Consider the Baire space $\mathbb{N}^\mathbb{N}$. A natural pre-order $\leq^*$ on $\mathbb{N}^\mathbb{N}$ is defined by $f\leq^*g$ if and only if $f(n)\leq g(n)$ for all but finitely many $n$. A subset $A$ of $\mathbb{N}^\mathbb{N}$ is said to be bounded if there is a $g\in\mathbb{N}^\mathbb{N}$ such that $f\leq^*g$ for all $f\in A$. Let $\mathfrak{b}$ be the minimum cardinality of an unbounded subset of $\mathbb{N}^\mathbb{N}$.

Let $\mathcal{I}$ be an ideal on $\mathbb{N}$. For any $f,g\in\mathbb{N}^\mathbb{N}$, we write $f\leq_\mathcal{I} g$ if and only if $\{n\in\mathbb{N} : f(n)>g(n)\}\in\mathcal{I}$. Note that $\leq_{\bf{Fin}}=\leq^*$, where $\bf{Fin}$ is the ideal on $\mathbb N$ containing all finite subsets of $\mathbb N$. A subset $A$ of $\mathbb{N}^\mathbb{N}$ is said to be $\mathcal{I}$-bounded if there is a $g\in\mathbb{N}^\mathbb{N}$ such that $f\leq_\mathcal{I} g$ for all $f\in A$. Let $\mathfrak{b}(\mathcal{I})$ be the minimum cardinality of a $\leq_\mathcal{I}$-unbounded subset of $\mathbb{N}^\mathbb{N}$.

Is there any class of ideals for which $\mathfrak{b}(\mathcal I)\neq\mathfrak{b}$?