Okay. Welcome back to electrodynamics and its applications Lecture 13. Today, I'm with my teaching assistant Melodie Glasser and my name is Professor Seungbum Hong. So, today we're going to discuss about the vector potential which is analogous to electrostatic potential in electrostatics, and this vector potential will be useful in magnetostatics. So, as written in this bullet point, we will continue our discussion of magnetostatics. So, let me ask Melodie what magnetostatics is. So, let's recapture the concept of magnetostatics. So, magnetostatics is when the magnetic field is constant, and that occurs when there is a flow of charge. However that is steady-state, so there is a consistent flow of charges. Exactly. So, we have a steady-state current that will close a loop, that is called circuit. There will be no change in charge density, therefore, there will be no change in electric field, but still we will have existence of magnetic field due to the current as shown in this Maxwell's equations. So, if we revisit Maxwell's third equation and fourth equation, we will see the fact that the divergence of the magnetic field is zero is always true, but the curl of magnetic field that only depends on the current will only be true for magnetostatics. Now, we want to solve these equations mathematically in a general way, that is to say without requiring any special symmetry or intuitive guessing. In electrostatics, one simply works out the scalar potential Phi by taking an integral of the charge and obtain the electric field from the derivatives of Phi. So, you may remember that the electric field, E is equal to minus of gradient of Phi. So, if I know Phi, I can just take the derivatives in space and get the electric field. Now, there is a corresponding procedure for finding magnetic field B if we know the current density J of all moving charges, and that will be the main focus of this chapter. So, we will denote the vector potential as A and the divergence of B is always zero, and this means that we can always represent B as the Curl of another vector field, A, which we call vector potential. Probably, we can recall the theorem that we learned in the previous chapter. That mean that there's multiple vector fields that could produce the same electric or magnetic field. That's exactly the case, and we will cover that in the following slides. So, as Melodie just mentioned, in order to have B, we can express B in terms of the curl of a vector field, but this is not unique. Okay? Before getting to that point, let's take a look at the definition of the curl operation. As you can see, we have three components x, and y, and z to represent the magnitude along each axis. As you can see, if we only look at one component here, the Del cross A sub x is equal to Round A_z over Round y, minus Round A_y over Round z. You need to do this for two other directions as well. You can see it's more complicated than electrostatic field. So, now we get to the point about what Melodie just mentioned. Adding a constant to a vector potential. So, you may remember that scalar potential Phi was not completely specified by its definition, as you can always find another potential Phi prime that is equally good by adding constant C. So, if I take a gradient of this function, they will produce the same electric field. So, likewise, the new potential Phi prime gives the same electric field since the gradient C is zero. Similarly, we can have different vector potentials, A which give the same magnetic fields. So, we can add to A any field which is the gradient of some scalar field without changing the physics. This is well-described in this box with equations, but I'm going to ask my teaching assistant Melodie if she can explain to you in details. Okay. So, basically what we're looking at here is we know the difference between A prime, and A is simply the gradient of some scalar field. If you take the curl of the gradient of a scalar field, we know that is going to be zero because you can't have circulation of a gradient as a gradient is always increasing or decreasing. Good. Very good. So, let's take a look. So, if the curl of a vector is zero, it must be the gradient of some scalar field. That was another theorem that we learned in the previous lecture. That is to say, the difference between those two vector field, A prime and A will be Del Psi is arbitrary scalar function or a scalar field. So, if we rearrange this equations, we can say A prime is equal to A plus Del Psi. That means that if A is a satisfactory vector potential for a problem, then for any Psi at all, A prime will be on equally satisfactory vector potential leading to the same field B. Now, so, we have some freedom of choice. Right? So, by suitable choice of Psi, we can make the divergence of A prime anything we wish. So, we will do some mathematical trick. We're going to put divergence operator in front of both of these terms, and as we can see the divergence of A prime will be equal to divergence of A plus divergence of gradient is Laplacian. So it'll be Laplacian of Psi. So, for magnetostatics, we will make a simple choice. Later when we take up electrodynamics, we will change our choice for our convenience. But for now, we will only use this one where the divergence a is zero, and you will see in the later slides why this handy and useful. Okay? So, let's take a look at the vector potential of a uniform magnetic field. This will be the simplest case, where you have only one direction of your vector and they are the same no matter where you are. In that case, we can imagine the x and y field is zero, and only we have z, z-field. If we equate them with the Curl of the vector potential, you can see only the last equation will have B Naught and the other will be zero. Now, by inspection, we see that possible solution include, but are not limited to these three examples. So, let's take a look at the first example. A_y is equal to xB Naught, A_x is equal to zero, A_z is equal to zero. So, if you put them into these equations, you will see they all satisfy the equations that are here. Now, let's take a look at the last examples here where we have minus half yB Naught, and A_y we have half xB Naught, and A_z is equal to zero. If you put them into this equation, again see that they satisfy this equation. But this is a typical example for a rotating field around the current carrying wire, and we will see that example in the following slide. So, we will see rotating vector potential for a uniform B-field, and the third solution has some interesting properties. Since the x component is proportional to minus y, and the y components is proportional to plus x, A must be at right angles to the vector from the z-axis as you can see. So, why are they at right angles? As you can see at any arbitrary point, the slope of your r vector is y over x. While the slope of your A vector is minus x over y. If you multiply two slopes and becomes minus one, they are at right angles. Right? That's the case here. So, you can understand that. Now, if you take a look at these three components, you can neatly arranged them into a vector form, which is written here. A is equal to one over two B Cross r prime. So, I'm going to ask Melodie if the vector potential A is equal to one over half B Cross r prime, which direction will A pointing to? So, A will be pointing perpendicular to both the magnetic field and the electric field. Exactly. Because this is a cross product, this will make this vector perpendicular or orthogonal to both of them. So, take a look at this picture. So, you have r and A is perpendicular to that, and B is coming out of this plane. So, it's also perpendicular to that as well. The vector potential A has the magnitude Br prime over two as you can inspect from this figure, and it rotates about the z-axis in counter clockwise, and that you can understand with right-hand rule. So, if you have B-field to my direction, you will have a counterclockwise rotation of A-field. Now, if the B-field is the actual field inside a solenoid, then the vector potential circulates in the same sense as do the currents of the solenoid. You will remember, inside the solenoid, you have uniform B-field. So, you can use that analogy to understand the vector potential as well as magnetic field, right? So continuing the discussion about the vector potential for a uniform field, the circulation of A on any closed loop gamma, can be related to the surface integral of the curl of A by Stoke's theorem. At this point, I'm going to ask Melodie what Stoke's theorem was about? Okay, so we learned about that in the prior lecture with the potato chip analogy. So basically, what we learned is that the line integral of the vector field on the outside is going to be equal to the surface integral of the curl of the vector field. Perfect. So that is now expressed in mathematical terms here, what Melodie has just explained to you. If you take a look at del cross A here, what does it reminds you of? I think we had that formula earlier, and it should be equivalent to the magnetic field. Exactly. So we can replace the cross product here, or the curl of the vector field A, vector potential A, by b. Once we do that, you see it becomes a flux of a magnetic field through the area of interest which is bounded by gamma. So, now we have this equation. So, if we take a circular loop of radius R prime in a plane perpendicular to a uniform field B, the flux is just pi R prime squared B because you have a symmetry, right. If we choose our origin on an axis of symmetry, so that we can take A as circumferential and a function of only R prime, the circulation will be, this one will be two pi r prime which is periphery of your circle times a is equal to the area of the circle which is pi r prime square times B. Therefore, A is equal to b r prime over two. So again you can see Br prime over two will be the magnitude of the vector potential. So, now we will move on to a little bit more deeper subjects. The vector potential of known's currents. Now, we have related vector potential with a magnetic field and we know magnetic field is related to current from the force equation. Now we are going to relate the current directly with the vector potential. So, let's see how that works. Mathematically identical equations for vector and scalar potential can be found when we rewrite our equations in terms of vector potential A and we can correlate that with the equations for electrostatic potential phi. Now, this is the fourth Maxwell force equations and we just replaced by del cross A and you remember this formula, right. So, del cross del cross A, is equal to gradient of del.A minus laplacian of A and you'll remember also that we chose the boundary condition or the specific number for our vector potential to have no divergence del.A is equals 0. Now you see why that's convenient because then you can make one term disappear. So now you'll only have what Laplacian. Laplacian of A will be equal to minus j over epsilon naught c squared. If you compare that with the Poisson's equation, laplacian of phi is equal to minus rho dot epsilon naught you have similarity. Similarity. You just replace A by phi by A and rho by j over c square, right. You know this has three components. So, you have AX AY AZ. That's different from phi. Phi is a scalar potential so you don't have one but you have three siblings here, right. So, then you can just do the math and see that you can have three independent equations that look exactly the same. Just the differences you have different subscript x and y and z, okay. So, we are discussing about amazing analogy between phi and A all we have learned about solving for potential when rho is known can be now used for solving for each component of A, when J is known. So, this is perfect strategy because we use limited resource to know a lot of knowledge, right. So, let's take a look at this picture, so you remember from an arbitrary distribution of charge in space if we know exactly the density function, then we can know at an arbitrary point what the electrostatic potential will be due to the set of charge that we were just thinking of, all right. This is the equation. Phi of one, which is the 0.1 where your observer is located, is equal to one over four pi epsilon not integral of rho two. Rho two is where the charges are located dV2 is also the volume integral around them over R12, which is the distance between the region of interest to the observation point. Now, replacing phi by A and replacing rho by j over C square. Now we are turning this into a current problem and magnetic field at the point of interest and you can see one-to-one corresponding equation can give you the general equation like this. As you can see, as long as the divergence of vector field A potential A is zero, the divergence of current density j will also be zero as you can see from here, alright. So that means it is good for steady-state current. So, we're going to now focus on a straight wire example, where you can see you have uniform current along one axis and we're creating a rotating magnetic field. So the current density vector J has only a z component with the magnitude of J sub Z is equal to i over PA squared. PA squared will be the area of the cross section of our wire, which is depicted in the picture. Since J sub x and j sub y are both zero, we can understand A sub z and A sub y will be zero. So, to get A sub z, we can use our solution for the electrostatic potential phi of a wire with a uniform charge density rho is equal to j z over c squared. We're now going to think the situation where we have wire full of charge which is like a linear charge that we learned before using the Gaussian surface of cylindrical surface with the symmetry that will allow us only to have radial field. So, Melodie, can you recap the Gaussian surface and the electric field dependence on r. When we were talking about it earlier, we talked about how if the charge is continuing in a cylinder then there will only be a field out in the radial direction and then next step is that we talked about the Gaussian surface. So regardless of the volume you take outside of the wire, the field will decrease by one over R because of the radial symmetry. Exactly. So if I compliment her explanation, even though you increase the radius of the Gaussian surface, the amount of charge you are containing inside the volume will be the same. So, the electric field will have one over r dependence. They will decrease one over r, which is less drastic than one over r square from a point charge. Since electric field is the gradient of electrostatic potential, if I want to reverse that equation, I have to integrate the electric field with respect to r. One over r if you integrate it becomes log r. That's why we have log R prime dependence for electrostatic potential and this is the equation. So Phi is equal to minus lambda which is the linear density of the charge on the wire over two pi epsilon naught times log r prime. Now, we're going to replace it mechanically, mechanistically one-by-one phi by A sub z and lambda, in case of lambda, we're going to replace that by pi A square j z over c square. You can understand that if you think about the geometry and the current. Once you understand this then you can understand that A sub z is equal to minus A because J sub z pi r squared this is current, i over two pi epsilon naught c squared times log r prime. So the magnetic field of a straight current carrying wire we just learned that before, but we can also use our approach to use the equations that relates to B field with the vector potential. So, if we put that into perspective or into this equations, then we will come up with the B field is equal to one over four pi epsilon naught c squared times two i two r prime. So magnetic field will depend on the current flowing in the wire as well as the distance from the wire. It will have one over one prime dependence. Like the electric field from a charged wire. Now, we will add complexity to our problem. So we will think about a long solenoid. So, we will now consider again the infinitely long solenoid with a circumferential current on the surface. So, then because we have a lot of rings with current carrying the total current will be n times i, i is the current through one ring and if we have n rings per unit length, which is here, the n turns per unit length then it is N i per unit length. So, just as we have defined a surface charge density sigma, we define here a surface current density. So you can imagine a long or on the surface of the cylinder we have sets of currents that forms a surface current density which is equal to current per unit length on the surface of a solenoid. Now, we can now understand, because this is rotating, we can put that as a function of phi, the angular position minus j sine phi and j cosine phi. Then this will express the rotating current. First we wish to find A sub x or points outside the solenoid and the result is the same as the Electrostatic Potential outside a cylinder with a surface charge sigma equals sigma naught sine phi. So, you may recall the example when we had our sphere with charge plus charges and minus charges and displace it a little bit, then the surface charge density followed cosine rule. Likewise, if you've shift charge in the lateral direction for cylinders, you will have the same kind of rule would be sigma naught becomes j over c squared. So this is analogy, okay. So, if we superimpose two charge cylinders, you will see in the pictures next to Melodie. Then from the top view, you will see it is something like displaced spheres. But in this case is displaced cylinders. So, this charge distribution is equivalent to two solid cylinders of charge one positive, one negative with slight relative displacement of their axis in y-direction. For the other one will be along x-direction. So, we will have two equations. The potential of such pair of cylinders are proportional to the derivative with respect to y of the potential of a single uniformly charged cylinder. So, we learned that perspective from our earlier chapter when we dealt with the dipole as a total energy due to the electric field by the test charge on that dipole. So, we can use that equation to understand this situation. Let me ask Melodie one thing. What was the magnetic field outside infinitely long solenoid. We discussed that was negligible or zero. Yeah. It was negligible or zero. However, the interesting thing is, even though magnetic field is zero outside the solenoid, we will prove we will have non-zero vector potential. So, potential of a cylinder of a charge is proportional to lnr prime and the potential of the pair is then phi is proportional to round lnr prime over round y, and if I do that is y over r prime squared. So, A_x is equal to minus K, this is coefficient. Y over r prime square, and A_y will be K x over r prime square. So, if you just rotate it 90 degrees with polarity change, you can understand this. So, although there is no magnetic field outside the solenoid as Melodie mentioned, there is A field that circulates around the z-axis. Now, then you may wonder, if I had vector potential, shouldn't I have B field? How does that work? Let me tell you this, even if you have potential outside something, if there is no change in potential, you don't have any electric field. Likewise, some rules, which is the rule of curl. If the curl of vector field is zero, even though you have vector field, you will have no magnetic field. We're going to prove that the curl of this vector potential outside solenoid will give you zero. Let's take a look at here. So, B sub z is equal to round or round x times K x over r prime squared minus round over round y times parenthesis minus K y over r prime squared. If you do the mathematical expansion, and do the work, then you will find it is zero. We have vector potential outside solenoid, but we don't have magnetic field. That is also manifested by bomb's experiment, which is a very important experiment for quantum mechanics. We will see A can influence something while we have no B. So, vector potential outside the solenoid. We can check our result against something else we know and double-check. The circulation vector potential around the solenoid should be equal to the flux of B inside the coil. That is what we have just discussed. That we know. This is one Stoke's' theorem, and we use the definition of magnetic field, that is the curl of a vector potential. So, we know this is true. Then we also know the magnetic field, B naught, is equal to nI over epsilon naught c square. Then we also know A_x is equal to minus K y over r prime square, and A_y is equal to K x over r prime square. So, the magnitude of this using Pythagorean Theorem, will be, A is equal to K over r prime. Now, the circulation is, 2pi r prime A, 2pi r prime A, because A is uniform, and this should be, as you can see A is K over r prime 2pi K. So, 2pi K, which is on the left side, 2pi K is equal to pi a square, this is the area of the solenoid, times the B-naught field nI over epsilon c square. So, K becomes nIa square over 2 epsilon naught c square. As a result, you have A is equal to nIa square over 2 epsilon naught c squared 1 over r prime. So, this is exactly the same as the one we proved in the previous lecture. Now, the homework is written here. How about the vector potential inside the solenoid? So, outside this 1 over r prime, but inside, you will be able to prove that is depending on r prime instead of 1 over r prime. You can prove it yourself and see how it works. Now, we will discuss an interesting example, where we can build up an angular velocity meter, where you can measure the angular velocity with high precision with very small materials. Very cheap materials. So, let's think about the example of retaining a long cylinder with Q on surface. If we have a thin cylindrical shell of radius, a, with surface charge density sigma, retaining a cylinder makes the surface current J is equal to sigma v, where v is equal to, a omega. You can see omega is the angular velocity. So, suppose we put a short piece of wire W, perpendicular to the axis of the cylinder, extending from the axis out to the surface and fasten it to cylinders so that it rotates. Now, if I have surface charge, and if I connect that with the grounded axis through this piece of wire, what happens? If it's a metal then the surface will also become grounded. So, you need to put some kind of insulator in between to make sure there's no grounding. Exactly. We will put some glue here. Gorilla glue maybe. By that, we are preventing the surface charge to flow into the ground. But if you do that now and we start to rotate it, we know we are creating surface current. Rotating current like solenoid, and that retaining current will create magnetic field perpendicular to this area. That magnetic field, will interact with the charges in the wire where we have the same number of minus and plus charges. But as you know, if I rotated in the same direction, it will create opposite current for minus and plus. Therefore, the force, the lowest force acting on each charge, will force them to separate to different extreme ends. So, you will have minus close to the axis of rotation and plus close to the surface. You are now creating a dipole. This dipole moment will be depending on the rotating velocity. If you can have a volt meter to measure that voltage, then you will be able to measure the angular velocity or rotation speed of the system. So, I'm going to ask Melodie. Where can we use this gadget? So, I think you might be able to use it to prove that there's rotational motion somewhere. So maybe like the rotation of the Earth. Yeah. Like Galileo proved that Earth is rotating around the Sun or rotating about itself, we can use this simple gadget to prove and also measure the angular velocity of Earth. Now, let's take a little further this discussion to the point which be a little bit difficult, but let's think about the frame of reference and what we mean by that, and also inertial coordinate system. So, inertial coordinate system doesn't change the inertia. In order to preserve the inertia, you need to have no force. In other words, you need to have no acceleration. In that case, it means either being stationary or moving with constant speed. What if I put myself in the frame of reference of the cylinder? So, imagine you are rotating herself inside a cylinder, then everything becomes stationary. Your wire doesn't rotate, so there is no velocity moment for Lorentz force. So, there's no way to separate this charges in the wire. However, as I just mentioned, that there's just a charged cylinder at rest and I know that the latter static equation, say there will be no electric field inside. So, there will be no force pushing the charge to the center. But there is no relativity rotation, especially for special relativity, meaning a rotating system is not an inertial frame. Why is not? Why is that the case Melodie? Because of acceleration. Exactly. If you start to rotate, you have acceleration at each moment you're changing the velocity vector. Although you're not changing the magnitude, you're changing the direction, and that also results in acceleration. We must be sure to use equations of electromagnetism only with respect to inertial coordinate systems. It would be nice to measure the absolute rotation of earth with such a charged cylinder as Melodie mentioned, but the effect is too small to observe with the most delicate instruments. But this was in 1960's. Maybe in 2020, where we're living right now, we can do it. So, why don't you challenge this and measure yourself. So, the homework here is to calculate the voltage difference between the cylinder and axis when using the cylinder and measure the absolute rotation of the Earth. Assume we use sigma equals 0.1 microcoulomb per square centimeter, which is the case where 99 percent screened unipolar ferroelectric surface.
