sawer said:
Thank you. I know they are equivalent. But I just want to know which one came first in the history?
Field lines?
I don't understand the connection between biot-savart, ampere and field lines.
I must admit I'm not familiar with the early history of electromagnetism. Note that this year we celebrate the 150th anniversary of Maxwell's equations. So there should come up some material about the history of electromagnetism.
Anyway, to answer your questions above. First of all one should note that the fundamental laws from a modern point of view are the differential form of Maxwell's equations for microscopic electrodynamics, i.e., (in Heaviside-Lorentz units which are the most natural units for electromagnetics):
$$\vec{\nabla} \times \vec{E}+\frac{1}{c} \partial_t \vec{B}=0, \quad \vec{\nabla} \cdot \vec{B}=0,$$
$$\vec{\nabla} \times \vec{B} - \frac{1}{c} \partial_t \vec{E}=\frac{1}{c} \vec{j}, \quad \vec{\nabla} \cdot \vec{E}=\rho.$$
Here ##\vec{E}## and ##\vec{B}## are the electric and magnetic components of the electromagnetic field, ##\rho## is the charge density, ##\vec{j}## the current density, and ##c## the speed of light in vacuo.
Ampere's Law and Biot-Savart's Law are now special cases for time-independent fields. Then the electric and magnetic components decouple, and for the magnetic components we deal with the two equations
$$\vec{\nabla} \cdot \vec{B}=0, \quad \vec{\nabla} \times \vec{B}=\frac{1}{c} \vec{j}.$$
The first equation implies that there's a vector potential, which is determined up to the gradient of an arbitrary scalar field. Thus we can introduce the vector potential via
$$\vec{B}=\vec{\nabla} \times \vec{A}$$
and constrain it by an arbitrary "gauge-fixing condition". As we'll see in a moment the most convenient choice is the Coulomb-gauge condition in this case:
$$\vec{\nabla} \cdot \vec{A}=0.$$
Now it is of utmost imporance to keep in mind that from Ampere's law (the 2nd equation above) it necessarily follows that the current must be source-free, i.e.,
$$\vec{\nabla} \cdot \vec{j}=0.$$
Otherwise there's no solution the equation. But now we use the vector potential in Ampere's Law and use the Coulomb-gauge condition, leading to
$$\vec{\nabla} \times (\vec{\nabla} \times \vec{A})=\vec{\nabla} (\vec{\nabla} \cdot \vec{A})-\Delta \vec{A}=-\Delta \vec{A}=\frac{1}{c} \vec{j}.$$
Now we know the solution of this Poisson equation from electrostatics. It's the same equation but now for the three (Cartesian!) vector components of the vector potential rather than the scalar potential in the electrostatic case. Thus you get
$$\vec{A}(\vec{x})=\frac{1}{4 \pi c} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' \frac{\vec{j}(\vec{x}')}{|\vec{x}-\vec{x}'|}.$$
Now taking the curl leads to
$$\vec{B}(\vec{x})=\frac{1}{4 \pi c} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' \vec{j}(\vec{x}') \frac{\vec{x}-\vec{x}'}{|\vec{x}-\vec{x}'|^3},$$
which is Biot-Savart's Law.
