A consideration on the last Maxwell equations: doubt

Question

If I consider the last equation of Maxwell,

$$\oint_\gamma \mathbf{B}\cdot d\boldsymbol{\ell}=\mu_0\left(I_C+\epsilon_0\frac{d\Phi(\mathbf{E})}{dt}\right) \tag 1$$

where $I_C$ indicates the conduction current generated by a potential difference, and $I_S$ it is the displacement current

$$I_S=\epsilon_0\frac{d\Phi(\mathbf{E})}{dt}$$

if I don't have sources the $(1)$ becomes:

$$\oint_\gamma \mathbf{B}\cdot d\boldsymbol{\ell}=\mu_0\epsilon_0\frac{d\Phi(\mathbf{E})}{dt}\tag 2$$

Is this term provided by the right side of the $(2)$ is due to the fact that even if I don't have a potential difference the charges in a wire always move in any direction?

Answer 1

When interpreting Maxwell's last equation $$\oint_\gamma \mathbf{B}\cdot d\boldsymbol{\ell}=\mu_0\left(I_C+\epsilon_0\frac{d\Phi(\mathbf{E})}{dt}\right) \tag 1$$ there is no need to bring potential difference into the game.
_{Actually the concept of electric potential, which is so important in electrostatics, is of limited use in electrodynamics. As said in Electric_potential - Generalization to electrodynamics: When time-varying magnetic fields are present (...), it is not possible to describe the electric field simply in terms of a scalar potential $V$ because the electric field is no longer conservative.}

$I_C$ is the conduction current (i.e. moving charges) regardless why the charges are moving. This current creates a magnetic field $\mathbf{B}$ wrapping around it.

The second term on the right side $\left(\epsilon_0\frac{d\Phi(\mathbf{E})}{dt}\right)$ is due to an electric field $\mathbf{E}$ which changes with time $t$ regardless where this electric field comes from. It has nothing to do with charges (even though its unit is also Ampère like it is for $I_C$). This changing electric field also creates a magnetic field $\mathbf{B}$ wrapping around it.

Answer 2

"Is this term provided by the right side of the (2) [is] due to the fact that even if I don't have a potential difference the charges in a wire always move in any direction?"

I think you have a misunderstanding. Your last equation needs no wire. It works everywhere, including in a vacuum.

The equation is an integrated form of the Maxwell equation itself, which, leaving out the source term, is $$\text {curl} \vec B = \mu_0 \epsilon_0 \frac{\partial \vec E}{dt}\ \ \ \text{that is}\ \ \ \nabla \times \vec B=\mu_0 \epsilon_0 \frac{\partial \vec E}{dt}$$ Some people think of this as telling you that a changing electric field acts like moving electric charges and generates a magnetic field around it. I prefer to think of the equation as relating two aspects of the electromagnetic field at a point, $\vec B$ and $\vec E$.