Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Apparent impossibility of uniform magnetic field

  1. Dec 7, 2005 #1
    Hi All,

    I would like to comment an idea and would appreciate having critics on that, as it appears to be a fallacy.

    According to the law for the magnetic force, if a charge is at rest relatively to some magnetic field then no net force act on it.
    However, if a charge is moving (no matter in which direction) in an uniform magnetic field then, due to simmetry, it will experience a field with the same structure as if it had remained at rest. So, the charge would not suffer magnetic action (force) in any possible state on motion. As we know it is not experimentally observed in the situations in which we try to produce uniform magnetic field, then the conclusion would be that what we call uniform magnetic field is in reality a field with small non uniformities.


    Thank you in advance for the comments,

    Sincerely

    DaTario
     
  2. jcsd
  3. Dec 7, 2005 #2

    NateTG

    User Avatar
    Science Advisor
    Homework Helper

    You may be confusing the notion of field with the notion of potential.
     
  4. Dec 7, 2005 #3

    Tide

    User Avatar
    Science Advisor
    Homework Helper

    We know that a force acts on charge moving in a magnetic field [itex]\vec F = q \vec v \times \vec B[/itex] so I think you're asking what does the charged particle see in its own rest frame. It does not see a simple magnetic field. It also sees an electric field perpendicular to its direction of motion and perpendicular to the magnetic field.

    The electric field arises out of the so-called Lorentz transformation which Einstein discovered describes the transformation of electric and magnetic fields from one reference frame to another.
     
  5. Dec 8, 2005 #4
    Correct so far if there is no electric field in this frame of reference.
    Your term "experience a field" is unfamiliar to me. I think you're thinking of something else hence the confusion here. A moving charged particle may experience a field if the particle's velocity has a component perpendicular to a magnetic field line. If the charge is moving parallel to the field then there will be no force on the charged particle.
    That is incorrect. The magnetic force is a velocity dependant one. The magnetic field makes its presence known by charges moving with respect to the field.
    I know of no observation that would make that true.

    Pete
     
  6. Dec 12, 2005 #5
    Ok, most of your answers pointed to what I agree to be the correct direction. But none of you addressed the question of translation simmetry.
    Maxwell's equation tells us that varying magnetic field imply observation of rotational electric field. That's ok.
    But in the situation I described, the charged particle is travelling with a given velocity in a region of uniform magnetic field.
    So, imagine the particle's perspective if it was at rest inside this region of constant magnetic field . Now imagine the particle's perspective if it was travelling with constant velocity inside this same region. As there is no observable difference between these two situations, one is tempted to conclude that no force will be experienced by the particle in any of these cases.

    DaTario
     
  7. Dec 12, 2005 #6
    I guess you are right. Thanks.

    But how the particle will get the knowledge that it is moving with velocity v relatively to a magnetic field which is uniform?

    It seems to be neccessary at least that all magnetic field be non uniform in some regions of space so as to allow one to infer its own movement.
     
    Last edited: Dec 12, 2005
  8. Dec 12, 2005 #7
    This example is actually my favorite method to demonstrate that the fields must transform when you transform between refrence frames. In the frame where the field is entirely magnetic and the charge is in motion you know that there must be a force acting on the charge, yet if you transform into the charge's refrence frame there is no velocity. There must still be a force, however, and thus there must be an electric field in the charge's reference frame to cause that force.
     
  9. Dec 12, 2005 #8
    You didn't ask a question of translation symmetry. This is how your question went
    That is true if and only if there is no electric field in the frame of reference that you're talking about.
    This is poorly stated. It will feel a force if and only if a component of the field is perpendicular to the direction of the magentic field. Warning: Do not make the mistake of thinking that you can divide the force by the charge and get an electric field. There is only a magnetic field present and only a magnetic force in this frame of reference. If the velocity of the particle is perpendicilar to the magnetic field then the particle will move in a circle. There is no problem with symmetry here. Here is a derivation

    http://www.geocities.com/physics_world/sr/cyclotron.htm

    What you're refering to is the relative existance of an electric field. In S, the "rest" frame (e.g. the one where there is only a uniform B field) there is zero electric field. Now change to a frame S' moving relative to S. In S' there is an electric field. This can be shown using Einstein's special theory of relativity.

    Pete
     
  10. Dec 14, 2005 #9
    The point I am trying to address is that of simmetry of translation inside a region where there is only an uniform magnetic field. The simmetry would imply no force at all, whatever the particle's state of motion.

    I agree that Lorentz tranformation seems to introduce the correct claim for electric field in such comparisons between different reference frame's perspectives. But I still have the impression that the issue of translation simmetry is not being touched.

    Nevertheless, I feel glad and grateful for your interesting comments.

    DaTario
     
  11. Dec 14, 2005 #10

    Tide

    User Avatar
    Science Advisor
    Homework Helper

    In that case, you must validate your assertion that the situation is symmetric. I suggest the symmetry only exists in the laboratory frame. From the perspective of the moving charge, there are more magnetic field lines approaching it from the front than receding from the back. The symmetry is broken and more careful analysis is required [see above!].
     
  12. Dec 15, 2005 #11
    I'm not sure exactly what you mean by translation symmetry, however I feel that this case, while it is certainly fairly symmetric, is perhaps not symmetric enough, as the case may be.

    It might be helpful, in this case, to consider the origins of the magnetic field. In general this approach will get you into trouble since things like permanant magnets and even simple electromagnets are not so easily dealt with, but in this specific circumstance some insight can be gained I think. It's long to write out, but the physical situation is really fairly simple once you get past my long winded descriptions :wink:

    So, how do we create an infinite uniform magnetic field? With an infinite sheet of current. The magnetic field lines will be parallel to the sheet and constant going out to infinity.

    Consider, for example, an infinite sheet in the x-y plane with a current in the x direction. The magnetic field will then point in the positive y direction, assuming you stay above the sheet of current. Also consider a positively charged particle moving through this field, relative to the sheet, in the positive x direction.

    In the rest frame of the sheet, where the magnetic field is constant, we say that there is a magnetic force [itex]q \vec{v} \times \vec{B}[/itex] which acts on the particle and pushes it in the negative z direction, towards the sheet. Now transform into the frame of the particle, and consider the charges in the sheet which had produced the current.

    In the rest frame of the sheet our standard model for the current says that positive charges are flowing along the x direction and negative charges are at rest in the sheet, spaced such that the sheet is charge neutral. In reality it's the other way around, but that's not important in this context.

    When we transform into the frame which is moving along the positive x direction we must consider two effects. First, since we are moving relative to the rest frame of the negative charges in the sheet they become more densely packed together. Second, since the positive charges flowing in the sheet are now moving with a smaller velocity they are less densely packed. These two effects combine to result in a net negative charge on the sheet in the frame of the moving observer. This negative charge causes an electric field pointing in the negative z direction which in turn produces a downwards force on our moving test particle, in agreement with what we had calculated as the magnetic force in the frame which was at rest with the sheet.
     
  13. Dec 15, 2005 #12

    vanesch

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    This is not true, for the following reason: the magnetic field vector (and the electric field vector) do NOT transform as galilean 3-vectors when you change reference frames. If you transform a magnetic field under a lorentz transformation, you get an electric field. So you can do a lorentz transformation to the rest frame of your particle, but in that case you also have to transform the B-field, which will result in a B and E field in that frame. The resulting E-field will accelerate the charge (in its rest frame).
     
  14. Dec 27, 2005 #13

    Thank you for the exposition. In fact I was already aware of this, as it can be found in Berkley's physics book on electromagnetism.
    My idea was only to produce a small theoretical result, using very basic ideas on symmetry, but I understood, I guess, all the points you all have allegated.

    Thank you once more.

    Best Wishes

    DaTario
     
  15. Dec 27, 2005 #14
    Did anyone ever read any of Feynmann's stories about when he went to Brazil? Just thought it was appropriate here.
     
  16. Jan 22, 2006 #15
    How so?

    Sincerely

    DaTario
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?