# Ampère's law from Biot-Savart law for linear currents with multivariate calculus

My book, W.E. Gettys's Physics, starts from the Biot-Savart law $d\mathbf{B}=\frac{\mu_0}{4\pi}\frac{Id\boldsymbol{\ell}\times\hat{\mathbf{r}}}{r^2}$, i.e.$$\mathbf{B}(\mathbf{x})=\frac{\mu_0}{4\pi}\int_a^b I\boldsymbol{\ell}'(t)\times\frac{\mathbf{x}-\boldsymbol{\ell}(t)}{\|\mathbf{x}-\boldsymbol{\ell}(t)\|^3}dt$$where $\boldsymbol{\ell}:[a,b]\to\mathbb{R}^3$ is a parametrisation of the current's path, to show that the magnetic field $\mathbf{B}$ at a distance $R$ from an infinite straight electric wire carrying a current $I$ has norm $B=\frac{\mu_0 I}{2\pi R}$ and direction and orientation as shown in the figure

$\hskip2in$ Then the book derives, from such an expression of the magnetic field produced by an infinite straight wire carrying current $I_{\text{linked}}$, that, for a closed path $\gamma$, Ampère's circuital law holds in the form $$\oint_{\gamma}\mathbf{B}\cdot d\mathbf{x}=\mu_0 I_{\text{linked}}$$ and then states, without proving it, that such a formula is valid for any current, not only flowing in an infinite straight line.

I have searched very much in the web and in this site in particular, but I only find derivations of Ampère's from the Biot-Savart law using integrations of the Dirac $\delta$, which I only knew in the context of functional analysis in the monodimensional case where $\int_{-\infty}^{\infty}\delta(x-a)\varphi(x)dx=\varphi(a)$. Is it possible, for the particular case of linear, monodimensional, current flows (as the current flow parametrised by $\boldsymbol{\ell}$ in the expression of $\mathbf{B}$ above), to prove Ampère's law, in the form $\oint_{\gamma}\mathbf{B}\cdot d\mathbf{x}=\mu_0 I_{\text{linked}}$, or in the form $\nabla\times\mathbf{B}=\mu_0\mathbf{J}$ where $\mathbf{J}$ is the current density from which I would derive the first expression by using Stokes' theorem, without using the Dirac $\delta$, only by using, say, the tools of multivariate calculus and elementary differential geometry? I heartily thank anybody posting or linking such a proof.

• For a line current, obviously the differential form of Ampère's law cannot be proved without using Dirac deltas, since the current density itself involves delta functions. – Emilio Pisanty Feb 7 '16 at 17:16
• @EmilioPisanty Thank you so much! The answer below uses Ampère's circuital law in the form $\oint_{\gamma}\mathbf{B}\cdot d\mathbf{x}=\mu_0 I_{\text{linked}}$. – Self-teaching worker Feb 7 '16 at 17:22
• Exactly. But the question also asks for delta-free proofs of the differential version under a line current, which is evidently impossible. – Emilio Pisanty Feb 7 '16 at 17:40
• @EmilioPisanty $\infty$ thanks for the precisation! – Self-teaching worker Feb 7 '16 at 17:42

Do you want a proof of Ampere's Law? Some book really follows the way you said. I think it is just an example rather than a proof.

For the proof of Ampere Law, there is no need to use the delta function, although this method is more simple in my opinion. Some geometry calculation is enough, but it is more tricky to use this method.

L1 is the source current. $$P$$ is a field point at $$\boldsymbol{r}_2$$ whose magnetic field we are interested in, then we have, $$\boldsymbol{B}(P)$$ according to the Biot-Savart law Then we calculate the line integral along $$L_2$$ passing through $$P$$.

$$\boldsymbol{B}(\boldsymbol{r}_2)\cdot\mathrm{d}\boldsymbol{l}_2=\frac{\mu_0I}{4\pi}\oint_\limits{(L_1)} \frac{\mathrm{d}\boldsymbol{l}_2\cdot (\mathrm{d}\boldsymbol{l}_1\times\hat{\boldsymbol{r}}_{12})}{r_{12}^2}=\frac{\mu_0I}{4\pi}\oint_\limits{(L_1)} \frac{(\mathrm{d}\boldsymbol{l}_2\times\mathrm{d}\boldsymbol{l}_1)\cdot\hat{\boldsymbol{r}}_{12}}{r_{12}^2}$$$$=\frac{\mu_0 I}{4\pi}\oint_\limits{(L_1)}\mathrm{d}\omega=\frac{\mu_0 I}{4\pi}\omega$$

Usually it takes at least 20 minutes to make it clear in class. I wish I could tell you the name of the book I used. But unfortunately, it is writen in Chinese.

I present you the main points of the demonstration, and I think it would be clear to you if you are familiar with the integral and vector analysis. Just be clear that the -dl2×dl1 can be treated as the area between the souce L1 and the L1' which is of a small displacement dl2 relative to L1

Ok, now we are calculating $$B(\vec{r_2})\cdot{d\vec{l_2}}$$, where $$d\vec{l_2}$$ is a small displacement in the line integral $$\oint_{(L_2)}$$. Now we have

$$B(\vec{r_2})\cdot{d\vec{l_2}}=\frac{\mu_0}{4\pi}\oint_{(L_1)}\frac{(-d\boldsymbol{l}_2\times d\boldsymbol{l}_1)\cdot\hat{\boldsymbol{r}}_{21}}{r_{21}^2}(1)$$

$$(-d\boldsymbol{l}_2\times d\boldsymbol{l}_1)$$ is just the area between line segment $$-d\boldsymbol{l}_2$$ and $$d\boldsymbol{l}_1$$. So if we consider the line integral in (1), $$\oint_{(L_1)}(-d\boldsymbol{l}_2\times d\boldsymbol{l}_1)$$ is the area between two 'circle', $$L_1$$ and $$L_1'$$ (see the first figure of my first answer), where $$L_1'$$ is another circle with a displacement of $$-d\boldsymbol{l}_2$$ from $$L_1$$. But don't forget there is also $$\hat{r}_{21}\over {r_{21}^2}$$ in the line integral which gives the solid angle with respect to point $${\vec{P}}$$.

Ok, now we are calculating $$B(\vec{r_2})\cdot{d\vec{l_2}}$$, where $$d\vec{l_2}$$ is a small displacement in the line integral $$\oint_{(L_2)}$$. Now we have

$$B(\vec{r_2})\cdot{d\vec{l_2}}=\frac{\mu_0}{4\pi}\oint_{(L_1)}\frac{(-d\boldsymbol{l}_2\times d\boldsymbol{l}_1)\cdot\hat{\boldsymbol{r}}_{21}}{r_{21}^2}(1)$$

$$(-d\boldsymbol{l}_2\times d\boldsymbol{l}_1)$$ is just the area between line segment $$-d\boldsymbol{l}_2$$ and $$d\boldsymbol{l}_1$$. So if we consider the line integral in (1), $$\oint_{(L_1)}(-d\boldsymbol{l}_2\times d\boldsymbol{l}_1)$$ is the area between two 'circle', $$L_1$$ and $$L_1'$$ (see the first figure of my first answer), where $$L_1'$$ is another circle with a displacement of $$-d\boldsymbol{l}_2$$ from $$L_1$$. But don't forget there is also $$\hat{r}_{21}\over {r_{21}^2}$$ in the line integral which gives the solid angle with respect to point $${\vec{P}}$$.

Can you understand what I wrote this time? Then there is not much left for us to move on.

• This site supports mathjax. – DanielSank Jan 5 '16 at 2:07
• No no, don't delete it. I'm just trying to help you learn the practices of this site, since you're new. – DanielSank Jan 5 '16 at 2:24
• I have edited the first answer. – FaDA Jan 5 '16 at 17:15
• So basically, I think you have followed all the stuff that I wrote in my second answer. We should move on. Now we calculate $\oint_{(L_2)}\boldsymbol{B}\cdot d\boldsymbol{l}_2$, the integration will just be the area of the surface generate by moving $L_1$ along a trace parallel to $L_2$, – FaDA Jan 6 '16 at 0:28
• Remember that $\oint_{(L_1)}(-d\boldsymbol{l}_2\times d\boldsymbol{l}_1)$ gives the surface of the area between $L_1$ and $L_1'$, $L_1$ and $L_1$ travels $-d\boldsymbol{l}_2$.So $\oint_{(L_2)}\oint_{(L_1)}(-d\boldsymbol{l}_2\times d\boldsymbol{l}_1)$ is the area between $L_1$ travels the entire $L_2$, which will form a close surface. I hope you can get to this, then we are very close to the final results. – FaDA Jan 6 '16 at 0:33