# Derivation of the Biot-Savart Law [duplicate]

Can someone provide a derivation of the Biot-Savart law for electromagnetic induction? To be clear, $$d\vec{B}~=~\frac{\mu_0}{4\pi}\frac{I d\vec{\ell}\times \vec{r}}{r^3}.$$

Is there a simple way to compute the magnetic field at a point between two Helmholtz coils, if the radii of the coils are the same and the current through each coil is the same?

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## marked as duplicate by Qmechanic♦May 10 '14 at 16:26

What should the law be derived from? – David Z May 16 '13 at 3:24
@DavidZaslavsky, From Ampere's Law, preferably. – Reuben Stern May 16 '13 at 18:15
I consistant derivation would be interesting to see. This is a good conceptual question, I absolutely dont seevwhy 2 closevoters think this should be closed. – Dilaton Jun 9 '13 at 7:33
I dont understand why has this question been marked as aduplicate when the question given in the link is actually a duplicate of this question! – Karan Singh Jul 24 '15 at 16:18

It is an experimental law not derivable from other more basic laws

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I think this is wrong. You can derive it directly from Maxwell equations, which are more fundamental from my point of view. – Noldig May 16 '13 at 7:16
And how do you think Maxwell equations have been obtained? I think from Coulomb and Biot-Savart laws! – richard May 16 '13 at 8:29
ok, historically you are correct. Let's say they follow from each other, or they are consistent. – Noldig May 16 '13 at 8:50
OK,people! Historically, Biot-Savart's Law -> Maxwell Equations -> Special Relativity But, I think now, you'll all agree that Special Relativity is the most fundamental of them all. So, why not use Special Relativity to give derivation of Biot-Savart's Law? I think I might write an answer – Cheeku Jul 11 '13 at 23:14

In the static case you can solve Maxwell equations using a vector potential via the poisson equatuion for the magnetic potential.

$\Delta \vec A(\vec r)=-\mu_0 \vec J(\vec r)$

Using the Greens function for the Laplace operator yields the solution of this differential equation.

$\vec A(\vec r)=\frac{\mu_0}{4 \pi}\int d^3r' \frac{\vec J(\vec r')}{|{\vec r-\vec r'}|}$

Now we can calculate the B field via $\vec B = \vec\nabla \times \vec A$ and use the identity $\vec\nabla\times(\phi\vec A)=\phi(\vec\nabla\times(\vec A))-\vec A\times\vec\nabla\phi$. Additionally we have to calculate the gradient of the scalar function 1/|(r-r')|. This gives the Bio Savart law.

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What is '$rot$'? Furthermore, I am not convinced that you can do what you have stated. I want to see the entire thing! – Killercam May 16 '13 at 8:05
rot is the german word of curl, I'm sorry for that. Which point is not clear or what do you think I can not do? – Noldig May 16 '13 at 8:16
I don't see how you can derive the full expression from what you have said... – Killercam May 16 '13 at 8:23
@Killercam: you can get to the Biot-Savart law from Noldig's answer for the vector potential by using explicitely cartesian coordinates and vector basis. – gatsu May 16 '13 at 9:14