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Why do we use $\mu_0$ during the derivation of $\vec{B}=\mu_0(\vec{H} + \vec{M})$ where $\vec{M}$ is magnetization?

The derivation given in Sadiku's Elements of Electromagnetics:
Let $\vec{J_f}$ be free volume current density, $\vec{J_b}$ be bound volume current density, \begin{align*} \nabla \, \times \left( \frac{\vec{B}}{\mu_0} \right) &= \vec{J_f} + \vec{J_b} = \vec{J} \\ &= \nabla \times \vec{H} \, + \nabla \times \vec{M} \\ &= \nabla \times (\vec{H} + \vec{M}) \\ \vec{B} &=\mu_0(\vec{H} + \vec{M}) \quad \blacksquare \end{align*}

I don't understand why we should use $\mu_0$ in the first place. Why don't we use $\mu$ instead? In free space, $\vec{M} = 0$ and \begin{align*} \nabla \times \vec{H} &= \vec{J_f} \\ \nabla \times \left( \frac{\vec{B}}{\mu_0} \right) &= \vec{J_f} \end{align*} then naturally we'd like to still have $\nabla \times \vec{H} = \vec{J} = \vec{J_f} + \vec{J_b}$ when $\vec{M} \neq 0$, so we could just change the $\mu_0$ to some constant $\mu$, so that \begin{align*} \nabla \times \vec{H} &= \vec{J} \\ \nabla \times \left( \frac{\vec{B}}{\mu} \right) &= \vec{J} = \vec{J_f} + \vec{J_b} \end{align*} but in the correct derivation, \begin{align*} \nabla \times \left( \frac{\vec{B}}{\mu_0} \right) &= \vec{J} = \vec{J_f} + \vec{J_b} \end{align*}

What is it that forces us to use $\mu_0$?

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  • $\begingroup$ My another question helped me to demystify this question. Basically, I was confused by the book defining $\oint {\bf H} \cdot d {\bf \ell} = I_{enc}$ while in fact it's not clearly explained in the text that $I_{enc} = I_{free}$. So after applying Stoke's theorem, I mistakenly treated $\nabla \times {\bf H} = {\bf J_f} + {\bf J_b}$ while it should be $\nabla \times {\bf H} = {\bf J_f}$ only, it's just definition and no point to add ${\bf J_b}$ to it. $\endgroup$ – TED May 19 at 6:18
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${\bf B}=\mu_0({\bf H}+{\bf M})$ is effectively just the definition of the auxiliary field ${\bf H}$. ${\bf B}$ is defined as the quantity that generates the velocity-dependent force in the Lorentz Force Law, and the magnetization ${\bf M}$ is the magnetic moment per unit volume. ${\bf H}$ is then defined in terms of the other two.

For the currents, the bound current ${\bf J}_{b}$ is defined as the curl of ${\bf M}$, and then the free current is whatever is left over, ${\bf J}_{f}={\bf J}-{\bf J}_{b}$. There is thus no room to change the constant $\mu_{0}$ to something else.

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