# Ratio of $B/H$ in a ferromagnet

On the page https://en.wikipedia.org/wiki/Saturation_(magnetic), it is stated that

The relation between the magnetizing field H and the magnetic field B can also be expressed as the magnetic permeability: $$\mu =B/H$$ or the relative permeability $$\mu/\mu _{0}=\mu _{r}$$, where $$\mu _{0}$$ is the vacuum permeability. The permeability of ferromagnetic materials is not constant, but depends on $$H$$. In saturable materials the relative permeability increases with $$H$$ to a maximum, then as it approaches saturation inverts and decreases toward one.

So the page says that for a ferromagnet, as we increase $$H$$, the value of $$\mu_r = \mu/\mu_o =\frac{B}{\mu_oH}$$ first increases to a maximum and then decreases towards one. I thought that makes sense: we have $$\boldsymbol B = \mu_o(\boldsymbol H + \boldsymbol M)$$. As we increase $$\boldsymbol H$$ at some point the material saturates and $$M$$ stops growing at $$M_{sat}$$. Increasing $$\boldsymbol H$$ further, at some point surely we can achieve $$H\gg M_{sat}$$ (since $$M$$ is now constant) and so we can approximate $$\boldsymbol B \simeq \mu_o\boldsymbol H$$, and so $$B/H \simeq \mu_o$$ for large $$H$$ and indeed $$\color{blue}{\mu_r\rightarrow 1}$$.

But then, the $$\boldsymbol B$$ field inside also saturates at some point and reaches a magnitude $$B_{sat}$$, right? So the ratio $$B/H = B_{sat}/H$$ should go to zero - since $$B_{sat}$$ is just a constant and $$H$$ goes arbitrarily large - hence $$\color{blue}{\mu_r\rightarrow 0}$$.

So which reasoning is correct? I know this is probably a silly question, but I'm missing something obvious and I'd be thankful if someone could clear this up for me.

EDIT: the experimental setup I have in mind would be a ferromagnetic coil with wire wound around it. By changing the current in the wire we control the external field $$\boldsymbol H$$.

• Also see this question, with no satisfactory answer yet: physics.stackexchange.com/questions/301080/… Mar 25, 2020 at 10:49
• What does it have to do with a bar magnet? What if the situation is just a wire wrapped around a toroidal coil, and we modify $H$ field by increasing current in the wire?
– user89768
Mar 25, 2020 at 11:00
• OK, I see. You are not talking about permanent magnets. Mar 25, 2020 at 12:06

Your first reasoning is correct and the second one is wrong. For high magnetic fields $$\mathbf{H}$$, the magnetization $$\mathbf{M}$$ will saturate but the magnetic induction $$\mathbf{B}$$ does not saturate as it is linearly proportional with the external magnetic field $$\mathbf{H}$$. This is commonly made a mistake when people plot the hysteresis curve (if you google "hysteresis curve magnet" you find multiple figures were the B-field is indicated on the y-axis and saturates for high $$\mathbf{H}$$ values - this is wrong!).
The picture below (taken from Physics behind the magnetic hysteresis loop—a survey of misconceptions in magnetism literature) is correct. The magnetization saturates and the magnetic induction approaches a linear behavior for high $$\mathbf{H}$$.
There are two "H" fields. There is the applied field $$H_a$$ that is formed by external agents, currents and permanent magnets, that create a vacuum $$H_a$$ such that would be the same when you place the magnetizable body that you wish to measure into the applied field. Neither inside nor outside the body is the applied field $$H_a$$ the same in the presence of the magnetizable body as the field $$H$$ except in very special circumstances. One such approximate case is the ferromagnetic toroid where the material properties, surrounding coil and geometry are such that essentially all flux lines stay within the toroid and hence $$H\approx H_a$$. This setup allows the measurement of the functional relationship between $$M$$ and $$H$$ by actually plotting $$M=M(H_a)$$. Another possibility is to shape the specimen as an ellipsoid for which it can be shown then if it is placed in a homogeneous applied field within the ellipsoidal body will also be homogeneous and there will be a tensorial relationship between $$\mathbf{H_a}$$ and $$\mathbf{H}$$, and also $$\mathbf{M}$$. The cause of the change from $$H_a$$ to $$H$$ is the appearance of the so-called de-magnetizing poles that accumulate on the surface of the specimen. These magnetic surface charges are in fact the divergence of the $$M$$ and their induced field is in opposition to $$H_a$$ unless the material is shaped so that hey cannot accumulate, an example for the latter is a toroid.
Regarding your question how could we have $$\mu_r \to 0$$ while calling $$B_{sat}$$ as $$H_a \to \infty$$ assume that $$H \approx H_a$$. Then you do get to a fixed saturated value $$M_{sat}$$ when all domains are parallel with $$H_a$$ and then $$B \approx \mu_0 H_a \approx \mu_0 H$$ and we say the $$B_{sat} = \mu_0 H$$ so that by definition $$\mu_{sat} = \mu_0$$ but as the quote from Brown says in [1] is not a useful concept.