Finding the Magnetic Field in a System with Intersecting Current-Carrying Pieces

[cdummie]

Homework Statement

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.

Homework Equations

Biot-Savart law

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?

 

[SammyS]

Homework Statement

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.
View attachment 86570

Homework Equations

Biot-Savart law

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?

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.

 

[cdummie]

In looking at you image enlarged, it appears that the current flows in opposite directions in the two shaded regions.
View attachment 86583

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.

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.

 

[SammyS]

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.

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.

...

Homework Equations

Biot-Savart law

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?

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.

 

[cdummie]

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.

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.

 

[SammyS]

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.

Yes.

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

 

[cdummie]

Yes.

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

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?

 

[SammyS]

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?

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.

 

[cdummie]

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.

Using Ampere's law, for the P1 i got ∫Bdl=μ₀∫J*dS

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

for the right part: μ₀∫J*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?

 

[SammyS]

Using Ampere's law, for the P1 i got ∫Bdl=μ₀∫J*dS

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

for the right part: μ₀∫J*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?

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.​

.

 

[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.

 

[SammyS]

View attachment 86957

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.

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

 

[cdummie]

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

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.

 

[SammyS]

In this case, left side (of the equation) 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.

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

 

[cdummie]

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

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.

 

[SammyS]

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.

Check your algebra in finding B.

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

 

[cdummie]

Check your algebra in finding B.

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

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:

 

[SammyS]

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:
View attachment 86999

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.

 

[cdummie]

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.

Thanks a lot.

 

[SammyS]

Thanks a lot.

How are you progressing with finding B at P₂ ?

 

[cdummie]

How are you progressing with finding B at P₂ ?

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?

 

[SammyS]

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?

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

 

[cdummie]

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

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?

 

[SammyS]

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?

What do you gt for the vertical component?

 

[cdummie]

What do you gt for the vertical component?

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

 

[SammyS]

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

Can you express cos(θ) in terms of R and a ?

 

[cdummie]

Can you express cos(θ) in terms of R and a ?

Well, if θ is angle between B_vertical and B then angle between B_horizontal and R (R to the point P1) is also θ since the sides of those two angles are orthogonal so cosθ can also be a/2R so B_vertical=B*cosθ=(μ₀JR/2)*(a/2R) so B_vertical=μ₀J*a/4
and if i multiply this by i have the solution for P2 since i should get same solution for the second part so B=μ₀J*a/2. Correct me if i am wrong.

 

[SammyS]

Well, if θ is angle between B_vertical and B then angle between B_horizontal and R (R to the point P1) is also θ since the sides of those two angles are orthogonal so cosθ can also be a/2R so B_vertical=B*cosθ=(μ₀JR/2)*(a/2R) so B_vertical=μ₀J*a/4
and if i multiply this by i have the solution for P2 since i should get same solution for the second part so B=μ₀J*a/2. Correct me if i am wrong.

You should use parentheses adequately. (a/2)/(R) is what you apparently intended.

a/2R = (aR)/(2) ≠ (a/2)/(R)​

(a/2)/(R) can also be written as a/(2R) .

More importantly, that's not the correct expression for cosθ. That's sinθ, not cosθ .

.

 

[cdummie]

You should use parentheses adequately. (a/2)/(R) is what you apparently intended.

a/2R = (aR)/(2) ≠ (a/2)/(R)​

(a/2)/(R) can also be written as a/(2R) .

More importantly, that's not the correct expression for cosθ. That's sinθ, not cosθ .

.

The way i see it, i don't understand how that is sinθ, which means i maybe made a mistake and i don't see where, so here's the picture (as for the parentheses part, i see, thanks that's what i meant):

 

[SammyS]

The way i see it, i don't understand how that is sinθ, which means i maybe made a mistake and i don't see where, so here's the picture (as for the parentheses part, i see, thanks that's what i meant):

View attachment 87124

B is tangent to the circle, thus B is perpendicular to the radius line.

 

[cdummie]

B is tangent to the circle.

I know it is, i just drew it wrong, but anyway, what's the deal with angle, i still don't understand how it is sine and not cosine?

 

[SammyS]

I know it is, i just drew it wrong, but anyway, what's the deal with angle, i still don't understand how it is sine and not cosine?

OK!

Yes, of course, you're correct!

 

[cdummie]

OK.OK!

Yes, of course, you're correct!

Thanks again for help on this one!

 

[SammyS]

Thanks again for help on this one!

Don't forget to double each result.