Note
In cylindrical coordinates, the curl with only azimuthal field goes to:
$$
\nabla \times \mathbf{B} = - \hat{\mathbf{r}} \frac{ \partial B_{\theta} }{ \partial z } + \hat{\mathbf{z}} \frac{ 1 }{ r } \frac{ \partial }{ \partial r } \left( r B_{\theta} \right) \tag{1a}
$$
Since we can assume that the field will not vary along $\hat{\mathbf{z}}$, the only term remaining is the second. If $B_{\theta}$ is constant with respect to radial distance, i.e., no local sources, then Equation 1a reduces to:
$$
\nabla \times \mathbf{B} = \hat{\mathbf{z}} \frac{ B_{\theta} }{ r } \tag{1b}
$$
Flux Freezing
The time variation of the flux can be written as:
$$
\begin{align}
\frac{ \partial \psi }{ \partial t_{1} } & = \int_{S} \ dA \ \hat{\mathbf{n}} \cdot \frac{ \partial \mathbf{B} }{ \partial t } \tag{2a} \\
& = - \int_{S} \ dA \ \hat{\mathbf{n}} \cdot \left( \nabla \times \mathbf{E} \right) \tag{2b}
\end{align}
$$
where $\hat{\mathbf{n}}$ is the outward unit normal of the surface $S$.
Suppose our surface $S$ was defined by a closed contour, $C$, with infinitesimal length, $d\mathbf{l}$. If this contour moves along with an arbitrary fluid element at velocity $\mathbf{V}$, then the area swept out per unit time by $C$ would be given by $\mathbf{V} \times d\mathbf{l}$. Therefore, we can write another part of the flux given by:
$$
\frac{ \partial \psi }{ \partial t_{2} } = \oint_{C} \ \mathbf{B} \cdot \left( \mathbf{V} \times d\mathbf{l} \right) = \oint_{C} \ \left( \mathbf{B} \times \mathbf{V} \right) \cdot d\mathbf{l} \tag{3}
$$
Note that the $\partial_{t}$ was pulled in to define the $\mathbf{V}$ from the area here.
Using Stokes' theorem we can rewrite Equation 3 as:
$$
\frac{ \partial \psi }{ \partial t_{2} } = \int_{S} \ dA \ \hat{\mathbf{n}} \cdot \nabla \times \left( \mathbf{B} \times \mathbf{V} \right) \tag{4}
$$
Then if we combine Equations 2b and 4 we have the total time rate of change of the flux, given by:
$$
\frac{ d \psi }{ dt } = - \int_{S} \ dA \ \hat{\mathbf{n}} \cdot \left[ \nabla \times \left( \mathbf{E} + \mathbf{V} \times \mathbf{B} \right) \right] \tag{5}
$$
The frozen-in condition requires that the following holds:
$$
\mathbf{E} + \mathbf{V} \times \mathbf{B} = 0 \tag{6}
$$
for a constant flux. If $\mathbf{V}$ is parallel to $\mathbf{B}$ then $\mathbf{E} = 0$ and the curl of $\mathbf{V} \times \mathbf{B}$ goes to (using standard vector calculus identities):
$$
\nabla \times \left( \mathbf{V} \times \mathbf{B} \right) = - \left( \mathbf{V} \cdot \nabla \right) \mathbf{B} \tag{7}
$$
because all other terms go to zero from geometry/symmetry and Maxwell's equations. The $\theta$ term of the gradient is given by:
$$
\left( \nabla f \right)_{\theta} = \frac{ 1 }{ r } \frac{ \partial f }{ \partial \theta } \tag{8}
$$
The rest is just symbol gymnastics to show that $\tfrac{B_{\theta}}{r}$ is a constant.
