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Electric field in magnetostatics?

  1. Mar 18, 2012 #1

    In magnetostatics theory, there exists a current of charges. So in this situation charges are not stationary, and the Coulomb's law, and all the relations derived from it, are not valid. My question is how can we obtain electric field when dealing with steady currents (within magnetostatics theory)?

    sorry for bad english
  2. jcsd
  3. Mar 18, 2012 #2


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    In magnetostatics, i.e., for time-dependent fields, charge distributions and currents, the equations for the electric field decouple, and for the electric field the rules of electrostatics still hold. You see this by writing Maxwell's Equations for this special case (in Heaviside-Lorentz units):

    [tex]\vec{\nabla} \times \vec{E}=0, \quad \vec{\nabla} \cdot (\epsilon \vec{E})=\rho, \quad \vec{\nabla} \cdot \vec{B}=0, \quad \vec{\nabla} \times \frac{\vec{B}}{\mu} = \frac{1}{c} \vec{j}.[/tex]

    Here, I've already worked in the usual consitutive equations for an isotropic medium in nonrelavistic approximation, [itex]\vec{B}=\mu \vec{H}[/itex] and [itex]\vec{D}=\epsilon \vec{E}[/itex]. Finally you need Ohm's Law (here also in its nonrelativistic approximation) [itex]\vec{j}=\sigma \vec{E}[/itex].
  4. Mar 18, 2012 #3
    Thanks for your good and clear answer. The only thing I'd like to correct is that in magnetostatics fields are NOT time-dependent.
  5. Mar 19, 2012 #4
    Hi Arham,

    For me, current is different from a motion of free charges. In case of current, the net charge in every differential volume is zero. Is this kind of related to your question?
  6. Mar 19, 2012 #5

    Consider a beam of electrons. It is a non-neutral current of charges. You can assign a current density to it. In magnetostatics, currents are steady, so [itex]\partial\rho/\partial t=0[/itex], and the continuity equation becomes: [itex]\nabla.J=0[/itex], where J denotes the current density.
  7. Mar 19, 2012 #6


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    Sure, that's a typo. I wanted to write "time-independent fields"...
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