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

  • #1
890
38

Homework Statement



upload_2017-8-22_13-25-2.png

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 ?
 

Answers and Replies

  • #2
Charles Link
Homework Helper
Insights Author
Gold Member
4,597
1,970
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.
 
Last edited:
  • #3
Charles Link
Homework Helper
Insights Author
Gold Member
4,597
1,970
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).
 
Last edited:
  • #4
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?
 
  • Like
Likes Charles Link
  • #5
890
38
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?
 
  • #6
Charles Link
Homework Helper
Insights Author
Gold Member
4,597
1,970
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.
 
  • Like
Likes Pushoam
  • #7
890
38
I got the answer.
Thanks for the guidance.
 
  • Like
Likes Charles Link
  • #8
Charles Link
Homework Helper
Insights Author
Gold Member
4,597
1,970
@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. :)
 
Last edited:
  • #9
890
38
Yes, it didn't cause the confusion. I got what you wanted to say.
Thanks.
 
  • Like
Likes Charles Link

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

Replies
4
Views
669
Replies
5
Views
3K
Replies
8
Views
8K
Replies
28
Views
6K
Replies
5
Views
3K
Replies
3
Views
1K
Replies
0
Views
1K
Top