Question

Derivation of Biot-savart's law from vector potential​

Answer 1

Answer:

The Biot-Savart law is a consequence of Maxwell's equations.

We assume Maxwell's equations and choose the Coulomb gauge, ∇⋅A=0. Then

∇×B=∇×(∇×A)=∇(∇⋅A)−∇2A=−∇2A.

But

∇×B−1c2∂E∂t=μ0J.

In the steady state this implies

∇2A=−μ0J.

Thus, we have Poisson's equation for each component of the above equation. The solution is

A(r)=μ04π∫J(r′)|r−r′|d3r′.

Now we need only calculate B=∇×A. But

∇×J(r′)|r−r′|=J(r′)×(r−r′)|r−r′|3

and so

B(r)=μ04π∫J(r′)×(r−r′)|r−r′|3d3r′.

This is the Biot-Savart law for a wire of finite thickness. For a thin wire this reduces to

B(r)=μ04π∫Idl×(r−r′)|r−r′|3.