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The tridimensional version of the Biot-Savart law says that the magnetic field generated at the point $\boldsymbol{r}\in\mathbb{R}^3$ by a tridimensional distribution of current defined by the current density $\boldsymbol{J}$ is$$\boldsymbol{B}(\boldsymbol{r})=\frac{\mu_0}{4\pi}\int_V\frac{\boldsymbol{J}(\boldsymbol{x}) \times(\boldsymbol{r}-\boldsymbol{x})}{\|\boldsymbol{r}-\boldsymbol{x}\|^3}d^3x$$where $V\subset\mathbb{R}^3$ is the region where the current is distributed.

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 $\boldsymbol{B}$ 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 heartily thank any answerer.

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