Hey, guys. So for this video, we're going to talk about a concept called the self-inductance. Now if you've seen our video on mutual inductance, this is going to be very, very similar to that discussion with a few small differences. And in case you haven't seen that video yet, pay attention because it's going to come up later. Let's check it out. So the whole idea of self-inductance and the whole thing here is that a current-carrying wire has the ability to induce an EMF on itself by changing its magnetic flux. 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 solenoid 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. That flux is just ba ∙ cosθ . What happens is that our magnetic field points off to the right and our area vector is going to 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 cosθ term just goes to 1. Alright. So what happens is we have the flux that depends on the current or sorry, we have the flux that depends on this magnetic field, but what happens is that's only just for one of the turns. If we wanted to figure out what the total amount of flux is, that would depend on n, which is the number of turns, and the magnetic field and 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 B 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 current. 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 see that φB is proportional to the currents. And all I'm saying here is that there is sort of like a mathematical 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 of the constant that goes out here is called L. And it's called the, it's called the self-inductance. And basically what the self-inductance represents is it's the ability for a current-carrying wire to generate or induce an emf on itself by changing its magnetic flux. That's really what it just represents. And the equation for that is pretty straightforward. We can actually get it from this, formula right here. So it's just going to be n ∙ βB ÷ I. And the units for that are given as Henrys. So if you've seen the video on mutual inductance, again, it's going to be the same exact thing. And in terms of more fundamental units, that's actually going to be a Weber per ampere. So we have flux in Webers, and we have amperes and currents on the bottom. Okay? So the next really important thing that you need to know is that it depends only 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 going to use this equation to calculate the self-inductance, 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 an example, the last thing I want to point out is that now we actually have a different formula we can use for the self-induced EMF. Emf. So we know that Faraday's law tells us that we can relate an emf with the number of turns and the change in the magnetic flux over the change in time. And we are perfectly okay to use that equation. But we can also write the self-induced EMF in terms of this self-inductance term that we just found out, and that's just going to be negative L ∙ ΔIΔt. 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 we wanted to figure out how each one of these things was changing with time, remember that n ∙ ΔβBΔt is the definition of what EMF is. And so both of these things, end up being equal to each other. So we just have a different expression that we can use for this EMF. Right? So if you have the change in flux over change in time, you can figure out what the EMF is. But now using this self-inductance, if you have the change in currents over change in time, you could also figure out what the EMF is. Alright? So that's basically it. Let's go ahead and check out an example. In this example, we're going to be calculating what the expression is for the self-inductance of a single current-carrying loop of wire. So if we want the self-inductance, remember that is going to be L. So L is our target variable here, and now we have 2 equations that involve L. 1 of them involves L and the self-induced EMF. That's going to be negative L ∙ ΔIΔt. But the thing is is that I don't have any information about the self EMF, and I have no information about how the current is changing over time. So this is not the equation I'm going to use. Instead, the other equation is that L = n ∙ βB ÷ I. Alright? 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 n = 1. So that means that sort of this just becomes 1 right here. And so L = βB ÷ I. So let's 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, then that means that this flux right here, this φB is equal to BA ∙ cosθ. But remember, because those things point in the same direction, this is just going to be 1. 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 what those equations are. So we have so Bloop × A. So remember that Bloop is going to be μ_{0} I ÷ 2 × little r where that's the radius. And then the area, the cross-sectional area right here is just the area of the circle, which is going to be π × r2. So we can just do some canceling out right here. We have an r2 on top and an r on the bottom, so we just cancel that out and leave 1 R remaining. And that means that our flux here, φB, is just equal to μ_{0} I × π × r ÷ 2. So now that we have an expression for the flux, in order to find out what the self-inductance is, we just need to plug this formula back inside of this. So that means that our self-inductance L is just going to be μ_{0} I × π × R ÷ 2, 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 currents will cancel out from the top and the bottom exactly like we said it would. And so this L basically just is μ_{0} I π oh, sorry. There's no I anymore. So it's just μ_{0} × π r ÷ 2. So it only basically just represents or it's only dependent on the number of coils or in the number of turns, which in this case was 1, and the shape of the coil itself. So we have that π × the r. It sort of is from, like, the circular shape of the loop. Okay? So this is the self-inductance for a coil of wire, and it's sort of like a physical property that just depends on that shape of the coil. Alright, guys. That's it for this one. Let me know if you have any questions.

# Self Inductance - Online Tutor, Practice Problems & Exam Prep

### Self Inductance

#### Video transcript

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?

^{−4}A/s

### Self-Inductance of a Toroid

#### Video transcript

Hey, guys. So let's check out this example. We're gonna work this one out together of calculating the self-inductance of a toroidal solenoid. So try and say that 5 times fast. Given some information about the geometry of a toroid, in the first part of the question, we're supposed to figure out what is the self-inductance of a toroid. And then in the second part, we're gonna figure out what the induced EMF is, assuming that the current is changing. Alright. So I want to take this actually really slow because remember that toroids can be pretty complicated shapes. But just remember that they're basically just big donuts of slinkies that have been wrapped up inside of each other. So that's actually pretty bad. Let's see if I can get this right on the 3rd try. There we go. So we've got this doughnut. And remember that a toroid 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. And I'm gonna try to do this as best as I can, sort of like that. That's pretty bad, but it's going to have to do. Alright? So we have the self-inductance. Let's see. So in part a, if we want to figure out what the self-inductance is, remember that it 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 of one of the little loops that it makes of one of the slinky turns. Right? So that's the cross-sectional area. And then the mean radius of 4 centimeters. Well, that's actually remember that the mean radius is not the radius in between the slinky like that. It's not the radius of the slinky itself. It is basically from the center of the donut out to the midpoint of that slinky. That is actually the mean radius. That's little r. And so we know that that little r is equal to 0 points, that's actually equal to 4 centimeters. So first, what I'm gonna do is I'm just gonna convert these. This is actually 6.25 times 10 to the minus 4 meters squared, because we have to apply that conversion from centimeters to meters twice. And then this is just equal to 0.04. Okay, cool. So now that we're done with the diagram and labeling all of our variables and all that stuff, what is the equation that's going to relate the self-inductance with all of these variables about the geometry of the coil? Well, remember that we can sort of relate all of these variables in together using the self-inductance 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 inductance always basically cancels out that current term in there. So all we have to do is just relate the φ, or just figure out an expression for the magnetic flux, which is BA cos(θ). So just a refresher, what does the magnetic field look like inside of a toroid? Well, it's basically kind of like a solenoid, but it always sort of points along the midline of that slinky. So it always basically goes around the center like that. So that B 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 because these things point along the same direction always. So that means that the magnetic flux right here, that's φ_{B}, is going to be Remember that this is going to be the magnetic field This is going to be the magnetic field of a toroid, and this is just the cross-sectional area which we're just given right here. Well, remember that for a toroid, you might have to look in your notes for this equation. The magnetic field is μ_{0} n i / 2π r, and then we just have the area right here. Okay? So what I can do is I can basically just now plug this whole expression for the flux back into this equation for the self-inductance. So that means that the self-inductance here is going to be n / i times this whole entire flux formula right here, which is equal to μ_{0} n ia / 2π r. I know it's kind of, kind of confusing there. There we go. And so we have this n 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-inductance here, in terms of actual numbers, is going to be 4π times 10 to the minus 7. That's the μ_{0} term. Then this n term actually gets multiplied twice because there's 2 of them. So I have 500 squared. Now the cross-sectional area is 6.25 times 10 to the minus 4, right? And then I have divided by, let's see, that's gonna be 2π times let's see, I actually don't need that parentheses. 2π times the mean radius, which is 0.04. And so if you go ahead and work this out in your calculators, you're gonna get a self-inductance of 7.81 times 10 to the minus 4 henries. Alright. So now that's the self-inductance of this toroidal solenoid. So you just relate it to the flux, the current term will cancel out, and it's just sort of like a property of this toroid. Cool. So now in the second part here, now that we know what this self-inductance is, if we have the current that is constantly decreasing, how can we relate this to the induced EMF of the coil? So, basically, what we're being asked for in this second part is what is epsilon of this L right here. So, sometimes you'll see this L for an inductor. What is ε_induced? Okay. So we know that the ΔI, the change in current is gonna be I_final - I_initial, which is going to be 2 amps - 5 amps. So in other words, the change in current was just equal to negative 3 amps. And we had the δ_time. So in other words, the change occurred over 3 milliseconds. So that's 0.003 seconds. Okay. So how do we get the inductance Or sorry. How do we get the self-induced EMF from the change in current over change in time? Well, remember that these variables here are related to the equation ε = -L ΔI/Δt. So let's see. We're trying to figure out what this induced EMF is, so we're trying to figure out what ε is. We know what the self-inductance is because we just calculated that in the last part. And now we have what the change in current over change in time is. So we have all of these variables. So that means that my ε_induced is going to be negative. Now, I have 7.81 times 10 to the minus 4. And now, I have the ΔI/Δt. So in other words, this is gonna be negative 3 divided by 0.003. What we'll see is that the two negative signs will actually cancel, and you should get an answer that is 0.78 volts. And that's our answer for the self-induced EMF. Alright, guys. That wraps this up. Let me know if you have any questions.

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