PhoenixFire
  • PhoenixFire
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.
Physics
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chestercat
  • chestercat
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Kainui
  • Kainui
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
  • PhoenixFire
@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
  • PhoenixFire
@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.

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Kainui
  • Kainui
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
  • Kainui
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
  • PhoenixFire
@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
  • PhoenixFire
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
  • PhoenixFire
But I just can't get the math to work out with this.
Kainui
  • Kainui
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.
PhoenixFire
  • PhoenixFire
|dw:1376126122650:dw| Here the forces all cancel each other out.
Kainui
  • Kainui
Wait, can they be like this: |dw:1376126177191:dw| I'm ridiculously tired, I feel like I just reasoned out that that couldn't be the case but now I'm second guessing myself.
PhoenixFire
  • PhoenixFire
|dw:1376126262056:dw| Yes, this will produce the same as my solution.
Kainui
  • Kainui
Sorry we posted at the same time, yeah definitely. Seems like it should work lol. So obvious, I'm pretty sure this is exactly what I saw in my mind when I first saw this problem lol.
PhoenixFire
  • PhoenixFire
haha, No problem. You can go sleep, come back to it when you can think. I have this drawing on my whiteboard... it makes sense... But the MATH JUST WON'T WORK!! ARGH.
Kainui
  • Kainui
Wait, no, it doesn't make sense **sigh** I'm sorry but this "obvious" solution is obviously wrong, I hate to say it.
PhoenixFire
  • PhoenixFire
Lol. Well, actually, I know half the answer now because I gave up and just inputted stuff until it told me. The correct answer is to the Left of Wire 1, and current flowing Downwards. Now I just have to find the Magnitude of the current and the distance from wire 1.
Kainui
  • Kainui
\[F_{12}=-F_{21}\]\[F_{13}=-F_{31}\]\[F_{23}=-F_{32}\]\[F_{21}+F_{31}=0\]\[F_{12}+F_{32}=0\]\[F_{21}+F_{31}=0\] Combining these equations yields: \[F_{12}=F_{23}=F_{31}\] That means these three forces are equal: |dw:1376126879222:dw| If those three forces are equal, then that means the vectors in the opposite direction are equal too. This can't be right though, since: Hell, I don't know, maybe that's right hahaha
PhoenixFire
  • PhoenixFire
I think I'm going to come back tomorrow after some thoughtful dreaming of physics. I'll do some more checking and reasoning then. Right now... this isn't working. I'm going to give you a medal just for sticking around and trying. Thanks @Kainui
Kainui
  • Kainui
Actually for some reason I was assuming current 1 and 2 had the same value but different direction for no reason. The two outside wires have the same current and the one on the inside must have the weaker current since to produce the same current across a further distance you must have a higher current attracting the further away ones. HERE: \[F_{12}=F_{13}\] I HAVE THE ANSWER
Kainui
  • Kainui
So here's a picture |dw:1376127747283:dw| I just reasoned out above that the forces are all equal in magnitude, equate the forces now for force of 1 on 2 with force of 1 on 3. \[\frac{ L \mu_0 }{ 2 \pi }\frac{ I_1 I_2 }{ d }=\frac{ L \mu_0 }{ 2 \pi }\frac{ I_1 I_3 }{ 2d }\]
Kainui
  • Kainui
Divide out everything. \[I_2 =\frac{ 1 }{ 2 }I_3\]
Kainui
  • Kainui
Wow ok that was simple I'm almost ashamed but hey not bad at 4:45 am now I can sleep without this question bothering my brain.
PhoenixFire
  • PhoenixFire
@Kainui Sadly when I used the numbers I have for the problem, it doesn't work. Since I've got the question wrong now and don't know the answer, here are the numbers I have. \(I_1=1.4\) A \(I_2=4\) A \(d=0.2\) m (Between Wire 1 and Wire 2) What I do know (through guessing and it telling me it's right) is that Wire 3 is to the LEFT of Wire 1 and the current in Wire 3 \(I_3\) is going downwards. |dw:1376174589678:dw| So, now with numbers try figure it out. Because every solution we came up with resulting in wrong values.
Kainui
  • Kainui
|dw:1376175212498:dw| I think this is the standard way to label forces, at the very least this is how I think of them. So now you have d, I1, I2, but you need to know D and I3. You can equate these forces:
Kainui
  • Kainui
\[F_{13}=F_{23}\]\[\frac{ L \mu_0 }{ 2\pi }\frac{ I_1 I_3 }{ D }=\frac{ L \mu_0 }{ 2\pi }\frac{ I_2I_3 }{ D+d}\]\[I_1 (D+d)=I_2D\]\[D(I_1-I_2)=-I_1d\] etc...Now go through a similar process (you can make 2 other equations) to finish solving for D and I3. I can help FOR REAL this time.
PhoenixFire
  • PhoenixFire
I thought if it as, force on 1 due to 2 = \(F_{12}\), but i'll use your way. With that equation \(D=0.107\) then rearranging \(F_{21}=F_{31}\) for \(I_3\), I get \(I_3 = 2.154\). And then the last one works.
Kainui
  • Kainui
I got D=.108, if it will really bother you about rounding that badly, if that's what you mean.
PhoenixFire
  • PhoenixFire
You what is funny... I had those values last night, but I was rounding too much and when I put them in the website said it was wrong. I had one submission left for distance, and put it 0.107 and it was correct. So I can only guess that the 2.154 A is correct too.
Kainui
  • Kainui
lol does that mean this nightmare is over haha
PhoenixFire
  • PhoenixFire
I guess that's what happens when you try do physics when you're tired. -sigh- THANK YOU for your help! The nightmare is indeed over.
Kainui
  • Kainui
Haha it really wasn't that bad, it was just us XD.

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