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

## Homework Statement ## 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 ?

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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.

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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).

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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?

• 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?

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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
Thanks for the guidance.

• Homework Helper
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@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. :)

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

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