Deriving Ampere's law from the Biot-Savart law

  • Thread starter fayled
  • Start date
  • #1
fayled
177
0
If we write the Biot Savart law as
B(r)=μ0/4π∫(J(r')xn/n2)dV'
where B is the magnetic field which depends on r=(x,y,z), a fixed point, J is the volume current density depending on r'=(x',y',z'), and n=r-r', a vector from the volume element dV' at r' to the point r. Note we integrate over the primed coordinates as J, the source of the current varies with these.

Then take the curl, making use of curl(AxB)=A(.B)+(B.)A-(A.)B-B(.A), and noting .J=0 (it depends on the primed coordinates) and that (n/n2.)J=0 for the same reason, we get
xB0/4π∫J(.n/n2)dV'-μ0/4π∫(J.)n/n2dV'.

Now term two can be shown to integrate to zero, which I understand (incidentally, the book says this second term is integrated over a volume enclosing all current, as I suspected - this is related to my problem below), but I have a problem with term 1. We get
xB0/4π∫J(.n/n2)dV'
and .n/n2=4πδ3(r-r'). So
xB0/4π∫4πJ(r'3(r-r')dV'
xB0J(r'3(r-r')dV'
Now according to my book, this reduces to
xB0J(r), which makes sense in a way because we can say
xB0J(r3(r-r')dV'
because only the value of J at the 'spike' is actually useful anyway. This is constant so
xB0J(r)∫δ3(r-r')dV' and then this integral is simply one giving the desired result.

However, we're integrating over (x',y',z') and so varying r', keeping r fixed. As our integral only need cover all of our current, and r could be outside of our current distribution, (according to my brain) we need not even integrate over the 'spike' of the delta function at r'=r, which makes the integral zero.

What is it I am misunderstanding? Thanks.

Edit: If the answer is something along the lines of 'we may as well integrate over all space because there's no current anywhere else anyway', I ask, if there's no current anywhere else anyway, why do we get two different answers by integrating/not integrating over all space.
 
Last edited:

Answers and Replies

  • #2
jambaugh
Science Advisor
Insights Author
Gold Member
2,335
313
You are not wrong!

As our integral only need cover all of our current, and r could be outside of our current distribution, (according to my brain) we need not even integrate over the 'spike' of the delta function at r'=r, which makes the integral zero.
You are not wrong... just considering a special case. IF you pick a point outside the current distribution the curl of B is zero as your mathematical intuition indicates. You are deriving Ampere's law -after all- which states that the curl of B at a point is proportional to the current density at that point. But r need not be outside the current distribution in which case you DO need to integrate over the spike and wherein B and J are not zero but still proportional.

All is right with the world.
 
  • #3
fayled
177
0
You are not wrong... just considering a special case. IF you pick a point outside the current distribution the curl of B is zero as your mathematical intuition indicates. You are deriving Ampere's law -after all- which states that the curl of B at a point is proportional to the current density at that point. But r need not be outside the current distribution in which case you DO need to integrate over the spike and wherein B and J are not zero but still proportional.

All is right with the world.

Oh dear, that is pretty obvious now :redface: Thankyou :)
 
  • #4
vanhees71
Science Advisor
Insights Author
Gold Member
2021 Award
21,592
12,408
Well, and if [itex]\vec{r}[/itex] is at a point outside the region where an electric current is flowing, your equation is correct too (for stationary currents only, of course!), because there [itex]\vec{j}(\vec{r})=0[/itex]. The local Maxwell Equations are always right!
 

Suggested for: Deriving Ampere's law from the Biot-Savart law

Replies
23
Views
839
Replies
12
Views
1K
Replies
5
Views
1K
  • Last Post
Replies
8
Views
775
  • Last Post
Replies
1
Views
805
Replies
5
Views
919
  • Last Post
Replies
1
Views
472
  • Last Post
Replies
5
Views
1K
Replies
9
Views
639
Top