- #1

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I intuitively suspect that, if ##V## is a cylinder of height ##h=b-a## and radius ##R##, whose symmetry axis is parallel to the unit vector ##\mathbf{k}##, flown through by a current having constant density ##\boldsymbol{J}\equiv J\mathbf{k}## on ##V## (and ##\boldsymbol{J}\equiv\mathbf{0}## elsewhere), this last expression of ##\boldsymbol{B}(\boldsymbol{r})##, with ##\boldsymbol{r}\in\mathbb{R}^3\setminus V##, equates that of a magnetic field generated, according to the linear version of the Biot-Savart law, by the current flowing through a straight wire, carrying the same current ##\pi R^2 J##, placed where the cylinder's axis is, i.e. $$\frac{\mu_0}{4\pi}\int_V\frac{\boldsymbol{J}(\boldsymbol{x}) \times(\boldsymbol{r}-\boldsymbol{x})}{\|\boldsymbol{r}-\boldsymbol{x}\|^3}d^3x=\frac{\mu_0}{4\pi}\int_a^b\frac{\pi R^2 J \mathbf{k}\times(\boldsymbol{r}-z\mathbf{k})}{\|\boldsymbol{r}-z\mathbf{k}\|^3}dz.$$Is that so, either for a cylinder of finite (##a,b\in\mathbb{R}##) or infinite (##a=-\infty##, ##b=+\infty##) length? If it is, how can it be proved?

I have tried to explicitly use cylindrical coordinates, so that the field can be expressed as$$\boldsymbol{B}(\boldsymbol{r}) =\frac{\mu_0}{4\pi}\int_a^b\int_0^{2\pi}\int_0^R\frac{J\mathbf{k} \times(\boldsymbol{r}-(\rho\cos\theta\mathbf{i}+\rho\sin\theta\mathbf{j}+z\mathbf{k}))}{\|\boldsymbol{r}-(\rho\cos\theta\mathbf{i}+\rho\sin\theta\mathbf{j}+z\mathbf{k})\|^3}\rho\, d\rho d\theta dz$$ but I am not able to handle the denominator. I do not want to use Ampère's law because all the proofs I have seen (like this) use ##\boldsymbol{J}\in C^2(\mathbb{R}^3)## while here we have a discontinuous density.

I ##\infty##-ly thank any answerer!