# Magnetic flux density at the center of uniformly magnetized sphere using general form of the Biot-Savart law

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

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