Given a Theory and a Model, you can define the Morley Rank of formulas with parameters from the model.
I'd like you to give me an example of a formula (with theory and model) with infinite Morley Rank.
In the linked page of Wiki, there's an example of it, but I can't seem to be able to prove it formally.
Here is a simple one.
In the theory $Th(\mathbb{Q},<)$ the formula $x=x$ has rank $\infty$.
First, let $a^1_n,a^2_n$ be an infinite sequence of 2-tuples of $\mathbb{Q}$ where $a^1_n < a^2_n <a^1_{n+1}<a^2_{n+1+}$.
Now, consider the formula (with parameters) $\varphi_n(x)\equiv (a^1_n <x<a^2_n)$. Note that $\varphi_n(\mathbb{Q})=\aleph_0$ as well as $\varphi_n(\mathbb{Q})\cap\varphi_m(\mathbb{Q})=\emptyset$ for $n \neq m$. Finally, realizing that each $\varphi_n(\mathbb{Q})\cong \mathbb{Q}$, we can repeat the processes indefinitely, and so we can construct arbitrarily large trees.
However, since $|\mathbb{Q}|=\aleph_0$, the argument above only works morally. What you actually do is just build arbitrarily large trees. You can do this by just picking large saturated models (say of size $\kappa$) and doing exactly the same above. You will construct a tree of size $\kappa$.
Therefore, the rank of $x=x$ has unbounded ($\infty$) rank in $(\mathbb{Q},<)$.