30. Induction and Inductance
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Hey, guys. So we saw how a current carrying wire can a deuce and current inside of another coil. That's just Faraday's law. When some problems, you're gonna need to know what the Mutual induct Ince's between two coils that we're gonna cover in this video. Now, just in case you've seen the video on self inductions, this is going to be very, very similar to that. If you haven't, that's perfectly OK, because you're gonna see it later. Let's check it out. So basically, the mutual induct INTs is just that for two conducting coils that are close to each other. A current through one coil current that's changing through one coil is gonna induce an E m F one occurrence on the other. And that z all this thing is now the coil with the changing current is often referred to as the primary coil. So that's the one with the current that's changing, and the other one is going to be the secondary coil. All right, so let's take a look at this diagram here that I have imagine that we have a coil like this and we have the current and I know this is sort of hard to visualize here, but the current is coming out towards you. So if use your right hand rule, you should be able to figure out the magnetic field goes to the right. In any case, don't let the direction of the magnetic field really trip you up. This is more of like a conceptual thing. So imagine this magnetic field points off in this direction. The fact that it goes through some area in coil to means that there is some magnetic flux and that magnetic flux is just equal to B times, a times the cosine of theta. Now what happens is I have two coils here, Coyle one and coil too. So I want to start labeling things properly. This is the magnetic field that is produced by coil one because it has a current going through it. And this is the area of coil to right. So what happens is that the total amount of flux through this second coil here depends on the number of turns, which I have right here, and it also depends on the magnetic field. Now, this magnetic field depends on the current that is going through coil one, so What happens is this magnetic field right here is actually proportional to the current that's going through coil one. So what we say is that the total amount of flux if it depends on the magnetic field and the magnetic field, which depends on the currents we just say that the total flux is proportional to call current one. So all we're saying here is that the total amount of flux through the second coil is actually proportional to the amount of current that is going through coil one. So whenever we have a proportionality, there could be multiple Constance that go out here all right. It's just a proportionality constant. And so what we say is that the total amount of turns times the total amount of flux through the second coil is actually equal to some proportionality constant called m times, the currents in that first coil. And that's what mutual inducting is this mutual inductions. M This letter M here is just a proportionality constant, which is called the mutual induct INTs. It's sort of just describes the relationship between two coils in which you change your current in one, and it induces a current and the other. Now we could get from this equation here that if you just if you just move over this current to the other side, we get that. The equation for mutual induct INTs is end to times the flux into divided by the currents and the units For this mutual induct, INTs are given as Henry's, which is just this capital letter h here on more fundamental units. Um, it's just the flux, which is given as a Weber, remember? And then we have the current on the bottom, which is just an ampere. And the most important thing that you need to know about the mutual induct INTs is that it depends on Lee on the number of turns and the shape of the coils. So what I'm saying is that it doesn't depend on the current Now, I know this sounds crazy because I had clearly have a current right here in this formula, but we're going to see that the currents this I one that's on the denominator will always cancel out. So let's go ahead and check out an example in which we have to solo noise, were given some letters and some some information about those solar noise. We need to find out what the mutual induct Ince's. Alright, so remember, If we're trying to find out the mutual inductions formula, that's always just gonna be this end to fight to over. I won. So we're just gonna be the number of turns times the total amount of flux divided by the total amount of current. So this is just a number right here, this end to and that's just given to us right here. By the way, we're just working with letters here. There's no numbers. So what happens is we have this soul annoyed that sort of wrapped up inside of another one. So this short solenoid right here is actually coiled, too. And this long solenoid over here is gonna be coil one. So let's say we had some sort of current that is going through this soul annoyed, and it's producing some magnetic fields. Now again, the direction of this magnetic field doesn't really matter. That's not the important thing. So let's just imagine that the magnetic field goes this way through this soul annoyed, and that's RB field. That's kind of hard to see. So you have some Byfield like this. Now, if you vary the current in I one, it's going to change and induced a current in coil to because the flux is changing. Right. So let's see this flux This given this flux over here is just given as let's see, we have end to and now we have the Byfield produced by coil one, the area through coil to and then the coastline of the angle. Now we can see here is that the magnetic field goes in this direction sort of down the solenoid and the area vectors of each of these things also point in the same exact direction. So what that means is that this coastline area goes away. This cosine theta term goes away because data is equal to zero degrees, right? Okay. And then we have to just divide this by the currents. Now, what we have to dio is we have end to We have the magnetic field, the area and the current. So we have to relate this to the length variables and and one. So remember, we're gonna have to get an expression here for this magnetic field. And this is the magnetic field generated by a soul annoyed. So the magnetic field that is generated by a solenoid We have an equation for that. And I know this is just this is a couple of videos back. We're gonna remember that the magnetic field that's generated by a solenoid is mu not times, little end times the current. But remember that this is the current in I one just the current that's traveling through the primary coil. So we actually have a different expression that we can use for this magnetic field because this little N is the number of turns per length. So we actually write this rewrite this as mu not times and one capital and one divided by the length times the current. So all we have to dio is just plug this expression back inside for this magnetic flux equation we can keep on. We can keep sort of eliminating some variables. So we have the mutual induct INTs is going to be, and two and now we have to plug in this expression over here. This is gonna be mu knots and one over l. A times I won and now we have the area of the second one this coastline term went away and then we just have to divide it by the current in one. Now we can see here is that the current I one will cancel out with the current I one that's in the denominator. And if you rewrite this, we're gonna get the mutual induct. INTs is equal to, Let's see mutual inducting is gonna be you knots and one and two times a two over l. Because remember this Ellis still on the bottom here, Now we can see is that the mutual induct INTs depends on on Lee the properties of the coils like the number of turns and the shape. It doesn't depend on the current, which is exactly what we said over here. This I won will always cancel out because the magnetic field B one depends on the current that's going through it, right, So this is sort of just like a physical property of the coils themselves. Now the last thing I want to mention here is that the IMF that's generated on the secondary coil can be written now in terms of the mutual induct INTs. Now we know from Faraday's law that the Epsilon is equal to n times Delta Phi over Delta T. And this negative sign really just represents lenses law, which is that it does the opposite of whatever the changes. Well, this can actually be written in terms of the mutual induct INTs by negative m times, Delta I over Delta T. And the way that we get this equation just really quickly is just from the equation that we had at the top of the page, where the total flux and to fi two is equal to m times I. So all we do is if this flux is changing and we just divided by Delta t, then this just becomes end two times Delta Phi over Delta t. But remember, if we're gonna dio if we're gonna divide the left side by Delta T, then we have to divide the right side by Delta t. So this just becomes Delta I over Delta T. And then these two things are basically just the same. So now it happens is that we have another equation that we can use to relate. The mutual induct INTs to the induced EMF on a secondary coil. And that's all that's all mutual inducted is It's just a property that describes relationship of changing currents and induced EMFs in a primary and secondary coil. All right, let's just do another example right here just to sort of drive everything home. So we have a solenoid of 25 turns and some area right here with another soul annoyed. So it's basically the same exact set up that we have here. So if it's some time, the current through the 10 centimeter solenoid is 100.5 amps and it's changing at 50 amps per second. Then what is the induced E M f on the 25 turn solenoid? Okay, so we have this, uh, this solenoid of 25 turns. That's this guy right here. It's wound around a 10 centimeter solenoid with 50 turns. So we know this is the larger one. That's gonna be the 50 turns, and we need to figure out what the induced E. M f is. So we have our induced CMF equation is gonna be negative n times Delta Phi over Delta T. Or we can relate this to the mutual induct INTs, which is negative. M times Delta I over Delta t. So let's take a look at our variables right here. So let's see, We have no information about the change in flux, but we do have an information about the amount of current that's changing. This current that's changing is actually the 50 Miller amps per second. So that means that this delta over Delta T is actually this guy right here. So we're gonna use the sort of right side off this equation instead of the normal Faraday's law. So we just have that the mutual, the induced e m f is gonna be negative. Negative M times Delta I over Delta T. Now all we have to do is figure out what the mutual inducting is. But we have to solo noise that we're working with, and we actually know what the mutual inducted is. We actually just came up with that expression right over here so we could just plug that in. So the induced EMF is going to be remember that this M term this mutual inductions between two solo noise is equal to mu knots and one and two area two divided by l. So that means that it's mu knots, which is equal to four pi times 10 to the minus seven now, and one is going to be this. 50 turns here now, and two is gonna be 25. And now we just have to figure out what the area is off that second solenoid. Well, the area of the second solenoid here is actually just given to us. And that's equal 2.5 Um, right there and then we have two divided by the length off that longer. Soledad, remember, the length of l is the length of the longer one. So that means this l here is going to be 10 centimeters, which is equal to 0.1. So that means that this mutual induct INTs is just equal to 7 times 10 to the minus five. And that's in Henry's. So now we can do is plug this expression back into our induced EMF equation. So this is just going to be negative 7.85 times 10 to the minus five. That's Henry's. And then we have to divide Multiply this by the changing current over change in time. That's 50 mg per second. So that means that is 0.5 amps per one second right So that's just what that expression means. That's Delta I over Delta T. So that means our induced E m f is going to be 3.93 times 10 to the minus six, and that is in volts. Alright, so that is our answer for the induced MF. And that's what this mutual inducting stuff is all about. Let me know. If you guys have any questions, we'll take a look. A couple more practice problems.
An outer solenoid with 30 turns is wound tightly around an inner coil 25cm long with a diameter of 4cm and 300 turns. The current in the inner solenoid is 0.12 A and is increasing at a rate of 1.75×103A/s. a) What is the average magnetic flux through each turn of the outer coil? b) If the resistance of the outer coil is 20mΩ, what is the magnitude of the induced current through the outer coil?
(Note:for the multiple choice, select the correct answer for part (b) only.)
(b) iind = 0.17 A
(b) iind = 5.0 A
(b) iind = 20 A
(b) iind = 150 A