 The goal of this video is to use Ampere's Maxwell's law to calculate the magnetic field at point P and Q in this case of charging capacitor. Now before we begin, just a quick recap of what this equation is all about. But remember this part is just the Ampere's law. And we explored in the previous video it was not sufficient to calculate magnetic fields in all places. So it was in sense incomplete. And that's where Maxwell stepped in and he completed that equation by adding this additional term called the displacement current. And now this is a complete equation and we can use this to calculate magnetic fields anywhere. Now if you are not familiar with that additional term what this displacement current is or why did he include that? It would be a great idea to go back and watch our previous videos on displacement current and Ampere Maxwell's law. So if you're fine with this let's go ahead. So like I said in this video we're going to put this equation to the test and calculate the magnetic field at point P and point Q. So let's start by calculating magnetic field at point P. How do I do that? Well just like what we should do for Ampere's law the first step is to choose an Amperean loop, a closed loop basically. So I want to choose a closed loop that passes through point P because this is where I want the magnetic field. And the loop I'm going to choose is circle because it's nice and circular symmetric. So we're going to choose a circular loop over here. And in fact that's the reason why we have chosen circular plate capacitor. So it just becomes easier for us to do these calculations. Alright next what's the next? Well now I have to walk across that loop and calculate what b.dl is. Now before I do that I have to choose which direction should I walk across. And this is something we've done before so we're doing it one more time. So how do I walk across? Well because I know that the current is to the right from the right hand thumb rule if you imagine our four fingers, if you imagine your right hand, oops, if you imagine your right hand thumb to be towards the right, then your four fingers I hope you can imagine is going to circle this way. And so I know my magnetic field is going to be in this direction. It's going to come out of the screen, it's going to go this way. And because I want the dot product to disappear I will also walk in the same direction. So I'm going to choose to walk this way. So now if I calculate b.dl at every point b and dl are in the same direction and so b.dl just becomes bdl. And if this looks familiar because this is the same thing we have done before when we were just using Ampere's law. So let's do that. So the left hand side would just become the integral of bdl. Remember our goal is to calculate the magnetic field bdl. b.dl just becomes bdl. And now I can say that everywhere the magnetic field is the same because of circular symmetry. So I can pull that b out. And what is the integral of dl? The integral of dl is just adding all these tiny, tiny dl vectors. And when you do that, you just get the length of this. We also circumference and circumference would be 2 pi r. So this would be 2 pi r. That's our left hand side. Now the right hand side, let's see what's that going to be equal to. Mu naught times, what is this? What is ion closed? Ion closed is the current that is punching through a surface attached to this loop. Remember to use Ampere's law or Ampere Maxwell's law, the first thing is you have to attach a surface. You can imagine dipping this loop in a soap solution and then a soap solution gets attached, a film gets attached. So you can have that as a flat surface attached. And if I do that, now notice the current that's punching through it, that becomes the current enclosed. And the current punching through it is just i. So our current enclosed is just i. Plus, now what is this part? The displacement current over here. To calculate the displacement current, I have to figure out what is the flux through this particular loop. Just let's quickly recall, what is the electric flux? How do we calculate electric flux? Remember electric flux is the product of the electric field at any point multiplied by the perpendicular component of the area. Now where would be the electric field over here? There is an electric field inside the wire. But in our case, if you consider this to be a very good conductor, then hardly any resistance, so hardly any potential difference, therefore hardly any electric field. So in our example, the electric field would be very tiny. And even the area would be just the thickness of the wire, which is going to be very tiny. So the flux in this case is going to be almost, negligibly almost zero. But even if the electric field was high enough, and that's possible if you have incredible amount of currents, and it's changing very quickly, it's possible, even if that was very high, notice our displacement current has an epsilon naught in it. And do you remember what's the value of epsilon naught? Again, a lot of things to recall over here. 1 over 4 pi epsilon naught you might remember is 9 times 10 to the power 9. So the epsilon naught value would be 1 by 4 pi times 10 power 9. So you can imagine epsilon naught would be a very tiny number. So if you were to put the values over here, then you would get numbers which is like 10 to the power minus 9, minus 10, minus 12. And this conduction current, this current, would be of the order of 1 or 2 amperes. And so you can completely neglect this term. And so we can say this is almost zero. And so what we see is that the displacement current inside the wire in almost all cases, all practical cases, would be zero. And in fact, that was the reason why Ampere's law worked when we are dealing with just wires, because the displacement current was almost negligible. All right. So how much do you get now? B would be mu naught times I divided by 2 pi r. And again, this is something that we have derived before just by using Ampere's law. Now we are using Ampere Maxwell's law and we get the same answer, which is nice. So this is the magnetic field at P. Now let's see what's magnetic field at Q. And if you remember from our previous video, we said that experimentally the magnetic field here should be the same as over here. Now would be a good time to pause the video and sort of challenge yourself. See if you can use the Ampere Maxwell's law to figure out the magnetic field at point Q is. So give it a shot. It's okay if you go wrong, but give it a shot. All right. So just like before, we'll choose a circular Amperian loop and we'll walk in the same direction. And since everything is the same as over here, for the left-hand side, we're going to get the same answer. The magnetic field and DL will be in the same direction. So we'll walk in this direction. B and DL will be in the same direction. The circumference is exactly the same. So left-hand side, which I'll write directly, will be exactly the same. So B times 2 pi r. Oops. I used the wrong color. So B times 2 pi r. All right. What would be the right-hand side? Okay. What about the enclosed current? Well, again, I have to attach a surface, attach a flat surface to it. And now what's the current punching through this surface? There is no current punching through this surface. It's between the plates of the capacitor. There's vacuum. No charge is flowing. And so the conduction current is zero. So you get zero here. So now, what's the displacement current? There has to be some displacement current. Would it be negligibly small like before? Let's find out. So to calculate displacement current, the first thing is I have to calculate the flux. So let me do that over here. So what will be the electric flux through this surface? Well, electric flux is electric field times the area. So over here, where is the electric field? Who's creating the electric field? Here, the electric field is between the plates of the capacitor. And because the capacitor plates are charged, there will be some electric field here. With this way, some electric field. But how much will that electric field be? How do I know that? We've studied this. We've seen this long, long time before. Electric field between two parallel plates. If you remember the expression for that, we derived it using Causs's law. It happened to be sigma divided by epsilon naught. And again, if you are unfamiliar with this, I recommend going back and watching our videos on Causs's law, where we applied that to calculate this expression. So that's our electric field. Now comes a question. What is the area? And again, if you hadn't tried this before, now would be a great idea to pause and think about what would be the area over here? Well, my first answer would be area of this loop. So pi r square. But let's be very careful. Area represents the area that's perpendicular to the electric field. Let me zoom in over here just so we can clearly see. We can see that there is no electric field over here or here. The electric field only exists within the space of the capacitor. Which means when we are considering the area, we should only consider this area where the electric field exists. Only this much of it. I don't know if you can see that properly. And so we should not consider the entire area of this loop. We should only consider the area of actually the capacitor plates. Does that make sense? That's the area because that's where the electric field is. Let's zoom back out. So over here, the area would be pi capital R square. Not small r square. Does that make sense? Because over here, there is no electric field. All right. So how much would this be? Well, sigma is the charge on the capacitor. Sorry, charge density of the capacitor. Charge per area. And pi r square is the area of the capacitor plate. So when you multiply, what do you get? This is charge per area times area. You get the charge. So this numerator is now representing the charge on the capacitor divided by epsilon naught. Now that we know what the flux is, we'll let's substitute. So if you calculate displacement current, we get epsilon naught times d phi over dt. What do you get if you differentiate this? Well, epsilon naught is a constant. So that stays the same. Times dq over dt, that gives you the current. So you get multiplied by i. And so notice our displacement current happens to be numerically equal to the current in the wire. Now, don't get me wrong. I'm not saying that there is a current flowing over here. What we are finding is that this value, displacement current, numerically ends up being the same value as this. And therefore, you can now see the value of b is exactly the same as before. You get mu naught times i divided by 2 pi r. And so indeed, this expression gives us the right answer. If we had just used Ampere's law and if we had neglected this term, we would have gotten 0. Ampere says this would be 0 because it's not enclosing any current. But Maxwell rightly said it's not just the current that matters. So this flux also induces magnetic fields. And so it's the displacement current that is producing the magnetic field over here.