[Physics] How to derive Biot-Savart’s law for the magnetic field of a surface charge

Question

The Biot-Savart law gives the magnetic field of a steady line current:
$$
B(r)=\frac{\mu_0}{4\pi}\int\frac{I\times\hat r}{r^2}dl.
$$
Now according to Griffiths, for surface currents, the Biot-Savart law becomes
$$
B(r)=\frac{\mu_0}{4\pi}\int\frac{K(r)\times\hat r}{r^2}da,
$$
where $K$ is the surface current density, given by
$$
K=\frac{dI}{dl_\perp}=\sigma v.
$$
While the version of the formula for surface currents makes sense to me intuitively (you simply consider the current for each infinitesimal area through an infinitesimal width $dl_\perp$ that lies perpendicular to the flow), I would still like to show it using calculus, at least to some extent. I have seen how to “convert” the magnetic field for a moving charge to the magnetic field of a surface current:
$$
F_\text{mag}=\int(v\times B)\sigma\,da=\int(K\times B)\,da.
$$
I was hoping something similar to this would also work for the formula for the magnetic field. Is there any intermediate [calculus] step thinkable, to justify the formula for the magnetic field of a surface charge?

Answer 1

You almost already have your answer!

If you wanted to "derive" $$B(r)=\frac{\mu_0}{4\pi} \int \frac{K(r)\times \hat{r}}{r^2}da$$ You simply look at the infinitesimal field due to an infinitesimal current carrying "Strip" or "wire". By the very definition of $K(r)$, you know that $$dI=Kdl_{p}$$ Where I have used $dl_{p}$ to mean the the length element perpendicular to our infinitesimal current element. Putting this in Biot Savart's law you obtain:
$$dB(r)=\frac{\mu_0}{4\pi} \int \frac{K dl_{p}\times \hat{r}}{r^2}dl$$ To get the full field at the point $r$ you simply integrate over $dl_{p}$ $$B(r)=\frac{\mu_0}{4\pi} \int \int \frac{K \times \hat{r}}{r^2}dl dl_{p}$$ It is precisely $dl dl_{p}$ that is identified with $da$