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Why does moving electron produce magnetic field?

  1. Sep 14, 2007 #1
    Why does moving electron produce magnetic field around it? Thank you.
  2. jcsd
  3. Sep 14, 2007 #2


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  4. May 20, 2011 #3
    i think what he's asking in lamon's terms is how the photons/laybons bounce off eachother to create a gravitational field, and i was wondering the same thing. is it because moving through space/time gives the particle new posatively charged particles to react with when it looses them?
  5. May 20, 2011 #4
    A moving electron alone actually does not produce a magnetic field. It requires electrons and protons to produce a magnetic field. As the electrons move relative to protons (ions) there is a relativistic charge per unit volume difference between the positive and the negative charges. This causes any external charges to feel a force we know as the magnetic field.

    The magnetic field is a relativistic correction to the electrostatic field.
  6. May 20, 2011 #5


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  7. May 20, 2011 #6
    That was a pretty clever derivation. It's interesting that the electric field is strongest perpendicular to the moving charge, just like the case for acceleration where the radiated power goes as [tex]sin^2 \theta [/tex] (Larmor formula), and hence the field as [tex]E \; \alpha \; sin \theta [/tex].

    I like formula (1539):
    [tex]B=\frac{v \times E}{c^2} [/tex]

    as it shows how weak the magnetic field is compared to the electric field. They're using SI units, so for meaningful comparison, you'd have to multiply the magnetic field by c:
    [tex]B=\frac{v \times E}{c} [/tex]

    To detect the magnetic field of a single charge is even worse, since the Lorentz force has a u/c x B, so you'd get a proportionality to uv/c2.

    I read somewhere that this asymmetry between electricity and magnetism is the result of symmetry breaking at low energy, but that makes no sense to me. Even if the particle is traveling at light speed, although the magnitudes of the two fields is about the same, that's just about all they have in common. What makes more sense to me is this: of course an electric charge/current will favor electric fields over magnetic fields. A magnetic charge/current, if they existed, would favor magnetic fields over electric fields.
  8. May 21, 2011 #7


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  9. May 23, 2011 #8
  10. May 23, 2011 #9


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    Yes, a very good derivation. In words, a charge sitting at rest produces a Coulomb field. When you view it from a moving rest frame, the E field becomes partly a B field. That's why we call it Electromagnetism.

    But then we'd have to call it... Magnetoelectrism!
    Last edited by a moderator: May 5, 2017
  11. May 23, 2011 #10
    "Becomes"? I don't see where this part is derived.

    The only relativistic derivation of the magnetic field I have ever seen was dependent on a positive charge as part of the system.
  12. May 23, 2011 #11
    There are at least two ways different from the website that you can use to derive the magnetic field of a constant-velocity point charge. One is to just use a Lorentz transform - just plug and chug. The second way is that most books already solve the Maxwell equations in terms of the sources, and there's an equation called the Feynman-Heaviside equation that tells you the field of a point particle - just set the acceleration to zero in those equations. The website does it yet a 3rd way, inserting the particle 4-velocity as way of introducing frame dependence. One possible point of confusion is the pre-factor [tex]\frac{1}{r'^3} [/tex] in front of the tensor. If r' is equal to say 5, then in the new frame that factor would still be numerically equal to 1/53, although written in terms of variables of the new frame: [tex]\frac{1}{r'(r,t)^3}[/tex] where [tex]r'=r'(r,t) [/tex] It's everything to the right of the pre-factor that produces the correct structure for tensor transformation. So if you tensor transformation that expression, don't replace [tex]\frac{1}{r'^3} [/tex] by [tex]\frac{1}{r^3} [/tex], which you could do only if the pre-factor were a scalar. But rather use [tex]r'=r'(r,t) [/tex]
    Last edited: May 23, 2011
  13. May 23, 2011 #12


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    Einstein derived it in his original OEMB paper.
  14. May 24, 2011 #13
    As far as the Einstein OEMB paper is concerned I can only find a part where he Lorenz transforms existing magnetic fields in his chapter:
    I looked up the Feynman-Heaviside equation and as far as I can see it is also derived starting with existing electrical and magnetic fields.

    As for the site mentioned in post #5 my maths is not up to scratch but if it refers to equations 1512 – 1515 : then as usual it refers back to existing magnetic fields. However it would be nice for me and others if someone could simplify and see how this works for a straight conductor, ie can someone work out B parallel with a conductor and also B at a right angle with this conductor. (with his site in mind).

    I still would like to see the transformation of a electrical field into a magnetic one. I have never ever seen a satisfactory solution.
  15. May 24, 2011 #14


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    Einstein derived the full expression for transforming arbitrary electromagnetic fields between reference frames. Then the link above took the known electric field (Coulomb's law - magnetic field = 0) in the frame where the charge is stationary and then simply calculated the magnetic field in a reference frame where the charge is moving.

    I don't understand your underlined "existing" emphasis. If you have no EM field in one frame then you have no EM field in any frame. You have to have something existing in one frame in order to transform it into another frame. In this case you have a purely electric field in one frame and in all other frames you will have a mixed electric and magnetic field. But you cannot have no EM fields in one frame and some EM fields in another.
  16. May 24, 2011 #15
    Just set B=0 in eqns 1512-1515. Then you can transform any purely electro-static problem (where B is equal to zero) into a electro- and magneto- static problem (where B is not equal to zero) .
  17. May 25, 2011 #16
    I found the writing that I last read on the relative correction to the electrostatic field. It depends on the fact that both positive and negative charges be present. Perhaps this is now out of date information. Here it is though, very easy to understand explanation.

  18. May 25, 2011 #17
    Hi DaleSpam can you give me the link to that paper? Yesterday I read one of his papers but either missed it or read the wrong paper.

    I did underline because I wanted to see how a purely electrical field transforms into a magnetic one.

    @ RedX. Ok put B=0 into 1515 that’s exactly what I’m after!

    The copy/past doesn’t work here so I’ll write out: B’ perpendicular = gamma x -V x E/c^2.
    Lets work out gamma x V/c^2. This is: (V/c^2+V^3/2c^4+…..) now it is customary to ignore higher powers of c^2 (but if you really want you can keep them in).
    The result is: B= - V x E/c^2 + …. And this to me is a bit of a revelation ie: a magnetic field can be caused by a travelling E field, where V is subject (in a first degree) to the every day Galilean relativity, that is relativity without Lorenz contraction.
    Also this formula shows me the absence of the 2 in the denominator which you would get when working out B, using Lorentz contraction.
    In short: this formula makes more sense to me then any other explanation.
  19. May 25, 2011 #18
    So I am lost. Did someone just determine that pure relative motion causes a charge to be accelerated perpendicular to the direction of travel?
  20. May 25, 2011 #19


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    Here it is, see section 6.


    He uses odd notation, but for a purely electric field in the unprimed frame simply set L=M=N=0 in the last set of equations in section 6. You can immediately see that L'=0, but M' and N' are non-zero. So a purely electric field in one frame is an electric and magnetic field in any other frame.

    [nitpick] I wouldn't call it a travelling E field. Fields don't have velocity. In one frame it is purely an E field, and in a frame which is travelling wrt that frame it is both an E and a B field. But there really isn't any sense in which the field itself has a velocity in either frame. [/nitpick]
  21. May 26, 2011 #20
    Well no not exactly. We have to work out E prime first. E’=gamma E, but that’s all the Lorentz contraction done. Then B= - (E’xV)/C^2.
    It’s a shame am going to be away for a week but later on I hope to be able to rewrite that paper you mentioned in post #16. (If someone already has done so please let me know).

    Thanks @ Dale
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