I have seen in some places the following expression for current in an electric conductor (for example Peter Monk's Finite Element Methods for Maxwell's Equations, equation (1.8)):-

\begin{equation} \vec{{\bf J}}=\sigma \vec{\bf E} + \vec{{\bf J}}_s \end{equation}

where $\vec{{\bf J}}$ is the electric current density, $\vec{\bf E}$ the electric field, $\sigma$ the conductivity, and $ \vec{{\bf J}}_s$ the "source current" density.

Supposing that Ohm's law is valid in the material, what does this additional term $\vec{{\bf J}}_s$ really mean? Isn't all the current caused by the electric field at the point?

  • $\begingroup$ "Isn't all the current caused by the electric field at the point?" I thought 'yes'. $\endgroup$
    – SRS
    Mar 10 '18 at 12:08
  • $\begingroup$ What about an electrical circuit with source, resistor and wire with zero resistance. Same current flowing in the resistor and wire but zero voltage drop along the wire hence no electric field along the wire. $\endgroup$
    – npojo
    Mar 10 '18 at 15:24
  • $\begingroup$ Thanks for both of your comments. @npojo you are right. Probably they may mean that when $\sigma$ is finite $\vec{{\bf J}}=\sigma \vec{\bf E}$ and when $\sigma$ is infinite (perfect conductor) $\vec{{\bf J}}=\vec{{\bf J}}_s$. $\endgroup$
    – praveen kr
    Mar 10 '18 at 21:47

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