Permeability constant in Ampère's circuital law must be vacuum permeability $\mu_0$?

Question

Just to be sure, is the permeability constant in Ampère's circuital law always equal to $\mu_0$, regardless of which medium the Amperian loop is placed in? That is, $\oint {\bf B} \cdot d {\bf \ell} = \mu_0 I$ and never equal to $\mu I$.

My reasoning is that if $\mu \neq \mu_0$, then using Stoke's theorem, \begin{align*} \oint {\bf B} \cdot d {\bf \ell} &= \mu I \\ \int (\nabla \times {\bf B}) \cdot d{\bf S} &= \mu \int {\bf J \cdot} d{\bf S} \\ \nabla \times {\bf B} &= \mu {\bf J} \\ &= \mu ({\bf J}_f + {\bf J}_b)\\ &= \mu (\nabla \times {\bf H} + \nabla \times {\bf M}) \\ &= \mu [ \nabla \times ({\bf H} + {\bf M}) ] \\ {\bf B} &= \mu({\bf H} + {\bf M}) \end{align*} But this contradicts with ${\bf B} = \mu_0({\bf H} + {\bf M})$.

Therefore, the $\mu$ must be equal to $\mu_0$.

Corollary:
Some books define Ampère's circuital law as $\oint {\bf H} \cdot d {\bf \ell} = I$. This is true if we are dealing with ${\bf B}$ in free space (or if $I=I_f$, see comment below). That is, we place the Amperian loop in free space such that \begin{align*} \oint {\bf B} \cdot d {\bf \ell} &= \mu_0 I\\ \oint \frac{\bf B}{\mu_0} \cdot d{\bf \ell}&= I \\ \oint {\bf H}\cdot d{\bf \ell}&= I \end{align*} If ${\bf B}$ is not in free space then $\frac{\bf B}{\mu_0} \neq {\bf H}$ and thus $\oint {\bf H}\cdot d{\bf \ell} \neq I$ (unless $I=I_f$, see comment below).

Answer 1

Yes, one of the correct forms of the law is $\oint {\bf B} \cdot d {\bf \ell} = \mu_0 I $, (not$ \mu$)and $I$ is all the currents included in the loop (free current in conductors, bound current due to spins, current due to orbital motion of electrons, everything).

Inside a linear medium (with permeability being constant inside it), one can use the formula $\oint {\bf B} \cdot d {\bf \ell} = \mu I_{free}$, where $ \mu = \mu_r \mu_0$, but this may not work if the loop of integration passes through multiple mediums.

Note that the above can be derived from $\oint {\bf H} \cdot d {\bf \ell} = I_{free}$ (in your question you have written $I$, but it should be $I_{free}$, then everything fits together and it is valid in any medium), which is another form of Ampere's law. In this form only free currents are included.