Okay, this is question twenty eight point seventy eight. We have a pair of long, rigid metal rods year each of length point five meters, which are parallel to each other on a frictionless table, and they're connected by these two springs. Now, the springs have unstructured length l zero, which I've actually just relabeled here J because it's hard to see the elves Hero looks a little like a ten here, so I'm going to say on stretched length l zero or ten, um, and fourth constant K. And we're told we have these two data points were told that when a current I runs through the circuit, the springs stretch by a distance of point oh, four meter or point four centimeters, which have converted to meet every serious a point Oh, four meters. And when a current of thirteen point one amps run through this, we get a stretch here of point Oh, eight meters point eight centimeters. So if we express this total distance D here between the two rods, then we can express this total D by jay or l zero. That's this one stretched length plus X, which is the amount that it's going to stretch, So this is what we're going to use. So in order to solve this, what we're going to need to do is think about the force between these two rods due to the magnetic field from the current, that that force is going to be pushing the two rods away. Since these currents are running in opposite directions on DH, then the force of the spring is going to be pulling the two rods together. And so what we want. Since we know that this whole system isn't moving, we want to set the force from the magnetic field equal to the force on the spring coming from the springs. So that will mean that our net force is zero, which is the state that our system isn't here. So let's first take a look at the force between two wires. This is the general expression that we have here for the force per unit length of two wires with current I won and I to let's just take a look here. We want just the force. We don't want force for unit length, so let's start by dusting, multiplying by the length on both sides. That's just going to bring this over to the numerator here. Now I want an eye too. I actually both just equal to ay because the same current is running through these two wires so I could turn this eye one times I to just into I squared up here and now our is the only other thing we don't know On our is thie distance between the two wires. So the distance between these two rods here is going to be able to d. So we're going to just set be there. And now the force from a spring, um, the force in general is going to be k times D. This is the force of a spring with constant K stretching over a distance. D But we have two of these springs acting on each rod, so I'm going to multiply that by too. So we have two k d is going to be the force pulling these two rods together, and this is the magnetic force pushing them apart. So now we want to do is set these equal. So I'm going Tio, come over here and set this speak will to ke ni. And now we just want to rearrange this and simplify this a bit. So let's keep our mu Not I squared over here and multiply by two Heidi on both sides. That's going to give us a for pi k b squared. And let's all this for a d So if we take me squared touching this, the order here is going to ring. We're going to have this in the numerator divided by our for Katie, our four pi K. I'm sorry. Yeah. Ah, in the denominator. And so now if we want Teo solve for D, we're going to need to just take the square root of both sides When we take the square root of this right side We see we haven't I squared so we can pull one factor of eye out and we'Ll be left with this Just on a second I'm going to draw the square root right here. So this whole thing taking this group So now that we have this Ah, all right. Remember that r D is equal to J plus specs and we want to solve for J. J. Is Ariel not here. So I'm going to put in J plus X in for De and that's equal toe I. And at this point, I was going to use the power of ah one half instead of writing out this hole Bare root sign taking the square root is he goto raising to a power of one half. So I'm just going to write this, uh, just like this. So now let's take a look at what we have, um, for each value that we're told here were given I and were given X. So let me just highlight this here. So we're given X and I for some, given each of these points and what we want to solve for is l not or J and K thes air the two values that part is asking us to find. So what we're going to do is we're gonna plug each of these values in, and then that's going to give us two equations and we'LL have two unknowns R J and R K. And we're going to have to solve that system of equations. So I'm that's going to do that down here. I'm just gonna plug in the two values that we know so that we can we have our equations to solve So we have our j r l not. And that's plus Oh, point out. Oh, for looking at this first equation And the current and the first question Big region are the first Great. A point up here is eight point O five. Now, at this point, let's just plug in the other thing really know, too. So we know that the length is point five meters. So I'm just gonna plug that in here sewing everything explicitly. And now we have everything up here except for kay and J, which is what we want to solve for. And then this putting the next day the point we have Oh, Pete for X and thirteen point one Here and again, we'LL put in point five. So these are the two greatest. We're going assaults where it's ah, do that on a new tab up here. Right, we have. Okay, it's just copy these. Go over here. Okay. And remember, our jay is really our l not all this put that I'm here. And so we remember. Um, okay, so now we have two equations and two unknown variables J and K. There much of different ways that we can solve a system of equations. One way would be substitution solve for one of these and plug it into the second equation. I'm going to do this by the subtraction method where I'm gonna write the It's just going to put this one above. And I'm going to subtract the second equation here from the first equation because there is one factor of Jae in each of these. And what that's going to do is cancel out our factor of J living us just with K. So we consult for Kay, play that Back in and sulfur J. So again, this isn't the only method to solve a system Immigration's. This is one that's going to be pretty easy to use here. So basically, I'm running this all out like one big ah, subtraction here. So I'm taking this minus this with us. Kanai, give us is the factors of j. R going to cancel out? So we're gonna have point oh, eight on the side, minus point oh four. And then our right side is going to be this whole thing minus this whole thing and that is that our factors here this is the same in both. So what we're gonna have out here. It's thirteen point one minus a point o five. So hopefully this makes sense. Where to? Subtracting breaths from this. And this is all a constant in both of these two cases. So we can simplify this out a little bit. We're going to have fine. Oh, for here. Um, the left side is going to be pretty much the same from what we have the thirteen point one minus eight point Oh five over here. So we'll have a factor of five point O five out front and this is the equation we're gonna solve. So this you can just you Not as a constant so we can just rearrange this square both sides to get rid of this multiplied by four pi k. I'm not going to do out all the algebra, but when you solve this equation, you should get K is equal to o point. Oh, wait. So that's our spring constant. Now we want to find r j r i'll not so to do that, I'm just going to use Just pick one of these equations. It doesn't matter which one I'll pick the first one up here and what we want to do is just plug in the value that we just found for Kay. So point out eight and solve this for Jay or for l not. So we'LL say i'll not. People's J equals o point o two four meters or point before centimeters. This is a result I'm just plugging the sin and solving for a j here. So this is the answer to part a. So now, um, looking at B B s if I have able to twelve amps What distance? Excellent age, spring stretch. So to solve this, we want to go all the way back to our original equation with everything. Actually, General all leased. This one of here, he's gonna just anyone over here. So this is our general equation. But now we know Ajay is through there. I'll not So point o two four and we also know it Pay is now, so we know that Kay, it is a point. Okay, so now we have an equation where we can plug in I and get X. Just rearrange the stencil for X or alternately, we could play connects and sulfur I because we only have these two unknown variables So we're told in this case that I is twelve camps. Well, we just want to Dio is plug twelve ants here into this equation and solve for X Odus Offer extra you need to disobey Tracked this on both sides and this gave us this whole expression for X. So again, I'm not going to actually like the sun. We can display this directly into a calculator. We also know our ill here, fellas, break five. So we have everything here is that we need to solve for X and we should get X equals O point oh, seven one meters or point seven one. Senators, that's our answer for part feed now for part C were asked what current is required for each spring to stretch one centimeter. So this is just the reverse case of this where, instead of having eye and selling Rex, we're told X and we want to solve her eye. So starting with this sama creation a grand. But now we know X is oh, point o one meters because that one centimeter So we just want Teo plug in our expression here point o two four plus oh, point o one and that's going to be equal to everything Here on the left side, on the right side. I'm sorry. Eso again in order to solve this. What? We're going to want to dio divide both sides by this whole thing. But these are all just numbers. And so this Can this be ah, plugged in right into the calculator we should get That eye is equal to fifteen point seven twins.