Net force between three long wires (undefined length) must equal zero.
Wire 1: Current going Up $I_1$
Wire 2: Current going Down $I_2$
Distance between wire 1 and wire 2 = $d_{12}$
How would you go about finding the location of Wire 3 relative to Wire 1 and the current $I_3$ of Wire 3 such that $F_{net}=0$ for each wire?

There's equations for two wires, force per unit length $\frac{F}{L}=\frac{\mu_0 I_1 I_2}{2\pi d}$

But I can't seem to figure out how to use it. Or if there's something else.

Question

Net force between three long wires (undefined length) must equal zero.
Wire 1: Current going Up $I_1$
Wire 2: Current going Down $I_2$
Distance between wire 1 and wire 2 = $d_{12}$
How would you go about finding the location of Wire 3 relative to Wire 1 and the current $I_3$ of Wire 3 such that $F_{net}=0$ for each wire?

There's equations for two wires, force per unit length $\frac{F}{L}=\frac{\mu_0 I_1 I_2}{2\pi d}$

But I can't seem to figure out how to use it. Or if there's something else.

kainui · Answer

I'm assuming the wires are parallel and lie in the same plane and are basically assumed to be infinite length.

The best metaphor to describe this scenario would be thinking in terms of point charges. You've probably already done this similar problem where you have 3 charges all in a line and have one charge between the two. Where do you place the charge between the other two in order to make the net force on that charge = 0? Same kind of idea here basically.

phoenixfire · Answer

@Kainui Yeah sorry, They're all parallel and hanging downwards. The question doesn't state a length, just "long".

I actually haven't done a question such as that. 
Since the net force has to be 0 on all the wires would I start with: 
$F_{1net}=0=F_{12}+F_{13}$
$F_{2net}=0=F_{21}+F_{23}$
$F_{3net}=0=F_{31}+F_{32}$
And some how find the distance $d_{13}$ and current $I_3$ that satisfies all three of these?

phoenixfire · Answer

@Kainui Also the problem I'm having is that the force on EACH wire must be 0. Not just on the wire that's being added in, which is slightly different than the question you said.

kainui · Answer

Sure, so we know it's a similar question to the charges because it is impossible for them to not be in the same plane. Let's look at all these wires from a top-view (since they're all dangling straight down) and we're only seeing each wire as a little circle cross section. There are only two ways they can possibly be:

|dw:1376122503206:dw|

So when the three wires aren't hanging in the same plane, there is no possible way you can cancel out the net forces of each other.

|dw:1376122653013:dw|

When all 3 wires are in the same plane, now you have the possibility of 2 wires acting equally in magnitude and opposite in direction on one of the other wires. And they can each do this to each other.

This takes a little bit of reasoning to understand I suppose, but hopefully looking at these wires from this perspective makes it easier to understand. Since they have a uniform current density, you can neglect the whole idea of them really being wires.

--

You have 3 equations that are good, and don't forget:

$$F_{12}=-F_{21}$$
and so on...

I think it's getting late because I started trying to solve this problem and feel like information is missing.

kainui · Answer

The only possible way I can see any of this happening is if the current is zero in all the wires. Maybe I need to go to bed or something, I remember looking at this earlier and thinking there was an answer but now I'm not so sure haha. Good luck.

phoenixfire · Answer

@Kainui 
This is actually all the information given. 
3 long wires hanging vertically. distance between 1 and 2 is $d_{12}$. on the left, wire 1 has current $I_1$ going upwards. and to the right, wire 2 has a current $I_2$ downwards. Wire 3 is located such that when it carries a certain current, each wire experiences no net force. 
Find the position of wire 3 in the form of "distance - right/left of wire 1" (eg. 20cm - left of wire 1).
Find the current in form of "amps - down/up". (eg 2A - Downwards)

That's all the information. I removed the numbers because I thought a generalized solution would help my understanding. 
Also I forgot that $F_{12}=-F_{21}$ but I just tried that and it still didn't work. I'll look at it more tomorrow though. Can't think straight right now.

phoenixfire · Answer

And I understand what you were saying about being in the same plane. 
So thinking about it, the current on this third wire should be downwards, and to the left of wire 1.

|dw:1376125426012:dw|

1 and 2 repel - opposite current
3 and 1 repel - opposite current
so the force on 3 due to 2 should attract 
and force on 2 due to 3 should attract.
all cancelling out.

phoenixfire · Answer

But I just can't get the math to work out with this.

kainui · Answer

Well I'm thinking if you combine these six equations you basically get that:

$$F_{12}=F_{23}=F_{31}$$

which seems wrong to me when I draw out a picture.|dw:1376125408688:dw|

So if you look at the possible positions, I've included the forces from wires 1 and 2 on each other to show they're repelling, they always will (UNLESS one of the wires has a negative current, which is basically contrary to the conditions they're telling us)

Top case: It seems like there's no way you can place the third one in the center because it will always be contrary to one or the other, so you'll have two only repelling the other one without a force to attract it back.

Bottom case: The two are repelling each other with equal magnitude so no matter what force you put on the outside to push one back, the force to attract the other one into equilibrium will never be strong enough to pull it back in.

I must be doing something wrong in my reasoning, this is bugging the pellet out of me.