Let $\Sigma$ be a sphere of radius $R$ charged with uniform surface density $\sigma$. Supposing $\Sigma$ rotates with constant angular velocity $\omega$, calculate the magnetic field at the center of the sphere.
Suppose $\boldsymbol \omega = \omega \mathbf{\hat z}$. We have a surface current
$$\mathbf K(\mathbf r') = \sigma \mathbf v(\mathbf r') = \sigma \boldsymbol \omega \times \boldsymbol \rho$$
where $\boldsymbol \rho$ is the vector separating $\mathbf r'$ from the axis of rotation (the $z$ axis). Since in spherical coordinate ($\theta$ longitude, $\varphi$ latitude) we can write $\boldsymbol \rho = R \cos \varphi \mathbf{\hat r}$, we have
$$\mathbf K(\mathbf r') = \sigma \mathbf v(\mathbf r') = \sigma\omega R\cos\varphi \boldsymbol{\hat \theta}$$
The magnetic field at the origin is given by
$$\begin{split}
\mathbf B(\mathbf 0) &= \frac{\mu_0}{4 \pi} \int_\Sigma \frac{\mathbf K(\mathbf r') \times (-R \mathbf{\hat r})}{R^3} \mathop\!\mathrm{d}a' = \frac{\mu_0}{4\pi} \int_{-\pi/2}^{+\pi/2} \int_0^{2\pi} \frac{\sigma\omega R^2\cos\varphi'\ \mathbf{\hat z}}{R^3} R^2\cos \varphi' \mathop\!\mathrm{d}\theta' \mathop\!\mathrm{d}\varphi' \\
&= \frac{\mu_0R\sigma\omega}{2}\int_{-\pi/2}^{+\pi/2} \cos^2\varphi' \mathop\!\mathrm{d}\varphi'\ \mathbf{\hat z} = \frac{\pi}{4} \mu_0 R \sigma \omega \ \mathbf{\hat z}
\end{split}$$
However, I am told that the answer should be
$$ \mathbf B(\mathbf 0) = \frac 2 3 \mu_0 R \sigma \omega\ \mathbf{\hat z}$$
What did I get wrong?
EDIT: Thanks to secavara's comments and answer, I figured out my mistake: mixing up cylindrical coordinates with spherical integration. Indeed $\boldsymbol{\hat \theta} \times \mathbf{\hat r}$ is not $\mathbf{\hat z}$: we have
$$\boldsymbol{\hat \theta} \times \mathbf{\hat r} = – \boldsymbol{\hat \varphi} = – (\cos\varphi \mathbf{\hat z} – \sin \varphi \mathbf{\hat u})$$
if we denote with $\mathbf{\hat u}$ the radial unit vector in cylindrical coordinates. So the integration should go
$$\begin{split}
\mathbf B(\mathbf 0) &= \frac{\mu_0}{4 \pi} \int_\Sigma \frac{\mathbf K(\mathbf r') \times (-R \mathbf{\hat r})}{R^3} \mathop\!\mathrm{d}a' = \frac{\mu_0}{4\pi} \int_{-\pi/2}^{+\pi/2} \int_0^{2\pi} \frac{\sigma\omega R^2\cos\varphi'\ \color{red}{\boldsymbol{\hat \varphi}}}{R^3} R^2\cos \varphi' \mathop\!\mathrm{d}\theta' \mathop\!\mathrm{d}\varphi' \\
&= \frac{\mu_0}{4\pi} \int_{-\pi/2}^{+\pi/2} \int_0^{2\pi} \frac{\sigma\omega R^2\cos\varphi'\ (\cos\varphi' \mathbf{\hat z} – \sin \varphi' \mathbf{\hat u})}{R^3} R^2\cos \varphi' \mathop\!\mathrm{d}\theta' \mathop\!\mathrm{d}\varphi' \\
&= \frac{\mu_0\sigma\omega R}{2} \left(\int_{-\pi/2}^{+\pi/2} \cos^3\varphi' \mathop\!\mathrm{d}\varphi'\ \mathbf{\hat z} – \int_{-\pi/2}^{+\pi/2} \sin \varphi' \cos^2\varphi' \mathop\!\mathrm{d}\varphi'\ \mathbf{\hat u} \right) \\
&= \frac{\mu_0\sigma\omega R}{2} \left(\frac 4 3\ \mathbf{\hat z} – 0\ \mathbf{\hat u}\right) = \frac 2 3 \mu_0\sigma\omega R\ \mathbf{\hat z}
\end{split}$$
and this is the correct answer. Thanks!
Best Answer
Ok, I think I finally got it. I am so used to my convention that I had to do it in mine and then translate it to yours. You are missing an additional factor of $\cos \phi$ in the integrand: you have one coming from the current, one from the volume element and one from the cross product between $\hat{\theta}$ and $\hat{r}$.