Problem with the Ampere-Maxwell Law [closed]

We know that $$J=\sigma E$$ where $J=dI/dS$ is the current density, $\sigma$ the conductivity, and $E$ the electric field.

So, if we put that into the Ampere-Maxwell law, then the law transforms from $\nabla\times B=\mu J +$ (second part) to $\nabla\times B =\mu\sigma Ε$ + (second part).

Is that consistent with the law? Because that states that the magnetic field comes from both the electric field that "pierces" the area of the amperian loop AND the rate of change of that electric field.

closed as unclear what you're asking by ACuriousMind♦, Kyle Kanos, Danu, Ali, JamalSMay 14 '15 at 9:06

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• What do you mean - is it consistent? You have described how Ampere's law is formulated inside a conducting medium. – Rob Jeffries May 12 '15 at 12:20
• So if you have say,an infinite wire with current that changes with time,then the magnetic field around it will be proportional to the magnitude of I at time t but also proportional to the rate of change of I at time t? – TheQuantumMan May 12 '15 at 12:35
• Why would that be a problem anyway? – Kyle Kanos May 12 '15 at 17:04
• i am just asking IF there is a problem.In other words,if my transformation is correct. – TheQuantumMan May 14 '15 at 17:35

Of course it is consistent. As commented, this is exactly the equation for a conducting medium. $$\nabla\times\mathbf B = \mu\sigma\mathbf E + \mu\epsilon\frac{\partial\mathbf E}{\partial t}$$
For instance, this is exactly the equation used to derive the wave equation in a conducting medium with no charge density: $$\nabla^2\mathbf E = \mu\epsilon\frac{\partial^2\mathbf E}{\partial t^2} + \sigma\mu\frac{\partial\mathbf E}{\partial t}$$