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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).

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  • $\begingroup$ Why do you think so? $\endgroup$
    – Dvij D.C.
    May 19 '19 at 4:02
  • $\begingroup$ @DvijMankad Because never have I seen any source defining $\oint {\bf B} \cdot d {\bf \ell} = \mu I$. It seems like the $\mu$ must be equal to $\mu_0$ in order for the law to work. $\endgroup$
    – endeneer
    May 19 '19 at 4:05
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    $\begingroup$ $\oint {\bf H} \cdot d {\bf \ell} = I_{free}$. $\endgroup$ May 19 '19 at 4:55
  • $\begingroup$ @ArchismanPanigrahi Yes, thanks for mentioning. $\nabla \times {\bf H} = {\bf J_f}$, $I_f = \int {\bf J}_f \cdot d{\bf S} = \int \nabla \times {\bf H} \cdot d{\bf S} = \oint {\bf H} \cdot d {\bf \ell} $. $\endgroup$
    – endeneer
    May 19 '19 at 5:08
  • $\begingroup$ @ArchismanPanigrahi Thank you. $\endgroup$
    – endeneer
    May 19 '19 at 5:13
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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.

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  • $\begingroup$ How would one proceed if the loop of integration passes through multiple mediums? For an example visit my question which got marked as "duplicate". $\endgroup$
    – ludz
    Aug 20 at 12:13
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    $\begingroup$ @ludz The two equations will remain as is. For $\oint {\bf B} \cdot d {\bf \ell} = \mu_0 I$, one needs to consider all currents including currents in the medium. For linear mediums, $\mu I_{free} = \mu_0 I_{total}$. The other equation $\oint {\bf H} \cdot d {\bf \ell} = I_{free}$ is for free currents, and will remain the same. $\endgroup$ Aug 20 at 12:18
  • $\begingroup$ @ludz It is always $\mu_0$, but you need to consider the total current (there can be bound currents in a medium, and you need to consider them also). $\endgroup$ Aug 20 at 12:26
  • $\begingroup$ If I understood you correctly then the equation for my example with two wires with different magnetic permeability is still $\oint \vec B \cdot \vec{dl}=\mu_0 I_{tot}$ but $I_{tot}$ is not necessarily equal to $2I$? $I_{tot}=free current + bound current$? $\endgroup$
    – ludz
    Aug 20 at 12:30
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    $\begingroup$ Okay thank you! Much appreciated for taking your time to answer even though this question was asked 2 years ago. $\endgroup$
    – ludz
    Aug 20 at 12:37

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