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Concept #1: Self Inductance

**Transcript**

Hey guys. In this video we're going to be talking about the concept of self inductance, which is related to mutual inductance. Alright, as we'll see, let's get to it. Now, when we consider two coils brought close together, if the current was changing in one coil it induces an EMF on second coil, however current running through a single coil, if that current is changing can change the magnetic flux through the single coil, which can induce an EMF on itself, this is known as a self induced EMF and this process is known as self induction, okay? Because it occurs on itself. Alright, we're going to define that the self inductance or the coefficient of self inductance exactly like, we defined the coefficient of mutual inductance. Alright, except there are no longer two coils to worry about it's all one coil, we're going to say the total flux through the one coil, which is just the number of turns times the magnetic flux through each turn divided by the current carried by that single coil, okay? The units here are still Henrys because this is still flux divided by current, okay? Now, the self induced EMF can then be described in terms of this self inductance, which is sometimes just referred to as inductance, okay? And, this is done exactly like, we did it with mutual inductance, except in this case we're saying that the self inductance times the rate at which the current is changing. Alright, no big deal, a very, very similar concept of mutual induction.

Let's do a couple examples, what is the self inductance of a coil of wire with N turns and a radius R, okay? So, self inductance as we know is the number of terms times the flux through each loop divided by the current in that circuit, okay? We know N, I is a variable, we're assuming and it's going to cancel anyway. Remember, that these inductance values have to be physical properties of the inductor, of the objects that undergo the inductance themselves, okay? So, the current should cancel, all we then need to figure out is, what the heck the magnetic flux is through each loop, okay? And the magnetic flux is just going to be the magnetic field times the area, okay? The area is easy, if the radius is R, the area is pi, R squared but what is the magnetic field? Well, the magnetic fields due to a single loop of wire is going to be mu naught times the current divided by two times the radius but we're talking about n loops of wire, each of them stacked on top of each other contributes a quantity to the magnetic field of mu naught, I over 2R, so the total magnetic field is actually in mu naught I over 2R to account for that in the number of loops that each contributes a magnetic field. So, this tells us then that the flux is n, mu naught I over 2R times pi, R squared, okay? And we can lose a factor of R and this can just become 2, I'm sorry, N mu naught, I over 2 times pi, R, okay? Now, that we know the flux, we can plug that back in to our equation for the self inductance. So, this whole thing leads us to say that the self inductance is just N times N mu naught I over 2, pi, R, right? That's the flux, the magnetic field times the area, divided by the current I, okay? And notice exactly, what we need the current cancels. Now, it's just a matter of prettifying it however you want, okay? All the variables are there, just represent it in the way that seems to make the most sense to you I'm just going to write it as mu naught, N squared, pi, R over 2, okay? No big deal, it's a straight up application of this formula exactly like we did before with mutual inductance, okay? Let's do one last example. A solenoid is 500 turns with the current of half an amp producing an average flux from each coil of 0.005 Weathers, if the self induced EMF on the solenoid is 10 millivolts, how quickly must the current be changing, okay? So, here what we're talking about is self induced EMF and we want to know how quickly must the current be changing, we know that self induction gives us an equation to relate the two, it says that the EMF induced on itself, the self induced EMF is going to be the inductance times the rate at which the current is changing. So, what we want to find out is the rate at which the current is changing right, sorry, let me scroll down a little bit, and that is just going to be the self induced EMF over the inductance, the self induced EMF were given, we're told it's 10 millivolts but we do not know the inductance, the self inductance, that we have to calculate. So, we'll use the formula that we were given, the number of loops times the flux through each loop divided by the current, okay? Now, the number of loops is 500, right? It says, the solenoid is 500 turns, the average flux through each loop is 0.005 Webers and the current in the circuit is half an amp, sorry, not 0.05 just 0.5 amps. So, this whole thing gives us an inductance of five Henrys. Now, we can take this inductance and we can plug it into this equation to figure out how quickly the current is changing. Remember, once again, the self induced EMF is 10 millivolts or 0.01 volts divided by 5 henrys, which is the inductance that equals to, sorry, instead equals 9.002 amps per second, okay? Now, we can just move the decimal place. So, we can call this 2 milliamps per second either answer is completely okay. Alright guys, that wraps up our discussion on self inductance, thanks for watching.

Practice: A circular coil of wire with 20 turns has a current changing at a rate of 0.12 A/s. If the radius of the coil is 20 cm, what is the induced EMF on the coil?

0 of 2 completed

A long straight solenoid has 800 turns. When the current in the solenoid is increasing at a constant rate of +0.20 A/s the emf induced in the coil is 0.060 V. What is the self-inductance of the solenoid?

A long straight solenoid has 800 turns. When the current in the solenoid is increasing at a constant rate of +0.20 A/s the emf induced in the coil is 0.060 V. When the current in the solenoid is 15.0 A, what is the average magnetic flux through each turn of the solenoid?

At an instant when the current through an inductor is I = 4.0 A and the current through the inductor is increasing at a rate of Δi/Δt = +0.50 A/s, the voltage across the inductor is 8.0 V. The resistance of the inductor is negligible. What is the self-inductance of the inductor?
(a) 2.0 H
(b) 32.0 H
(c) 16.0 H
(d) 8.0 H
(e) 4.0 H
(f) none of these answers

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