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Magnetic field of charge moving at constant velocity

  1. Jul 29, 2011 #1
    Can anyone show me how to calculate the magnetic field of a single charge moving at a constant velocity - including both the current and displacement current terms?

    Also, what is the significance if any of this result to Galilean vs Lorentz transformations? It seems like a problem that would have interested Maxwell and yet I cannot find any reference to it in his biographies.
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  3. Jul 29, 2011 #2


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    Your idea of using Lorentz transformations is correct (using Galileo transformations is wrong of course).

    The solution of the problem of a point charge at rest is of course the electric Coulomb field and vanishing magnetic field. Then you go to a reference frame, where the charge moves with a given velocity, by using a Lorentz boost, and you obtain the electromagnetic field. To that end you arrange the electric and vanishing magnetic field into the antisymmetric 2nd-rank field-strength tensor, which has well defined transformation properties or you introduce the four potential which transforms as a four vector (modulo gauge transformations).

    The same holds for the four-dimensional current, which transforms as a four vector.
  4. Jul 29, 2011 #3
    Thanks for your answer, but I hadn't asked my question clearly enough. What I wanted to know is what's the solution in a non-relativistic framework. In particular, is there a net magnetic field, or does the current "charge x speed" get cancelled by the displacement current leaving zero field. It's of interest because in Maxwell's day this type of problem was known and I'm interested to know whether it would have produced "anomolous" results in the sense of a magnetic field appearing when speed goes from zero to constant.
  5. Jul 29, 2011 #4


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    The question is, what you mean by "non-relativistic framework". Maxwellian electromagnetics is a relativistic theory, and thus you have to modify it to make it consistent with Galilei invariance.

    Such questions are addressed, e.g., here:

    Le Bellac M and Levy-Leblond J M 1973 Galilean electromagnetism Nuov. Cim. B 14 217-233.
  6. Jul 29, 2011 #5
    By "non-relativistic" I mean the equations that Maxwell originally formulated rather than the re-formulation on a relativistic basis.

    Getting back to my original question, which I'd love someone to answer, what is the magnetic field of a single charge moving at a constant velocity? It clearly has an electrostatic field E but does it or doesn't it have a field B?
  7. Jul 29, 2011 #6
    To clarify this, there seem to be 3 distinct situations that are widely studied:
    1) currents in wires - zero net charge, no electrostic field - Biot-Savart only
    2) particle beam - net charge, field is static so no displacement - Biot-Savart only
    3) moving charge - net charge, dynamic field, displacement current - so what's the solution?
  8. Jul 29, 2011 #7


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    A Google search on "magnetic field of a moving charge" leads to this as the first hit:


    Searching instead for "magnetic field of a moving point charge" leads to this as the first hit:


    These are for v << c (the non-relativistic limit). You can find the exact version, good for all v < c, at

    http://farside.ph.utexas.edu/teaching/em/lectures/node125.html (equations 1538 and 1539)
  9. Jul 29, 2011 #8
    Maxwell's equations as originally formulated are relativistic. Try doing a Galilean transformation on them and it doesn't work.

    Moving charges create magnetic fields, even if you have a single charge. If the charge is moving at a constant velocity, you will have a static magnetic field similar to that created by a very small length of wire carrying a current: the magnetic field will curl radially around the path the electron takes and the magnetic field strength will be proportional to the inverse square of the distance between the observation point and the charge.
  10. Jul 29, 2011 #9
    For item (3), if the charge is moving with a constant velocity (or very close to constant) it will create an electrostatic field according to Coulomb's law and a magnetostatic field according the Bio-Savart law the two fields will be independent.

    If the charge is has a non-constant velocity (is accelerated), then the electric and magnetic fields are non longer static, but change in time and couple to each other. You need all 4 Maxwell equations to describe this. The most intuitive approach is to reduce Maxwell's equations down to inhomogenous wave equations, which basically tells you that accelerating charges emit (and absorb) electromagnetic waves.
  11. Jul 29, 2011 #10


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    It does.

    They are called the Lienard Wiechert potentials:

    Also, vanhees71 and chrisbaird are correct, the Lienard Wiechert potentials come right from Maxwells equations and are fully relativistic. They work for any speed.
    Last edited: Jul 29, 2011
  12. Jul 30, 2011 #11
    On Lorentz invariance I stand corrected, thanks all of you for putting me right.

    Thank you also for drawing my attention to Lienard Wiechert which is very helpful for following through the calculation.

    Can anyone give a simple explanation of why for the steadily moving particle, whose electric displacement is dynamic and therefore generating a displacement current (DP), nevertheless the contribution of the DP to the magnetic field B is zero, and the field B is entirely the Biot Savart field. Or is this answer simply a low-velocity approximation?
  13. Jul 30, 2011 #12


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    It is not a low velocity approximation.

    I don't understand your displacement current comment. Why would you think it is zero?
  14. Jul 30, 2011 #13
    Another way of asking the question is why, for a moving point charge, is the Lienard Wiechert solution to the two vector equations :
     x B = J + dE/dt and  x E = -dB/dt =>
    => B = Q VxR/4πrrr (rrr=r cubed as symbols are unavailable, please excuse the poor symbols, the best I could do)
    the same result as the Biot Savart solution to Ampere's law on its own:
     x B = J =>
    => B = Q VxR/4πrrr
    Is there an insightful explanation of this, or is it simply a coincidence?
  15. Jul 30, 2011 #14


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    The Biot-Savart law for a point charge at constant velocity must be the same as the Lienard Wiechert solution for a point charge at constant velocity. They are both derived from the same equations (Maxwell's equations) for the same boundary conditions.
  16. Jul 30, 2011 #15
    Surely there is some significance that when you solve just Ampere's law with the point charge (and so just one equation  x B = J) you get the same answer as when you solve the full maxwell's equations (ie two vector equations  x B = J + dE/dt and  x E = -dB/dt). Why is the older formulation that pre-dates Maxwell (Biot-Savart was derived in 1820 on the basis of Ampere's simpler formulation) giving the same answer as Maxwell? After all the equations are different!

    What strikes me as odd about the moving charge case is that in the Ampere's circuital equation there is no reference to an electric field and the magnetic field depends only on the charge-current.

    Yet in the Maxwell formulation the Electric Field explicitly enters the equations, it is clearly changing so dE/dt is non-zero and yet the Electric Field does not effect the solution in any way; then also the Magnetic Field is changing too and so gives rise to curl x E. So in some sense the changing Electric Field coupled with the changing Magnetic Field ends up leaving the solution the same as if there had been no Electric Field, and I'm curious why this is.
    Last edited: Jul 30, 2011
  17. Jul 30, 2011 #16


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    If the equations are different then the answer is not the same. Perhaps you are thinking of the low-velocity approximation to Biot-Savart:
    which is not at all the same as the Lienard Wiechert potential:

    If you use the same equations you get the same answer. If you use different equations then you get different answers. You seem to be saying that you think you use different equations but get the same answer.
  18. Jul 30, 2011 #17
    Ampere's law is:

    [tex]B*2 \pi R=\frac{1}{c^2}\frac{d}{dt} \int \frac{q}{4 \pi \epsilon_0 r^2} \hat{e_r}\cdot d\vec{A} [/tex]

    where if you imagine the charge moving along a line, and r is the point you want to calculate the field, then R is the perpendicular distance from r to the line. The surface of integration is a disc of radius R whose center goes through the line. Just to set it up a little bit more:

    [tex]B*2 \pi R=\frac{1}{c^2}\frac{d}{dt} \int \frac{q}{4 \pi \epsilon_0 (L^2+x^2)} \frac{L}{\sqrt{L^2+x^2}} [2\pi x dx] [/tex]

    where L is the distance to the point on the field you want to calculate along the line, the factor in [...] is the measure, and the factor next to that is the cosine the angle the field makes with the surface. The integral over x goes from 0 to R.

    I just did this calculation with the help of an online integrator and differentiator (because I'm lazy), and you get the Biot-Savart magnetic field.

    edit: just to be more clear, L=L(t), so when you take the timed derivative of L, you get the velocity v.
    Last edited: Jul 30, 2011
  19. Jul 31, 2011 #18
    I agree. Now if you solve the multiple equations of Maxwell, you will also get Biot Savart. So two different sets of equations give you the same answer.

    What I'm asking is not about solving equations (for the result could be a coincidence) but what is the deeper insight into why the two approaches give the same result. They don't for oscillating charges. So what is the general insight?
  20. Jul 31, 2011 #19


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    I don't understand what you mean by different approaches to solve for the full Maxwell equations. If you mean the Biot-Savart Law from magneto statics, of course it's not applicable to the magnetic field of accelerated charges since it works only in the special case of magnetostatics, i.e., stationary currents.

    In any case, you get the right answer by using the retarded Green's function (Lienard-Wiechert potentials for an arbitrarily moving point charge). In the case of a uniformly moving charge you get the fields also from Lorentz boosting the Coulomb field for the charge at rest. Of course, both solutions must be identical since Maxwell's electromagnetics is a relativistically covariant classical field theory.
  21. Jul 31, 2011 #20


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    This is not correct, as shown by the links posted above.
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