Since the curl of E is the time derivative of B,
$\nabla \times \vec{E} = -\frac{\partial B}{\partial t}$
Do changing magnetic fields always produce solenoidal electric fields? For instance a changing magnetic field in time in the z direction produces an electric field in the $\phi$ direction. Which makes sense since it causes a current in a wire.
$ \renewcommand{\div}{\vec{\nabla} \cdot} \renewcommand{\curl}{\vec{\nabla} \times} \newcommand{\e}{\vec{E}} \newcommand{\b}{\vec{B}} $The answer is yes. We know that $\div \e=\rho$ and $\curl \e = -\dot{\b}$. The the helmholtz decomposition theorem tells us $\e$ can be written as the sum of a irrotational piece $\e_\mathrm{ir}$ and a solenoidal piece $\e_\mathrm{sol}$. These pieces satisfy
$$\e_\mathrm{ir} + \e_\mathrm{sol}=\e$$
$$\div \e_\mathrm{ir} = \rho$$
$$\curl \e_\mathrm{ir} = \vec{0}$$
$$\div \e_\mathrm{sol} = 0$$
$$\curl \e_\mathrm{sol} = -\dot{\b}$$
Now if we say that the changing magnetic field is only responsible for producing $\e_\mathrm{sol}$ and not $\e_\mathrm{ir}$, which is a reasonable way of thinking about it, then your statement is correct: changing magnetic fields can only produce solenoidal electric fields.
