Shrinking Loop. A circular loop of flexible iron wire has an init...

Question

Shrinking Loop. A circular loop of flexible iron wire has an initial circumference of 165.0 cm, but its circumference is decreasing at a constant rate of 12.0 cm/s due to a tangential pull on the wire. The loop is in a constant, uniform magnetic field oriented perpendicular to the plane of the loop and with magnitude 0.500 T. (a) Find the emf induced in the loop at the instant when 9.0 s have passed.

Accepted Answer

Everyone today, we are going to determine the induced E. M. F. In the loop after 11 seconds. Where in this particular practice problem, we will have a very stretchable and high conductivity conducting polymer based wire where the wire is stretched into a circular loop of circumference two m. And it's placed where it's plane perpendicular to a uniform to 0.2 Tesla magnetic field. And when released the circumference of the loop actually will start to shrink at a uniform rate of 15 centimeters per second. So we want to determine to induce E. M. F. In the loop after 11 seconds. So the way we want to calculate this is two by utilizing Faraday's law. So by utilizing Faraday's law, we can actually calculate the E. M. F. Both the magnitude and the direction. Because we know in this particular problem that we will have an induced E. M. F. Because the flux actually changes because the circumference or the area of the loop is also changing. So recall that um I'm just gonna absolute value everything here just because we don't really care about the direction. But recall that you uh with Faraday's law, the E. M. F. Really close to the change in the magnetic flux over D. T. And also recall that magnetic flux can also be calculated by multiplying the magnetic field with its area. So, by substituting this formula into this, then we will get d be multiplied by a over D. T. Well, in this practice or in this problem we know that the magnetic field is going to be uniform and concept 0.2 Tesla. So that is a constant value that we can just bring out. So this will just then be be multiplied by D. A. Over D. T. Just like so okay, so we know how to actually calculate the induced E. M. F. Using Faraday's law. And we are actually given that the circumference is changing. So what we need to do now is to relate this D. A. Over D. T. We need to relate the A. Over D. T. With the change that we are experiencing. Which is the change in circumference to D. C. Over D. T. With C being the circumference. So now we know that the A. Is just pi R. Squared. And we also know that the circumference is just to buy our, what we want to do is to actually express the area in terms of C. The way we want to do that is to buy, substitute this are here with an equation of our here that contains a C. So using this formula here, we know that the R. Is going to be C. Over two pi just through rearrangements. So then we can actually Um substitute this formula here to this equation and get an area of pi multiplied by c. over two pi squared. And therefore from this we can conclude that the A. Is going to then be C. Square over four pi. And that will be the area. In terms of circumference here. Now what we need to realize D over DT so we want to plug this into D over DT formula. So the A over D. T. Will equals two D. Over DT times C squared over four pi. We know that the four pi is constant, so we can pull it out, this is going to be 1/4 pi D. C. The over D. T multiplied by C squared. So using the chain rule and the derivative um formula that we have previously discovered, we want to express this just so that we have this formula here in terms of D. C. Over D. T. So this will then be see Multiple by 2/4 pi multiplied by D. C. Over D. T. Just like, so, so simplifying everything, we can get the over D. T. To equal C. Over two pi multiplied by D. C. Over D. T. Just like, so, okay, so now that we find this D. A. Over D. T. In terms of D. C. Over D. T. We can plot this value into the induced E. M. F. Formula here to get the induced E. M. F value. So the nuc M F. Is being multiplied by D. A. Over D. T. And that will equals two, essentially be multiplied by C over two pi multiplied by the C. Over D. T. And that is essentially just plugging in our D. C. Over D. T. So from here, we want to then revisit our problem statement where it says that we want to determine to induce EMF in the loop after exactly 11 seconds. So when D equals 11 seconds, does your conference then we should actually calculate that because we want to know what this D C over D D value is. So the circumference is going to be our initial minus time passed, Which is 11 seconds, multiplied by the rate of the shrink. And that will be our initial, which is two m two m. Our initial state conference minus the time pass, which is 11 seconds, multiplied by the rate, which is 0.15 m per seconds. And these are all given in the problem statement. So at the equals 11 seconds, the circumference will then be um 0. m So we can actually plug all this value into our equation here. Be multiplied by C over two pi multiplied by D C over D T equals. This is going to be 0.20 Tesla given in the problem statement, the C. Is the one that we just found which is 0.35 m over two pi. The D C over D. T. Is given, which is 0.15 m per second. And therefore the induced EMF is then going to be 1.67 million volt or 0. fold. Just like. So, so that will be the answer to our problem, which is 0.0167 fold, or 1.67 million fold, which actually corresponds to option D. So D will be our answer to this problem. And if you guys have any sort of confusion still in this particular problem, make sure to check out our other lesson videos regarding this topic. And that will be all for this particular problem. Thank you.

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