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Ampère's law is stated as

$$\nabla\times\vec{B}=\mu_0\vec{J}.$$

I am told this only works in steady currents and not with time varying ones.

However Maxwell's addition of $+ \mu_0\varepsilon_0 \frac{\partial\vec E}{\partial t}$ means that this works for a time varying current.

Why in the case of a steady current does this mean that $\frac{\partial\vec E}{\partial t}$ is zero, giving us the original Ampère law, but for a time varying current $\frac{\partial\vec E}{\partial t}$ is needed and is non zero?

I understand conceptually that a time varying current means that there is a propagation delay and Ampère's law (non-corrected) is instantaneous, but from the Maxwell-Ampère law alone I cannot figure out why.

As in steady current there are electrons moving (but cancelled out by protons) but in time varying the E field around wire should also be zero? What am I missing?

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According to Ampere's law, $\nabla\times\vec{B}=\mu_0\vec{J}.$

If we take divergence both side and recall divergence of curl is zero, we get, $$\nabla.\vec{J}=0$$ Which means $\vec J$ is solenoidal. Therefore, at each cross-section all the entering current also leaves, so the value of current is not changing with time.

Now, the equation of continuity tells us, $$\nabla.\vec{J} + \frac{\partial \rho}{\partial t}=0$$ Therefore we conclude that $\frac{\partial \rho}{\partial t}=0$

But what does it mean? It means there is no source or sink for charge density, which actually results in steady current.

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In a circuit with a capacitor, and variable current, there is a magnetic field around the component, the same way as around other points of the wire. Even if there is no flow of charges between the plates. That means: the changing electric field between the plates plays the same role of the current in the wire.

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  • $\begingroup$ Yes, but in the case of a circuit without a capacitor, when there is no time varying current, the j term is another way of saying de/dt. but when the current changes direction (time varying) why is de/dt needed when it can just be j? $\endgroup$ – jensen paull Oct 24 '20 at 23:14
  • $\begingroup$ The equation is valid anyway, because in this case $E = 0$ and $dE/dt$ vanishes. $\endgroup$ – Claudio Saspinski Oct 25 '20 at 0:53
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$\vec{J}=\sigma \vec{E}$, so if the current (density) is changing, then so is the electric field.

Note that unless the conductivity $\sigma$ is infinite then even a steady current requires there to be an electric field in order to "push" the electrons through the wire. Since the boundary conditions for the electric field demand that it is continuous tangential to an interface, then an electric field exists outside the wire too.

Or are you just asking why the displacement current ($\partial \vec{E}/\partial t$) term is required at all? The answer is that without it, Ampere's law does not work in situations with time-varying electric fields, because the curl of the B-field can be non-zero in regions where there is no conduction current density.

The region outside a wire carrying a time-varying current is an example of that. Without the displacement current term the curl of the B field would be zero and there would be no electromagnetic waves.

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  • $\begingroup$ I understand that for a moving point charge at each point in space there is a (changing current density?)/changing e field at each point in space which means that there is a curl in the magnetic field not just at the point where the charge is located at a specific point in time. i also understand that the magneyic field genersted by this point charge moving causes a curl in the E field. But for the case of a steady current there is no change in current density at each point in the wire and no change in the electric field around the wire according to amperes law for steady current, $\endgroup$ – jensen paull Oct 25 '20 at 14:48
  • $\begingroup$ but for a time varying current the current density changes, but only because the velocity of charges decrease not because there is less charge density, so is the reason the de/dt for non steady currents is non zero because the electric field around a moving charge is impacted by its velocity via its curl? $\endgroup$ – jensen paull Oct 25 '20 at 14:52
  • $\begingroup$ @jensenpaull J can change either because of a change in charge velocity or charge density. Either requires a change in E. The microscopic version of Ohms law is $J =\sigma E$. So $dJ/dt = \sigma dE/dt$. An E field is required to make a current flow except in superconductors. $\endgroup$ – ProfRob Oct 25 '20 at 16:56
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For a stationary case $\vec E$ is not zero outside the wire, but it is rotation free and $\vec \nabla \cdot \vec E = \rho /\epsilon_0$. For a time varying case $\vec E$ is no longer rotation free but still you have $\vec \nabla \cdot \vec E = \rho /\epsilon_0$. What may confuse you is that the rotation of E originates from a current, not from a charge.

The deeper reason is that originally E and B were created to describe static electricity and magnetism, where the distinction between the two is clear. We kept using these fields to describe non-static situations and here a clear distinction no longer exists. Welcome to electromagnetism.

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