32. Electromagnetic Waves

Displacement Current and Maxwell's Equations

# Displacement Current and Maxwell's Equations

Patrick Ford

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Hey, guys, in this video, I want to talk about something called Displacement Current and Maxwell's Equations. Okay, most of electromagnetism was done throughout the early 18 hundreds and in near the 18 sixties or so. I don't remember exactly when there was a guy named James Clerk Maxwell, who sort of put everything together. Gas is Law, Faraday's Law and Peers Law. Everything that we've seen so far. But he did make one small change toe amperes law, and that is the invention or not the invention because he didn't create physics but the discovery and the inclusion of the displacement current. Okay, so that's what we want to talk about in this video. Let's get to it. Okay, now, amperes law as we know it says that the integral around some sort of closed loop this should technically read this. Okay, but you guys know what it says is on Lee, dependent upon what I have called I subsea the conducting, the current A current forms by physical charges moving. Okay, there's actually a second current. I specifically put in air quotes because it's not a real current. It's not a current composed of conducting charges, but it carries the same unit as current, and it acts like a current. Okay, The second current can also produce a magnetic field. According to amperes law, this second current is known as the displacement current. Okay, And this is what James Clerk Maxwell was known for. The displacement current is given by the permitted Vitti of free space times the rate at which the electric flux is changing with time. So it's very similar to Faraday's law, right? The rate at which the magnetic flux was changing with time. So everything that we learned when we were talking about Faraday's law, how to find the magnetic flux, how to find the change in the magnetic flux with time, we're gonna apply that to the displacement current. But now it is for the electric field. Okay, so let's do a quick example. This was the sort of classic, like, quintessential problem where they realized that there was this sort of imaginary current that didn't exist that was producing a magnetic field. If you have an actual conducting current going to a capacitor plates and an actual conducting current leaving a capacitor plate, what is the magnet? The magnitude of the magnetic field produced between the two capacitor plates, as shown below. Okay, so as the conducting current comes towards that, uh, capacitor plate, it's depositing positive charges, right? And as the current leaves the other capacitor plate, it's carrying positive charges away, and it's leaving it negative. Obviously, I said this backwards. Okay, that's from the view of the positive charges, which we know is wrong. Okay, But either way, as positive charges buildup on one plate and negative charges build up on the other, an electric field is growing between these plates. Let me draw this one here a little bit differently. An electric field is growing between these plates. That electric field is definitely producing a changing magnetic flux through some arbitrary ring. Whatever you decide to measure the magnetic flux through. Okay, so we're going to apply amperes law using this imaginary displacement current, okay? And we're going to say that this integral around an imaginary loop of radius r equals mu, not times the displacement current, which is mu not times epsilon, not times the rate at which the electric flux is changing with time. Okay, so welcome. First we have to figure out how quickly is the electric flux changing with time. Before you can even do that, you need to know what you're measuring the flux through all of this And NPR's law deals with that imaginary imperial loop that I drew in green. That imaginary imperial loop which has a radius. Little are that is the surface that you're measuring the flux through. Okay, so first, let's measure the electric flux. This is going to be the electric field times the area of our imperial loop. Okay, Now, what is the area of that loop? What is the electric field between the parallel play capacitors? Alright, I'm gonna remind myself for this the area of the loop is pretty easy. The radius of the loop is little r squared. Do not put capital, our capital ours what I reserved for the radius of the capacitor plates. We're considering the flux through this imaginary imperial loop, just like we would consider how much current was contained in the imaginary imperial loop. Okay. And the next thing is, what's the electric field? While the electric fields between any two parallel plates is the voltage across them divided by the distance, the voltage in a capacitor as we know is always the charge on that capacitor divided by the capacitance. This is the charge divided by the capacitance times the distance. Now, the capacitance of a parallel plate capacitor is absolutely not times the area of that capacitor over deep. So I'm just gonna plug that in. This is Q divided by excellent. Not this is a of the capacitor now times d divided by D. Okay, that's the area of the capacitor. So that's going to depend upon capital are not little. Are those DS we're gonna cancel and this is going to become Q over. Absolutely not. Pie capital R squared. Okay, let me give myself a little bit of space here so I can sort of put this all together. I don't have a whole lot of space, so I'm gonna move over to the left. The flux. Now, the total electric flux is going to just be the multiplication off that electric field at times, the area of the imperial loop. So that's gonna be Q over. Absolutely not. Pie capital R squared times pi little r squared. Okay. And the pies they're going to cancel now. What is the rate at which the electric flux is changing. Well, looking here at our equation for the electric flux. Three. Only thing that's changing is the amount of charge on the capacitor. That's it. Not even little are is changing. We're choosing this for a particular fixed radius for the imperial loop, so that is a constant. Everything is a constant in this, except for Q. The charge on the capacitor that's increasing as the current dunks charge onto the capacitor plate. So this is going to be de que d t times are squared over Esplanade Capital R squared. Okay, now what is Dick Udt? How quickly is charged building up on the plate that's actually equal to the actual conduction current, right? So this is actually equal to the conduction current because the conduction current that's going into the plate is what's dropping charges onto that plate. So this becomes R squared for excellent capital R squared times that induction current finally plugging this into and piers law, we'll get our answer. Okay, Sorry. This is about to get a little bit squished here. Something that we have to assume is that the magnetic field is constant around the loop, so that's just gonna be the magnetic field times the length of the loop or the circumference, which is two pi times. Little are because little are remember is the radius of the imperial loop. And that equals Epsilon times the rate at which the electric flux is changing. Look at this right here. The rate at which the electric flux is changing It has an excellent not in the denominator that's going to cancel with the absolute not in the numerator. And this is just gonna be little r squared over capital r squared. I see if I want to divide this whole two pi r Over here, the final answer for my magnetic field is going to be our over two pi capital R squared. I see. I am actually missing a mu Not okay. Really quickly. Do not forget this Mu Not like I did that. Mu not there. The mu not never appeared here. Okay, so we do need, um you not there. That is my bad guys. Mu not mu. Not me or not. Okay, so this is the magnetic field, and look, it depends upon an actual current, as it should. It should definitely depend upon a physical current because the physical current is what produces a magnetic field. But notice the trick here and amperes law. When you draw on imperial loop, you Onley include currents within that loop, looking at the image. What current is between that loop? Nothing. There's no current. They're all that's There is an electric field that is changing. That changing electric field is causing a changing electric flux through the ampere Ian Loop. And that's producing the magnetic field. Okay, it definitely depends upon a real current, as it should, but it's not produced by the rial current. It's produced by this imaginary thing called the Displacement Current. Okay, and that was Maxwell's contribution. Okay, so the three laws that rep that form Maxwell's equations gas is law for Electric Field, God's Law for Magnetic Field and Faraday's Law. Those stay the same, but the Final One and Peers law that one gets a tune up because he realized during his time working on electromagnetism that NPR's law could not predict or sorry could not explain the existence of a measurable magnetic fields between capacitor plates. So we do have to add this extra term right here, dis displacement current term. Okay, so you say that NPR's law depends upon the total current, the conduction current, which is produced by riel charges moving and the displacement current, which is produced by a changing electric flux. Alright, guys, that wraps up our discussion on the displacement current and how it changed Maxwell's equations. Alright, thanks for watching.

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