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?