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Will the Lorentz force expression be valid for a magnetic field created by a magnetic monopole?

I haven't seen any derivation of Lorentz force expression yet and I don't know whether it was derived applying Lorentz transformations to field created by charge particles.

But if this is the case then magnetic field created by moving charges of course has zero divergence ,as we know (not same as magnetic field of a monopole) and a magnetic monopole can create magnetic field even when it is at rest. So should the Lorentz force expression be applied in the case of magnetic field by a monopole?

Edit:I was talking about the force on a charge particle due to magnetic field created by a monopole.

$$F=q_e(v \times B)$$

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Yeah, that would remain the same, but with an analogous part for the electric field, like that:

$$F=q_e(E+v \times B)+q_b(v \times E +B)$$

but Maxwells formulas would be a little bit different, the divergent of the magnetic field would be the analogous for the electric, giving the charge density. And Faraday's law would include a term of magnetic charge current like the one in Ampere's equation.

That is possible because a magnetic charge would only create a divergent of the magnetic field, but since as a field B and E are analogous, all the formulas would change according to this simetry.

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  • $\begingroup$ How do you know 'that' will remain same? $\endgroup$
    – Paul
    Commented Mar 14, 2015 at 18:11
  • $\begingroup$ It actually depends, if you are talking about the force felt by an electric charge, then it would remain the same, but not for a magnetic charge, which is given by the formula above. $\endgroup$
    – Ivan Lerner
    Commented Mar 14, 2015 at 18:13
  • $\begingroup$ I was talking about force on a charge particle due to magnetic field created by monopole? $\endgroup$
    – Paul
    Commented Mar 15, 2015 at 3:17
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    $\begingroup$ Because $E$ has the same units as $vB$, your formula of the force law should be modified as $\mathbf{F}=q_e[\mathbf{E}+(\mathbf{v}\times \mathbf{B}]+q_m[\mathbf{B}-\frac{1}{c^2}(\mathbf{v}\times \mathbf{E})]$. $\endgroup$
    – Wang Yun
    Commented Dec 17, 2015 at 12:29
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    $\begingroup$ I believe there is a sign error in your magnetic Lorentz force term. $\endgroup$
    – ProfRob
    Commented Nov 9, 2017 at 23:39

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