by Patrick Ford

Hey, guys. So let's check out this example. We're gonna work this one out together of calculating the self induct INTs of a two royal solenoid. So try saying that five times fast. So given some information about the geometry of a tor oId in the first part of the question, we're supposed to figure out what is the self inducted of it Tor oId. And in the second part, we're gonna figure out what the induced E. M. F is assuming that that's, um, current is changing. All right, so I want to take this actually really slow because one of the Tories could be pretty complicated shapes, which remember that they're basically just big doughnuts of slink ease that sort of been wrapped up inside of each other. So that's actually pretty bad. Um, let's see if I could get this right on the third try. There we go. So you've got this doughnut. Remember that? The Tauride is basically like if you took a slinky with a bunch of coils and you sort of wrapped it around itself. So you have a bunch of coils like this. I'm gonna try to do this as best as I can sort of like, that's That's pretty bad, but it's gonna have to dio All right, so we have the self induct INTs. Uh, let's see. So in part A. If we want to figure out what the self inducted is, remember that is L. And we have some information, like the number of turns, which is 500. The cross sectional area, which is 6.25 centimeters squared. So we have that we have n number of turns, which is 500. And then the cross sectional area represents the area off one of the little loops that it makes of one of the little slinky turns. Right? So that's the cross sectional area, and then the mean radius of four centimeters. Well, that's actually remember that The mean radius is not the radius in between The Slinky like that's not the radius of the slinky itself. It is basically from the center of the doughnut, out to the midpoint of that slinky that is actually the mean radius. That's little are, and so we know that that little R is equal to zero points. That's like equal to four centimeters. So first, what I'm gonna do is. I'm just gonna convert these. This is actually 6.2 25 times 10 to the minus 4 m meters squared because we have to apply that conversion from centimeters 2 m twice. And then this is just equal to 0.4 Okay, cool. So now, now that we're done with the diagram and labeling all of our variables and all that stuff, what if the equation that's going to relate the self induct INTs with all of these variables about the geometry of the coil? Or remember that we can sort of relate all of these variables and together using self inductions formula, which is the number of turns times the flux divided by the currents. So we know what the number of turns is. And we know that this current, even though if we don't have it in our equation or in our problem, it's going to cancel out because the induct INTs always basically cancels out that current term in there. So all we have to do is just relates the fi or just figure out an expression for the magnetic flux, which is be a times the co sign off data so just a refresher. What is the magnetic field look like inside of a tour? OId Well, it's basically kind of like a soul annoyed, but it always sort of points along the midline off that slinky. So it always basically goes around the center like that so that that be term would just go around like this. And because we're talking about the area, the area is going to be the area of the cross sections of one of the loops right here. So that means that always at all times, the area vector always points along with the magnetic field. So they always basically just go follow each other like that. So what that means is that the cosine term will go to just one, because these things points along the same direction, always. So that means that the magnetic flux right here that's fi B is going to be Or remember that this is gonna be the magnetic field is gonna be the magnetic field of a Tor oId. And this is just the cross sectional area which were just given right here. We actually know what that cross sectional area is. Well, remember that for a tor oId you might have to look in your notes for this equation. The magnetic field is mu not times the number of turns times the current divided by two pi times the mean radius on. Then we just have the area right here. Okay, So what I can do is I could basically just now plug this whole expression for the flux back into this equation for the self induct INTs. So that means that the self induct its here is gonna be n divided by I times this whole tire flux formula right here, which is equal to mu knots times n i over two pi r and then I have a right here. Right. So I have, you know, I over two pi r you know, that's kind of ah, kind of confusing there. There we go. And so we have this end that pops up twice, So it's gonna pick up a squared, and just like we expected, the current term will go away because it's in the top and the bottom. So that means that the self inductions here, in terms of actual numbers, is going to be four pi times 10 to the minus seven That's the mu, not term. Then this end term actually gets multiplied twice because there's two of them. So I have 500 squared now. The cross sectional area is 6.2 25 times 10 to the minus four, right? And then I have divided by Let's see, that's gonna be two pi times. Let's see, actually, don't need that parentheses. Two pi times the mean radius, which is 20.4 And so, if you go ahead and work this out in your calculators, you're gonna get a self induct. INTs off 7.81 times. 10 to the minus four. Henry's right. So now that's the self induct INTs off this to royal soul. Annoyed. So you just related to the flux. The cancel the current term will cancel out. And it's just sort of like a property off this tour. OId cool. So now in the second part here, now that we know what this self inducted is, if we have the current that is constantly decreasing, how can relate this to the induced E m f of the coil? So basically what we're being asked for in this second part is what is Epsilon off this l right here. So sometimes you'll see this Alfred and Dr what is Epsilon induced? Okay, so we know that the Delta I the change in current is gonna be I final minus I initial, which is going to be two amps, minus five amps. In other words, the changing current was just equal to negative three amps and we have the number of the delta time. So, in other words, the change occurred over three milliseconds, So that 0.3 seconds Okay, so how do we get the induct INTs or Sorry, How do we get this self induced E m f from the change in current over change in time? Remember that these these variables here are related to the equation. Epsilon is equal to negative. L times Delta II over Delta T. So let's see, we're trying to figure out what this induced e m f is. We're trying to figure out what Absalon is. We know what the self inducted is because we just calculated that in the last part, and now we have what the changing current over changing time is. So we have all of these variables so that means that my epsilon induced is going to be negative. Now I have 7. 81 times 10 to the minus four, and now I have the Delta I over Delta T. So, in other words, this is gonna be negative. Three divided by 0.3 What we'll see is that the two negative signs will actually canceled, and you should get an answer. That is 0.78 volts. And that's our answer for the self induced E M F our guys that wraps this up, let me know if you have any questions.

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