# What is the $\mathbf{H}$-field inside and outside the magnet?

A permanent cylindrical magnet with diameter $$D$$ and length $$L$$ possesses an uniform magnetization $$\mathbf{M}$$ going in the $$z$$-direction in a cylindrical coordinate system. The magnet is placed with its centre in the origin, with its length going along the $$z$$-axis.

Problem

Given the expression for the magnetic field along the $$z$$-axis $$\mathbf{B}(z)= \mathbf{a_z} \cdot \frac{\mu_0M}{2} \Bigg[ \frac{z+ \frac{L}{2}}{\sqrt{\big(z+\frac{L}{2} \big)^2+ \big(\frac{D}{2} \big)^2}}- \frac{z- \frac{L}{2}}{\sqrt{\big(z-\frac{L}{2} \big)^2+ \big(\frac{D}{2} \big)^2}} \Bigg]$$ arrive at an expression for the $$\mathbf{H}$$-field along the $$z$$-axis. What is the direction of $$\mathbf{H}$$ inside and outside the magnet?

My attempt

Given the simple relation $$\mathbf{H}=\frac{\mathbf{B}}{\mu_0}-\mathbf{M}$$ we get

$$\mathbf{H}(z)=\mathbf{a_z} \cdot \frac{M}{2} \Bigg[ \frac{z+ \frac{L}{2}}{\sqrt{\big(z+\frac{L}{2} \big)^2+ \big(\frac{D}{2} \big)^2}}- \frac{z- \frac{L}{2}}{\sqrt{\big(z-\frac{L}{2} \big)^2+ \big(\frac{D}{2} \big)^2}} \Bigg]-M \cdot \mathbf{a_z}$$

But I'm not sure how to answer what direction $$\mathbf{H}$$ has inside and outside the magnet.

Edit

I found this example in my book, that revolves around the exact same scenario, a cylindrical magnet with uniform magnetization $$\mathbf{a_z}M_0$$. In the example, the author writes about the relationship about $$\mathbf{H}$$ and $$\mathbf{B}$$. He also talks about their direction, which I have marked with yellow. So if the scenario is the same, doesn't the example answer also answer my problem completely?

• Oct 27, 2020 at 19:50
• Well, judgning by the answers in those two links it seems the example included in my edited question is exactly what I need to fully solve my problem. @RobJeffries
– Carl
Oct 27, 2020 at 20:10
• What book are you using? Oct 30, 2021 at 9:01
• @user1700890 Field and wave electromagnetics, by Cheng I think.
– Carl
Oct 30, 2021 at 13:53

I will give a more qualitative answer than just formulas. Considering the given proposal one has to be very careful since $$\mathbf{H}$$ not only has to fulfill Maxwell-equations and material equations, but also the boundary conditions which are not the same for $$\mathbf{B}$$ and $$\mathbf{H}$$.

The Maxwell equations in case of a permanent magnet without any currents are those of magnetostatics:

$$\nabla \times \mathbf{H}= 0 \quad\quad \text{and} \quad\quad \nabla\cdot \mathbf{B}=0$$

and there is the relevant material equation (assuming cgi-units): $$\mathbf{B} = \mathbf{H} + 4\pi \mathbf{M}$$.

As we want to know $$\mathbf{H}$$ we eliminate $$\mathbf{B}$$ and get:

$$\nabla \cdot \mathbf{H} = -4\pi \nabla \cdot\mathbf{M} \quad {and} \quad \nabla \times \mathbf{H}= 0$$

This system of equations resembles to the equations of electrostatics:

$$\nabla \cdot \mathbf{E} = 4\pi \rho \quad {and} \quad \nabla \times \mathbf{E}= 0$$

So the solution will resemble an already well-known case of electrostatics.

In the following we also consider that the permanent magnet is directed along the z-axis, the only non-zero component of the magnetisation is $$M_z$$.

The only locations where the divergence of the magnetisation is non-zero are the ends of the permanent magnet. At the borders along the magnet, the z-component of the magnetisation $$M_z$$ also suffers an abrupt change, but in radial direction, so it does not matter since changes of $$M_z$$ in radial direction do not contribute to the divergence (only $$\frac{\partial M_z}{\partial z}$$ contributes).

So the permanent magnet resembles to two plates of opposite electrical charges at each end of the magnet, i.e. the $$\mathbf{H}$$-field looks like that of a cylindrical electrical condensator.

The equivalent charge of such a system is: $$\rho = -\frac{\partial M_z}{\partial z}$$ and assuming that the abrupt change at each end of the magnet is the same (respectively symmetrical) both charge densities $$\rho_1 = - \rho_2$$ as it is for a condensator. If the diameter $$D \rightarrow 0$$, the field line picture would resemble to that of two opposed point charges $$+$$ and $$-$$. So in order to know the $$\mathbf{H}$$ field in space as a function of coordinates just refer to the $$\mathbf{E}$$ field in space for a condensator.

• Hi Frederic, thank you for your answer but I am not really sure how to follow it from there. However, I made an edit to the question, would you mind looking at it?
– Carl
Oct 27, 2020 at 16:16
• I accepted your answer, in hopes of getting an additional response. @FredericThomas
– Carl
Oct 28, 2020 at 9:52
• @Carl: Thank you for the green check mark. In the comments there are references to posts which explain even better what I said. In particular you find pics for the directions of the field lines of B and H, so I wonder what you still need more. A formula ? Anyway if I gave a formula it would be one based on the electric condensator picture. Oct 28, 2020 at 9:57
• I saw the other posts that were linked and they also helped me, but I just want to be sure that I understand it correctly. Can you confirm that, in my case that: 1. Outside the magnet the H and B-field have the same direction and are proportional. 2. Inside the magnet H and B point in opposite direction and are related by the equation $\mathbf{H}=\mathbf{B}/\mu_0 -\mathbf{M}$?
– Carl
Oct 28, 2020 at 10:10
• @Carl: Locally this relation is correct. Inside the magnet it is like that $\mathbf{H}=\frac{\mathbf{B}}{\mu_0} -\mathbf{M}$. But outside of the magnet $\mathbf{M}=0$. So the formula given in your post only "holds" (I did not check the part concerning the $\mathbf{B}$-field though) inside the magnet (well outside one can set M=0 of course). Oct 28, 2020 at 11:46