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The square loop in Figure $\mathrm{P} 20.62$ is made of wires with a total series resistance of 10.0$\Omega$ It is placed in a uniform $0.100-\mathrm{T}$ magnetic field directed per- pendicular into the plane of the paper. The loop, which is the paper. The loop, which is hinged at each corner, is pulled as shown until the separation between points $A$ and $B$ is 3.00 $\mathrm{m} .$ If this process takes $0.100 \mathrm{s},$ what is the average current generated in the loop? What is the direction of the current?

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So for this problem, we're asked to use figure p 20.26 which is made of wires that have a total resistance are of 10 homes. It's in a uniform magnetic field. So be is 0.1 Tesla and it's perpendicular into the plane of the paper that magnetic field is so the loop, which is the paper is hinged at some corner and is pulled has shown until the separation points between A and B is three meters. So I drew a little picture here, Um and what they mean by a M. B. So this is point A here and this point up here is point B and we're told that this is three meters. So then I call that distance here Dee and I say that that's equal to three meters and ah, we're told that it takes 30.1 seconds and it wants us to find the current generated in this loop in the direction of that current. So Delta T here is 0.1 seconds the distance here from B to the origin point at the center. I called that distance l that's gonna be equal to 1.5 meters. Okay, and then using Pythagorean theorem, we can then find side s. So according to stagnant here, um s would be equal to the square root, uh, D squared minus l squared. Playing those values into this expression, we find that this is equal to 2.6 meters. Okay? And again, we're asked to find the current. So to find the current, we could use homes law where I is equal to the IMF and you see a meth divided by the resistance. So therefore, we need to find absolutely to find that induced e m f. And we can do that using the expression where the induced ium F is equal to the change in the magnetic flux belt. If I divided by the change in time. Okay, so therefore, using this expression, we can find the induced E m f. And from the induced E m f. We can go ahead and find the current because we know the resistance. So let's start a new page here to make sure we have no firm to work this out. So the inducing and meth is going to be equal to minus the change of magnetic flux. The changing magnetic flux is equal to the magnetic field times the area. That's the flux divided by Delta T change in time, which we know cable. The magnetic field is constant. It's not changing. So this is minus magnetic field. The only thing that's changing is the area Delta, eh? Divided by Delta T. Okay, so now we just need to find the change in area because this is going to be equal to minus magnetic field multiplied by initial area minus final area. And then this is all divided by delta T. Okay, so then you just find the initial area in the final area, plugged those values into this expression when we found the induced M f. And then we can go ahead and find the current from that. So the final area is equal to 1/2 times D times two times the distance s that we found from the diagram zero. Plugging those values into this expression, we find this is equal to 7.8 square meters and of course, the initial here, it's just going to be equal to, you know, just the sense. It's a square D squared which is equal to nine meters square so plugging those values into this expression for the induced E m f as well as the values for Delta T In the magnetic field, we find that the induced you met is equal to 1.2. But the units here, volts. Okay, so now I'm going back to this page to plug in that 1.2 volts for that current, um, it's actually go ahead and do it over here, so we have enough for him to do that. We find that the current I again which is equal to that induced e m f, which we just found, divided by the resistance in the wires, which we were told us 10 homes. So that's 1.2 volts divided by 10 homes, which comes out to be zero 0.12 when the units of current are amperes. So we can box set into the solution to our question.