Can someone please explain why I'm not getting the known answer, $(2/3)\mu_0M$, using this setup for a uniformly magnetized sphere of radius $b$ and magnetization $M$:

starting with $$dB = \frac{\mu_0I}{4\pi}\frac{(dl \times R)}{R^3}$$

given that volume current density = 0 and a surface current density due to magnetization $J = M\sin(\theta)$

Plugging in:

$$I = Mb\sin(\theta)d\theta$$

$$dl = b\sin(\theta)d\phi$$

$$R = b$$

When I integrate, I consistently get $(\pi/4)\mu_0M$ instead of $(2/3)\mu_0M$, due to having a $\sin^2$ in the integral instead of the desired $\sin^3$.

It seems like I might be going wrong with the $R = b$ part. I know it is supposed to be a vector pointing from each dl to the field point, but the dl's are all on the surface of the sphere and the field point is the center, so every $R$ should just be the radius $b$, right?

I know there are methods to solving this starting from other versions of the the Biot-Savart Law, but I want to understand where I'm going wrong with this approach specifically. Thanks


1 Answer 1


It seems you have set up the sphere so that the current is parallel to the z axis. We know from symmetry that the magnetic field at the centre of the sphere (and anywhere else along the z-axis) will also be parallel to the z-axis, so we are only interested in the z-component of it, which is obtained by taking the dot product with $\hat{z}$:

$$ B_z(0) = \frac{\mu_0}{4\pi} \int \hat{z} \cdot \frac{\mathrm{d}K \times (-\hat{r})}{r^2} $$

where $\mathrm{d}K = Mb^2 \sin^2 \theta \,\hat{\varphi} \, \mathrm{d}\varphi \, \mathrm{d}\theta$.

When you take the cross product of $\hat\varphi$ with $-\hat{r}$, you get $-\hat\theta$, and when you take the dot product of that with $\hat z$, you get $\sin \theta$, so that's where the third factor of $\sin\theta$ comes from, leaving us with

$$ B_z(0) = \frac{\mu_0}{4\pi} \int M \sin^3 \theta \, \mathrm{d}\varphi \, \mathrm{d}\theta = \frac{2\mu_0 M}{3} $$

  • $\begingroup$ Over the course of the past two weeks, I've literally spent close to 10 hours trying to figure this out. I was actually close to coming to this conclusion yesterday, but couldn't quite work it out. This clears it up for me. I kept thinking -theta hat dot z was cos(theta), when it's really cos(90-theta) which equals sin(theta). Thank you very much $\endgroup$ Apr 4, 2023 at 11:27

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