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Acopper rod of length 0.79 m is lying on a frictionless table (see the drawing). Each end ofthe rod is attached to afixed wire by an unstretched spring that has a s...

Question

Acopper rod of length 0.79 m is lying on a frictionless table (see the drawing). Each end ofthe rod is attached to afixed wire by an unstretched spring that has a spring constant of k=73 Nlm.A magnetic field with a strength of 0.28 T is oriented perpendicular to the surface of the table: (a) What must be the direction of the current in the copper rod that causes the springs to stretch? (b) If the current is 11A by how much does each spring stretch? Table (top view)Fied wires(out of paper)Copper

Acopper rod of length 0.79 m is lying on a frictionless table (see the drawing). Each end ofthe rod is attached to afixed wire by an unstretched spring that has a spring constant of k=73 Nlm.A magnetic field with a strength of 0.28 T is oriented perpendicular to the surface of the table: (a) What must be the direction of the current in the copper rod that causes the springs to stretch? (b) If the current is 11A by how much does each spring stretch? Table (top view) Fied wires (out of paper) Copper rod (a) Direction (b) Number Units



Answers

A copper rod of length 0.85 $\mathrm{m}$ is lying on a frictionless table (see the drawing). Each end of the rod is attached to a fixed wire by an unstretched spring that has a spring constant of $k=75 \mathrm{N} / \mathrm{m}$ . A magnetic field with a strength of 0.16 $\mathrm{T}$ is oriented perpendicular to the surface of the table. (a) What must be the direction of the current in the copper rod that causes the springs to stretch? (b) If the current is $12 \mathrm{A},$ by how much does each spring stretch?

We have. The force equals I Omega. Be not sign 90 days one So you don't write it. Forces to K X. It was I, Omega. Be not so. We have the changing lent of the spring. X equals Iomega. Be not over.

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.

Hi in the given problem here this is U shaped that garlic rails over which there is a conductor create conductor allowed to move towards right with the velocity V. Within a magnetic field which is directed into the plane of okay like this a uniform magnetic field directed into the plane of paper the length of this conductor which is allowed to move over the rails. It's given us 50.0 centimeter. Cool. And market here as L equals to 50.0 centimeter or 0.50 m magnetically uniform magnetic field is 0.0 Tesla. In the first part of the problem, the speed given to destroy towards right is we is equal to 7.50 m per second. Works right? So the emotionally um of indios will be given by the expression be into V into L. Since this is 0.0 Tesla multiplied by V which is 7.50 meter per second. Multiplied by the length which is 0.50 m. The CMF induced here comes out to be 3.0 volt, which becomes an answer for the first part of this problem. No. In the second part of the problem, we want to know the direction of burden induced in it. In this conductor, A. B. Cool using lemmings right and rule the direction of current in dues will be from and B two and A. It is the answer for the second part of the problem. Now, in the third part of the problem, resistance of the circuit has been given as 1.50 home. The current passing through this conductor will be given by, you know, if induced divided by its resistance means this is three by 1.5 which comes out to be too. And here. So a magnetic force experienced by this conductor will be given by B into I into L. Means this is 0.0 multiplied by two empty and multiplied by 0.50 m length of the conductor. So this force comes out to be 0.80 new done magnetic force acting on this conductor. No, if you look for its direction as the current is from the tweet, so direction of this force will be left, so keep on moving the conductor with the same speed. We need apply the same force 0.80 newton worse right. Which is answered for the third part of this problem. Thank you.

Here, we've got two parallel wires hooked to a circuit so that they both are hooked in series. And the current runs also through connected springs at both ends of the wires. These are identical springs. What will be using is some data to data points on the current that flows through the wires versus the amount that the springs stretch. And we see that they stretch because as the wires uh current through them, they actually repel each other. And so those two springs will stretch as the wires repel and they will become a length L zero Plus X. What will be trying to do with that data is see if we can use the data to experimentally determine both the unstructured length of these springs and the spring constant of these springs. The identical springs. There are some principles that we're going to use. There's a number of them, but the first one is that the springs will stretch until they balance the magnetic force of repulsion from the current with the spring force. So that is the main principle that will be using. Um So we're after um the two parameters in the spring and we'll be using some known things about the magnetic force to evaluate um those constants along with the data. Um So what that means is we'll need to break apart each of those two forces and the force of the spring, I'll write down its magnitude is the effective spring constant times L zero Plus X. This is the magnitude only the direction of course will be trying to push pull back against the magnetic force. We can kind of show that yeah with some arrows there and the magnetic force will be equal and opposite on both wires. So we really need to just consider say the top wire. It's and why I wrote it as effective spring constant is I do note that there are two springs, they're working in parallel. Um And so the spring stiffness effective is going to be bigger than just the single springs. And since they're identical it is twice uh the individual spring constant. Okay, there are a couple things that I need to evaluate the magnetic force. First of all, let's just evaluate the force on the top wire. Call that Wire one. Um It's the force is going to come from the current in the wire Times its length which is 0.5 m times the magnetic field that it's sitting in. So bag magnetic field is coming from the bottom wire. We'll call that wire to and furthermore that magnetic field can be evaluated using amperes Law for a long straight wire. Um You're not I over two pi times the distance between them, which is L not plus X. So we can put that all together. You not I squared length over two pi L not plus X. Okay. So we have our two expressions now we have the magnetic force and we have the spring force and now let's equate them and see what we can do with our data to try to evaluate um the stiffness and unstructured. Hte length. Okay. So we're ready to put everything together. So we have let's go ahead and put it in red. You know what I squared L over two pi L not plus X equals Okay, effective times a lot plus X. And now I want to think about um the data points that we're going to use. Um What are the things that depend on that data? There's the current and there is the X. So what I'm going to do is I'm going to regroup this equation to put all the constants On one side of the equation and all the variables together on the other and the constants should be the same regardless of which data point you have. Um So kind of rearranging let's put the constants on the I don't know the right and the variables on the left. I hope I don't mess up all these constants. I tend to drop a few here and there. Yeah that I do believe that looks like we have all of our constants together. Yeah. Um Okay so um what this says is that no matter what the data point is um So let's call the these two data points maybe one and 2. I know I've got a lot of number ones and number twos but His data .1 and data point to we know that they must agree because They concur in the constant on the right hand side. So what that means is I for data .1 squared L not plus X one quantity squared equals I too quantity squared. I will not plus X two quantity squared. Okay. And so notice there's only one unknown in that equation. We've got two data points that we can put in if we wanted to we could take the square root of both sides. That might make things a little bit easier. And so we have I won over L not plus X one equals I too Over L. Plus X two. And let's see two. I want to put I'll put things in terms of mm. We can then solve this equation for L. Not. Okay. And yeah it's kind of a Uh huh ugly looking thing. I will kind of show some intermediate steps. Um So let's see I too over I one. That might be a good thing. Show Is 13.1 Over 8.5. 1.63. And so if I cross multiply, what I wind up with is let's see l not plus eight millimeters is equal to 1.63 Times L. Plus four. And one more intermediate step .63. A lot Is 1.434, 8 times 10 to the -3. And I will come up with my L. Not in meters if I've used meters for my X. So wanna known dale that feels good. Um And let's see what I can do with that now uh nicely is I can use one of my data points and solve it for the K effective. So I can go back substitute in my known a lot and solve for K effective and hence get the spring constant of a single spring. Yes. Yeah. Yeah. Okay. And I don't think it matters which um, data point you actually use. It should work with both of them. So yeah, it'll do a little bit of algebra. Um, but using either data point and do not is four pi times 10 to the minus seventh. Tesla meters per amp L is .5 m. And use any data point that you want Either the 8.5 Or the 13.1 current and either the four or the um, eight millimeters. But to use the L not that you have found in meters and I am working out a very small spring constant. So it's they're very squishy springs eight point oh three Times 10 to the -2 newtons per meter. And what this means is that each of these springs is one half of that stiffness. So even sloppier very weak springs. Um And now we can go on and do some what ifs. So we have an equation that is basically governing the a current and the extension and this this equation governs the current and the amount of extension. Actually, yeah. Um So if you put in the new current and a new X. And set it equal to one of the known values, you should be able to solve for either the unknown current or the unknown X. So let's just kind of write out an equation for um it dependent and independent variable. Okay, equation that relates hi your necks and it looks like it will be a linear equation. So on one side you put the unknown X on the other, you use one of your you're known data points. Let's see. That's the eight. And this will give an equation for X in meters and current in amps. Let's see. Mhm. Okay. Mhm Cross multiply. Let's see. Right. Yeah. Yeah. Okay. So patients, while I work out my nice linear equation, it won't take I put a little bit more work to work out the slope and the intercept. So we would expect this to be a linear relationship between the current and the extension. Okay, so let's take a just look at a couple of cases. But certainly we have everything we need to predict. A new current. Let's suppose that we had a current of 12 amps. And the question is, what is X? Um So you would use the equation 12- too tight. 2.97 Divided by one point. Check that number. That looks a little messy. At 1.27 Types tend to the 3rd is equal to X in meters. And we get about uh seven .1 times 10 to the -3, which makes sense. So this credit is a little bit less Than the 13.1 Amps And the X is a little bit less than the eight. And finally one more case, suppose X is equal to 10 or Your .1 m. We would be asking what current would produce that. So again, we can just use our nice linear relationship to predict that current and that should come out to be in amps. And we get about 15.7 amps, Which again makes sense. 10 is just a little bit more than the eight kilometers. So we would expect the current just a little over the 13 amps.


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