So, now we're going to discuss about the field of a small loop, which is the smallest unit of a "magnetic dipole" and we will think about the magnetic dipole as written in detail. So, let's use the vector potential method, to find the magnetic field or small loop of current where r which is the distance between the loop and the observer is much larger than the dimension or the length scale of the loop itself. It will turn out that any small loop is a magnetic dipole which produces a magnetic field like the electric field from an electric dipole. So, we will compare electric dipole with magnetic dipole to enhance our knowledge in this new field. Okay. So, have you thought about this? You have a ring. Right? Somehow. If my body creates a current that will have magnetic field perpendicular to my ring, then you can create a circulating current around your ring. Depending on the materials, the conductivity you can change this dynamics. So, you can have some interactions with your ring and your body, right? Sometimes that might not be insignificant. That can also influence probably your circulation of blood around your limbs. Anyway that's a distraction. So, we're going to think about the vector potential of a magnetic dipole, as we just promised. We take first a rectangular loop, because rectangular loop is perfectly fine to understand in Cartesian coordinate. I just learned Cartesian is coming from Descartes, the geometry. Anyway. So, there are no currents in the z direction so a sub z is zero. There are currents in the x-direction on the two sides of length a and in each leg the current density and current is uniform. So, we have aa here and bb there, is a rectangle. So, the solution for a sub x, is just like the electrostatic potential from two charged rods, as you can see here. So, if you have two charged rods, how does it look like from a distance? A dipole moment. Exactly. A dipole moment with point charge at each end. Right. So, the potential of the dipole moment is well-known, we just learned that before. One over four pi epsilon naught times p. this is the unit vector to your observer e sub r over r square, where p is equal to lambda ab. Lambda is the linear density of the charge. So, if I multiply lambda by a this becomes the total charge and b is the distance between those two charges. So, that's why the dipole moment can be written as lambda ab where p is the dipole moment of charge distribution. Now, the dipole moment points always from where to where? Melody. From the negative to the positive. Yes. From the negative to positive as depicted in the picture. So, in conclusion, the positive wire is pointing this way and my dipole is pointing out away, so it's negative y direction. So, let's take a look at the equation a little bit further. So phi is one over four pi epsilon naught p.e sub r over r square. We replaced them by one over four pi naught lambda ab over r-squared over times y over r. That becomes minus lambda ab over four pi epsilon naught y over r cube. Now, we're going to use some analogy and exchange some parameters. So, lambda is rho which is volume density of charge times area because this is linear density. I, current is equal to current density times the area. So, if I can replace lambda by j I can replace rho by j. I can replace lambda by i. Right? So, if I replace phi by a sub x then it becomes a sub x is equal to minus i ab over four pi r c squared times y over r cube. The same goes to y with different polarity because now we have a dipole along positive x direction. So, minus becomes plus and we just replace y by x or x by y. So then, you can understand these two equations. Now, with these equations you understand a sub y is proportional to x and a sub x is proportional to minus y. So, the vector potential at large distances goes in circles around the axis. It's like, when your head, when you're dizzy or confused, you have birds rotating around your head. Right? So, you can see goes in a circle around z axis circle in the same sense as I end the loop. So, they are like mimicking each other. A is mimicking the circulation of the current. So, it's easy to understand or memorize. Here, you see iab, we can put it as mu, which is the magnetic dipole moment. For electric dipole moment we have q times D to Q times distance but here we have current times area. So now, I want you to prove that the vector potential of a small plane loop of any shape is given by the same equation if we replace iab by ai times area of loop. Melody, can you give our students hint how to solve this homework? No. Maybe. Okay. So, let me give you some hint. Do you remember the potato chip or arbitrary shape? You can put it into a rectangular pieces and circulation of those rectangular pieces will add up to make the equivalent circulation of any arbitrary loop? Use that fact, to understand this problem. Okay. So, we can put our equation in vector form if we define the direction of the vector mu, to be the normal to the plane of the loop with a positive sense given by the right-hand rule. So, if I'm rotating like this, magnetic dipole will be upright. So, here you see the rotatings in counterclockwise we will make sure that your magnetic moment is in positive z direction. Let's take a look at the equation here. The vector potential a is equal to one over four pi epsilon not c squared, times mu the vector cross product r over r cubed, which is equal to one over four pi epsilon not c squared mu cross-border e sub r over r square. Using this equation and inserting it into bx by bz, you will be able to get all of these equations by yourself. Where, dot dot dot mean mu over four pi epsilon r c square. All right. So, let's take a look at these amazing table. So, we are comparing magnetic versus electric dipole moments. The components of the B-field behave exactly like those of the E-field for dipole oriented along the z-axis and take a look at this, B sub z, E sub z, and you see one over r cubed minus three z square over r to the fifth is the exactly the same, only the difference you see mu is replaced by b, epsilon not c square is replaced by epsilon not and the rest you can see them by yourself. So, let's discuss a little bit about the magnetic dipole moment. So Melody. Do we have to pause for a magnetic dipole? No. We don't have that, right? So, the word dipole is slightly misleading when applied to magnetic field because there are no magnetic poles, as Melody answered, corresponding to electric charges. So, the magnetic dipole field is not produced by two charges but by an elementary current loop. It is curious though that starting with completely different laws we can end up with the same kind of field as we saw from the comparison between the components of magnetic field and electric field from each typo. Why should that be? I think it has a lot to do with the math we're using to analyze it like the simplification to me in discussing the system. Exactly. So, we are not thinking of the higher-order terms and as Melody just mentioned, it is because the dipole fields appear only when we are far away from all the charges or currents meaning we are doing approximation. So, through most of the relevant space the equations for E and B are identical. So they give the same solutions. However, be careful to sources whose configuration we summarized by the dipole moments are fiscally quite different. But this is also strings. So, you can forget about the real physics, you can only focus on the math and then you can go get out to the physics and make sense of them. Okay. So, we're going to discuss the vector potential of a circuit. Circuit is our current loop where you have no starting point and ending point. So, we're often interested in the magnetic field produced by circuits of a wire in which diameter virus is very small, compared with the dimensions of the whole systems, like printed circuit system board or even the flexible circuit board that is using smartphones. So for thin wire, we can write our volume element as dv is equal to Sds where the capital S is the area of the cross-section of the wire or ds is a small length component of your wire along its length direction. Since the vector ds is in the same direction as j the current density we can write a vector of equation jdV is equal to jSds. As js is current i, in a wire, the vector potential becomes a of one is equal to one over four pi epsilon not c square times integral of i dx2 over r12. So now, we can use some measurable current into the equation to get the vector potential. So now, that leads us to the law of Biot Savart. So, installing electrostatics we found that electric field of a known charge distribution could be obtained directly by an integral that is depicted here, e of one is equal to one over four pi epsilon not times integral of rho of two times E sub one to dv2 over r12 squared. This is an arbitrary loop that we just made a small loop inside there. You see next to Melody, there's an observation point one. So, the distance between those two will be the r12 to many times it takes much easier to do the integral for the potential and that's great. That was the case for let us statics. It is not always the case for magneto statics. Right. We see there's a similar integral which relates to magnetic field B to the current I. As you can see B of one is equal to del cross a of one which is the curl of vector potential, is equal to del cross and this is the equation we know. We will transform that equation, to use the current. So, the curl operator means taking the derivatives of a's of one that is to say it operates only on the coordinates x1, y1 and z1 and if you look at this equation only the denominator has subscript one, denominator doesn't have subscript one. So, we have for the x-component of B, b sub X is equal to round az over round y1 minus round ay over round z1. If you lay this out expand it, you will be able to prove that can be arranged in this neat form of the vector form. So, the integral gives me directly in terms of the known currents so you can translate this into this one. Where you just have to know the current along the circuit and do the integration to the get the B-field at the observation point. So now, another homework will be, what is the advantage of the vector potential if we can find B directly with a vector integral? All right. That's it and I hope to see you again in the next lecture. Thank you and have a nice day.