# How is the surface integral of $\vec J$ carried out in Ampere's law?

In Ampere's law, the right hand side consists of the surface integral $$\int_s\vec j\cdot ds$$ which is supposed to represent the current passing through the surface of the enclosed volume. However, $$\vec j$$ leaving the surface will contribute a positive value to the integral while that entering will contribute a negative value. Hence the integral should give a zero value, rather than the current $$I$$

• The article says it's not a closed surface Commented Nov 16, 2019 at 19:19

## 1 Answer

The right integral is: $$i_{\Sigma}(t)=-\iint_{\Sigma}{\mathbf j (\mathbf r,t)\bullet \mathbf n(\mathbf r) \ d\Sigma}$$ where $$\mathbf n$$ is the versor of the surface. If you have a closed surface: $$i_{\Sigma}(t)=-\iint_{\Sigma}{\mathbf j (\mathbf r,t)\bullet \mathbf n(\mathbf r) \ d\Sigma}=-\iiint_{V}div \mathbf j(\mathbf r, t) \ dv$$

• If you could clarify the mathematics, it would be really nice. Commented May 4, 2020 at 15:14