I Maxwell’s equations with varying charge but constant current

1. Sep 13, 2017

phyzguy

Well, let's look at it. If we look at the curl B Maxwell's equation in integral form (Ampere's law), we can write:
$$\int B \cdot dL = \mu_0 [\int J \cdot dS + \epsilon_0 \frac{\partial}{\partial t} \int E \cdot dS]$$
In a circuit, we can write $J = \sigma E$ or $E = \rho J$, where rho is the resistivity. Then we can write:
$$\int B \cdot dL = \mu_0 [\int J \cdot dS + \epsilon_0 \rho \frac{\partial}{\partial t} \int J \cdot dS] = \mu_0 [I + \epsilon_0 \rho \frac{\partial I}{\partial t} ]$$
or, for some frequency ω:
$$\int B \cdot dL = \mu_0 [I + i \epsilon_0 \rho \omega I ]$$

So the question becomes what is the magnitude of the combination ερω compared to 1. I think if you put in some numbers you will find that in any normal circuit where a reasonable current is flowing that the second term is negligible. It is only when the resistivity is so high that no significant current is flowing that the second term matters. It is hard to call this case a "circuit".

2. Sep 13, 2017

mertcan

Thank you but also may I ask you to look at my second question in post 20. Because I am aware that curl of time dependent magnetic field should depend on current density (conductivity * electric field)and change in electric field *free space permittivity but when I applied curl of jefimenko equation for time dependent magnetic field I can not obtain the desired result (current density (conductivity * electric field)and change in electric field *free space permittivity ) could you help me about that providing mathematical demonstration? ?????????

3. Sep 13, 2017

phyzguy

I don't really understand your question. Maxwell's equations always hold, so we always have:

$$\nabla \times B = \mu_0 J + \frac{1}{c^2} \frac{\partial E}{\partial t}$$

whether or not B is time dependent. Given any distribution of currents and charges, you can always calculate E and B by calculating the retarded potentials, as described here. However, taking the curl of B as given there and showing that it reduces back to the right-hand side of Maxwell's equations may be a difficult exercise.