1

concept

## Self Inductance

7m

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Hey, guys, over this video, we're gonna talk about a concept called the self induct Inst. Now, if you've seen our video on mutual induct into this is gonna be very, very similar to that discussion with a few small differences. In case you haven't seen that video yet, pay attention because it's gonna come up later. Let's check it out. So the whole idea of self induct INTs on the whole, the whole thing here is that a current carrying wire has the ability to induce an IMF on itself by changing its magnetic flux. So let's take a look at how that happens here. I have a single coil of wire and it turns into a loop like this, right? So it turns into sort of like a solar like kind of thing. So if this solenoid looking thing has a current going through it, it generates a magnetic field. And if this magnetic field is passing through a surface, then that means that we have a magnetic flux and that flux is just be a times the coastline of data. And what happens is that our magnetic field points off to the right and our area vector is gonna be perpendicular to that surface. So that also points to the right like this. And because those two things point along the same exact direction that cosine theta term just goes toe one. All right, so what happens is we have the flux that depends on the currents, or so we have the flux. That depends on this magnetic field. But what happens is that Onley just for one of the turns, if we wanted to figure out what the total amount of flux is that would depend on end, which is the number of turns and the magnetic field in the area. The problem is that this magnetic field also depends on another variable. Remember that this magnetic field depends on the amount of current that passes through it. So be depends on I. So let's see what's happening here. We have the flux that depends on the magnetic field, but the magnetic field depends on the currents. So that means that there is a relationship between the flux and the current as well. So basically, what I'm just saying is that there is a proportionality. So we say that five B is proportional to the current and all I'm saying here is that there is sort of like a mathematical relationship between the flux and the currents. And whenever we have a proportion, there is one or more constants that could go out here in the front. And if you were to sort of write out an equation for this, this just turns out to be n on the left, right, so that's the total amount of flux is equal to I. But the proportionality, the content that goes out here is called L. And it's called the It's called the self induct INTs and basically with self inductions represents is it's the ability for a current carrying wire to generate or induce an E m F on itself by changing its magnetic flux. That's really what it just represents. The equation for that is pretty straightforward. We can actually get it from this formula right here, so it's just gonna be end times the magnetic flux divided by the currents and the units for that are given as Henry's. So if you've seen the video on mutual induct, it's again. It's gonna be the same exact thing. And in terms of more fundamental units, that's actually gonna be a Weber. Per and Pierce, we have flux and Webber's We have amperes and current on the bottom. Okay, so the next really important thing that you need to know is that it depends on Lee on the number of turns, which is n and it depends on the shape of the coil, which is going to give us that flux value. So what happens is we're going to see that this current value is always going to cancel out. So even though we're gonna use this equation to calculate the self induct INTs, we're going to see that the current will cancel out. And this l is only sort of like a physical property of the coil. Now, before we get into it, um, example the last thing I wanna point out is that now we actually have a different formula we can use for the self induced e m. F. So we know that Faraday's law tells us that we can relate an e m f with a number of turns and the change in the magnetic flux over the change in time. And we're perfectly okay to use that equation. But we can also write the self induced MF in terms of this self inducted term that we just found out and that's just gonna be negative. L times Delta I over Delta T And just in case you need to know where this equation comes from, it actually just comes from this relationship right here. So if we were to just divide both sides by Delta t So if you wanted to figure out how each one of these things was changing with time, remember that end times the number of end times the change influx over the change in time is the definition of what e. M. F is. And so both of these things end up being equal to each other. So we just have a different expression that you can use for this e m f Right. So if you have the change influx over change in time, you could figure out what the e. M f s. But now, using this self induct INTs, if you have the changing currents over change in time, you could also figure out what the MFS Alright, that's basically it. Let's go ahead and check out an example and this example we gonna be calculating what the expression is for the self induct INTs off a single current carrying loop of wire. So if you want the self induct INTs, remember, that is going to be l So l is our target variable here and now we have two equations that involved l One of them involves l and the self induced PMF. That's gonna be negative. L times Delta I over Delta T. But the thing is, is that I don't have any information about the self induced CMF and I have no information about how the current is changing over time. So this is not the equation I'm gonna use instead. The other equation is that l is equal to end times the magnetic flux divided by the current. All right, so let's go ahead and work with this. We have a single current carrying loop of wire, so if it's a single loop, that just means that end is equal to one. So that means that sort of this just becomes one right here. And so l is just equal to the magnetic flux divided by the currents. So take a look at here, Right, So we have a loop of wire that has a current that's going through it. So using our right hand rule, if you were to take your right fingers and curl them in the direction of this, your thumb should be pointing into the page. So that means that the magnetic field points in this direction. And if it's a loop of wire like this, then that means that the area vector also points in the same direction, sort of into the page. So that means that both of these things, because they sort of point along the same direction away from you into the page. That means that this flux right here, this fiber be is equal to be a times the co sign of data. But remember, because those things point in the same direction, this is just gonna be one. So that means the flux is just the magnetic field times the area. But for a loop of wire, we can actually figure out what that magnetic field is. So let's go ahead and write out. Those equations are So we have so be for a loop of wire. Someone writes, be loop times a. So remember that be for a loop of wire is going to be Munitz. I divided by two times. Little are where that's the radius and in the area. The cross sectional area right here is just the area of the circle, which is gonna be pi times little r squared so we could just do some canceling out right here. We have an R squared on the top and are on the bottom, so you just cancel that out and leave One are remaining. And that means that our flux here, Phoebe, is just equal to you, not I Times pi times are divided by two. So now that we have an expression for the flux in order to find out what the self inducted is, we just need to plug this formula back inside of this. So that means that our self inducted l is just going to be mu knots. I times pi times are over two. That's the self. That's the magnetic flux that we just found divided by the currents. So what we see here is that the current actually cancel out from the top and the bottom exactly like we said it would. And so this l basically just is immune not I pi Oh, sorry. There's no I anymore. So it's just Munitz times pi r divided by two. So it on Lee basically just represents or it's on Lee, dependent on the number of coils or in the number of turns, which in this case was one and the shape of the coil itself. So we have that pi times the are It's sort of this from, like the circular shape of the loop. Okay, so this is the self induct its for a coil of wire. And it's sort of like a physical property that just depends on that shape of the coil. Our guys. That's it for this one. Let me know if you have any questions.

2

Problem

A single loop of wire with a current of 0.3A produces a flux of 0.005 Wb. If the self-induced EMF on this loop is 10 mV, how quickly must the current be changing?

A

1.67×10

^{−4}A/sB

0.60 A/s

C

1.67 A/s

D

60 A/s

3

example

## Self-Inductance of a Toroid

7m

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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.

Additional resources for Self Inductance

PRACTICE PROBLEMS AND ACTIVITIES (4)

- A long, straight solenoid has 800 turns. When the current in the solenoid is 2.90 A, the average flux through ...
- At the instant when the current in an inductor is increasing at a rate of 0.0640 A/s, the magnitude of the sel...
- At the instant when the current in an inductor is increasing at a rate of 0.0640 A/s, the magnitude of the sel...
- A 2.50-mH toroidal solenoid has an average radius of 6.00 cm and a cross-sectional area of 2.00 cm^2. (a) How ...