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I've been trying to tackle this problem for more than a day now, and would appreciate some help. The problem is to prove that the magnetic dipole moment of a spherical ball of mass $m$ and charge $Q$, whose charge is distributed uniformly only on its surface, rotating about its center, is equal to $$ \vec{\mu} = \frac{5Q}{6m} \vec{L}.$$ My reasoning is this: We know the formula for the dipole moment in terms of the surface charge density $K$, which is $$\vec{\mu} = \frac{1}{2} \oint \vec{r} \, ' \times \vec{K} \, dA.$$ Ok, next we have to figure out these quantities. Well, by definition $\vec{K} \equiv \sigma \vec{v}$, where $\sigma$ is the surface charge density. Using this we can write it as $$ \begin{align} \vec{K} &= \sigma \vec{v} \\ &= \sigma \vec{\omega} \times \vec{r} \\ &= \sigma \omega R \sin \theta \hat{\phi}. \end{align} $$ Next, we know that $\vec{r} \, ' = R\hat{r},$ since what we are integrating only lies on the surface. Now, all we do is take the cross product: $$ \begin{align} \vec{\mu} &= \frac{1}{2} \oint R\hat{r} \times \sigma \omega R \sin \theta \hat{\phi} \, dA \\ &= \frac{\sigma \omega R^2}{2} \oint \sin \theta (-\hat{\theta}) \, dA.\\ \end{align} $$ We know that the dipole moment will be in the $z$ direction because that is the direction of the angular momentum, so we can rewrite $\hat{\theta}$ in our equation above, taking only the $z$ component. $\hat{\theta} = \cos \theta \cos \phi \hat{i} + \cos \theta \sin \phi \hat{j} - \sin \theta \hat{k}$ and so we get $$\vec{\mu} = \frac{\sigma \omega R^2}{2} \oint \sin^2 \theta \hat{k} \, dA$$ We know that $dA = R^2 \sin \theta \, d\theta \, d\phi$, and our limits of integration are from $0$ to $2 \pi$ for $\pi$ and from $0$ to $\phi$ for $\theta$, so our final result comes out to: $$ \begin{align} \vec{\mu} &= \frac{\sigma \omega R^2}{2} \int_{\phi = 0}^{2 \pi} \int_{\theta = 0}^{\pi} \sin^2 \theta \hat{k} (R^2 \sin \theta \, d\theta \, d\phi) \\ &= \frac{\sigma \omega R^4}{2} \int_{\phi = 0}^{2\pi} \int_{\theta = 0}^{\pi} \sin^3 \theta \hat{k} \, d\theta \, d\phi \\ &= \frac{\sigma \omega R^4}{2} \frac{4}{3} 2 \pi \hat{k} \\ &= \frac{4}{3} \sigma \omega \pi R^4 \hat{k}. \end{align} $$ Last but not least, we need to put this into the correct form as the problem asks us for. We know that $\sigma = \frac{Q}{4\pi R^2}$ and $\vec{L} = I_{sphere} \vec{\omega} = \frac{2}{3} mR^2 \omega \hat{k},$ since the angular velocity is directed in the $z$ direction. To finish it off, we plug in these values and get: $$ \begin{align} \vec{\mu} &= \frac{4}{3} \frac{Q}{4\pi R^2} \omega R^4 \pi \hat{k} \\ &= \frac{Q\omega R^2}{3} \hat{k} \\ &= \frac{Q}{3} \frac{3\vec{L}}{2m} \\ &= \boxed{\frac{Q}{2m} \vec{L}} \end{align} $$ oops... Once again, this is not the correct answer and I cannot find any faults, so help would be much appreciated! Thank you in advance as usual.

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  • $\begingroup$ I think your assumption that $\vec{r}' = R\hat{r}$ with constant $R$ is wrong. You are rotating around the $z$ axis, so your electrons circle around that axis with different $R(z)$: for $z = \pm R_0$, $R(z)=0$, for $z=0$, $R(z) = R_0$. Currently, you are calculating a rotating cylinder (or ring) instead of a ball (but for that, your result would be correct). $\endgroup$ – Solarflare Nov 15 '16 at 20:54
  • $\begingroup$ Hm okay, but what would $\vec{r} \, '$ be then? Would it be $\vec{r} \, ' = R_0 \sin \theta \hat{r}$, since at $ z = R_0$, $\vec{r} \, ' (R_0) = \vec{r} \, '(\theta = 0) = 0$ and at $z = 0$, $\vec{r} \, ' (0) = \vec{r} \, '(\theta = \pi) = R_0.$ $\endgroup$ – Josh Pilipovsky Nov 15 '16 at 21:57
  • $\begingroup$ Yes, although you have to make sure you have the right sign. Alternative: set $r(z)^2=R^2-z^2$, and use an integral over z (not sure what's easier). Also check if you maybe have a solid ball (for the mass of the ball and therefore I ), you might get the missing 5 there (or maybe the 5 comes from the integral $R^4dr$). $\endgroup$ – Solarflare Nov 15 '16 at 22:58
  • $\begingroup$ Duuuuuuude! That's the whole problem! The moment of inertia is supposed to be $\frac{2}{5} mR^2$, then we get the correct answer. What stupid mistake... $\endgroup$ – Josh Pilipovsky Nov 15 '16 at 23:07
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The moment of inertia is supposed to be $$I = \frac{2}{5} mR^2,$$ because it is a solid sphere, not a thing spherical shell. Plugging the correct moment of inertia yields the correct answer!

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