Why are the effects of a magnetized material neglected in this case?

[vcsharp2003]

Homework Statement

I came across the following text from my textbook as shown in screenshot, but it doesn't make sense to me.

A material inside a solenoid will affect the magnetic field inside a current carrying solenoid since the material will enhance the magnetic field due to the current in solenoid if the material is paramagnetic or ferromagnetic, and it will diminish the magnetic field due to the current in solenoid if the material is diamagnetic.

Also, I do not get what is meant by end effects of a magnetized material.

Relevant Equations

$\vec B = \mu_0 \vec H + \mu_0 \vec I$

According to what I know, the net magnetic field $\vec B$ inside a solenoid is given by $$\vec B = \mu_0 \vec H + \mu_0 \vec I$$ where $\vec B$ is the net magnetic field inside a current carrying solenoid, $\vec H$ is the magnetic intensity ( aka magnetic intensity or magnetizing field intensity), $\vec I$ is the intensity of magnetization (aka magnetization). The first term on right hand side of above equation is the magnetic field due to the current in the hollow solenoid, while the second term is the magnetic field due to magnetization of some material inside the solenoid So, if there is some material inside the solenoid i.e. the solenoid is not hollow, then the contribution of the material to $\vec B$ will be $\mu_0 \vec I$, and since $\vec I \neq 0$ therefore $\mu_0 \vec I \neq 0$.

According to the equation below, we cannot ignore $\mu_0 \vec I$ when determining $\vec H$.

$$\therefore \frac {\vec B - \mu_0 \vec I} { \mu_0 }= \vec H $$

 

[kuruman]

Maybe I misunderstood what you are asking, but Maxwell's equation in the static case $\mathbf{\nabla}\times\mathbf{H}=\mathbf{J}$ is pretty clear that the source of circulating magnetic intensity is the transport current $\mathbf{J}$. Hence the textbook's statement "The magnetic intensity in a material is then determined by the external sources only, even if the material is magnetized."

According to the equation below, we cannot ignore $\mu_0 \vec I$ when determining $\vec H$.

$$\therefore \frac {\vec B - \mu_0 \vec I} { \mu_0 }= \vec H $$

I think you got it backwards. One uses Maxwell's equation and $\mathbf{J}$ (if any) to determine $\mathbf{H}$ and then the known magnetization (if any) to determine $\mathbf{B}.$

 

[Charles Link]

The $\vec{H}$ can arises from two types of sources= the first is currents in conductors, and Biot-Savart or ampere's law are used to compute the $H$ from the free currents. The second source for $H$ is magnetic poles, where magnetic pole density $\rho_m=-\nabla \cdot M$. (I like to use the standard convention of $M$ rather than $I$ for magnetization.) For a uniform $M$, this gives by Gauss' law a magnetic surface charge density $\sigma_m=M \cdot \hat{n}$, e.g. You get magnetic poles on the end faces of a uniformly magnetized cylinder. The inverse square law is used to compute the $H$ from magnetic poles.

To respond to @kuruman above, the differential equation$\nabla \times H=J_{conductors}$ has a solution to $H$ of the Biot-Savart form, but this misses the homogeneous solution (where $\nabla \times H=0$)from the magnetic poles that must be included. See also https://www.physicsforums.com/threa...cs-and-solving-with-the-curl-operator.975978/

 

[Charles Link]

and a follow-on to post 3=notice for a long rod that is uniformly magnetized, the $H$ from the poles on the end faces will be small and perhaps negligible near the center of the rod. Meanwhile a magnetized torus has no end faces, and thereby no magnetic poles.( The surface magnetization charge density $\sigma_m= M \cdot \hat{n}=0$ for the toroid surfaces. There are no end faces to get a non-zero $M \cdot \hat{n}$).

 

[Charles Link]

and a second follow-on: There are two methods for solving these problems with magnetic materials, and they both get the exact same answer for the magnetic field $B$. One is the pole method, that is being worked above. It should be noted that $B$ is the real and only magnetic field, and while $H$ is given a name "magnetic field intensity", it ($H$) is a mathematical construction that sometimes turns out to be proportional to the magnetic field $B$, while other times it turns out to be just the pole term, which in a permanent magnet is a (negative) correction term for the $B=\mu_o M$ for geometries other than a torus or very long cylinder. In the pole method, once $H$ is computed, the magnetic field $B$ is then found using $B=\mu_o H +\mu_o M$. The magnetic field $B$ is the only real magnetic field, where $H$ is really just a necessary thing that needs to be computed to then calculate the magnetic field $B$. The $H$ in general is not a magnetic field.

The second method of solving these problems uses magnetic surface currents, and this method does not introduce an $H$, and it doesn't recognize any magnetic poles. Instead it uses magnetic current density $J_m=\nabla \times M$, (resulting in magnetic surface current per unit length at boundaries of $K_m=M \times \hat{n}$), and the magnetic field $B$ is computed by Biot-Savart or ampere's law from magnetic currents as well as currents in conductors. (Note also in the pole method above, magnetic surface currents are not recognized as sources of the magnetic field).

See also https://www.physicsforums.com/threads/a-magnetostatics-problem-of-interest-2.971045/

 

[Charles Link]

One additional comment is, upon reading the title of this post, the effects of the magnetic material really are not neglected. When the magnetic field $B$ is computed from the pole method, the calculation $B=\mu_o H +\mu_o M$ is performed. Without any magnetic poles for the two cases mentioned, the $H$ is very simple, but the material is in fact accounted for completely with the $\mu_o M$ term. There will be a $\mu_o H$ correction term for geometries other than the two that were mentioned. In addition, outside the material the magnetic field $B$ is precisely $B=\mu_o H$.

It is also of interest that the $\mu_o M$ term is what gets computed in the surface current method for $B$ for the long (uniformly magnetized) cylinder, and the $\mu_o H$ is precisely the (subtractive=e.g it often points opposite the $M$) correction term to $\mu_o M$ that gets employed for any other geometries.

Finally, if there are any currents in conductors, for both the pole method and the surface current method, the currents in conductors get treated the same way, using Biot-Savart in the computation of $B=B_{Biot-Savart \, conductors}$. In the pole method, the currents in conductors get a Biot-Savart type contribution to $H$, (call it $H_{conductors}$), so that $B_{Biot-Savart \, conductors}=\mu_o H_{conductors}$.

One more additional comment is that it is actually surprising that the magnetic pole method gives the exact same result for the magnetic field $B$ as the surface current calculation. It is somewhat of a mathematical coincidence that this is the case, and it can be very useful to take advantage of this in doing calculations with the magnetic pole method. e.g. with a torus with uniform magnetization, (in cylindrical coordinates in the $\phi$ direction), we can say immediately

[Edit: We have $\sigma_m=M \cdot \hat{n}=0$, with $M$ always parallel to the surface. Note we also need to check that $\rho_m=-\nabla \cdot M=0$. See below. Thereby the are no sources for $H$.]

that $H=0$ if there are no currents in conductors, and that $B=\mu_o M$, while a calculation using magnetic surface currents with $K_m=M \times \hat{n}$ and Biot-Savart would take considerable more effort, (including a $J_m=\nabla \times M$ term to get things exact).

Meanwhile with a google of $\nabla \cdot$ in cylindrical coordinates, the calculation gives for the pole method that $\rho_m=- \nabla \cdot M=0$, (also a necessary condition for $H$ to be precisely zero), in cylindrical coordinates with $M=M_{\phi} \hat{\phi}$.