Cutoff a divergence-free function while staying divergence-free

Question

This question is related to another question I've asked. Let us consider the two channels $A = \mathbb R \times[-1, 1]$, $B = \mathbb R \times[-2, 2]$ and the functions $\boldsymbol{\varphi}, \boldsymbol{\psi}: \Omega \subset \mathbb R^2 \to \mathbb R^2 \in C^\infty(\Omega) \cap L^2(\Omega)$ , for $\Omega$ an open set containing $A$ and $B$, which have free divergence. Assume that $\text{supp }\boldsymbol{\varphi}\subset A$ and that $$\|\boldsymbol{\varphi} - \boldsymbol{\psi}\|_{L^2(\Omega)} < \varepsilon.$$ I would like to show that, in this case there exists a divergence free $\boldsymbol{\tilde{\psi}} \in C^\infty(\Omega) \cap L^2(\Omega)$, equal to $\boldsymbol{\boldsymbol \psi}$ on $A$, such that $\text{supp }\boldsymbol{\tilde\psi}\subset B$ and $$\|\boldsymbol{\varphi} - \boldsymbol{\tilde\psi}\|_{L^2(\Omega)} < \varepsilon.$$ As suggested by @Calvin Khor in the comment of the previous question, I could define a cutoff function $$ \theta = \begin{cases} 1 & \text{in } A,\\ 0 & \text{in } B^c. \end{cases} $$ As $\boldsymbol{\psi}$ has a free divergence, there exists a smooth scalar function $\zeta$ such that $\boldsymbol{\psi} = \nabla \times (0,0,\zeta)$. If we set $\boldsymbol{\tilde\psi} = \nabla \times (0,0,\theta\zeta) = \zeta\nabla \theta + \theta \boldsymbol{\psi}$ we have \begin{align} \|\boldsymbol{\varphi} - \boldsymbol{\tilde\psi}\|_{L^2(\Omega)}^2 &= \|\boldsymbol{\varphi} - \boldsymbol{\psi}\|_{L^2(A)}^2 + \|\boldsymbol{\tilde\psi}\|^2_{L^2(B \backslash A)}\\ &= \|\boldsymbol{\varphi} - \boldsymbol{\psi}\|_{L^2(A)}^2 + \|\zeta\nabla \theta + \theta \boldsymbol{\psi}\|^2_{L^2(B \backslash A)} \end{align} Is there a way from here to show that $\|\zeta\nabla \theta + \theta \boldsymbol{\psi}\|^2_{L^2(B \backslash A)}$ must be lower than $\varepsilon~?$ I really don't see how to control this term..