Edit. As Ben Crowell points out in his comments below (and in the subsequent chat), my argument below is incomplete. In particular, I used a specific integral expression for the electric field (written in terms of the charge density) that is not generally valid. In my own mind, that integral expression was justified by writing the electric field as $\mathbf E = -\nabla\Phi$, but as Ben pointed out in the chat, this cannot be done in general; one must also include a vector potential term. As a result, my claim that the time-independence of the charge density leads to the time-independence of the electric field does not hold with the assumptions stated. To make something like my argument work, one really needs to ensure, either by assumption, or perhaps by some other formulation of the problem, that the time derivative of the electric field vanishes. Thanks to Ben for his time!
There is a somewhat simple, rigorous argument coming from applying Maxwell's equations. User sybtc refers to Maxwell's equations in his answer, but he provides nothing in the way of detail, and I don't think the result is just an obvious application of Gauss's law. Gauss's law shows that a static charge distribution will lead to a static electric field, but one needs to analyze the other equations to show that such a charge distribution also does not create a magnetic field. One can actually prove the following claim which I think is what you are looking for (although I'm not sure it's a "simple physical" result).
Claim. For any localized charge distribution $\rho$ for which the corresponding current $\mathbf J$ vanishes, the magnetic field is everywhere zero.
Proof. First, recall the continuity equation:
\begin{align}
\frac{\partial \rho}{\partial t} +\nabla\cdot\mathbf J = 0
\end{align}
The hypothesis $\mathbf J = \mathbf 0$ therefore implies that the charge distribution is static; $\partial\rho/\partial t = 0$. By Gauss's Law
\begin{align}
\nabla\cdot \mathbf E = \frac{\rho}{\epsilon_0},
\end{align}
the electric field of a localized charge distribution has the following integral expression:
\begin{align}
\mathbf E(t,\mathbf x) = \int_{\mathbb R^3}d^3x'\,\rho(t,\mathbf x')\frac{\mathbf x - \mathbf x'}{|\mathbf x - \mathbf x'|^3}
\end{align}
It follows that since $\rho$ is time-independent, then so is $\mathbf E$:
\begin{align}
\frac{\partial\mathbf E}{\partial t} = \mathbf 0
\end{align}
That fact, the fact that the current vanishes, and Ampere's Law
\begin{align}
\nabla\times \mathbf B = \mu_0\mathbf J +\mu_0\epsilon_0 \frac{\partial \mathbf E}{\partial t}
\end{align}
combine to require the magnetic field to have zero curl;
\begin{align}
\nabla\times\mathbf B = 0
\end{align}
But now recall that one of Maxwell's equations tells us that the divergence of the magnetic field is zero;
\begin{align}
\nabla\cdot \mathbf B = 0
\end{align}
It follows from the Helmholtz Decomposition, that the magnetic field must vanish everywhere provided it falls off sufficiently rapidly at infinity, a reasonable property that should be true of physical charge distributions.
