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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?

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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.

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  • $\begingroup$ Thank you very much also for your answer that I have appreciated (see also the comment to the previous user). $\endgroup$ – Sebastiano Jun 21 '20 at 10:30
  • $\begingroup$ Why is there a downvote? What is the reason? $\endgroup$ – Sebastiano Jun 21 '20 at 19:14
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"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 $\vec E$.

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  • $\begingroup$ Meanwhile thank you very much for your explanation that I have understood well. Perfect. I give the check mark to the user Thomas because he has fewer votes than you, but I'm always for honesty and kindness. $\endgroup$ – Sebastiano Jun 21 '20 at 10:29
  • $\begingroup$ Why is there a downvote? What is the reason? $\endgroup$ – Sebastiano Jun 21 '20 at 19:14

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