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I'm ok with the following expression for Biot-Savart: $$ {\mathbf B}({\mathbf r}) = \frac{{\mu _0}}{{4\pi }}\int\frac{{{\mathbf J}({\mathbf r'}) \times {(\mathbf{r-r'})}}}{{|\mathbf{r-r'}|^3 }}dV'. $$ Many sources give a slightly different formula for a curve C: $$ {\mathbf B}({\mathbf r}) = \frac{{\mu _0 I}}{{4\pi }}\int_C\frac{{d{\mathbf r'} \times {\mathbf{(r-r')}}}}{{|\mathbf{r-r'}|^3 }}. $$ That the general formula implies the one for a curve seems reasonable, and even intuitive. But how would one go about showing it properly?

My idea was to write the current density as a product of delta functions whose arguments have uncountably many roots (those on the curve C), thus the triple integral would be annihilated by the delta functions, while a new integral would arise because of the delta functions. However this is quite complicated, and it doesn't seem to work out. Anyone has any ideas?

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Yes, the strategy is right. If one has a one-dimensional current $I$ in the thin wire, the volume density of the current is $$ \vec J(\vec r) = I\cdot \delta^{(2)}(\vec r - \vec R_{\rm nearest}) \cdot \vec n $$ where $\vec R_{\rm nearest}$ is the point on the wire that is closest to the point $\vec r$. There are other ways to write the current but this is probably the simplest one. Also, $\vec n$ is the normal direction along the wire at the point $\vec R_{\rm nearest}$.

Now, substitute this $\vec J$ to the first formula which involves a volume integral. In some local coordinates, the two transverse integrals cancel against the two-dimensional delta-function, $I$ survives as a factor, and what's left from $\vec J \cdot dV$ except for the parts we have already accounted for is simply $\vec n\cdot d z$ where $z$ is a local coordinate along the wire, and that's what is called $d\vec r'$ in the second form of the integral. Because of the delta-function, the three-dimensional integral gets reduced to the one-dimensional integral along the wire because this is where the two-dimensional delta-function vanishes.

I've been a bit sloppy about another detail that would make the notation less comprehensible if I used it immediately. The two-dimensional delta-function should have an argument that is a two-vector. So the argument shouldn't be just $\vec r-\vec R_{\rm nearest}$ but the projection of this 3-vector to the 2-plane orthogonal to $\vec n$ (the wire).

It may be right to say that to write the exact delta-function-containing formula for $\vec J$ is more complicated than to guess the second form of the law immediately. But a virtue of this comment is the correct assertion that a formula for $\vec J$ that uses the delta-functions does exist and the second form may indeed be derived from the first if things are done right.

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