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Displacement Current and Maxwell's Equations | 12 mins | 0 completed | Learn |

Concept #1: Displacement Current and Maxwell's Equations

**Transcript**

Hey guys, in this video I want to talk about something called displacement current and Maxwells equations, most of electro magnetism was done throughout the eighteen hundreds and in near the eighteen sixty's or so I don't remember exactly when there was a guy named James Clerk Maxwell who sort of put everything together Gauss's law, Faraday's law, Ampere's law everything that we've seen so far but he did make one small change to Ampere's 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, so that's what we want to talk about in this video let's get to it. Now Ampere's law as we know it says that the integral around some sort of closed loop this should technically read this but you guys know what it says is only dependent upon what I have called Isub C the conducting the current, a current formed by physical charges moving, there is actually a second current I specifically put in 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, the second current can also produce a magnetic field according to Ampere's law this second current is known as the displacement current and this is what James Clerk Maxwell was known for the displacement current is given by the permitivity 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 fllux was changing with time flux, 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 going to apply that to the displacement current but now it is for the electric field.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 go into a capacitor plate and an actual conducting current leaving a capacitor plate what is the magnitude of the magnetic field produced between the two capacitor plates as shown below. So as the conducting current comes towards that 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 that backwards that's from the view of the positive charges which we know is wrong but either way as positive charges buildup on one plate and negative charges buildup 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, so we're going to apply Amperes law using this imaginary displacement current 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 near Mu not times epsilon not times the rate at which the electric flux is changing with time.

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 is an Amperes law deals with that imaginary amperian loop that I drew in green imaginary amperian loop which has a radius little R that is the surface that you're measuring the flux through. So first let's measure the electric flux this is going to be the electric field times the area of our amperian loop. Now what is the area of that loop what is the electric field between the parallel plate capacitors alright I'm going to minimise myself for this the area of the loop is pretty easy the radius of the loop is little R squared do not put capital R, capital R is what I reserved for the radius of the capacitor plate we are considering the flux through this imaginary amperian loop just like we would consider how much current was contained in the imaginary amperian loop and the next thing is what's the electric field well the electric field 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 epsilon not times the area of that capacitor over D. So I am just going to plug that in this is Q divided by epsilon not this is A of the capacitor now times D divided by D. That's the area of the capacitor so that's going to depend upon capital R not little R those Ds are going to cancel and this is going to become Q over epsilon not pi capital R squared. 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 going to move over to the left. The flux now the total electric flux is going to just be the multiplication of that electric field times the area of the amperian loop so that's going to be Q over epsilon not pie capital R squared times pi little R squared and the pies are 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 the only thing that's changing is the amount of charge on the capacitor that's it not even little R is changing we're choosing this for a particular fixed radius for the amperian loop so that is a constant everything's a constant in this except for Q the charge of the capacitor that's increasing as the current dumps charge on to the capacitor plate so this is going to be D Q D T times R squared over epsilon not capital R squared. Now what is D Q D T how quickly is charge 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 on to that plate so this becomes R squared epsilon not capital R squared times that induction current finally plugging this into Amperes law we'll get our answer, 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 going to be the magnetic field times the length of the loop or the circumference which is 2 pi times little R because little R remember is the radius of the amperiun loop and that equals 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 epsilon not in the denominator that's going to cancel with the epsilon in the numerator and this is just going to be little R squared over capital R squared I C if I want to divide this whole 2 pi R over here the final answer for my magnetic field is going to be R over 2 pi capital R squared I C I am actually missing Mu not really quickly do not forget this Mu not like I did that Mu not there the Mu not never appeared here. So we do need a Mu not there that is my bad guys Mu not, Mu not, Mu not.

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 a physical current is what produces a magnetic field but notice the trick here in Amperes law when you draw an amperian loop you only include currents within that loop looking at the image what current is between that loop nothing there's no current there all that's there is an electric field that is changing that changing electric field is causing a change in electric flux through the amperian loop and that's producing the magnetic field. It definitely depends upon a real current as it should but it's not produced by the real current It's produced by this imaginary thing called the displacement current and that was Maxwell's contribution so the three laws that form Maxwell's equations Gauss's law for electric field Gauss's Law for magnetic field and Faraday's law those stay the same but the final one Amperes law that one gets a tune up because he realized during his time working on electromagnetism that Amperes law could not predict or sorry could not explain the existence of a measurable magnetic field between capacitor plates so we do have to add this extra term right here this displacement current term. So you say that Amperes law depends upon the total current the conduction current which is produced by real charges moving and the displacement current which is produced by a change in the electric flux. Alright guys that wraps up our discussion on the displacement current and how it changed Maxwell's equations. Thanks for watching.

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Concept #1: Displacement Current and Maxwell's Equations

The figure shows a discharging capacitor. If looked at from the left, in the region between the capacitor's plates there is a magnetic field that is:
A) Pointing away from us
B) Clockwise
C) Pointing towards us
D) Uniformly zero
E) Counterclockwise

In an RC circuit, the capacitor begins to discharge. During the discharge, in the region or space between the plates of the capacitor, there is
a. An electric current but no magnetic field
b. A magnetic field but no electric field
c. Both electric and magnetic fields
d. No fields of any type

A constant current I is supplied for a brief time to charge a parallel plate capacitor. The capacitor has circular plates of radius R with gap d (d << R). Point 1 is at R + d from the wire, and point 2, is at a distance R + d from the center of the capacitor.
During the time interval that the constant current I is flowing through the capacitor, the magnetic field at point 2, B2 is:
a) B2 > B1
b) B2 = B1
c) 0 < B2 < B1
d) B2 = 0
e) not enough information to tell

A constant current I is supplied for a brief time to charge a parallel plate capacitor. The capacitor has circular plates of radius R with gap d (d << R). Point 1 is at R + d from the wire, and point 2, is at a distance R + d from the center of the capacitor.
After the capacitor is charged and the current I goes to zero
a) the electric flux and the magnetic field between the plates are both zero.
b) the electric flux between the plates is zero and magnetic field between the plates is non-zero.
c) the electric flux between the plates is non-zero and the magnetic field between the plates is zero.

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