# Finding the strobilization of the magnetization at sphrerical coordinates.

1. Mar 7, 2009

### vchris5

1. The problem statement, all variables and given/known data

If the magnetization of a sphere is:
$$\vec{M} = \hat{\phi }\frac {r}{R}sin\theta M_{o}$$

How much are the captive streams:
$$J_{b}=\vec{\bigtriangledown }\times \vec{M}$$
$$K_{b}=\vec{M}\times \hat{n}$$

3. The attempt at a solution
I find that the first is:
$$J_{b} = \frac {2M_{o}}{R}(cos\theta \hat{r} - sin\theta \hat{\theta })$$
but i don't know if that is correct and the point that i really want help is how to work the K.

Thank you a lot.

2. Mar 7, 2009

### gabbagabbahey

I've never heard the term "captive streams" before, in English $\vec{J}_b$ and $\vec{K}_b$ are usually called "bound currents"

Your solution for the bound volume current $\vec{J}_b$ is correct ....As for the bound surface current $\vec{K}_b$, what is the outward normal direction $\hat{n}$ for a spherical surface?

3. Mar 8, 2009

### vchris5

Hello gabbagabbahey,
You are right about the term "bound currents".
For the surface current $\vec{K}_b$ , the normal direction $\hat{n}$ is always vertical to the spherical surface.But how are the coordinates of $\hat{n}$ to put them to the type of strobilization..??
Thanks a lot for your interest.

4. Mar 8, 2009

### gabbagabbahey

I'm not sure what "strobilization" means.... are you translating this from another language?

Also, the surface normal is always perpendicular to the surface, not "vertical"....In spherical coordinates, $$\hat{n}=\hat{r}$$....do you see why?

5. Mar 8, 2009

### vchris5

With the word strobilization I mean the $$\vec{M}\times \hat{n}$$.
Sorry but my English are not so good with the physics terms..

Oh yes, i see. So the surface current bound must be:
$$K_{b}=\frac{r}{R}sin\theta M_{o}\hat{\theta }$$ ???

6. Mar 8, 2009

### gabbagabbahey

looks good to me

I guess by "strobilization", you maybe mean "vector cross product"?

7. Mar 8, 2009

### vchris5

gabbagabbahey first of all, thanks a lot.
And the next problem is how is the A(x) with these current bounds??

8. Mar 8, 2009

### gabbagabbahey

vchris, first of all, your welcome!

And next, use the integral equations for the vector potential A:

$$\vec{A}(\vec{r})=\frac{\mu_0}{4\pi} \int_{\mathcal{V}} \frac{\vec{J}_b}{|\vec{r}-\vec{r}'|}dV'+\frac{\mu_0}{4\pi} \oint_{\mathcal{S}} \frac{\vec{K}_b}{|\vec{r}-\vec{r}'|}dA'$$

9. Mar 8, 2009

### vchris5

I know this formula. I will try to find a solution and i will upload it to be able to compare...!!!!
Thanks again!!

Last edited: Mar 8, 2009
10. Mar 9, 2009

### vchris5

I found that the surface current bound gives:
$$\frac{\mu_oMr}3sin\theta \hat{e_\varphi}$$ for r<R
$$\frac{\mu_oMR^3}3\frac{1}{r^2}sin\theta \hat{e_\varphi }$$ for r>R
But i can't find a solution for the other current bound