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I want to compute the magnetic field due to a homogeneously charged, rotating sphere with radius $R$, angular velocity $\vec{\omega} = \omega\hat{z}$ and total charge $Q$. I want to use Biot-Savart law, so I first computed the current density $\vec{j}(r,\theta,\phi) = \frac{\omega Q}{4 \pi R} \delta(R-r) \cdot (-\sin\theta \sin\phi,\sin\theta\cos\phi,0)^T$. After taking out all the constants with respect to integration and removing the integral over the radial component with the $\delta$ function, what stays in the Biot-Savart integral is

$$\vec{B}(\vec{r}) \propto \iint_{\theta,\phi}^{\pi,2\pi}\frac{\vec{j}(\vec{y}) \times (\vec{r} - \vec{y})}{|\vec{r}-\vec{y}|^3}\sin{\theta}\;\text{d}\phi\;\text{d}\theta$$

where $\vec{r} = |\vec{r}|\cdot(\sin\theta_r\cos\phi_r, \sin\theta_r\sin\phi_r,\cos\theta_r)^T$ is the point where the $B$-field is evaluated, $\vec{y} = R\cdot(\sin\theta\cos\phi, \sin\theta\sin\phi,\cos\theta)$ is a point on the sphere. and the $\vec{j}$ is now missing the $\delta$ part, and some constants.

The problem I have is, that the vector inside the integral is

$$\begin{pmatrix} \sin\theta\cos\phi(|\vec{r}|\cos{\theta_r} - R\cos\theta )\\\sin\theta\sin\phi(|\vec{r}|\cos\theta_r-R\cos\theta)\\-|\vec{r}|\sin\theta\sin\phi\sin\theta_r\sin\phi_r-|\vec{r}|\sin\theta\cos\phi\sin\theta_r\cos\phi_r+R\sin^2\theta\end{pmatrix}.$$ Since all terms involving $\sin\phi$ or $\cos\phi$ will integrate to zero, this implies that I will get a $B$-field that is along the $z$ axis, no matter where it's evaluated. This can't be, because in the next part I'll show that far away the field looks like that of a dipole.

I have been sitting over this for hours, can someone help me clear this up? Where am I going wrong?

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The problem is with your current density $\vec{j}$, let's start with it's basic definition as current over area, $\vec{j} = \frac{Q\vec{v}}{4\pi R^2}$, now knowing that the angular velocity can be used to get the linear velocity of the charge on the spherical surface using $\vec{v}=\vec{\omega}\times R\hat{r} = \omega R \hat{z} \times \hat{r} = \omega R (\cos\theta\hat{r}-\sin\theta\hat{\theta})\times\hat{r} = \omega R\sin\theta \hat{\phi}$

So that your current density is:

$$\vec{j} = \frac{Q\omega}{4\pi R} \sin\theta \delta(r-R)\hat{\phi}$$

Notice that this makes sense since the charges should only be moving in the $\phi$ direction around the z-axis, you should never have had any $r$ or $\theta$ components before. Also, at the poles of the sphere the current density goes to zero as expected and with the maximal current density at the equator.

With this current density your integrand will have a $\sin ^2\theta$ term which will clear up your previous issues.

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  • $\begingroup$ Thanks for your effort. I'm going to compute everything out with this approach, and see what happens. However, I'm sceptic, because if this is going to work, then I feel that there should be a computational mistake in my working above, as the coordinate systems are supposed to be equivalent, and so my approach should be equally valid, right? Or why should this work, and the above not? Notice that my $\vec{j}$ doesn't have a $z$ component either, and is the same except that I wrote out the $\hat{\phi}$. $\endgroup$ – Pascal Engeler Mar 23 '14 at 11:06
  • $\begingroup$ I misunderstood what you did and wasn't paying careful enough attention. I didn't realize that you were expressing everything in cartesian. I still think it makes sense to use spherical coordinates and it should be computationally easier but clearly both approaches should agree. $\endgroup$ – Elvex Mar 23 '14 at 17:00
  • $\begingroup$ But how can this finite rotating sphere create a $\vec{B}$ field that only has a $z$ component at any point in space? My understanding of magnetostatics tells me that the field lines should be closed for this problem, and this seems to be violated by my result. I imagine that the field should look like a superposition of the fields created by small current-circles around the sphere, which (by my intuition) would look similar to the field created by only one current-circle. $\endgroup$ – Pascal Engeler Mar 23 '14 at 17:22

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