Let $ \Omega \subset \mathbb{R}^3 $ be a convex polyhedron with connected boundary $ \partial \Omega $. $ \Omega $ can be divided into finite subdomains. For simplicity, it is assumed that $ \Omega $ consists of only two subdomains, namely a conducting subdomain $ \Omega_c $ and an insulate subdomain $ \Omega_d = \Omega \setminus \overline{\Omega}_c $. $ \Omega_c $ is surrounded by $ \Omega_d $.
In subdomain $ \Omega_c $, the electric conductivity $ \sigma $ is positive and bounded, namely $ 0 < \sigma_* \le \sigma \le \sigma^* < \infty $. However, $ \sigma = 0 $ in $ \Omega_d $. The magnetic permeability $ \mu $ satisfies $ 0 < \mu_* \le \mu \le \mu^* < \infty $ in $ \Omega $.
Suppose $ \mathbf{J}_s \in L^2 \bigl( [0,T], \mathbf{L}^2 ( \Omega_d )\bigr) $.
The variational formulation reads as: Given $ \bigl( \mathbf{A} ( 0 ), \phi ( 0 ) \bigr) = ( \mathbf{A}_0, \phi_0 ) $, find $ \bigl( \mathbf{A} ( t ), \phi ( t ) \bigr) $, such that \begin{multline*} \bigl( \sigma \partial_t \bigl( \mathbf{A} ( t ) + \nabla \phi ( t ) \bigr), \mathbf{Q} + \nabla \psi \bigr)_{ \Omega_c } + \bigl( 1/\mu \nabla \times \mathbf{A} ( t ), \nabla \times \mathbf{Q} \bigr)_{ \Omega } + \bigl( 1/\mu \nabla \cdot \mathbf{A} ( t ), \nabla \cdot \mathbf{Q} )_{\Omega} \\ = \bigl( \mathbf{J}_s (t), \mathbf{Q} \bigr)_{\Omega_d}, \end{multline*} for a.a. $ t \in \left[ 0, T \right] $ and any $ ( \mathbf{Q}, \psi ) \in \mathbf{H}_N^1 ( \Omega ) \times H^1 ( \Omega_c ) / \mathbb{R} $, where $$ \mathbf{H}_N^1 ( \Omega ) = \{ \mathbf{u} \in \mathbf{H}^1 ( \Omega ) : \mathbf{u} \times \mathbf{n} = 0 \; \mathrm{on} \; \partial \Omega \}, $$ My question is: Is it right to define the norm of $ \mathbf{H}_N^1 ( \Omega ) \times H^1 ( \Omega_c ) / \mathbb{R} $ as $$ \| \mathbf{Q} + \nabla \psi \|_{\mathbf{L}^2 ( \Omega_c )} + \| \nabla \times \mathbf{Q} \|_{\mathbf{L}^2 ( \Omega )} + \| \nabla \cdot \mathbf{Q} \|_{\mathbf{L}^2 ( \Omega )}, $$ and what is the relationship between the above norm and $$ \| \mathbf{Q} \|_{\mathbf{H}^1 ( \Omega )} + \| \psi \|_{H^1 ( \Omega_c )}? $$ Thank you very much.
Remark:
This question is inspired by Biro's paper. I wonder whether the weak solution is well-posed.
As far as I understand, the norm should be $$ \| \mathbf{Q} + \nabla \psi \|_{\color{red}{\textbf{L}^2(\Omega)}} + \dots \qquad (*) $$
I do not have a rigorous arugument to prove that with your choice something goes wrong, but let me explain my perspective.
$\widehat{\mathbf{H}}_0^1 ( \Omega ) \times H^1_0 ( \Omega )$ is cartesian product of spaces, hence it is natural to endow it with a graph norm. It is well known that $\| \nabla (\cdot) \|_{\mathbf{L}^2}$ is a norm on $H^1_0 ( \Omega )$. Recall that in standard topological situations it holds the continuous embedding $$ \mathbf{H}(\text{div}; \Omega) \cap \{ \mathbf{u} \in \mathbf{H}(\text{curl};\Omega) : \mathbf{u} \times \mathbf{n} = \mathbf{0} \ \text{on} \ \partial \Omega \} \hookrightarrow \mathbf{H}^1(\Omega), $$ or in $\widehat{\mathbf{H}}_0^1 ( \Omega )$ if we want; it follows that the norms $$ \| \cdot \|_{\mathbf{L^2}} + \| \text{curl}(\cdot) \|_{\mathbf{L}^2} + \| \text{div}(\cdot) \|_{\text{L}^2} $$ and $$ \| \cdot \|_{\mathbf{H}^1} $$ are equivalent on $\widehat{\mathbf{H}}_0^1 ( \Omega )$, whence the (a) graph norm on $\widehat{\mathbf{H}}_0^1 ( \Omega ) \times H^1_0 ( \Omega )$ would read $$ \| \mathbf{Q} \|_{\mathbf{L}^2(\Omega)} + \| \text{curl}(\mathbf{Q}) \|_{\mathbf{L}^2(\Omega)} + \| \text{div}(\mathbf{Q}) \|_{\text{L}^2(\Omega)} + \| \nabla \psi \|_{\mathbf{L}^2(\Omega)}, \quad (\mathbf{Q}, \psi) \in \widehat{\mathbf{H}}_0^1 ( \Omega ) \times H^1_0 ( \Omega ), $$ which is indeed equivalent to the one in $(*)$.
I imagine that you choose $\mathbf{L}^2(\Omega_{c})$ because the weak formulation you wrote down only requires summability in $\Omega_c$ for $\mathbf{Q}, \nabla \psi$, but I am afraid that some functional analysis problems may arise: for instance, is $\widehat{\mathbf{H}}_0^1 ( \Omega )$ even a Banach space if the norm does not read also the behavior of functions in $\Omega_d$?
These considerations more or less contain the answer to the second question too. For what concernes the norms you wrote down, the second one is stronger than the first one. Instead, if we modify the first as in $(*)$, they turn out to be equivalent.