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Magnetic force in a moving inertial reference frame

  1. Apr 3, 2012 #1
    A charged particle is placed next to a current-carrying wire. The wire produces a magnetic field, but if the particle is at rest, the field exerts no force on it. However, in a different inertial reference frame moving at speed v parallel to the wire, the particle is seen to be in motion, and so there is a magnetic force perpendicular to the wire! Electrostatic force is the same in both cases. Where does the extra magnetic force come from?
  2. jcsd
  3. Apr 3, 2012 #2


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    Hi BucketOfFish! :smile:
    You're saying, the particle has zero acceleration, so how come the maths says there's a radial magnetic force on it (from the perpendicular magnetic field)?

    Because that extra force is balanced by the extra electric force from the moving electrons (i forget the exact argument :redface:, it has something to do with lorentz contraction making the electrons closer together than the positive charges, so that the negative electric field no longer balances the positive electric field). :wink:
  4. Apr 3, 2012 #3
    In the referential of the magnetic field source, the observer sees the moving particle deviated by the Lorentz force F=q*VxB (V and B are vectors).
    In the referential of the particle there is a transverse electric field E=VxB and the force onto the particle is F=q*E, the same.
    This is explainable by relativity. An observer in an inertial reference frame moving relative to the source of a magnetic field sees an electric field which is the Lorentz transform of the magnetic field. It is this electric field that exerts the transversal force from the viewpoint of the moving charge.
  5. Apr 3, 2012 #4
    Wow, this is crazy! How did people deal with this problem before relativity? Also, is all this generally explained in an undergrad EM course, or will I have to wait until grad school to get the full picture?
  6. Apr 3, 2012 #5


    Staff: Mentor

    Hi BucketOfFish,

    Here is a link that explains this concept at a good undergraduate level:

    Start with the "the talk itself" link and if you get stuck anywhere then look at the notes.
  7. Apr 3, 2012 #6
    Thanks, Dale! This looks really interesting!
  8. Apr 3, 2012 #7
    I had these same questions to arise also. Looking at what Dale posted, it talks about length contraction. So wait, if a wire then has a current through it, and a particle is said to be at rest as the OP said, there is a force on the particle even though the wire is said to be electrically neutral when current is not flowing? We just don't call it a magnetic force in this frame of reference? If so, that really makes me look at this differently!

    I originally assumed that if a wire carries current and a test charge is placed at rest (with respect to the frame that defines the current) that it experienced no force at all because the wire is neutral and v is zero making B also zero. But length contraction states that the moving current will be contracted relative to the rest test charge making for a higher charge density of the like, and thus an electric field. This means that the test charge particle will experience a force from the current carrying wire regardless if it is said to be at rest or not. The only thing that changes is whether you classify it as an electric force, a magnetic force, or a mix of the two which is dependent on your frame of reference.

    Is that correct?
  9. Apr 4, 2012 #8
    There is no force onto a test charge at rest near a current carrying wire at rest.
    "The separation between the electrons in the current carrying wire is not Lorentz contracted because there is only a finite number of electrons in the wire and charge is conserved. Thus the electron remains spread out over the entire wire.".
    Nevertheless the magnetic force can be explained by the length contraction of moving charges. This is well explained in the Bickerstaff's course, chapter 14.8.1 and 2, page 233:
    http://exvacuo.free.fr/div/Sciences/Cours/Relativite/R.Paul%20Bickerstaff%20-%20Claustrophobic%20physics.pdf [Broken]
    Last edited by a moderator: May 5, 2017
  10. Apr 4, 2012 #9


    Staff: Mentor

    Yes. If there is a net EM force on the test charge in one frame then there will be a net EM force on the test charge in all frames, but its "classification" depends on the frame. Similarly, if there is no net EM force on the charge in one frame then there will be no net EM force in all frames. In some frames that will be due to a cancellation betwen the electric and magnetic forces, and in others they will each simply be 0.
  11. Apr 5, 2012 #10
    There is still a finite amount of protons in the wire also, and yet if we say the electrons are stationary and the protons are moving, they do indeed contract and there is a force. What makes protons so special? I still don't see the difference. :)
    Last edited by a moderator: May 5, 2017
  12. Apr 5, 2012 #11


    Staff: Mentor

    What Exnihiloest said above is correct. If a test charge is at rest next to a neutral current-carrying wire it experiences no force.

    In another reference frame the wire still carrys current, but also is charged, the resulting electric force is exactly equal and opposite to the magnetic force, for 0 net force in all reference frames.

    I don't understand your comment about protons being special. Everything transforms the same way from the conditions given in the original frame.
  13. Apr 5, 2012 #12
    Yes, I believe I understand this. If in the lab frame where protons are at rest and the electrons are moving, and also given that in this frame the wire is said to be neutral - the test charge experiences no force. If you change frames by going to the frame of the flowing electrons, you then see the average separation of these electrons lengthen, and also in consequence, the protons contract. This results in a net positive charge and thus an electric field you would expect to act on the test charge. But - in this frame, the test charge is now seen to be moving with the protons, and thus there is a magnetic field that exactly cancels out the electric field in the given frame. Is this correct? If so, I think I understand this.

    What I do not understand, as you say, is the conditions given in the original frame. I don't understand why - given that the electrons have a velocity in the lab frame, that their average length is the same as the protons. I certainly see it possible, but I don't know what makes it the given. I have heard that charge is conserved, but that still doesn't solve the problem I mentioned above:
    If the protons are said to initially be moving, and the electrons + the test charge are said to be initially at rest, you could conclude with the original statement that charge is conserved and thus the average separation of the electrons and protons is the same. This initial condition is just like the before case except that the test charge is now at rest with the electrons instead of the protons.

    Now, those initial conditions set, I see that the test charge experiences no force. Yet when looking at the other initial conditions where electrons are said to be moving and the protons + the test charge are said to be at rest, when the test charge is then set into motion so that its speed is equal to the electrons, it experiences a force. But yet, at this given state, it seems identical to the initial conditions of the second case I mentioned at which the test charge was said to not experience a force. This either arrives me at a contradiction, or there is some reason for the chosen initial conditions.

    Does the way I am explaining this make sense? It seems very hard to write this (to many assumptions seem to be needed - hence why I believe I am not understanding what has already been said.)
  14. Apr 6, 2012 #13


    Staff: Mentor

    That is under experimental control. If the experimenter wanted to make the wire so that it was electrically charged he could do it simply by putting a very high voltage on both ends of the wire. Since the self-capacitance is very small the voltage would have to be very high on both ends of the wire in order to get any significant charge on it.

    Although it is not explicitly stated, the usual understanding for this scenario is that the experimenter is doing the usual thing of putting a small voltage across the wire to generate a large current, and not a large voltage on both ends of the wire to give it a small charge. In principle you could do both, i.e. a very high voltage on both ends with a small difference in voltage between the two sides, in order to get both a current and a net charge.
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