I always thought that a change in electric field induces a magnetic field and vice-versa. Moreover, I imagined that any current distribution will give rise to a magnetic field. But then I wrote this down: Maxwell's equations in absence of magnetic field.
\begin{align} \nabla \cdot \mathbf{E}(\mathbf{x},t) &= \frac {\rho(\mathbf{x},t)} {\varepsilon_0}\\ \nabla \times \mathbf{E}(\mathbf{x},t) &= 0\\ \frac{\partial \mathbf{E}(\mathbf{x},t)} {\partial t}&= -\frac{\mathbf{j}(\mathbf{x},t)}{\varepsilon_0} \end{align}
the second equation gives \begin{align} \mathbf{E}(\mathbf{x},t)=-\nabla \phi(\mathbf{x},t) \end{align} so that the rest becomes
\begin{align} \nabla^2 \phi(\mathbf{x},t) &= -\frac {\rho(\mathbf{x},t)} {\varepsilon_0}\\ \nabla \frac{\partial\phi(\mathbf{x},t)} {\partial t}&= \frac{\mathbf{j}(\mathbf{x},t)}{\varepsilon_0} \end{align} (Note: already we can see charge conservation, i.e. $\partial_t \rho+\nabla\cdot\mathbf{j}=0$)
Then we get the usual solution of the first equation \begin{align} \phi(\mathbf{x},t) = \iiint \frac{\rho(\mathbf{x}',t)}{4\pi\epsilon_0 |\mathbf{x} - \mathbf{x}'|}\, \mathrm{d}^3\! x', \end{align} which can be written as \begin{align} \partial_t\phi(\mathbf{x},t) = -\iiint \frac{\nabla_{\mathbf{x}'}\cdot\mathbf{j}(\mathbf{x}',t)}{4\pi\epsilon_0 |\mathbf{x} - \mathbf{x}'|}\, \mathrm{d}^3\! x', \end{align} and the second equation, due to gradient theorem, becomes \begin{align} \partial_t\phi(\mathbf{x},t)&= \phi(\mathbf{0},t) + \frac{1}{\varepsilon_0}\int_0^1 \mathbf{j}(\lambda\mathbf{x},t)\cdot\mathbf{x}\,\mathrm d\lambda \end{align} so that \begin{align} \phi(\mathbf{0},t) + \frac{1}{\varepsilon_0}\int_0^1 \mathbf{j}(\lambda\mathbf{x},t)\cdot\mathbf{x}\,\mathrm d\lambda=-\iiint \frac{\nabla_{\mathbf{x}'}\cdot\mathbf{j}(\mathbf{x}',t)}{4\pi\epsilon_0 |\mathbf{x} - \mathbf{x}'|}\, \mathrm{d}^3\! x' \end{align}
I guess what I'm trying to show here is $\mathbf{E}(\mathbf{x},t)=\mathbf{E}(\mathbf{x})$ so that $\mathbf{j}(\mathbf{x},t)=0$, but I might be wrong. Also I don't see an obvious way to continue with this derivation.
EDIT:
A current density doesn't always produce a magnetic field, so that configurations with $\mathbf{E}(\mathbf{r},t)$ and $\mathbf{B}=0$ do exist. The final question I'm wondering about is whether the last equation is some sort of a constraint on $\mathbf{j}(\mathbf{x},t)$ or is it an equality true in general?