B-field inside a linear magnetic sphere (in a uniform external B-field)

[Pushoam]

Homework Statement

Homework Equations

The Attempt at a Solution

Because of the external magnetic field $\vec B_0$ , a uniform magnetization will be in the direction of external magnetic field.
Because off this uniform magnetization, there will be a uniform magnetic fied in the direction of magnetization.
So, this implies that $\vec H$ will be uniform and in the direction of the external magnetic field.

Let's take direction of external magnetic field along z - axis.
Take a square loop such that its two sides is along the z - axis and the other two are along the x-axis.
let's take one z-axis side at infinity, then Ampere's law gives $\vec H = 0$, which gives $\vec B = \vec M = 0$.
Is this correct ?

 

[Charles Link]

See also https://www.physicsforums.com/threa...field-of-a-uniformly-polarized-sphere.877891/ There is a "pole method" of solving magnetostatic problems (as opposed to the surface current method) that is directly analogous to the electrostatic problem with $D,E,P$ replaced by $B,H,M$ respectively, and $\epsilon_o$ replaced by $\mu_o$. The one item to check in the units is that $B=\mu_o H+M$ is the equation that is used rather than $B=\mu_o H+\mu_o M$, because the analogous electrostatic equation is $D=\epsilon_o E+P$. Se especially post #9 of the "link" above. Instead of a uniform applied $E$ field, you have a uniformly applied $H_o$ field where $H_o=B_o/\mu_o$. And see also post #7 of the "link" for some of the details of the Legendre solution. $\\$ Another "link" that you also might find useful is https://www.physicsforums.com/threads/electric-field-of-a-charged-dielectric-sphere.890319/ , especially post #2 along with the complete discussion.

 

[Charles Link]

You may also wonder how the "magnetic pole" method could possibly work. That is actually what they taught us back in 1975-1980 as undergraduates as well as in graduate school, and they really weren't even teaching the surface current method at that time. Anyway, in about 2009-2010, I finally tied it all together. If you are interested in the details, here is a paper that I wrote up: It was my first attempt at using Latex, so the Latex isn't nearly as polished as it could be. https://www.overleaf.com/read/kdhnbkpypxfk You can click on the green arrow at the upper middle left to get the full page on the right. $\\$ Editing: One objection I had to the magnetic potential method as presented in the textbooks, as well as below by @NFuller , is that it appeared to lack complete mathematical rigor, in that although the potential satisfies $\nabla^2 \phi=0$ in the region where there aren't any poles, does the solution that is obtained with the boundary conditions with some distribution of poles, (outside of the $\nabla^2 \phi=0$ region), give the same result for the $B$ field as that computed from magnetic surface currents from the magnetization? Perhaps it necessarily works, but I wasn't completely convinced until working the calculations that I did in this paper. Basically, I questioned the equation $B=\mu_o H+M$, and it needed to be derived, rather than assuming that it works, just because $D=\epsilon_o E+P$. In addition the question is, is it correct to write $H_m(x)=\int \frac{\rho_m(x') (x-x') }{4 \pi \mu_o |x-x'|^3} d^3 x'$ where $\rho_m=-\nabla \cdot M$, analogous to the electrostatic $E$ field equation? $\\$ [Assuming $B=\mu_oH+M$, we have $\nabla \cdot B=0$, so that $\nabla \cdot H=-\nabla \cdot M/(\mu_o)$. This allows us to write the integral expression for $H_m$ assuming the equation $B=\mu_o H+M$ is correct. Note: With this expression, we also need to add solutions of the homogeneous equation $\nabla \cdot H=0$, resulting from any currents in conductors, besides the magnetic pole contribution to $H$ which is $H_m$]. $\\$ The answer is yes, but for me it wasn't completely obvious. ($x$ and $x'$ and $H_m$, $B$, and $M$ are vectors. The notation is shortened for brevity).

 

[NFuller]

Let's take direction of external magnetic field along z - axis.
Take a square loop such that its two sides is along the z - axis and the other two are along the x-axis.
let's take one z-axis side at infinity, then Ampere's law gives ⃗H=0H→=0 \vec H = 0 , which gives ⃗B=⃗M=0B→=M→=0 \vec B = \vec M = 0.
Is this correct ?

The general method for solving this problem is the use of a magnetic scalar potential. Have you studied this yet?
Since there are no currents or time-varying fields, the magnetic field can be written as
$$\mathbf{H}=-\nabla\phi$$
$$\nabla^{2}\phi=0$$
So the magnetic potential is a solution of the Laplace equation. Do you know the general solution of this equation in spherical coordinates?

 

[Pushoam]

Have you studied this yet?

Sorry, I haven't studied this yet.
But, I think I understood your method.
What I understood is:
Since the sphere is linearly magnetic, when the sphere is put in the presence of external uniform magnetic field, it gets uniform magnetization $\vec M$. Both $\vec M$ and $\vec B_0$ are in the same direction. Now , magnetic field due to this magnetization inside the sphere is also uniform and in the direction of magnetization. So,finally the net magnetic field and the magnetization and therefore, the $\vec H$ inside the sphere are uniform and in the same direction.
This allows me to take,
$\nabla \times \vec H = 0$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)
$\vec H = - \nabla \phi$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1.1)
$\nabla . \vec H = 0$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)
$\nabla^2 \phi = 0$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2.1)

Since the question asks to calculate $\vec B$ inside the sphere, I can do the same thing with $\vec B$ and get the corresponding eqns.
But, according to uniqueness theorem , I need to know the normal component ( to the spherical surface ) of $\vec B$ or $\vec H$ to specify it uniquely. How to get it?

Can't it be solved in the following way?
$\vec M = C \vec B \text{ magnetization due to magnetic field inside the sphere} \\ \vec B = \vec B_0 + D \vec M \text{ magnetic field as superposition of magnetic field due to magnetization and the applied magnetic field } \\ = \vec B_0 / (1-CD)\text { where C and D are appropriate constants}$

Since, there exists no magnetic pole, I want to know whether it is useful to learn magnetic pole method.

Let's take direction of external magnetic field along z - axis.
Take a square loop such that its two sides is along the z - axis and the other two are along the x-axis.
let's take one z-axis side at infinity, then Ampere's law gives ⃗H=0H→=0 \vec H = 0 , which gives ⃗B=⃗M=0B→=M→=0 \vec B = \vec M = 0.
Is this correct ?

Isn't the above argument correct?

 

[Charles Link]

Sorry, I haven't studied this yet.
But, I think I understood your method.
What I understood is:
Since the sphere is linearly magnetic, when the sphere is put in the presence of external uniform magnetic field, it gets uniform magnetization $\vec M$. Both $\vec M$ and $\vec B_0$ are in the same direction. Now , magnetic field due to this magnetization inside the sphere is also uniform and in the direction of magnetization. So,finally the net magnetic field and the magnetization and therefore, the $\vec H$ inside the sphere are uniform and in the same direction.
This allows me to take,
$\nabla \times \vec H = 0$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)
$\vec H = - \nabla \phi$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1.1)
$\nabla . \vec H = 0$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)
$\nabla^2 \phi = 0$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2.1)

Since the question asks to calculate $\vec B$ inside the sphere, I can do the same thing with $\vec B$ and get the corresponding eqns.
But, according to uniqueness theorem , I need to know the normal component ( to the spherical surface ) of $\vec B$ or $\vec H$ to specify it uniquely. How to get it?

Can't it be solved in the following way?
$\vec M = C \vec B \text{ magnetization due to magnetic field inside the sphere} \\ \vec B = \vec B_0 + D \vec M \text{ magnetic field as superposition of magnetic field due to magnetization and the applied magnetic field } \\ = \vec B_0 / (1-CD)\text { where C and D are appropriate constants}$

Since, there exists no magnetic pole, I want to know whether it is useful to learn magnetic pole method.

Your solution is close to being right.
In general, $B=\mu_o H+M$ , so you start with $B_i=\mu_o H_i+M_i$ and $B_o=\mu_o H_o$.
Inside the sphere $M_i=\chi_m H_i$
In addition by superposition, $H_i=H_o+H_m$ where $H_o$ is the applied field, and $H_m=-\frac{1}{3} M_i$ for the spherical geometry just like the corresponding electrostatic problem. (The demagnetizing factor $D=\frac{1}{3}$ for a sphere).
Solve for $H_i$ and $M_i$, and then $B_i=\mu_o H_i+M_i$.
$\\$ For a simpler problem connecting the surface current method to the pole method see https://www.physicsforums.com/threads/magnetic-field-of-a-ferromagnetic-cylinder.863066/ The magnetic pole method is a mathematical shortcut that can be very useful. This "link" is the case of a simple permanent magnet where the resulting magnetic field is shown to be computed both by pole method and surface current method in the posting. $\\$ A physics professor at the University of Illinois-Urbana who has taught E&M there for quite a number of years tells me that they currently don't present the magnetic pole method any more until graduate school. When I was a student there, they taught it to us in the advanced undergraduate E&M class.

 

[Pushoam]

I got the answer.
Thanks for the guidance.

 

[Charles Link]

@Pushoam A couple of minor corrections on post #6 (It gets a little tricky because there are a couple systems of units that get used=c.g.s. uses $B=H+4 \pi M$, and there are two types of MKS units: $B+\mu_o H+M$, and $B=\mu_o H+\mu_o M$). $\\$ For the $B=\mu_o H+M$ that I used, I should have written $M_i=\mu_o \chi_m H_i$ and also $H_m=-(1/3)(M_i/\mu_o)$. $\\$ What I wrote in post #6 works if you use $B=\mu_o H+\mu_o M$. Hopefully it didn't cause too much confusion. :)

 

[Pushoam]

Yes, it didn't cause the confusion. I got what you wanted to say.
Thanks.