# Find the magnetic field of a uniformly magnetized sphere.

1. Dec 7, 2013

### bfusco

1. The problem statement, all variables and given/known data
Find the magnetic field of a uniformly magnetized sphere. (this is all that was given in the problem)

3. The attempt at a solution
I chose the z axis in the same direction as M.
$$J_b=\nabla \times M=0$$
and
$$K_b=M \times \hat{n}=Msin\theta \hat{\phi}$$

Apparently, I can treat this problem as a sphere rotating spherical shell, is that because the surface current $K_b$ is in the $\hat{\phi}$ direction?

So, if i can treat it as a rotating sphere, $K=\sigma v$, where $v=R\omega$

And using the Biot Savart law, I get to the point
$$B=\frac{\mu_0}{4\pi}\int (\sigma R \omega)(\hat{\phi} \times \hat{r})/r^2 dArea$$

Where r is the vector pointing from the source of the field to the point in question. for spherical coordinates I use (s,$\theta$,$\phi$)

Im not entirely sure how to determine the direction of the cross product, because idk the direction of r, although i want to guess it is in the $\hat{\theta}$ direction because that is the only direction left for a sphere. I also am not entirely sure what $r^2$ is equal to. is it $r^2=R^2+s^2-2Rcos\theta$?

2. Dec 7, 2013

### WannabeNewton

Have you worked with the vector potential $\vec{A} = \frac{\mu_0}{4\pi}\int \frac{\vec{K}(\vec{r}')}{|\vec{r}-\vec{r}'|}dA'$ before? Here $\vec{r}'$ points from the origin of the chosen coordinate system to an arbitrary source point (i.e. an arbitrary point on the sphere) and $\vec{r}$ points from said origin to an arbitrary field point. It's much easier to calculate $\vec{A}$ for a uniformly rotating charged sphere (which is what the uniformly magnetized sphere is equivalent to as far as the magnetic field goes for exactly the reason you stated in our post: we have a bound surface current flowing in the azimuthal direction) and then get $\vec{B}$ from $\vec{B} = \vec{\nabla}\times \vec{A}$.

By the way, $\vec{K} = \sigma \vec{v}$ is fine but the relationship between $\vec{v}$ and $\vec{\omega}$ isn't as sample as you have written. The general relationship is $\vec{v} = \vec{\omega}\times \vec{r}'$. As you can probably tell from visualizing the system $\vec{r}'$ and $\vec{\omega}$ will not be perpendicular for all source points on the sphere; you have to actually carry out the cross product.

Also, just as a hint to get you started, orient your coordinate system so that $\vec{r}$ is along the $z$-axis and $\vec{\omega}$ lies at some angle in the $xz$-plane.

3. Dec 7, 2013

### bfusco

I have and that does seem easier, however I still come across an issue of determining the directions of all the given parameters. For example in $v=\omega \times r'$ how do i determine the direction of r'? I would image for the sake of simplicity i can choose it perpendicular to $\omega$, but im not sure if that is in the direction of $s,\theta$ or $\phi$, or if it has components

4. Dec 7, 2013

### WannabeNewton

First let's orient our coordinate system centered on the sphere so that $\vec{r}$ is along the $z$-axis and $\vec{\omega}$ lies in the $xz$-plane at some angle $\varphi$. If you're wondering why we can do this, assume that we initially had our coordinates oriented so that $\vec{\omega}$ pointing along the $z$-axis and $\vec{r}$ pointing in some arbitrary direction; now rotate the coordinate system about the $z$-axis until $\vec{r}$ is on the $xz$-plane and rotate within this plane until $\vec{r}$ is along the $z$-axis and $\vec{\omega}$ is at some angle $\varphi$ within this plane.

Back to your question, $\vec{r}'$ points from the origin to an arbitrary point on the sphere hence you can't choose it to be perpendicular to $\vec{\omega}$. However you can definitely write down a formula for it. $\vec{r}'$ points from the origin to some point on a sphere of some radius $R$ right? So how would you express $\vec{r}'$ in terms of cartesian unit vectors $\hat{x},\hat{y},\hat{z}$ and the $\theta,\phi$ angles in spherical coordinates?

5. Dec 7, 2013

### bfusco

so I would do the typical conversion, however to be sure I understand it and i am not just regurgitating it, I will describe the conversion.

Noting that $\phi$ is the angle from the projection of r into the x-y plane from the x axis, and $\theta$ is the angle from r to the z axis.

$r'_x=$Proj_r(to x-y plane)$cos\phi$ which is $Rsin\theta cos\phi$
$r'_y=$Proj_r(to x-y plane)$sin\phi$ which is $Rsin\theta sin\phi$
$r'_z=Rcos\theta$

Then i would have to figure out the components for $\omega$, which isn't difficult to express in terms of that angle $\rho$ you stated, but that $\rho$ isn't one of the spherical coordinate angles.

6. Dec 7, 2013

### WannabeNewton

That's perfect. You may want to put primes on the spherical coordinate angles when you perform the integral for $\vec{A}$ so that you don't confuse source points with field points (you don't have to if you don't want to it isn't really a big deal but sometimes it helps with book keeping).

Indeed the angle $\varphi$ that $\vec{\omega}$ makes within the $xz$-plane isn't a spherical coordinate angle but that's ok just work with it for now and you'll see later what to do with it. Go ahead and compute all the quantities you need to find $\vec{A}$ in this coordinate system.

7. Dec 7, 2013

### bfusco

Cool. Thank you, but before I finish the problem, there is something I never understood and would really appreciate if you could explain it to me (assuming you understand it). In my first post when I stated $r^2=R^2+r^2 -2Rrcos\theta$, why is r^2 equal to that?

8. Dec 7, 2013

### WannabeNewton

You have to be careful with that formula. If we have two vectors $\vec{r}$ and $\vec{r}'$ then $|\vec{r} - \vec{r}'|^2 = (\vec{r} - \vec{r}')\cdot(\vec{r} - \vec{r}') = r^2 + r'^2 - 2\vec{r}\cdot \vec{r}' = r^2 + r'^2 - 2rr'\cos\alpha_{rr'}$ where $\alpha_{rr'}$ is the angle between $\vec{r}$ and $\vec{r}'$. The angle between them will in general not be the spherical coordinate angle $\theta$, which only measures the angle vectors make with the $z$-axis.

However for this problem we oriented our coordinate system so that the field point $\vec{r}$ points along the $z$-axis so in this orientation, the angle between $\vec{r}$ and the source point $\vec{r}'$ will in fact be the spherical coordinate angle $\theta$ (the polar angle). This is actually why this orientation is so useful for this problem to begin with.