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Limit to strength of electron magnetic field

  1. Oct 28, 2007 #1
    if an electron is accelerated closer and closer to the speed of light, will its magnetic field grow forever or will it approach a limit?
  2. jcsd
  3. Oct 29, 2007 #2


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    Its magnetic field will grow forever.
  4. Oct 31, 2007 #3
    A point charge travelling at velocity v wrt to an observer will create 'rings' of magnetic field strength B centered along the axis of travel.

    The magnitude of B will be greatest when the particle is at nearest distance to the Observer (O)when it path is at a right angle to O.

    Using the Biot-Savart Law for a point charge:

    B= µ(0)/4π. qv/r² When at closest.

    The maximum velocity will tend to c but can never get to c so the B-field will only tend to this upper limit. I.e. In answer to the original question, there is an upper limit.

    Hence the case of two protons travelling past an observer at near c will show that the repulsion due to the +ve charges will be almost balanced by the attraction due to the Magnetic effects.

    Hence beams of like charges travelling at near c in say a particle accelerator will stay confined to a beam far longer than if you just calculated the Electrostatic repulsions, because the Magenetic attraction between the charges almost compensate when near c wrt to the lab.
  5. Nov 1, 2007 #4


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    "B= µ(0)/4π. qv/r² When at closest."

    Lienard, Wiechert, and I disagree. For a constant v, B at closest approach is

    It is true that F_E~F_B
  6. Nov 1, 2007 #5
    Yes. Definitely. The magnetic field of a moving charge is proportional to the electric field strength as measured in the charges rest frame, the particle's velocity and the value [itex]\gamma = (1 - \beta^2)^{-1/2}[/itex] which goes to infinity as v -> c.

  7. Nov 1, 2007 #6


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    http://www.phys.ufl.edu/~rfield/PHY2061/images/relativity_21.pdf gives an explicit formula for the magnetic field of a moving charge. Earlier and later webpages made by replacing the '21' with other numbers may also be of interest, though you have to type the URL in manually AFAIK. These earlier webpages include discussion and derivation and other topics on relativistic electrodynamics.

    The formula for the field as a function of angle is involved, but setting [itex]\theta=0[/itex] makes it simpler, one can see in that case that the field gets multiplied by a factor of gamma as other posters have remarked.

    Note that the electric field of a moving charge particle is given in http://www.phys.ufl.edu/~rfield/PHY2061/images/relativity_14.pdf and may not be intuitively obvious.
  8. Nov 4, 2007 #7
    so the magnetic field perpendicular to the particles motion increases forever but so does the electric field due to length contraction, and both by the same amount?

    does length contraction entirely explain the strength of the magnetic field everywhere? if so then that would mean that the magnetic field does have a limit but then gets contracted into a smaller space.
    Last edited: Nov 4, 2007
  9. Nov 4, 2007 #8


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    The Gauss law integral of the electric field around the charge must be a constant. So the integral of the normal field * surface area for any surface enclosing the charge must be a constant number, the enclosed charge.

    Your remarks about the field being squashed are generally correct, but I'm not sure I understand the exact question well enough to give a yes or no answer. Hopefully the Gauss law integral will clarify things for the electric field case.
  10. Nov 4, 2007 #9

    if i am reading this correctly, one first calculates the compressed electric field (which increases without limit in one direction) and then the magnetic field is simply the cross product of that and its velocity (which has a limit).

    therefore one could say that the magnetic field does have a limit but it and the electric field both get compressed (without limit)due to relativistic length contraction.
  11. Nov 5, 2007 #10


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    To me, saying the magnetic field "had a limit" would imply that there is some number M such that B is always less than M, and this is false.

    Similarly, there is no number M such that E is always less than M.

    Other than that, I think you have the right idea.
  12. Nov 5, 2007 #11
    i mean that the ratio of the strength of the magnetic field to the strength of the electric field has a limit.
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