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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?

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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 $da$

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  • $\begingroup$ Amazing, thank a lot! I was thinking in infinitesimal "patches" of the surface , not realising I should be thinking about infinitesimal strips. Then indeed it makes sense to use $dl$ first, and obtain $dl\,dl_p$ later. Thanks again! $\endgroup$ – Sha Vuklia May 19 '17 at 12:27

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