# Định luật Biot-Savard

Định luật Biot-Savard định nghỉa tương quan giửa Dòng điện và Từ trường . Định luật Biot-Savart Law cho rằng:

${\displaystyle d\mathbf {B} =K_{m}{\frac {Id\mathbf {l} \times \mathbf {\hat {r}} }{r^{2}}}}$

Với

${\displaystyle K_{m}={\frac {\mu _{0}}{4\pi }}\,}$, where ${\displaystyle \mu _{0}}$ is the magnetic constant
${\displaystyle I\mathbf {} }$ is the current, measured in amperes
${\displaystyle d\mathbf {l} }$ is the differential length vector of the current element
${\displaystyle \mathbf {\hat {r}} }$ is the unit displacement vector from the current element to the field point and
${\displaystyle r\mathbf {} }$ is the distance from the current element to the field point

Phương trình trên chỉ đúng với dòng điện ổn

Mật độ dòng điện
${\displaystyle \mathbf {B} =K_{m}\int {{\frac {\mathbf {j} \times \mathbf {\hat {r}} }{r^{2}}}dv}}$

Với

${\displaystyle \mathbf {\hat {r}} ={\mathbf {r} \over r}}$ is the unit vector in the direction of r.
${\displaystyle dv}$ = is the differential unit of volume.
Dòng điện ổn
${\displaystyle \mathbf {B} =K_{m}I\int {\frac {d\mathbf {l} \times \mathbf {\hat {r}} }{r^{2}}}}$
Điện tích điểm với vận tốc không đổi

In the special case of a charged point particle ${\displaystyle q\mathbf {} }$ moving at a constant velocity ${\displaystyle \mathbf {v} }$, then the equation above reduces to a magnetic field approximately of the form:

${\displaystyle \mathbf {B} =K_{m}{\frac {q\mathbf {v} \times \mathbf {\hat {r}} }{r^{2}}}}$

This formula, however, is wrong. This is because a point charge moving in a straight line does not constitute a steady current nor a constant charge distribution, both of which are essential to magnetostatics. In particular, in this case, there is a changing electric field and an induced magnetic field that must be added to correct the above formula. Even though the equation is not precisely correct, it is a very good approximation (unreasonably good, in fact).

Cực nhỏ

${\displaystyle \mathbf {H} =\epsilon \mathbf {v} \times \mathbf {E} }$

and hence,

${\displaystyle \mathbf {B} =\mathbf {v} \times {\frac {1}{c^{2}}}\mathbf {E} }$