# Homework Help: Magnetic induction vector?

1. Jul 30, 2015

### cdummie

1. The problem statement, all variables and given/known data
In the system shown in the picture, there's a current whose constant density is J=0,5A/mm^2. System contains two pieces as shown in the picture, in the area where two pieces intersect, there's no current. If R=1mm and a=1,25mm (a - distance between centers of the circles), find the vector B in at points P1 and P2.

2. Relevant equations
Biot-Savart law

3. The attempt at a solution
I think i should solve this using superposition, or even better, find the B for the first part and multiply it by two in both cases (P1 and P2) since i should get the same value for both parts of the system. Now, i could think of the first part as lot of lines with current flowing through them, then i could sum up (integrate) all lines in that area and find the value B. The problem is, how to find area?

2. Jul 30, 2015

### SammyS

Staff Emeritus
In looking at you image enlarged, it appears that the current flows in opposite directions in the two shaded regions.

In the region on the right J is into the page. On the left,J is out of the page.

Therefore, the net current flow is zero.

3. Jul 31, 2015

### cdummie

But, isn't direction of vector B defined by right hand rule, so it would have vertical direction for both left and right part in the point P1? Does it matters if net current is zero if those two parts are separated, like there in picture? Please explain, i don't think i understood it.

4. Jul 31, 2015

### SammyS

Staff Emeritus
In post #2, I was only referring to the current, not the B field.

You are correct about the B vector at point P1 being vertical for the left part and for the right part, being taking separately, so you can just double the result for either to get the overall resulting B.

Your description of your solution in Post #1 was somewhat vague - lacking detail.
Do you need to use the Biot-Savart law? If not use Ampere's law along with superposition to make the solution fairly easy.

Consider this as two overlapping circles (actually circular cylinders). Each has current density J, the left with J out of the page, the right with J into the page.

5. Aug 1, 2015

### cdummie

So basically, if i find B for both cylinders, final result will come up as if there's no current in part where they intersect because in that part i will have two currents of the same intensity but different direction, so summed up, it would be zero.

6. Aug 1, 2015

### SammyS

Staff Emeritus
Yes.

That does allow you to treat each cylinder independently, making B relatively easy to calculate for each.

7. Aug 3, 2015

### cdummie

Is there any difference while calculating B for two cylinders, i mean if center of one of the cylinders is 0 then center of another should be a, is that correct?

8. Aug 3, 2015

### SammyS

Staff Emeritus
Each point, P1 & P2, is equidistant from the axis of each cylinder, so for each point, the magnitude of B is the same for each cylinder. At P2, the direction of B from each differs.

9. Aug 6, 2015

### cdummie

Using Ampere's law, for the P1 i got ∫Bdl0J*dS

for the left part: ∫Bdl=B2rπ (r is the distance from center to the P1 -it's a/2)

for the right part: μ0J*dS= μ0*J∫2rπdr (limits of integration are 0 to r) =μ0*J*r2*π*2

so B=μ0*J*r2/2

Doing this for the second part i got the same solution just opposite direction (minus sign), which means B is zero, which is wrong, but i don't see any mistakes. What is incorrect here?

10. Aug 6, 2015

### SammyS

Staff Emeritus
For P1:
You had indicated previously that the contribution from the left conductor is in the same direction as that from the right conductor. Have you abandoned this?

Are you going to plug-in a/2 for r or not. sometimes you do sometimes not.

What are the integration limits for $\displaystyle \ \int \vec J \cdot \vec{dS} \$ what is r? In fact, since $\ \vec J \$ is constant, why use an integral at all?

You have made algebra errors in solving for B.​
.

11. Aug 6, 2015

### cdummie

This is how i did it for the P1, but i still don't understand how to find B for the second part, i know it should be in the same direction and it should have the same intensity, but how can i know that the part where they intersect is not included here, i mean if i go and calculate B for the second part i don't see how part where they intersect is not included, and if radius on the right side of the equation is a/2 again (and i think it is, because that is the distance from center to point whose B i am looking for), but if distance from center is a/2 that doesn't mean that the two parts are placed exactly like in the picture, i mean, there could be more (or less) intersected area and still these limits of integration and radius would remain the same.

12. Aug 6, 2015

### SammyS

Staff Emeritus
For Ampere' Law. how are the path for the line integral and the region for the surface integral related ?

13. Aug 7, 2015

### cdummie

In this case, left side is circumference of the circle and area of the same circle is the right side of the Ampere's law, correct me if i am wrong.

14. Aug 7, 2015

### SammyS

Staff Emeritus
Yes. So that's a/2, not R.

15. Aug 7, 2015

### cdummie

Oh, ok, so B=a*J but again, if distance from center is a/2 that doesn't mean that the two parts are placed exactly like in the picture, i mean, there could be more (or less) intersected area and still these limits of integration and radius would remain the same.

16. Aug 7, 2015

### SammyS

Staff Emeritus
Check your algebra in finding B.

For the rest of what you say: It's not clear to me what you mean.

17. Aug 7, 2015

### cdummie

I took the upper limit to be a instead of a/2.

I mean, while calculating B i integrated like i have a full circle, but there's intersection with no current, if i do the same thing for the second part, how can i be sure that intersected part is not included if i calculated both times as if i have a full circle, i mean, will intersected area remain the same if two cylinders are placed, for example, like this:

18. Aug 7, 2015

### SammyS

Staff Emeritus
That will change the direction of the B vector, but that's all.

Don't forget, you're using super-position to get the overall answer.

19. Aug 7, 2015

### cdummie

Thanks a lot.

20. Aug 7, 2015

### SammyS

Staff Emeritus
How are you progressing with finding B at P2 ?

21. Aug 7, 2015

### cdummie

Well, i have B=μ0JR/2 for the one part, but i am not sure, i mean should i express R using a? Because this way, this could be like i am finding B for two cylinders separately, they could be far away one from another, am i right?

22. Aug 7, 2015

### SammyS

Staff Emeritus
The B vectors from the two conductors have different direction.

Verify that only the vertical components remain.

This will bring ' a ' back into the mix

Last edited: Aug 7, 2015
23. Aug 8, 2015

### cdummie

Well i know, since they (horizontal components) have the same intensity but opposite directions their sum will be zero, but how that brings back a?

24. Aug 8, 2015

### SammyS

Staff Emeritus
What do you gt for the vertical component?

25. Aug 9, 2015

### cdummie

Well, since i got B=μ0JR/2 and if θ is angle between vertical component of vector B and vector B then Bvertical=Bcosθ , but i don't see how this can be related with a.