Magnetic Surface Currents and the Interaction of Magnetic Fields with Matter

[lorenz0]

Homework Statement

A solenoid $L = 10 cm$ long with radius $r << L$ consisting of $N=100$ coils with a current $i$ through it is completely filled with a cylinder of ferromagnetic material whith a curve of first magnetization as shown in the figure. Estimate the value of the magnetization field $M$ assuming that the magnetic working point is at $H = 4 kA / m$.
In this condition, calculate the current $i$ flowing through the solenoid and the total magnetization current present on the lateral surface of the ferromagnetic cylinder.

Relevant Equations

$\vec{H}=\mu_0 \frac{N}{L} i$, $\sigma_{m}=\vec{M}\cdot \hat{n}$

From the graph we see that at $H=4 kA/m,\ B=1.5T$.
We have that $M=\frac{B}{\mu_0}-H=\frac{1.5T}{\mu_0}-4kA/m$ and from Ampere's Law that $i=\frac{HL}{N}=\frac{4kA/m\cdot 0.1 m}{100}$ and the current (density on the surface is) $\sigma_{m}=M$.

Does this make sense? I am having difficulties in understanding the concept of magnetic working point and how $\vec{B},\ \vec{H},\ \vec{M}$ work together when magnetic fields interact with matter so I would be grateful for some feedback.

 

[Steve4Physics]

Hi @lorenz0. Not a speciality of mine but since there are no other replies yet, I’ll give it a go.

First, there are some problems.

There is no graph visible (to me anyway) on the post.

The use of the term ‘magnetic working point’ is ambiguous. The term is often used for a special point on a B-H curve (e.g. as explained here: https://www.hsmagnets.com/blog/magnetic-working-point/). But in this question the term probably just refers to an arbitrary point on a B-H curve – so what I presume you've done -reading-off the value of B from a graph - sounds fine.

In the equation $H = 4 kA / m$ presumably ‘k’ means ‘kilo’ (as opposed to some constant). $H = 4000A / m$ might be clearer. But maybe having the missing graph would clarify this.

$\vec{H}=\mu_0 \frac{N}{L} i$ is incorrect because $\mu_0$ is not required. Also the left side of the equation is a vector but the right side is a scalar. However it looks like you’ve used the correct formula later in your working.

$\sigma_{m}=\vec{M}\cdot \hat{n}$ should be $\vec {\sigma_{m}}=\vec{M}\times \hat{n}$ - we need the cross-product. Also it’s more usual (I think) to use $j_s$ for surface current.density.

For information (just in case you don’t already know), note that the surface current is not an actual physical flow of charge across the surface. It is an ‘equivalent’ current across the surface resulting from the internal magnetic dipoles. In this question, is corresponds to an (equivalent) current flowing around the circumference of the ferromagnetic cylinder. Its units are A/m so it is a sort of linear current density.

As far as I can tell, your working is OK. You have found the surface current (linear) density - do you see how to get the total surface current?

 

[Charles Link]

First note there are two systems of units that are very common: One that uses $\vec{B}=\mu_o \vec{H} +\mu_o \vec{M}$, and the other that uses $\vec{B}=\mu_o \vec{H}+\vec{M}$. You (the OP) are using the first one, which is ok, and will be used in the paragraph that follows.

There are two methods of solving problems with magnetization $M$ and magnetic fields $B$. (and they can be worked with both units above). One is the magnetic pole model, where there is a surface magnetization charge density $\sigma_m=\mu_o \vec{M} \cdot \hat{n}$. The second method is the magnetic surface current model, where magnetic surface current per unit length $\vec{K}_m=\vec{M} \times \hat{n}$. It is this second method that the problem is asking about, and the comparison is made between the current density of the currents in the conductor that are part of the solenoid, which is also the $\vec{H}$ that gets computed, vs. the magnetic surface current density $\vec{K}_m$, which has the same geometry as the solenoid currents.($\vec{K}_m$ is on the outer surface of the iron cylinder).

The formulas using the second set of units are $\sigma_m=\vec{M} \cdot \hat{n}$ and $\vec{K}_m=\vec{M} \times \hat{n}/\mu_o$. Including this here because you are likely to see it written both ways depending on which textbook you might encounter, because both units are in widespread use.

Edit: The problem has you compute the current $i$, but they would do well to have you then compute the current per unit length by computing $H=i N/L$ and compare it to magnetic surface current per unit length $K_m= M$, to compare apples to apples. (It is really a somewhat poor comparison to compare $i$ to $K_m L=ML$, which is what I believe the statement of the problem asks for, but perhaps it is of some interest. On the other hand, they are simply trying to give you some introduction to the material. It should be noted that $\vec{K}_m$ points along the currents, while $\vec{H}$, although computed using the current, points along the z-axis).

The magnetic surface currents enhance the magnetic field from the currents in the coil by a very large factor. You can compute $B$ with and without the iron cylinder, by including or omitting the $M$ in $B=\mu_o H + \mu_o M$.

For a couple of simple inputs, we have $H =4000$ amps/m and $N=100$ with $L=.1$ m. What does that give you for current $i$?
They give you $B =1.5$ T. Using $B=\mu_o H+\mu_o M$, can you assume $M \approx B/\mu_o$ ? (and ignore the $\mu_o H$).