A diamagnetic sphere in a uniform magnetic field

In summary, the solution part a) is to calculate the external magnetic field, which is done by using the Legendre methods. One additional input is that you really need to compute the resulting magnetization before you can do part "a".
  • #1
milkism
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Homework Statement
a) Find the induced surface current density
b) Find the M, B and H-fields
Relevant Equations
See solution.
Problem:
c34309ccc41761c2e590be707866b514.png

Solution part a)
6d28d5d25bc62c3f53b3de748320335e.png

where formula 6.14 is just M x n.

We need to do part b without seperation of variables, I'm quite stuck. Will B just be the magnetic field inside a solenoid? How can I find the other fields.
 
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  • #2
See https://www.physicsforums.com/threa...f-homogeneous-dielectric.940015/#post-5943369

The solution is similar for the analogous magnetic problem. There are Legendre methods for solving this, but it can help to start with a simpler approach.

There is ## D=\epsilon _o E+P ## and ## B=\mu_o H+M ##. You need to watch the units on ## M ## though when transposing the two, because some books write it as ## B=\mu_o H +\mu_o M ##.
 
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  • #3
One additional input: You really need to compute the resulting magnetization ## M ## before you can do part "a" which is to compute the magnetic surface current density.

additional item: to compute the magnetic field ## B ## outside the sphere, the most practical way seems to be to use Legendre methods, although I think Griffith's in his E&M book presents an alternative way that also works. See https://www.physicsforums.com/threa...ormly-polarized-cylinder.941830/#post-5956930
The above is for a cylinder, but the method would also work for a sphere.
 
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  • #4
Charles Link said:
One additional input: You really need to compute the resulting magnetization ## M ## before you can do part "a" which is to compute the magnetic surface current density.

additional item: to compute the magnetic field ## B ## outside the sphere, the most practical way seems to be to use Legendre methods, although I think Griffith's in his E&M book presents an alternative way that also works. See https://www.physicsforums.com/threa...ormly-polarized-cylinder.941830/#post-5956930
The above is for a cylinder, but the method would also work for a sphere.
We need the external magnetic field to calculate the external H-field? The M-field outside is zero.
 
  • #5
The ## E ## and ## H ## are corresponding fields when transposing the two. If you can follow how the ## E ## is calculated, you then convert it to ## H ## with ## \mu_o ## in place of ## \epsilon_o ##, and ## M ## in place of ## P ##.

and yes, for ## B_{outside} ##, the ## M_{outside}=0 ## so that ## B_{outside}=\mu_o H_{outside} ##.

One additional note: You might find this thread on magnetostatics to be good reading: https://www.physicsforums.com/threads/a-magnetostatics-problem-of-interest-2.971045/

and it pays to use units on ## M ## where ## B=\mu_o H+M ## in doing these problems with comparisons to the electric field, and convert afterwards, if your book uses an ## M' ## where ## B=\mu_o H+\mu_o M' ##.

See https://www.physicsforums.com/threa...iformly-polarized-sphere.877891/#post-5513730 post 7, to get ## E_{outside} ##. To this you need to add the applied field, but that part is simple. Once you have ## E_{outside} ## you can transpose to ## H_{outside} ##. (The uniform ## P ## (or ## M ##) that you get with an applied field should be a simple matter to compute, using the link of post 2 above with the demagnetizing factor ## D=1/3 ## for a sphere).
 
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  • #7
Charles Link said:
@milkism Did you get an answer for the ## M ## ?
Luckily we were allowed to use already calculated expressions for the internal- and external magnetic fields. This is what I got:
0d6a29e67c4f87316759acefefe40582.png

(The angular velocity is supposed to be bold, but I have dificulty bolding greek symbols. And "with our own given magnetic susceptibility" instead of out.).
 
  • #8
I don't agree completely with your answer. Since this is really not an Introductory Physics problem, let me show you what I got for ## B_{inside} ## and ## H_{inside} ##, etc. The first part I don't agree with that ## B_{inside}=(2/3) M ## using units where ## B=\mu_o H+M ## and ## M=\mu_o \chi H=-\mu_o |\chi| H ##, because ## H_i=H_a-M/(3 \mu_o) ##, so the ## H_a ## complicates matters somewhat.

Let me show you how the ## H_i ## is obtained from ## H_a ##, where you really need a ## B_a ## with a subscript for the applied ## B ##, where ## B_a=\mu_o H_a ##.
## E_i=E_a-P/(3 \epsilon_o) ## using the demagnetizing factor ## D=1/3 ## for a sphere.
Now ##P= P_i=\epsilon_o \chi E_i ## so ## E_i=E_a/(1+\frac{\chi}{3}) ##, so that ## H_i=H_a/(1-\frac{|\chi|}{3}) ##, and ## M=-\mu_o |\chi| H_i=-|\chi|B_a/(1-\frac{|\chi|}{3}) ##, and ## B_i=\mu_o(1+\chi )H_i=B_a(1-|\chi|)/(1-\frac{|\chi|}{3}) ##.

Note: ## M=\mu_o \sigma R \omega ##, but not every formula works for that solution, because of the ## B_a ##.

Once we get agreement with the inside, then we'll take a closer look at the outside. Please let me know if you were able to follow the calculations.
 
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  • #9
Charles Link said:
I don't agree completely with your answer. Since this is really not an Introductory Physics problem, let me show you what I got for ## B_{inside} ## and ## H_{inside} ##, etc. The first part I don't agree with that ## B_{inside}=(2/3) M ## using units where ## B=\mu_o H+M ## and ## M=\mu_o \chi H=-\mu_o |\chi| H ##, because ## H_i=H_a-M/(3 \mu_o) ##, so the ## H_a ## complicates matters somewhat.

Let me show you how the ## H_i ## is obtained from ## H_a ##, where you really need a ## B_a ## with a subscript for the applied ## B ##, where ## B_a=\mu_o H_a ##.
## E_i=E_a-P/(3 \epsilon_o) ## using the demagnetizing factor ## D=1/3 ## for a sphere.
Now ##P= P_i=\epsilon_o \chi E_i ## so ## E_i=E_a/(1+\frac{\chi}{3}) ##, so that ## H_i=H_a/(1-\frac{|\chi|}{3}) ##, and ## M=-\mu_o |\chi| H_i=-|\chi|B_a/(1-\frac{|\chi|}{3}) ##, and ## B_i=\mu_o(1+\chi )H_i=B_a(1-|\chi|)/(1-\frac{|\chi|}{3}) ##.

Note: ## M=\mu_o \sigma R \omega ##, but not every formula works for that solution, because of the ## B_a ##.

Once we get agreement with the inside, then we'll take a closer look at the outside. Please let me know if you were able to follow the calculations.
Yes I was able to follow the calculations.
Would B_outside be $$-\frac{M_0 R^3}{3 \mu_0 r^2} \left( \cos(\theta) \mathbf{\hat{r}} + \frac{\sin(\theta)}{r}\mathbf{\hat{\theta}} \right)$$?
 
  • #10
milkism said:
Would B_outside be
See the last link of post 5. It's post 7 of the "link". I would need to study this in detail=I don't have extra time at the moment, but don't forget to include ## B_a ## besides the contribution to ## B ## from the magnetization. For the magnetization, you of course need to use what I just computed in post 8.
 
  • #11
and a follow-on: For the ## \hat{a}_r ## component, when you take the derivative of ## 1/r^2 ##, you get ##-2/r^3 ##, (using the gradient per the link I mentioned above). I'm not going to check all of the signs, but what you have looks almost correct, but both terms are with a ## 1/r^3 ##. Don't forget to include the original ## B_a ## though.
 
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  • #12
Charles Link said:
and a follow-on: For the ## \hat{a}_r ## component, when you take the derivative of ## 1/r^2 ##, you get ##-2/r^3 ##, (using the gradient per the link I mentioned above). I'm not going to check all of the signs, but what you have looks almost correct, but both terms are with a ## 1/r^3 ##. Don't forget to include the original ## B_a ## though.
True, also forgot to take the negative when doing the gradient it should be this ##\frac{M_0 R^3}{3 \mu_0 r^3} \left( 2\cos(\theta) \mathbf{\hat{r} + \sin(\theta)}\mathbf{\hat{\theta}} \right)##. So we have ##B_i = B_a - \frac{M}{3 \mu_0} ##
 
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  • #13
Looks good for ## B_{outside} ##, but add ## B_a=B_o \hat{z} ##.

For ## B_i ##, what I have in post 8 is complete. (You don't need to add any ## B_a ## to that). The expression ## B_i=B_a-M/(3 \mu_o) ## is not correct though=we have ## H_i =H_a-M/(3 \mu_o) ##, but ## B_i=\mu_o H+M=\mu_o H_a-M/3+M=B_a+(2/3)M ##.
 
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  • #14
One additional item=I put the "link" in post 5 above, but I think this one is worth repeating=this "link" ties together the magnetic surface current method and the magnetic pole method with a simple example. You should find it good reading. See https://www.physicsforums.com/threads/a-magnetostatics-problem-of-interest-2.971045/
It should be noted that currents in conductors also contribute to ## H ## in the pole method, with a calculation using Biot-Savart divided by ## \mu_o ##. I may have mentioned this later in the thread, but I don't see it in the first couple of posts.
 
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