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Pan B Net ForceIf the rectangular loop of wire has resistance of 4,5 n determine the net force that exerted on the rectangular loop by the long straight wire Use co...

Question

Pan B Net ForceIf the rectangular loop of wire has resistance of 4,5 n determine the net force that exerted on the rectangular loop by the long straight wire Use coordinate system where orce the right positive and force the Ieft is negative.enter the value in units OF N.View Available Hint(s)AEdFpactSubmitPrevious Answers Request Answer

Pan B Net Force If the rectangular loop of wire has resistance of 4,5 n determine the net force that exerted on the rectangular loop by the long straight wire Use coordinate system where orce the right positive and force the Ieft is negative. enter the value in units OF N. View Available Hint(s) AEd Fpact Submit Previous Answers Request Answer



Answers

The long, straight wire $A B$ shown in Fig. $P 28.72$ carries a current of 14.0 A. The rectangular loop whose long edges are parallel to the wire carries a current of 5.00 A. Find the magnitude and direction
of the net force exerted on the loop by the magnetic field of the wire.

This problem covers the concept of the force exerted by a current carrying wire on another file. So to solve this problem, first ability to analyze the direction of the force on the section A. B. There to the straight fire. The direction of the magnetic field is into the plane and also on the sexual see to the direction is into the plane. So since the current is moving along the airport direction in the loo, the force on A. B. By the long straight wire is longer prior direction. And on the section C. D. The forces along the left direction by symmetry. Um, can say the net force on A. D. And B. C. Is zero. So they can buy it by symmetry. Yeah, the net force on B. C. And 80 is in Seattle. Now let's calculate the force on the section A. B. So the foods on the section A. B equals full by into tenders minus seven Tesla majors, foreign peel into the current in the straight fire That is 40 NPR into the current in the loop that is 20 ampere into the length of the section A. A. B. That is six centimeter. We can write 0.6 m upon To buy into the distance between the section every and the straight wire that is two cm. Or we can add 0.02 m. So the force on A B. A magnitude is 4.8 In to Tenders -4 notice. Now let's calculate the force on the section. Uh, city. So the fourth on sex and city equals four by Into tennis -7 islam meters for MPO, into 40 mpl into 20 mps, Into the length. That is 0.06 m upon to buy Into the distance will be four cm, therefore 0104 metre. So the fourth magnitude on the CDS 2.4 into 10 days minus four. Know them. The force on the section series along the left direction. Therefore, the net force on the loop is equals 4.8 in two tenders -4. Newton minus 2.4 Into. And this man is for Newton are the net force on the lupus, 2.4 into and this man is four Norton to the right.

Hey, this is question twenty eight point sixty four. We have one long wire appear with current that will call I won That has a current of fourteen amps. And then we have this wire loop down here with current I too, which I'll call. I'LL call this current night too. And I, too, is five amps. And we want to know what the force is on the wire on this loop of the wire. Ah, caused by this long wire up on the top, so did it out. We're gonna need to use our equation for the force, um, on two segments of wire and we want to know the actual absolute force. So let's multiply both sides here. Bye. Well, so that we, um, are just solving before the force directly. Now, um, since the's wires and the super separated by just a different distances, we're going to need to consider each side of the loop separately. So let's first think about the thes two horizontal lines over the two vertical parts of the loop. You think about it. The magnetic field exerted on both of these from this wire loop is going to be the same. But the current in either of these is in opposite directions. So when we add up the sum of the force on this wire and the sum of the force on this wire, we're going to get two values that are equal in magnitude but opposite in directions. So one of these is going to be negative, and one will be positive. But that means that they're essentially just gonna cancel out because they're equal in magnitude but opposite in direction. So we can just ignore these too vertical parts of the loop because they're not going to contribute anything to the force. They're just going to cancel. So now let's consider. We're gonna have to just add the forest on the top to the force on the bottom. And let's just think about this. Conceptually, we have the force on the top. We have two wires that are going in the same direction, with currents going in the same direction, and when we have two currents going in the same direction, they will attract each other, So this is going to be a an attractive force. But on this bottom wire, the currents are going in opposite directions from the original wire. So that means that these ones are going to repel. So the direction of the force exerted on the top is going to be the opposite of the force exerted on the bottom. So when we add these together, we're just going to want to make sure that will consider this to be positive. And then we want to subtract the value from this. So the fact that the currents are in opposite directions, mate, we're going to need to make sure that this term is negative. So now that we have established that, Ah, let's actually go ahead and do that. So let's consider the force on the top wire first. So we have our eye, one which we know is fourteen amps. Are I too, which we also know the length is going to be the length of this wire. So Well, I know that as well. That'LL be just twenty centimeters. Um and now we need two pi r. So the distance between the distance from this first, uh, wire to the top wires, that's two point six centimeters, which is going to be point o too. Six liters. So that's what we're gonna have ground here. So this is the forest on the top wire. And then remember, we wanted us to subtract me Magnitude Subtract the result of the force exerted on this bottom wire. So the bottom wire, everything is the same. We have the same current left again. That's going in opposite direction. Which is why we bring in this negative sign. And now the distance it is ten centimeter is so point. Oh, so point one meters, Um, when we can simplify this a bit before we actually go through and plugging everything to solve because we have this factor of ah knew not. I want I two times l divided by two pi in both of these terms. And then we just have went over a point. Oh, two six minus one over. Oh, point one. So this is what we're going to plug in here and again. We're just looking for a magnitude. So we just want to take the absolute value of our results here, and this is going to give us seven. Let's just make sure that we are sure that we know everything that's playing in. So I'm just gonna explicitly say that Earl here we're gonna have to plug in as point twenty centimetres, which is point two meters. So now we explicitly have everything here to plug into our equation. So now we plug in and solved. We'LL get the forest is equal to seven point nine seven times ten to the minus five britons again, um, we have, um we're told that, um these, uh we essentially have, um when we were considering the direction we know that this top part of the loop is closer to this wire. So since we have s o, the force on this top part is going to dominate. So ah, sense. The current's going in opposite er in the same direction attract thie. This, uh, Luke is going to experience a force upward because it's going to be net attracted to this top wire since the bottom wire is farther away from it, then this one. So this is going to be a force upward

So in this problem, we have a rectangular loop of wire here on the left, and then we have a long, just single wire on the right. Ah, so on the left, we have current I won on the right. We have current I too, and we have all of the relevant distances. And so we want to find in part a what? The Net force on this rectangular loop of wire is. So to start this label, these sides of this loop so we'll say side 112 three, and four. Um, so right away, because sides one in two are carrying the same magnitude of currents. Um, these are just gonna produce forces that cancel out. So from the right hand rule side, one is gonna be making a force in the upward direction side three, making a force in the downward direction so they'll cancel each other out because I have the same current through each other and at the same distance away from the wire. So we don't have to worry about those. So we can just say than the net force is gonna be the force from wire to plus the force from wire four and again with the right hand roll. We could see why or two will make a force going out like there and wire four Will won't make one out that way. So I just want to deal with only the magnitudes of these two forces. So I'm just going to kind of build into the equation that F two is negative and then f four is positive. So now we just need to know what form these forces take and how to solve for those. So we know that for a wire, the force on its due to a magnetic field is just gonna be that magnetic field times the current in the wire times the length of the wire. So our magnetic field is coming just from this big long wire here. And so we know that's always going to be you, not times its current over two pi r. So then adding in these other terms will then have I won, and then we'll have l. So this is just our general form for what the forces from wire two and four gonna take. Now we just plug in the specifics, Um, so it looks like we could factor out always, Um, you not eye to eye one. Oh, over to pie. So the only things that are gonna be changing here are the are So we said after his negative story gonna have a minus one over and our distance for wire to We have five centimeters plus seven centimetres. So points 112 meters. Oops, that's a point. And then we have plus riff for and four is only 57 years away. So points 05 So we plug in our length through these So 0.26 meters and then we have, um, I, too, as 100 amps And I want us to amps so plugging all those in and we get a positive 1.2 times 10 to the minus fifth or five Newton's And so that's in the X that direction So being pulled towards the big wire. Then question, um, her part be asked us to find the net torque on the wire. Um, but we know the net force is just in the rightward direction. So all there is is just a force going like that, Um, and more so. All these forces are in the same plane as each other. There's really no way to make a rotation acting here, at least because all these forces air acting like we can treat them as a big force. But really, it's like, uh, like a consistent, smooth bunch of little forces here. So there's no way there's gonna be a net, Tourky. There's no way these could be causing your rotation, at least if all these forces are in the same plane like this, so there's just no torque.

In this problem were asked to find out the magnetic forces that are acting on each side of a triangular shape that is made out of wire that carries a particular current. Now it's gonna look something like this where we have a triangle, a made out of wire. We have a constant current going through that train. Give a wire were also given an angle Phi here and the length of this side here. Now we're gonna label the corners A B and C So this four, therefore the side on the bottom here decide a B, and we can label it the length of Syed A B, and that is given as two meters. Now, we also know that there is a magnetic field directed directly to the right like this. And of course, that is gonna, uh, bring about the magnetic forces that are acting on the charges moving through this wire or because the current dislike the way. Now, if we're in order to figure out the forest on every particular side, we just have to follow our force equation. The magnitude of our force will be given by the current flowing through the particular part of our triangle. How long that side of the triangle is? The magnetic field magnitude in sine of the angle between the magnetic field in the current direction. So we just need to apply this equation for every side of this triangle and then just keep track of both the magnitudes of the forces on each side, on each wire each side of triangle, and think about the right hand rule to figure out the direction of that force. So let's first start with the bottom part of the triangle or what we call side A B. No, it was Think about how this force equation here works for this particular side of our triangle. So the magnitude of the magnetic force is going to be the current I going through that wire times the length of the wire length a B. We know that value. I believe it is two meters, and then we know the magnitude of the magnetic four magnetic field through which the wire is do the white that the wire suspended in. Sorry. And we need to figure out the angle that is between the current carrying direction in the magnetic field direction in the bottom part of this triangle. So just highlight. Decide right the bottom side of the triangle. We see that the current is heading to the left and that our magnetic field is heading directly to the right. So the angle between two completely opposite vectors so directors that in the opposite direction the angle between those two is 180 degrees and sign of 100 degrees is actually zero. And this makes sense because the current or the charge slow. The motion of the charges is directly anti parallel to the external magnetic field, and therefore there's gonna be no magnetic force acting on the bottom part of the wire. So that's one part. So it's only one part of our force. And, of course, since the Forces zero, that means really has no direction. Now let's think of the other side's Let's think about the other sides of this triangle. First, let's think about the vertical side, the vertical part war side A. C. And so that means we're gonna be thinking about this side right here for now. Just a highlight it. So it's right out our equation again, actually will quit before we read on equation. We notice right that we're gonna need the length, but we're only given the length 80 not the links a c So we need to use some trigonometry to figure that out. If you look at this triangle, we have the angle phi which is in the corner. Be so if we know this length a B as well. We can just use a tangent function to figure out the side the length a c So just doing a little strong trigonometry length a C divided by length A B gives us the ratio, the economic function of tangent. And we know the angle here is given as 55 degrees if you two years, 55 degrees. So we now know how to express the length of side A C in terms of length of Syed A B, which we know, and in terms of this angle phi. So when we go on right out the magnitude of the force that's acting on this vertical side, we get the current times the saut length of side a C times the magnetic field strength times sine of the angle between the current direction and the magnetic field direction now it was easy up top to the bottom side. The side of way just made it zero. We don't plug in any new numbers, but now we're gonna have to put those in. So our current value, the current flowing through the wire is 4.70 amps. The length of the side is gonna be the length of sight a B or two meters multiplied by tangent of 55 degrees multiplied by the magnitude of the magnetic field 1.8 Tesla's multiplied by sine of the angle between the current direction and the magnetic field direction. If we go up to our drawing, we'll see that the current is heading up in this part of the wire and the magnetic field direction is heading to the right. So the angle between those two is actually just 90 degrees. So we can plug a 90 degree in here and we remember 90 degrees turns into signing. Made agrees. It's one. So we just plug all this into our calculator and they end up getting that The magnitude of the force acting on the side. A sea of this triangle is 24.2 24.2. Munitz, The question is, is what is the direction of this force? So we have to use the right hand rule. So we put our pointer finger along the direction of the current like this, we twist our wrist to make sure our middle finger points in the direction, man, it field toe left. And when we do that, we see there a thumb is gonna be pointing. Well, you drive a bigger It's a circle with an accident. So it's into the screen. So the direction of this force is actually going to be into the screen. Okay, so what I'm gonna do is keep this diagram. I'm just going to real quickly a race, our work for the bottom part and replace that for our work for side A B, and then they can continue on with the problem. So we're gonna find the magnitude of the force on the slanted side or side A B. C. Now, again, we're gonna need it the length of side BC. So when I use a little bit of trigonometry, But since we know side, uh, a be let's use our coastline function because it's adjacent to our angle Phi and you want to find the value of our iPod news. So if we look at our trigonometry function, we confined the length of science a C by looking at this ratio here, sign right. So the length of side A C is going to be given by the length of side A be divided by co sign. So when we gonna go, will they go plug in our values for the force acting on this side, we take the current multiplied by length of side A C multiplied by the magnitude of making the field multiplied sine of the angle between the two. Now what's plug these things in that we know, right? So we know that the current is 4.7 amps. The length of side A C is the length of Syed A B or 2.2 meters divided by co sign of five. Then we're gonna have this, uh, multiplied by the value of the magnetic field or 1.8 Tesla's and multiplied by sine of the angle between the current the magnetic field. So if you look at that, that's gonna be this angle right here. This angle right here, which is actually the same as angle Phi. So we can actually just plug in the same angle phi into our sine function right here. So after we do a little bit of simplifying, make some room here. Since we already talked about this stuff do a little bit of simplifying and we see that the magnitude the force is gonna be given by 4.7 amps multiplied by the two meters. You see, we have a co son and the denominator and a sign in the numerator that becomes tangent of ah five, which is tangent of 55 degrees, again multiplied by 1.8 Kessel's. We see that mathematically, this ends up being the exact same statement as this. So the magnitude of the force on side B. C is the exact same and magnitude 2.4 mutants. But we have to figure out the direction. So if we put our point of using the right hand right, So if you put our pointer finger along the direction of the current, and then we put our middle finger along the direction of the magnetic field like so Oops, let me do that one direction. The magnetic field like this completely to the right. Then the direction of our thumb actually points outward. That's the only way that we can get those to be situated by rotating our wrist and our thumb points out. So the direction of the force on the side, the magnetic force on side B C, is actually out of the screen. The screen so we have no force on Side A. B affords the magnitude of 24.2 Newton's into the screen on site A. C and that force of magnitude 24.2 Newton's out of the screen, four side C B or B C. Whichever you want to say that. So the next part that was all for part A that was off a part of this problem. The next part actually deals with finding the Net force acting on this object, which we don't need the diagram for this anymore. So I'm gonna clear this out there. We're gonna talk about net forces, right? So remember, we have a force. The magnetic force Onside A B was actually being zero. The force onside A C was 24.2 newtons into the screen, and the force on Side B C was 24.2. Newton's out of the screen. So when we add all these together, add all these together to get the net force right, we add all these up. We have to think about the directions. While zero Force, of course, is gonna have no contribution, let's say let's call into the screen as negative. So we're gonna have a negative 24.2 Newton's being contributed from Side A C and then if into the screen is negative, will consider out of the screen is positive. And so we're gonna have a total of 24 positive. 24.2 Nunes being contributed from the side BC You'll notice that when we add all these together, we end up just getting a net force of zero, which is something that happens when we have a current carrying loop it suspended inside a uniform magnetic field. You'll see that we have a zero net force acting on this. That means this loop is not gonna translate in space anywhere. Move around. But you'll learn that there's gonna be a net Twerk on this thing. So generally current carrying loops in external magnetic magnetic fields like to twist around and spin because they have a non zero network. But they will generally not translator, move around in space linearly because they have no net force, and that is our answer. An explanation.


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