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Copper wire is 8.00 m long; and has cross-sectional area of [.00 10 4 m. This wire forms Iturn loop in the shape of square and is then connected to battery that app...

Question

Copper wire is 8.00 m long; and has cross-sectional area of [.00 10 4 m. This wire forms Iturn loop in the shape of square and is then connected to battery that applies potential difference of 0.100 V. Ifthe loop is placed in uniform magnetic field of magnitude 0.400 T, what is the maximum torque that can act 0n it? The resistivity of copper is 1.70 * 10 * M: Deline Resistivity(5 marks)

copper wire is 8.00 m long; and has cross-sectional area of [.00 10 4 m. This wire forms Iturn loop in the shape of square and is then connected to battery that applies potential difference of 0.100 V. Ifthe loop is placed in uniform magnetic field of magnitude 0.400 T, what is the maximum torque that can act 0n it? The resistivity of copper is 1.70 * 10 * M: Deline Resistivity (5 marks)



Answers

A copper wire is $8.00 \mathrm{~m}$ long and has a cross-sectional area of $1.00 \times 10^{-4} \mathrm{~m}^{2}$. The wire forms a one-turn loop in the shape of square and is then connected to a battery that applies a potential difference of $0.100 \mathrm{~V}$. If the loop is placed in a uniform magnetic field of magnitude $0.400 \mathrm{~T}$, what is the maximum torque that can act on it? The resistivity of copper is $1.70 \times 10^{-8} \Omega \cdot \mathrm{m}$.

So for this problem, for part A, we can first write down that the maximum torque is going to be you be which is equal to I a B. Now the only thing that's really going to be a changing factor in this problem is the area. So if we have a one meter lum, a wire of length one meter, then if we turn it into a square, each side is going to be 0.25 meters. And so therefore this becomes 0.25 squared times I times be and this gives us a maximum torque of 0.8 0.78 Newton meters. And so this is the answer to part A. Now for part B. This is the same exact thing. Except so now we have t is equal to i b and now the area Well, we know that the circumference is going to be equal to one meter and the circumference is equal to two pi r. So the radius is equal to 1/2 pi, and this means that the area is equal to pi one over to buy squared, which is equal to 1/4 pi and so that torque is equal to you. I over four pi times be and that maximum tour comes out to 0.1 zero noon meters. And so the maximum torque is actually greater if it's a circle. And now lastly, we have part C. What we first have to do is write down that the torque is equal to mu cross be and remember that mu is actually equal to i times a times the normal vector to the loop. And so this means that technically, the torque the magnitude of the torque is going to be I times a a times b times these sign of the angle between the normal, which I'll just included here. Just so, um, we don't forget it the angle between the normal vector of the loop and the magnetic field. Now what we're looking for is to set this equal to the torque, the maximum torque on the square coil. So that's from part a. The maximum torque on the square. Poile is equal to 2.25 squared times. I times be so immediately we can cross out the eyes and the bees. So remember this is for the square and This is for the circle. So that means that you have to plug in for our area the area of the circle which was 1/4 pi. So we have 1/4 pi times. The sign of data is equal 2.25 squared. And so the data is equal to the inverse sign of four pi times 40.25 squared and this angle comes out to 51.6 degrees. And so that is the answer to part C.

In this problem, we are given the length of a wire, and we know that it's made into a coil of one turn. And you know the current going through that wire and the magnetic field with which the wire is suspended and then were asked to find with a maximum torque is on that one turn circular coil. Well, if we look at the equation for the Twerk on a coil in a magnetic field, it's given by the number of turns times that current through that coil times a cross sectional area that coil times the magnitude magnetic field in which the quail is suspended both bye bye sine phi or phi is the angle between the normal of our coil. So the normal sticking out of the cross sectional area of our coil. Uh, between that vector and the vector direction of our magnetic field. Now, when the Twerk is a maximum, or I should say the tour will be a maximum when Phi is 90 degrees or sign of five. Societal 90 degrees is one right. Remember, sine function varies from 0 to 1, and therefore, if we want the maximum amount of work we want to solve this equation for when sign If I is one or sign of 90 degrees is one or sign If I is one and that just goes away, we have an expression for the maximum amount of Twerk. That is just an I A B now were given much of the information we need. We know that it is only a one turn coil. We want to We know the current going through that coil, But we don't know this cross sectional area. We only know the link of the wire of this coil. So we know though we have some wire of length, will call it little L and that this wire is made into a circle. Right? So this wire is made into a circle of from a wire of length L we know that the circumference of that circle must equal the length of that, uh, wire. So from this, we can try to figure out what the area will be not in terms of the radius of the circle, but in terms of the circumference of the circle. So let's just do a little bit exploration algebraic lee to figure that out. So how do we do this right? Well, we want expression for the area cross sectional area of the circle. And that, of course, is just gonna be pi r squared. So if we knew the radius of this circle, we could figure out the cross sectional area. Of course, we don't quite know the Radius, but we do know a relationship between the circumference, which we know is l and the circumference of a circle, which is two pi times are so we can use this equation to solve for R in terms of the length of our wire, which is the length divided by two pi. And then we can plug that into the equation for our area. So if we do that right here, the area is going to be pie are all squared. But art is l over two pi When we a square All this relief for the pi times l squared divided by four pi squared so the area can be expressed in terms of the length of this wire and it's given by the length squared divided by four pi. So that is the area now in terms of something that we know and let me draw my pie a little bit better here. Four pine. And so that's gonna be a relationship between the cross sectional area and the length of the wire and the length of the wires. Something you know, Cool. So now I have an expression for the maximum Twerk, which is what we're trying to find here and now I know how to express my area in terms of the links from repress replace a with l squared over four pi. And then, of course, I have my magnetic field term here. So I plug all the numbers in the night now because I do know the current. It's 4.3 amps. Remember? I plugged in one for my number of turns already multiplied by 7.0 times 10 to the negative two meters all squared. That's the length of our wire. Divide that by four pi and then multiplied by r 2.50 Tesla magnetic field. So what puggle? That's in my calculator. What do I get in the end, I get an answer of about 4.19 times 10 to the negative three meters. That is the magnitude of the Twerk that this coil is feeling the maximum work that this coil experiences in this magnet

Let us assume that the length of its side of square is A. So the perimeter of the land will be equal to the perimeter of the square. L. Will be close to pour A. From here. You get a Z. equals two L. Development. For let's put the value of L. Yeah, This is one divided food. Well this will come out to me zero point 125 We do. No. Now we can find the area of the square. Lou. Area A. is equals two he Smalley square. That is also 0.125. The square, it will come out to be zero point 01 5 6 Military Square. This is the area of the square. Look. No, let us write the maximum torque which the loop this loop can experience. So let Corby Town now will be and I A. B. This is the maximum tow And is the number of times. That is one multiplied where I is the is the uh current that is 12 ampere multiplied. Where is the area of the loose square loops, europe and 0156 victory Square. And the B is the magnetic coalition that is zero one to islam. After calculating this bigot, the maximum torque is zero zero double 24 newton meters, double too four newton meters. This is the required value of mack center.

So I have a little set up here of our situation. We have wear that were given the total length of that is made into a square and put into a magnetic field, and it's a motor. So we have this axle running down the middle, and then this is set to rotate when it's in the magnetic field. So it rotates around the axle. Okay, so now we know the total length of this way, er, and it's wrapped into a square, which means that each side of the square is 1/4 l or L, divided by four. But because this magnetic field is parallel to two sides, we have this topside here in the bottom side, are parallel to be, which means that it's not gonna have any force in this direction. So it's not gonna have any contribution to the torque. The total torque, because we know that magnetic field relative to current involves a sign of fi or fada. In the forces case. There's an England wall, by the way, and it has to be perpendicular to each other. If they're parallel, they don't contribute any force. So our area that we need to take into consideration. We only have two sides. So this is the hard part for this problem is figuring out what we need to use for our area. So our equation for torque, which is what we're looking for, is number of turns times the current times, the area times the magnetic field sign off by we want when our torque is at its max, which means that we want our I wonder if I hear to be 90 degrees because that maximizes this equation every other value of five less or other than 90 degrees. So when when Smith advise 90 degrees, we have sign of one which is the highest value for sign. Every other value is gonna be less than one which will lower the value of this total work. So when sign is 19 year when the data are five, here is 90 degrees. We have our maximum we can get, Which is why Dr Drone Bee, I've drawn be perpendicular to two sides. It can only be Max for a total of being perpendicular to a total of two sides, so we could draw up and down and reversed his axle. But this direction works fine. for our maximum torque, which is perpendicular to two sides. So we have We have be we have end actually want to calculate for one end to end which simply change it to two after calculating what one end is so we also have be We need to find this area though, so area is a question mark in the moment. So we have We have our total length and we have that one side of the quarter out. But we only have two sides that contribute. So that means if we take our area which would be the inside of this this square here. So typically, if all sides contributed if this magnetic field say was into the page out of the page so if it was coming out or in it would be perpendicular toe all four sides are area would be equal to one side of the square squid. So this number squared so l divided by four, which is one side times another side. So this this would be another l over for because it's a square. So they're all the same link. So we times this side by this side to get her area, But we only have two sides that contribute to our overall torque, which means we need to divide that by two. So it's typical area cut in half using this number, and we also are given the current were able to calculate torque now using sign of 90 degrees, which is one. So this whole, this whole part drops out. So we have actually have to do this twice. We have Thao the first case and is equal to one, so we don't need to take that into consideration. We have the current times are area, which is this. So you take your l divided by four square that and then divide that number by two. That's your area and the magnetic field. And then for part two for two coils, that just means we plug in a twofer and instead of one. So this this you take what you get for part one for one coil and just multiply that by two for to end. And that will give you what two loops I would give you for torque. And there you have it


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