For me the "physical intuition" is the fact that exclusion processes tend to clog up for large densities and hence you cannot expect a faster current than in the $\rho = \frac{1}{2}$ case, even when you add more particles.

For me the "physical intuition" is the fact that exclusion processes tend to clog up for large densities and hence you cannot expect a faster current than in the $\rho = \frac{1}{2}$ case, even when you add more particles.

PS: to answer your question in the comments, the phase transition should occur at $\rho = \frac{1}{2}$ by symmetry with respect to interchanging the notion of "particles" and "holes". Indeed (heuristically speaking) if the process has $\rho = \frac{1}{2}$, there are just enough holes that each particle basically behaves like a free particle. If you now increase the density, the front of particles still can only move as fast as the free particle and all the other ones block each other behind it until the first particle in front has moved away a bit. Only then the next one can follow and in this case again it moves like a free particle. This is (my) intuition why the particle current cannot be higher than in the "free" case.