# Non-constant Current Density of Magnetic Field

Tags:
1. Apr 18, 2015

### Aristotle

1. The problem statement, all variables and given/known data

2. Relevant equations
Ienclosed = ∫ JdA
∫B*ds = m0*I

3. The attempt at a solution
This is my take of finding
when r is greater than R--so, the magnetic field is m0*I/2pir, since
B(2pir) = m0*I
This is because the amperian circle covers the whole wire.

Can somebody verify that this is correct? (when r>R) I mean it doesnt matter that the density is nonunfirom correct?

2. Apr 18, 2015

### Delta²

Correct but still you got to calculate I which will not be the same as if the current density was uniform.

3. Apr 18, 2015

### Aristotle

Oh I understand that "I" will be different when r is INSIDE the circle. But "I" outside should just be "I" total without having to integrate right?

4. Apr 18, 2015

### Delta²

For an amperian circular loop that has radius any r>R , the current I will be the same "I total", but how you gonna find it without integrating the current density? it will be

$I=\int\limits_{0}^{R}\int\limits_{0}^{2\pi}\vec{J}\cdot d\vec{S}$

while for r<=R it will be
$I=\int\limits_{0}^{r}\int\limits_{0}^{2\pi}\vec{J}\cdot d\vec{S}$

Last edited: Apr 18, 2015
5. Apr 18, 2015

### Aristotle

Well originally I integrated it from limits R to r (r>R) for J*dA....and got an answer of a current equaling to I(r^5-R^5)/(R^5). And so B equaled m0*I(r^5-R^5)/(2piR^5*r)...so then I realized integrating for the amper circle outside would not be reasonable.

6. Apr 18, 2015

### Aristotle

I mean sure finding r inside the circle is a different current, but my main concern is outside.

7. Apr 18, 2015

### Aristotle

Doing it at 0 to R is the complete wire. It doesn't relate to the r of the amp. circle that is outside it, right?

8. Apr 18, 2015

### Delta²

You integrated current density from R to r??? But the current density is zero outside the wire isnt it?
when the radius r is >R then you integrate from 0 to R cause the loop encloses the whole wire.

9. Apr 18, 2015

### Aristotle

Oh yeah you make a great point, I just thought back on the section of Gauss and remembered that. Thanks!
But as I integrated from 0 to R...and plugged in the Jo for the current density...I got a result of just "I". (a good indication that it is correct).

So it wouldnt really be necessary at all to integrate right?--I mean everything cancels in the end to get a final "Itot". Or is it safe to just integrate anyways?

10. Apr 18, 2015

### Delta²

Well integrating from 0 to R i seem to get a total current $I=\frac{2}{5}\pi J_0R$

11. Apr 18, 2015

### Aristotle

Yup same here--when you find j0 and Im sure you could--you get 5I/2PiR :)...it then equals to I haha

12. Apr 18, 2015

### Aristotle

I guess integrating is a good way to show it, and indeed you're correct about the integration, but it would seem a tad redundant in having to do that since I that is enc equals the current of the wire in its entirety.

13. Apr 18, 2015

### Delta²

yes ok, it is just that the total current I is not given, instead you are given the current density equation, R and J_0.

14. Apr 18, 2015