A constant charge density does not imply a zero magnetic field. Even considering a set of isolated charges, suppose they were (mechanically) moved along a circular path. The charge density could remain the same but there would be a current flow. The curl of the magnetic field produced would be $\mu_0 \vec{J}$, where $\vec{J}$ is the current density.
If the charge density is static, all you can say (from the equation of charge continuity) is that the divergence $\nabla \cdot \vec{J} = 0$. This also does not imply time-independence. For instance, you could speed up the circular motion of the charges to get a larger current density and a larger magnetic field.
If you can have a time-dependent magnetic field then you will also get a time-dependent electric field. The time-independence of $\rho$ only tells you that the divergence of the electric field is time-independent.
Both of these follow from desirable properties of this hypothetical magnetic charge, namely:
- Magnetic charge is conserved.
- Magnetic field lines radiate outwards from positive magnetic charges.
- The net force between two magnetic charges moving at constant speed along parallel tracks is less than that between two stationary charges.
All three of these properties hold for electric charges. The last one may not be as familiar, but it basically works as follows: if we have a positive electric charge moving at constant velocity, it generates a magnetic field in addition to its electric field. A second positive electric charge moving parallel to the first one will therefore experience a magnetic force, and if you work out the directions, this force works out to be attractive. Thus, the net force between the two charges (electric and magnetic together) is less than the magnitude of the force they would exert on each other if they were at rest. [ASIDE: This can also be thought of in terms of the transformation properties of forces between different reference frames in special relativity, if you prefer to think of it that way.]
Now, the conservation of electric charge can be written in terms of the continuity equation:
$$
\vec{\nabla} \cdot \vec{j}_e + \frac{ \partial \rho_e}{\partial t} = 0
$$
Note that this can be derived from Ampère's Law and Gauss's Law ($\epsilon_0 \vec{\nabla} \cdot \vec{E} = \rho_e$), using the fact that the divergence of a curl is always zero:
$$
0 = \vec{\nabla} \cdot (\vec{\nabla} \times \vec{B}) = \mu_0 \vec{\nabla} \cdot \vec{j}_e + \mu_0 \frac{\partial ( \epsilon_0 \vec{\nabla} \cdot \vec{E})}{\partial t} = \mu_0 \left( \vec{\nabla} \cdot \vec{j}_e + \frac{\partial \rho_e}{\partial t} \right)
$$
If we want to extend Maxwell's equations to magnetic charges, we need to have a magnetic version of Gauss's Law and add in a magnetic current term to Faraday's Law:
$$
\vec{\nabla} \cdot \vec{B} = \alpha \rho_m \qquad \vec{\nabla} \times \vec{E} = \beta \vec{j}_m - \frac{\partial \vec{B}}{\partial t} ,
$$
where $\alpha$ and $\beta$ are arbitrary proportionality factors. But if we try to derive a continuity equation for magnetic charge from these two facts (as we did above for electric charge), we get
$$
\beta \vec{\nabla} \cdot \vec{j}_m - \alpha \frac{\partial \rho_m}{\partial t} = 0,
$$
and this is equivalent to the continuity equation if and only if $\alpha = - \beta$. Beyond this, the choice of $\alpha$ is to some degree arbitrary; different values correspond to different choices of which type of magnetic charge we call "positive", and what units we use to measure it. If we want to have magnetic field lines radiating away from "positive" magnetic charges, then we will want $\alpha > 0$; the usual choice in MKS units is to pick $\alpha = \mu_0$ (and $\beta = -\mu_0$), as you have in your equations above.
This negative sign in the magnetic current term Faraday's Law then implies that the electric field lines created by a moving magnetic charge will obey a "left-hand rule" instead of a "right-hand rule". In other words, the direction of $\vec{E}$ created by a moving magnetic charge would be opposite the direction of $\vec{B}$ created by a moving electric charge. If we still want two magnetic charges moving along parallel tracks to exhibit a lesser force than what they feel when at rest, then we must also flip the sign of the $\vec{v} \times \vec{E}$ term in the Lorentz force law to compensate for this flip.
Best Answer
Let us begin with the Maxwell's equations for electrostatics and magnetostatics.
For electrostatics: $$\nabla\cdot \mathbf{E}=\frac{\rho}{\epsilon_0} \\ \nabla\times \mathbf{E}=0 $$
For magnetostatics: $$\nabla\cdot \mathbf{B}=0 \\ \nabla\times \mathbf{B}=\frac{\mathbf{J}}{c^2\epsilon_0}$$
The equation $\nabla\cdot \mathbf{B}=0$ in magnetostatics says that the divergence of the magnetic field is always zero. Comparing this with the analogous equation $\nabla\cdot \mathbf{E}=\dfrac{\rho}{\epsilon_0}$ in electrostatics, we can conclude that there are no magnetic charges. This implies that the magnetic field lines neither starts nor ends (for exceptions, see this paper). Then what is the origin of the magnetic field?
Magnetic fields are always associated with electric currents, and the equation $\nabla\times \mathbf{B}=\dfrac{\mathbf{J}}{c^2\epsilon_0}$ says that the curl is proportional to the electric current density. So the magnetic field lines in seldom cases form closed loops around the current vectors (for the general case, see this post).
Now the question is: Why don't electric field lines form closed loops?
At first sight, one might think that electric field lines also form closed loops when $\rho=0$. But this is not the case. To understand this, look at the equation $\nabla\times \mathbf{E}=0$ which says that the curl of electric field is always zero. This simply means that electric field lines can never form closed loops. This fact is closely related to conservative nature of the electrostatic field.