Consider two examples of number fields:
$K= \mathbb Q(\sqrt{-13})$ and $L=\mathbb Q(\sqrt{-11})$, with respective rings of integers $\mathcal O_K = \mathbb Z[\sqrt{-13}]$ and $\mathcal O_L = \mathbb Z[\frac{1+\sqrt{-11}}{2}]$
Then suppose we have elements $a + b\sqrt{-13} \in \mathcal O_K$ and $x + y\sqrt{-11} \in \mathcal O_L$
Then $N(a+b\sqrt{-13}) = a^2 + 13b^2$ but is the norm of $x+y\sqrt{-11}$ equal to $x^2 + 11y^2$?
Yes, conceptually finding the norm for elements in $\mathbb{Q}(\sqrt{-11})$ is no different from finding norms in $\mathbb{Q}(\sqrt{-13})$.
I'm not sure what definition of the norm you are working with, but according to the definition using embeddings we have for $\alpha = x + y\sqrt{-11}$ in $\mathbb{Q}(\sqrt{-11})$ the norm of $\alpha$ is
$$ N(\alpha) = \sigma_1(\alpha)\sigma_2(\alpha) $$
where $\sigma_1$ and $\sigma_2$ are the two embeddings of $\mathbb{Q}(\sqrt{-11})$ into $\mathbb{C}$. Namely, $\sigma_1$ is the identity and $\sigma_2$ sends $\sqrt{-11}$ to $-\sqrt{-11}$. So you get
$$ N(\alpha) = (x + y\sqrt{-11})(x - y\sqrt{-11}) $$