Verify Ampere's law using Biot- Savart equation

Question

how can I verify Ampere's law using Biot- Savart equation on a circular current:

namely, to prove:

$$\oint\limits_{W}B\,\mathrm{d}w=\mu_0 I$$

what is the curve W we should integrate with?

Answer 1

Assuming the filament is in magnetostatic conditions, Ampère's law reads $$\oint_{\mathrm W}\,\mathbf B\cdot\mathrm d\boldsymbol w=\mu_0 I_{\text{enc}}$$ where $\mathrm d\boldsymbol w$ is tangent to the curve $\mathrm W$.

Let's then prove how Biot-Savart's law satisfies Ampère's law for any curve:

Substituting Biot-Savart's law we get that $$\oint_{\mathrm W}\oint_C\dfrac{\mathrm d\boldsymbol\ell\times(\boldsymbol{r-r'})}{|\boldsymbol{r-r'}|^3}\cdot\mathrm d\boldsymbol w=4\pi$$ has to hold. Let's compute that integral. First we'll employ $$\boldsymbol\nabla\times\left(\dfrac{\mathbf a}{\phi}\right)=\boldsymbol\nabla\dfrac{1}{\phi}\times\mathbf a+\dfrac{1}{\phi}\boldsymbol\nabla\times\mathbf a$$ where we identify $\mathbf a=\mathrm d\boldsymbol \ell$ and $\phi=|\boldsymbol{r-r'}|$. Hence, since $\mathrm d\boldsymbol\ell$ only depends on $\boldsymbol r'$, $$\oint_{\mathrm W}\oint_C\dfrac{\mathrm d\boldsymbol\ell\times(\boldsymbol{r-r'})}{|\boldsymbol{r-r'}|^3}\cdot\mathrm d\boldsymbol w=\oint_{\mathrm W}\boldsymbol\nabla\times\oint_C\dfrac{\mathrm d\boldsymbol\ell}{|\boldsymbol{r-r'}|}\cdot\mathrm d\boldsymbol w$$ Using Stokes' theorem, the integral becomes $$\iint_{S_\mathrm W}\boldsymbol\nabla\times\boldsymbol\nabla\times\oint_C\dfrac{\mathrm d\boldsymbol\ell}{|\boldsymbol{r-r'}|}\cdot\mathrm d\boldsymbol S_\mathrm W$$ where $$\boldsymbol\nabla\times\boldsymbol\nabla\times\oint_C\dfrac{\mathrm d\boldsymbol\ell}{|\boldsymbol{r-r'}|}=\boldsymbol\nabla\left(\oint_C\mathrm d\boldsymbol\ell\cdot\boldsymbol\nabla\dfrac{1}{|\boldsymbol{r-r'}|}\right)-\boldsymbol\nabla^2\oint_C\dfrac{\mathrm d\boldsymbol\ell}{|\boldsymbol{r-r'}|}$$ but $$\oint_C\mathrm d\boldsymbol\ell\cdot\boldsymbol\nabla\dfrac{1}{|\boldsymbol{r-r'}|}=-\oint_C\mathrm d\boldsymbol\ell\cdot\boldsymbol\nabla'\dfrac{1}{|\boldsymbol{r-r'}|}=-\oint_C \mathrm d\left(\dfrac{1}{|\boldsymbol{r-r'}|}\right)=0$$ and so the first term vanishes. Moreover, $$\boldsymbol\nabla^2\dfrac{1}{|\boldsymbol{r-r'}|}=4\pi\delta^{(3)}(\boldsymbol{r-r'}),$$ and thus, the original integral reduces to $$\begin{aligned} &\oint_{\mathrm W}\oint_C\dfrac{\mathrm d\boldsymbol\ell\times(\boldsymbol{r-r'})}{|\boldsymbol{r-r'}|^3}\cdot\mathrm d\boldsymbol w\\ &=-4\pi\iint_{S_{\mathrm W}}\oint_C \delta^{(3)}(\boldsymbol{r-r'})\,\mathrm d\boldsymbol\ell\,\mathrm d\boldsymbol S_{\mathrm W}\\ &=-4\pi\iint_{S_{\mathrm W}}[\delta(y-y')\delta(z-z')\boldsymbol i+\delta(x-x')\delta(z-z')\boldsymbol j+\delta(x-x')\delta(y-y')\boldsymbol k]\cdot\mathrm d\boldsymbol S_{\mathrm W}=4\pi\,\blacksquare \end{aligned}$$