Maybe quantum mechanics provides the answer:
The wave equation for a point particle in an external electromagnetic field given by the potentials (\Phi, \mathbf{A}) is:
<br />
i \, \hbar \, \frac{\partial \, \Psi}{\partial t} = \frac{1}{2 m} \, \left( \frac{\hbar}{i} \, \nabla - q \, \mathbf{A} \right)^{2} \, \Psi + q \, \Phi \, \Psi<br />
Taking the complex conjugate of this equation, we get:
<br />
-i \, \hbar \, \frac{\partial \, \Psi^{\ast}}{\partial t} = \frac{1}{2 m} \, \left( -\frac{\hbar}{i} \, \nabla - q \, \mathbf{A} \right)^{2} \, \Psi^{\ast} + q \, \Phi \, \Psi^{\ast}<br />
Multiplying the first one by \Psi^{\ast} and the second one by \Psi and adding them up, we get:
<br />
i \, \hbar \, \left( \Psi^{\ast} \, \frac{\partial \Psi}{\partial t} + \frac{\partial \Psi^{\ast}}{\partial t} \, \Psi \right) = \frac{1}{2 m} \, \left\{ \Psi^{\ast} \, \left(\frac{\hbar}{i} \, \nabla - q \, \mathbf{A} \right)^{2} \, \Psi - \left[\left(-\frac{\hbar}{i} \, \nabla - q \, \mathbf{A} \right)^{2} \, \Psi^{\ast}\right] \, \Psi \right\}<br />
It is obvious that the term in the large parenthesis on the lhs is:
<br />
\frac{\partial}{\partial t} \left(\Psi^{\ast} \, \Psi \right)<br />
What is not so obvious, but can be checked by straightforward calculation using nabla calculus rules is that the term in the braces on the rhs is:
<br />
-\nabla \cdot \left[ \hbar^{2} \, \left(\Psi^{\ast} \, \nabla \Psi - \nabla \Psi^{\ast} \, \Psi \right) + \frac{2 \, q \, \hbar}{i} \, \mathbf{A} \, \Psi^{\ast} \, \Psi \right]<br />
Recalling that \rho(\mathbf{r}, t) = \Psi^{\ast}(\mathbf{r}, t) \, \Psi(\mathbf{r}, t) is the probability density for finding the particle at position \mathbf{r} at time t, we see that the probability current density for a particle in an external electromagnetic field is given by:
<br />
\mathbf{J} = \frac{\hbar}{2 m i} \, \left(\Psi^{\ast} \, \nabla \Psi - \nabla \Psi^{\ast} \, \Psi \right) - \frac{q}{m} \, \mathbf{A} \, \Psi^{\ast} \, \Psi<br />
The first term is present even in the absence of an electromagnetic field. The second term is of the form \mathbf{J} = \rho \, \mathbf{v}, where \mathbf{v} = -q \, \mathbf{A}/m.
If I can dare to say, it is as if the presence of a vector potential \mathbf{A}(\mathbf{r}) leads to dragging of charged particles with the background velocity that depends both on their charge and on their mass. The presence of a magnetic field \mathbf{B} produces a vorticity of this background velocity field:
<br />
\mathbf{\omega} \equiv \nabla \times \mathbf{v} = -\frac{q}{m} \, \left(\nabla \times \mathbf{A}\right) = -\frac{q \, \mathbf{B}}{m}<br />
If we look at the formula for the vorticity of the velocity field of a rigid body rotating with angular velocity \mathbf{\Omega} (\mathbf{v} = \mathbf{\Omega} \times \mathbf{r} \Rightarrow \mathbf{\omega} = \nabla \times \mathbf{v} = 2 \, \mathbf{\Omega}), we see that the vorticity of the "dragging" velocity field for the particle in a magnetic field is the same as if there is a rotation with half the cyclotron frequency of the particle (\omega_{c} \equiv |q| |B|/m) and with the direction of the angular velocity in the opposite (same) direction as the direction of the magnetic field if the particle is positive (negative).
