4
$\begingroup$

Biot-Savart law for a linear current distribution is:

$\displaystyle \vec{B}=\frac{\mu I}{4\pi}\int\frac{\vec{dl}\times \vec{r}}{r^{3}}$.

In the book that my professor uses says that if we have current flowing in a wire of very small cross section $S$, in a length $\vec{dl}$ of the wire we have $dV=Sdl$ so we can write that $\vec{J}dV=\vec{J}Sdl=\vec{I}dL=I\vec{dl}$.

Then we can obtain Biot-Savart law for volume distribution. How can that happen? In the linear distribution we have one integral and in the surface distribution we have triple integral! How is that possible? I don't mean why we use triple integral for volume distribution. I mean how is that possible to go from one to another.

$\endgroup$
1
  • $\begingroup$ In this case the current density is proportional to a Dirac delta function. (The current density diverges on the wire and is zero everywhere else.) Upon integration this reduces to a line integral. A good E&M text should supply the details. $\endgroup$ – user26872 Mar 24 '15 at 18:22
4
$\begingroup$

Remember that $\vec{j}$ is a current density, which is a current per unit area. This should be sufficient to observe that dimensionally, the correspondence between $Idl$ and $jdV$ retains proper units. In 3D, you have the current density $\vec{j}$ which gives a direction to the local current and gives you information about how that current is distributed over a cross sectional area whose surface normal is in the direction of the current. Two of the integrals in the volume integral will integrate that current density over the surface, and the third one will integrate over the path of current, just like the $dl$ integral.

EDIT: To be more clear, I hope, your main problem is why $Idl$ is a 1D integral yet $jdV$ is a 3D integral. You should not view the current density $\vec{j}$ as being a surface density like construction, to see this think of some generic volume which has current flowing through it. Current is the flow of charge and at any location in the volume the current must have some direction. There are only two other directions that are orthogonal to this direction, which form a cross-sectional area. You have to allow for the current to be distributed over this cross sectional area in a general way, i.e. have some functional dependence on the two orthogonal coordinates that can be integrated over.

There is also a surface current, usually denoted by $\vec{K}$, which is a current per length orthogonal to the direction of current flow. If you have a 2D surface with current running through it, the current flow has some direction to it and then the current may be distributed along a line orthogonal to this direction. The corresponding form for a surface current is now $KdA$.

So $Idl$ is a one dimensional current problem, $KdA$ is for current distributed over a surface, and $jdV$ is for a volume that contains current flow. 1D, 2D, and 3D integrals.

$\endgroup$
2
  • 1
    $\begingroup$ Excuse me but I don't quite understand. Can you explain that mathematically? $\endgroup$ – Adam Mar 25 '14 at 0:38
  • $\begingroup$ Thanks but I know that. $\endgroup$ – Adam Mar 25 '14 at 16:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.