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

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).

• Why do you think so? May 19 '19 at 4:02
• @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. May 19 '19 at 4:05
• $\oint {\bf H} \cdot d {\bf \ell} = I_{free}$. May 19 '19 at 4:55
• @ArchismanPanigrahi Yes, thanks for mentioning. $\nabla \times {\bf H} = {\bf J_f}$, $I_f = \int {\bf J}_f \cdot d{\bf S} = \int \nabla \times {\bf H} \cdot d{\bf S} = \oint {\bf H} \cdot d {\bf \ell}$. May 19 '19 at 5:08
• @ArchismanPanigrahi Thank you. May 19 '19 at 5:13

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.

• How would one proceed if the loop of integration passes through multiple mediums? For an example visit my question which got marked as "duplicate".
– ludz
Aug 20 at 12:13
• @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. Aug 20 at 12:18
• @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). Aug 20 at 12:26
• 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 $2I$? $I_{tot}=free current + bound current$?
– ludz
Aug 20 at 12:30
• Okay thank you! Much appreciated for taking your time to answer even though this question was asked 2 years ago.
– ludz
Aug 20 at 12:37