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!

  • 1
    $\begingroup$ Be careful about the transformation between unit vectors of distinct coordinate systems. I am not sure if I understand your convention for the angles, but does it make sense that the the current goes in the $\hat{\theta}$ direction? $\endgroup$ – secavara Jan 13 '18 at 15:22
  • $\begingroup$ I don't see why it shouldn't: the sphere is spinning about the $z$ axis, so it drives the charge density in the $\boldsymbol{\hat \theta}$ direction (a "speck of charge" draws a circle centered on the vertical axis in the anticlockwise sense). We should expect the current to vanish at the poles, and indeed it does due to che cosine term in the second equation $\endgroup$ – giobrach Jan 13 '18 at 15:30
  • $\begingroup$ Oh your $\theta$ is the azimuthal angle (with a range of $2\pi$) and your $\phi$ is your polar angle (with a range of $\pi$), Ok, I find this convention confusing, but that's fine. Is $\hat{r}$ the spherical radial unit vector or the cylindrical unit radial vector? $\endgroup$ – secavara Jan 13 '18 at 15:42
  • $\begingroup$ Spherical. My convention for the components is $$\begin{cases} x = r\cos\varphi\cos\theta \\ y = r\cos\varphi\sin\theta \\ z = r\sin\varphi \end{cases}$$ $\endgroup$ – giobrach Jan 13 '18 at 15:44

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 $\hat{r}$.

  • $\begingroup$ Why does the third cosine term pop out? Isn't the triple $\mathbf{\hat r}, \boldsymbol{\hat \theta}, \mathbf{\hat z}$ orthonormal? $\endgroup$ – giobrach Jan 13 '18 at 16:41
  • $\begingroup$ Nope. In your convention, the orthonormal triples are $\hat{\theta},\hat{z},\hat{\rho}$ (cylindrical) or $\hat{\theta},\hat{\phi},\hat{r}$ (spherical). $\endgroup$ – secavara Jan 13 '18 at 16:44
  • 1
    $\begingroup$ Ah! That was silly on my behalf. Thank you for your help! $\endgroup$ – giobrach Jan 13 '18 at 16:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.