# $B$ vs $H$, Ampere's Law, $\mu \ne \mu_0$, why is the field weaker outside a high permeability core?

Apologies in advance for the basic question - I haven't taken physics in a while.

In trying to understand an electromagnet from "first principles", I'm left wondering why magnetic field lines should "warp" due to a high permeability material. So for example, if I have an electromagnet with a high permeability core, my understanding is that the magnetic field lines would be stronger through the core and weaker outside the core (as compared to the case without the core). But why would the magnetic field outside the core be any different if I remove the core?

Specifically, Ampere's law as I understand it states (assuming steady state)

$$\nabla \times H = J$$ and $$B = \mu H .$$

Regardless of what $$\mu$$ is anywhere, shouldn't the $$H$$ field be identical because the differential equation fully defines $$H$$? And in that case, why would either $$B$$ or $$H$$ change outside the material (where $$\mu=\mu_0$$) if I remove the core?

• What are you fixing when you insert or remove the core? Are you just considering a long solenoid arrangement where you fill the coil with a high $\mu_r$ material whilst keeping the coil current constant? Oct 13, 2023 at 9:30
• The differential equation does not define $H$, it only defines $\nabla \times H$. Same $\nabla\times H$ is compatible with different $H$'s. Oct 13, 2023 at 10:23

In order to obtain a solution of a differential equation apart from the equation also the boundary conditions are needed. They are modified by the core.

Let's go into the details. First of all, in case of areas with non-homogeneous distribution of magnetic material --- needless to say we assume that it is ferromagnetic ---, it is much more useful to write:

$$\mathbf{H} = \frac{1}{\mu_0} \mathbf{B} - \mathbf{M}$$

where $$\mathbf{H}$$ are the magnetic field, $$\mathbf{B}$$ is the magnetic flux and $$\mathbf{M}$$ the magnetization of the material. We will use another important equation for magnetic phenomena which is

$$\nabla \cdot \mathbf{B} =0$$

since it will provide us with additional information on $$\mathbf{H}$$:

$$\nabla\cdot \mathbf{H} = -\nabla\cdot \mathbf{M}$$

The only location where we expect a change of the magnetization and thereby a non-zero $$\nabla \cdot \mathbf{M}$$ is the border of the core (denoted as $$\partial core$$). It jumps there from 0 to the magnetization value of the core. Of course if there is no core at all, then $$\nabla \cdot \mathbf{M} =0$$ everywhere. In this way one can see that the insertion of the core matters, because if it is there it provides additional boundary conditions:

$$\nabla\cdot \mathbf{H}|_{\partial core} = -\nabla\cdot \mathbf{M}|_{\partial core}$$

If it comes to $$\mathbf{B}$$, one solves first for $$\mathbf{H}$$, then $$\mathbf{B}$$ is determined by

$$\mu_0(\mathbf{H} + \mathbf{M}) = \mathbf{B} \quad\text{in the core}$$

and

$$\mu_0\mathbf{H} = \mathbf{B} \quad\text{outside the core}$$

Yes, $$\mathbf{B}$$ also changes in the outside proximity of the core (inside the core it changes too).

• Ohh, so is it roughly correct to say that everything is always $\mu_0$ but $\mu$ is just an approximation for $1 + M/H$ for dia-/para- magnetic materials (ignoring the vector division hackery)? I didn't realize M != 0 as I thought M was only for "permanent" magnets Oct 13, 2023 at 10:28
• Well first if you have a core in your electromagnet, it has a magnetization. The formula $B = \mu H$ is only valid if $M=cH$ everywhere. But for a ferromagnet this is not valid. Of course I assumed that the core of the electromagnet is ferromagnetic. Otherwise why would one put a non-ferromagnetical material inside ? Oct 13, 2023 at 11:21

The solenoidal (no monopole) law for magnetic field is $$\nabla \cdot {\bf B} = 0$$. This says that lines of magnetic field can neither begin or end since there are no magnetic monopoles.

There is no such law for the H-field and for a linear material where $${\bf B} = \mu {\bf H}$$, the no monopole law tells us that the divergence of the H-field is minus the divergence of the permeability. The permeability has a non-zero divergence at and near the boundary of the core material.

If we consider an electromagnet consisting of a long coil and insert a cylinder of material (whilst keeping the coil current constant), the parallel lines of magnetic field must be continuous across the boundary of the coil material to ensure their divergence is zero. i.e. The B-field is the same immediately either side of the boundary of the material. However, since $$\vec{H} = {\vec B}/\mu$$, the H-field does change, because the ends of the magnetic material act as sources and sinks of H-field.

Note that the fields at the boundary would need to be derived as the vector sum of the fields due to the different parts of the coiL, since the assumption of a uniform, infinitely long solenoid would no longer be met.

Finally, the differential equation $$\nabla \times {\bf H} = {\bf J}$$ does not uniquely define $${\bf H}$$, only its curl. In particular $$\nabla \times ({\bf H} + \nabla \Phi) = \nabla \times {\bf H} = {\bf J}\ ,$$ where $$\Phi$$ is any scalar function.