Why is $\nabla \times B$ perpendicular to $B$? I'm considering a magnetic field inside a load capacitor where
$$\boldsymbol{\nabla}\times\textbf{B} =\mu_{0} \varepsilon_{0} \frac{\partial\textbf{E}}{\partial t}\\
\textbf{j}_{c}=0.$$
Why is $\boldsymbol{\nabla}\times\textbf{B}$ is perpendicular to $\textbf{B}$?
 A: The condition that the magnetic field and its curl be orthogonal is satisfied, generally, for plane-wave situations where the curl is given simply by $\nabla\times\mathbf B = i\mathbf k\times \mathbf B$, as well as a number of similarly 'nice' geometries, but it is not a general property of electromagnetic fields.
In the specific case you refer to ─ the magnetic field induced by the displacement current inside a charging capacitor ─ is a bit tricky because what you have is a differential equation for $\mathbf B$,
$$
\nabla\times\mathbf{B} =\mu_{0} \varepsilon_{0} \frac{\partial\mathbf{E}}{\partial t},
\tag1
$$
where the source $\frac{\partial\mathbf{E}}{\partial t}$ is assumed to be known and homogeneous over space (possibly with some time dependence), and this differential equation does not have unique solutions. This means that we must put in additional constraints to mold it to our needs: we normally choose the solution
$$
\mathbf B = (Cy,-Cx,0)
$$
for $C$ a constant and with $\frac{\partial\mathbf{E}}{\partial t}$ along $z$, because it's the most useful one (but note that it has an unphysical dependence on the origin!). However, the magnetic field
$$
\mathbf B = (Cy,-Cx,B_0)
$$
is also a solution of $(1)$, and it no longer satisfies $\mathbf B \cdot(\nabla\times\mathbf B)=0$, so the property is not inherent to the statement of Ampère's law ─ it comes from the external conditions we impose on the problem. In other words, in that problem $\mathbf B \cdot(\nabla\times\mathbf B)=0$ because we want it to.
