm looking at J depends on v. The equation being J = pv.

https://en.m.wikipedia.org/wiki/Current_density
When considering just two particles, I still see no way for the v in context of "vB" to be a different v than the one found in "J=pv"
 
  • #29
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From what I'm looking at J depends on v. The equation being J = pv.
I am not sure what you are “looking at”. In a specific case that may be true, but in general J is not necessarily related to v. Be careful not to over-generalize if you are making claims based on specific cases.

My comments above are general. They are not limited to specific cases.

When considering just two particles, I still see no way for the v in context of "vB" to be a different v than the one found in "J=pv"
Since ##v## is already defined as the relative velocity between the primed and unprimed frame and since the charge may have any velocity ##u## in the unprimed frame then there are an infinite number of possibilities where ##u\ne v## and ##J=\rho u##
 
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  • #30
PhDnotForMe
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I am not sure what you are “looking at”.
Oops. I was looking at the link I referenced following the statement. I will put it below this message as well.

https://en.m.wikipedia.org/wiki/Current_density
in general J is not necessarily related to v.

Can you give me an example of a case where J is not related to v?
 
  • #31
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Can you give me an example of a case where J is not related to v?
Any case where ##\rho## is not at rest in either frame.

I am going to make a guess as to what is happening here. You are looking at two pages, the “Classical Electromagnetism and Special Relativity” page and the “Current Density” page. Both pages use the symbol ##v##, so you are trying to put the equations together. The problem is that the ##v## on each page has a different meaning. On the first page it says “The primed frame is moving relative to the unprimed frame at velocity ##v##”. On the second page it says that ##v## is “the velocity of the charges”. Those two quantities are different in general.

You cannot go simply looking at symbols and mixing equations together. You need to understand what the symbol means in each context. The text and figures around the symbol usually explain, but from some of your questions it seems that you are either not reading or not paying attention to that important information.
 
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  • #32
PhDnotForMe
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Any case where ##\rho## is not at rest in either frame.

I am going to make a guess as to what is happening here. You are looking at two pages, the “Classical Electromagnetism and Special Relativity” page and the “Current Density” page. Both pages use the symbol ##v##, so you are trying to put the equations together. The problem is that the ##v## on each page has a different meaning. On the first page it says “The primed frame is moving relative to the unprimed frame at velocity ##v##”. On the second page it says that ##v## is “the velocity of the charges”. Those two quantities are different in general.

You cannot go simply looking at symbols and mixing equations together. You need to understand what the symbol means in each context. The text and figures around the symbol usually explain, but from some of your questions it seems that you are either not reading or not paying attention to that important information.
Okay. So if I see a stationary proton and that proton begins to accelerate faster and faster in some direction, the B-field created will increase as it moves faster. Because the B-field changes, the E-field must change too. Will the transverse E-field of the proton increase or decrease?
 
  • #33
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Okay. So if I see a stationary proton and that proton begins to accelerate faster and faster in some direction, the B-field created will increase as it moves faster. Because the B-field changes, the E-field must change too. Will the transverse E-field of the proton increase or decrease?
See https://en.wikipedia.org/wiki/Liénard–Wiechert_potential in the section "Corresponding values of electric and magnetic fields". I don't know the answer to your question but that formula is how to calculate it.
 
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