Introduction to Vector Potential
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- 1.1 Vector Potential
- 1.2 Exercise - Finding the Current Density
- 1.3 Poisson_s Equation for Vector Potential
- 1.4 Exercise - Vector Potential of Finite Wire
- 1.5 Calculating the Vector Potential of a Given Magnetic Field
- 1.6 Exercise - Infinite Spring
- 1.7 Exercise - Infinite Cylinder
- 1.8 Boundary Conditions for Vector Potential
- 1.9 Multipole Expansion and Equation for Magnetic Dipole
- 1.10 Exercise – Thick Plane with Uniform Current Density

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[{"Name":"Introduction to Vector Potential","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"1.1 Vector Potential","Duration":"14m 35s","ChapterTopicVideoID":21381,"CourseChapterTopicPlaylistID":99495,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/21381.jpeg","UploadDate":"2020-04-06T22:16:29.4270000","DurationForVideoObject":"PT14M35S","Description":null,"MetaTitle":"1.1 Vector Potential: Video + Workbook | Proprep","MetaDescription":"Vector Potential - Introduction to Vector Potential. Watch the video made by an expert in the field. Download the workbook and maximize your learning.","Canonical":"https://www.proprep.uk/general-modules/all/physics-2-electricity-and-magnetism/vector-potential/introduction-to-vector-potential/vid21452","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.085","Text":"Hello, in this lesson,"},{"Start":"00:02.085 ","End":"00:05.850","Text":"we\u0027re going to learn about vector potential."},{"Start":"00:05.850 ","End":"00:08.430","Text":"In the past in electricity,"},{"Start":"00:08.430 ","End":"00:16.575","Text":"we\u0027ve seen that if we take the curl of the electric field and if it is equal to 0,"},{"Start":"00:16.575 ","End":"00:19.125","Text":"then that means that E,"},{"Start":"00:19.125 ","End":"00:23.175","Text":"the electric field, is conservative."},{"Start":"00:23.175 ","End":"00:30.045","Text":"If it\u0027s conservative then what we get is the E field"},{"Start":"00:30.045 ","End":"00:37.150","Text":"is equal to the negative gradient of some function Phi."},{"Start":"00:38.060 ","End":"00:43.730","Text":"Every time we have that the curl of the E field is equal to 0,"},{"Start":"00:43.730 ","End":"00:47.630","Text":"then we can say that the electric field is conservative,"},{"Start":"00:47.630 ","End":"00:50.465","Text":"and therefore we can give the electric field"},{"Start":"00:50.465 ","End":"00:56.310","Text":"this value as being the negative gradient of a function."},{"Start":"00:56.810 ","End":"01:02.280","Text":"The negative gradient, and this is of course a scalar function."},{"Start":"01:03.260 ","End":"01:11.970","Text":"This Phi is of course the potential of the electric field and its scalar, as we said."},{"Start":"01:13.100 ","End":"01:15.470","Text":"With the magnetic fields,"},{"Start":"01:15.470 ","End":"01:22.235","Text":"we\u0027ve already seen from Maxwell\u0027s second equation that div B,"},{"Start":"01:22.235 ","End":"01:23.780","Text":"or the divergence of B,"},{"Start":"01:23.780 ","End":"01:27.450","Text":"the magnetic field, is equal to 0."},{"Start":"01:27.450 ","End":"01:36.575","Text":"From here we get that B is equal to the rotor of A,"},{"Start":"01:36.575 ","End":"01:40.800","Text":"where A is some vector."},{"Start":"01:41.300 ","End":"01:47.660","Text":"Also in general, whenever the divergence of some vector field is equal to 0,"},{"Start":"01:47.660 ","End":"01:51.635","Text":"we can write out the equation where the curl"},{"Start":"01:51.635 ","End":"01:56.820","Text":"of a different vector field will give us this original vector field."},{"Start":"01:58.060 ","End":"02:00.515","Text":"So in this equation over here,"},{"Start":"02:00.515 ","End":"02:04.245","Text":"we had that Phi was the scalar potential."},{"Start":"02:04.245 ","End":"02:07.430","Text":"Similarly in this equation,"},{"Start":"02:07.430 ","End":"02:10.835","Text":"A is going to be the vector potential,"},{"Start":"02:10.835 ","End":"02:14.130","Text":"which is the topic of our lesson."},{"Start":"02:14.570 ","End":"02:18.200","Text":"The vector potential is useful as"},{"Start":"02:18.200 ","End":"02:21.950","Text":"sometimes it\u0027s easier to calculate the vector potential and"},{"Start":"02:21.950 ","End":"02:24.935","Text":"from this to calculate the magnetic field"},{"Start":"02:24.935 ","End":"02:29.320","Text":"rather than to calculate the magnetic field using other methods."},{"Start":"02:29.320 ","End":"02:36.290","Text":"This is, of course an equation to put into your equation sheets so please"},{"Start":"02:36.290 ","End":"02:39.335","Text":"do and now what we\u0027re going to do is we\u0027re going to look at"},{"Start":"02:39.335 ","End":"02:44.280","Text":"Ampere\u0027s law given this vector potential."},{"Start":"02:44.660 ","End":"02:52.850","Text":"Ampere\u0027s law is equal to the rotor of the magnetic field,"},{"Start":"02:52.850 ","End":"02:57.495","Text":"which is equal to Mu naught J."},{"Start":"02:57.495 ","End":"03:01.265","Text":"Now instead of the rotor of B,"},{"Start":"03:01.265 ","End":"03:08.480","Text":"we\u0027ll place n this over here because B is equal to the rotor of A."},{"Start":"03:08.480 ","End":"03:13.895","Text":"What we have is the rotor of B,"},{"Start":"03:13.895 ","End":"03:23.295","Text":"which is the rotor of A and the right side stays exactly the same."},{"Start":"03:23.295 ","End":"03:32.345","Text":"We can use identity which says that the rotor of the rotor of a vector field is equal to"},{"Start":"03:32.345 ","End":"03:40.010","Text":"this Nabla of the divergence of A"},{"Start":"03:40.010 ","End":"03:47.625","Text":"or the gradient of the divergence of A minus Nabla squared A,"},{"Start":"03:47.625 ","End":"03:52.035","Text":"where this is Nabla squared A is the Laplacian of A,"},{"Start":"03:52.035 ","End":"03:56.525","Text":"and this is again equal to the same right side, Mu naught J."},{"Start":"03:56.525 ","End":"03:58.895","Text":"Now a note on this previously,"},{"Start":"03:58.895 ","End":"04:05.810","Text":"we saw that if we take Nabla squared of a scalar function,"},{"Start":"04:05.810 ","End":"04:09.395","Text":"some Phi, let\u0027s say of the potential."},{"Start":"04:09.395 ","End":"04:16.380","Text":"We saw that this was equal to negative Rho divided by Epsilon naught."},{"Start":"04:16.880 ","End":"04:22.820","Text":"We actually saw that this, the Laplacian,"},{"Start":"04:22.820 ","End":"04:29.555","Text":"Nabla squared of this scalar function was equal to"},{"Start":"04:29.555 ","End":"04:37.930","Text":"Nabla of the divergence of this scalar function."},{"Start":"04:37.930 ","End":"04:42.559","Text":"However, when we\u0027re dealing like over here with a vector function,"},{"Start":"04:42.559 ","End":"04:44.390","Text":"this is not the case."},{"Start":"04:44.390 ","End":"04:47.435","Text":"When we\u0027re dealing with A vector,"},{"Start":"04:47.435 ","End":"04:51.769","Text":"if we have the Laplacian of A vector,"},{"Start":"04:51.769 ","End":"04:57.547","Text":"this does not equal this"},{"Start":"04:57.547 ","End":"05:04.100","Text":"and that is exactly why this over here does not equal 0."},{"Start":"05:04.100 ","End":"05:08.210","Text":"It actually equals some value."},{"Start":"05:08.210 ","End":"05:14.975","Text":"This is only correct when we have a scalar but over here,"},{"Start":"05:14.975 ","End":"05:18.304","Text":"when we have a vector field,"},{"Start":"05:18.304 ","End":"05:20.930","Text":"then they are not equal to one another."},{"Start":"05:20.930 ","End":"05:23.870","Text":"So please don\u0027t get confused."},{"Start":"05:23.870 ","End":"05:29.225","Text":"Let\u0027s just go over what this Laplacian of A is,"},{"Start":"05:29.225 ","End":"05:31.310","Text":"or Nabla square of A."},{"Start":"05:31.310 ","End":"05:33.380","Text":"We\u0027ll write it over here."},{"Start":"05:33.380 ","End":"05:37.470","Text":"If we have Nabla squared of A,"},{"Start":"05:37.470 ","End":"05:44.475","Text":"what we\u0027re doing is Nabla squared of each component of A,"},{"Start":"05:44.475 ","End":"05:47.040","Text":"so it\u0027s x, y, and z component."},{"Start":"05:47.040 ","End":"05:51.730","Text":"That means that we have Nabla squared of A_x,"},{"Start":"05:51.730 ","End":"05:55.500","Text":"the x component of A in the x direction."},{"Start":"05:55.500 ","End":"05:59.990","Text":"Of course, we get a vector after doing this operation,"},{"Start":"05:59.990 ","End":"06:04.075","Text":"plus Nabla squared of A_y,"},{"Start":"06:04.075 ","End":"06:06.915","Text":"the y component in the y direction,"},{"Start":"06:06.915 ","End":"06:17.550","Text":"plus Nabla squared of the z component in the z-direction."},{"Start":"06:17.550 ","End":"06:21.270","Text":"Then we can break up each one of these."},{"Start":"06:21.270 ","End":"06:26.175","Text":"What does Nabla squared of the x-component mean?"},{"Start":"06:26.175 ","End":"06:27.975","Text":"Let\u0027s just write this out."},{"Start":"06:27.975 ","End":"06:34.680","Text":"Nabla squared of A_x is equal to"},{"Start":"06:34.680 ","End":"06:40.380","Text":"d_2 by dx^2 of A_x"},{"Start":"06:40.380 ","End":"06:46.950","Text":"plus d_2 by d_y^2 of A_x,"},{"Start":"06:46.950 ","End":"06:54.675","Text":"plus d_2 by d_z^2 by A_x."},{"Start":"06:54.675 ","End":"07:00.770","Text":"Of course, we can do this exact same thing to A_y and A_z to each component."},{"Start":"07:00.770 ","End":"07:08.193","Text":"Just here we\u0027d switch with A_y or A_z."},{"Start":"07:08.193 ","End":"07:11.740","Text":"In other words, for the x component,"},{"Start":"07:11.740 ","End":"07:14.095","Text":"we take the second derivative."},{"Start":"07:14.095 ","End":"07:19.846","Text":"We take the derivative twice of the x component with respect to x,"},{"Start":"07:19.846 ","End":"07:22.660","Text":"and then we add it to"},{"Start":"07:22.660 ","End":"07:28.930","Text":"the second derivative of the same components just with respect to y."},{"Start":"07:28.930 ","End":"07:36.595","Text":"We add all of that onto the second derivative of the same component with respect to z."},{"Start":"07:36.595 ","End":"07:40.120","Text":"Then all of that is in the x-direction."},{"Start":"07:40.120 ","End":"07:45.050","Text":"Then we do the same exact thing with A_y and A_z."},{"Start":"07:46.470 ","End":"07:49.405","Text":"Back to this. Now we can see,"},{"Start":"07:49.405 ","End":"07:53.590","Text":"this equation over here is a little bit complicated."},{"Start":"07:53.590 ","End":"07:59.935","Text":"We\u0027re going to try and write it in a simpler fashion."},{"Start":"07:59.935 ","End":"08:04.426","Text":"Just as we\u0027ve seen before when dealing with a scalar field,"},{"Start":"08:04.426 ","End":"08:09.160","Text":"let\u0027s say we have this scalar potential and we add"},{"Start":"08:09.160 ","End":"08:16.055","Text":"some constant C. This is equal to some potential phi tag."},{"Start":"08:16.055 ","End":"08:26.300","Text":"We\u0027ve seen that the electric field is equal to the gradient of this potential phi,"},{"Start":"08:26.300 ","End":"08:31.555","Text":"and it\u0027s also equal to the gradient of phi tag."},{"Start":"08:31.555 ","End":"08:35.245","Text":"Why is this? Because if we take the gradient, of course,"},{"Start":"08:35.245 ","End":"08:37.285","Text":"of either phi or phi tag,"},{"Start":"08:37.285 ","End":"08:41.530","Text":"this constant just disappears because it\u0027s a constant."},{"Start":"08:41.530 ","End":"08:44.155","Text":"What\u0027s important is that our electric field is"},{"Start":"08:44.155 ","End":"08:47.469","Text":"equal to both because when we add the constant,"},{"Start":"08:47.469 ","End":"08:50.050","Text":"when we take the derivative,"},{"Start":"08:50.050 ","End":"08:53.210","Text":"our equation doesn\u0027t change."},{"Start":"08:53.820 ","End":"09:00.280","Text":"The electric field has to be equal to the same thing over here,"},{"Start":"09:00.280 ","End":"09:03.055","Text":"either to this potential or to this potential,"},{"Start":"09:03.055 ","End":"09:08.300","Text":"which have to be equal when using either potential."},{"Start":"09:08.550 ","End":"09:10.825","Text":"In a similar way,"},{"Start":"09:10.825 ","End":"09:15.040","Text":"we want to have the same effect when using a vector field."},{"Start":"09:15.040 ","End":"09:20.650","Text":"We want to take some vector field a and add some constant."},{"Start":"09:20.650 ","End":"09:25.870","Text":"Of course, it\u0027s going to be some scalar function."},{"Start":"09:25.870 ","End":"09:29.065","Text":"We can take the divergence of"},{"Start":"09:29.065 ","End":"09:34.210","Text":"a scalar function f. This can be equal to vector field A tag."},{"Start":"09:34.210 ","End":"09:35.740","Text":"In the same way, we want,"},{"Start":"09:35.740 ","End":"09:43.660","Text":"let\u0027s say our magnetic field to be equal to so if we take the rotor of A,"},{"Start":"09:43.660 ","End":"09:54.010","Text":"we want this to be the exact same thing as if we would take the rotor of A tag."},{"Start":"09:54.010 ","End":"09:59.950","Text":"Because when we take the rotor of f of this nabla F,"},{"Start":"09:59.950 ","End":"10:02.680","Text":"it will be equal to 0."},{"Start":"10:02.680 ","End":"10:05.335","Text":"What we get over here,"},{"Start":"10:05.335 ","End":"10:13.000","Text":"this will be equal to the rotor"},{"Start":"10:13.000 ","End":"10:22.360","Text":"of A plus this,"},{"Start":"10:22.360 ","End":"10:30.655","Text":"which is just equal to the rotor of A plus"},{"Start":"10:30.655 ","End":"10:38.335","Text":"the rotor of nabla"},{"Start":"10:38.335 ","End":"10:46.078","Text":"F. There is an identity that says that this is always equal to 0,"},{"Start":"10:46.078 ","End":"10:49.210","Text":"so this will cancel out."},{"Start":"10:49.210 ","End":"10:56.470","Text":"The rotor of the divergence of a scalar function is always equal to 0."},{"Start":"10:56.470 ","End":"11:03.382","Text":"Then we get that this is just equal to the rotor of A,"},{"Start":"11:03.382 ","End":"11:07.370","Text":"and this is of course equal to this."},{"Start":"11:11.820 ","End":"11:15.309","Text":"This is what is important over here,"},{"Start":"11:15.309 ","End":"11:19.840","Text":"to see that we can add this constant like"},{"Start":"11:19.840 ","End":"11:24.460","Text":"in the scalar field and that we get that they are equal."},{"Start":"11:24.460 ","End":"11:29.845","Text":"We get the same field by either the electric field or the magnetic field,"},{"Start":"11:29.845 ","End":"11:34.970","Text":"whether we add the constant or not."},{"Start":"11:36.480 ","End":"11:39.760","Text":"Using what we have over here,"},{"Start":"11:39.760 ","End":"11:43.075","Text":"what we just saw, scroll down,"},{"Start":"11:43.075 ","End":"11:49.135","Text":"so what we can see is that we can always choose"},{"Start":"11:49.135 ","End":"11:52.780","Text":"this vector field A such that div"},{"Start":"11:52.780 ","End":"12:00.050","Text":"A will be equal to 0 and our magnetic field will not change."},{"Start":"12:00.660 ","End":"12:07.810","Text":"Seeing how we can have that these are the same,"},{"Start":"12:07.810 ","End":"12:10.210","Text":"we can always therefore choose this"},{"Start":"12:10.210 ","End":"12:15.295","Text":"a vector such that the divergence of A will equal to 0,"},{"Start":"12:15.295 ","End":"12:19.930","Text":"and therefore our magnetic field won\u0027t change."},{"Start":"12:19.930 ","End":"12:22.255","Text":"How can we find such an A?"},{"Start":"12:22.255 ","End":"12:28.758","Text":"We find which scalar function where we do the divergence to it,"},{"Start":"12:28.758 ","End":"12:33.260","Text":"it will cancel out this A vector."},{"Start":"12:33.630 ","End":"12:40.008","Text":"We find such a function where this cancels out with this,"},{"Start":"12:40.008 ","End":"12:46.885","Text":"and then we can find the A vector such that its divergence will be 0,"},{"Start":"12:46.885 ","End":"12:51.320","Text":"which means that its magnetic field will not change."},{"Start":"12:51.510 ","End":"12:59.000","Text":"Sorry, what I meant to say is that this div F will cancel out with div of this A."},{"Start":"13:02.190 ","End":"13:05.260","Text":"Given this over here,"},{"Start":"13:05.260 ","End":"13:07.630","Text":"if we look at our equation again,"},{"Start":"13:07.630 ","End":"13:12.100","Text":"if we can choose an A vector such that its divergence is"},{"Start":"13:12.100 ","End":"13:16.660","Text":"equal to 0 and we still get the same magnetic field,"},{"Start":"13:16.660 ","End":"13:21.325","Text":"then we can choose this A such that this over here is equal to 0."},{"Start":"13:21.325 ","End":"13:26.230","Text":"We can say that this is equal to 0 because of this."},{"Start":"13:26.230 ","End":"13:32.080","Text":"In that case, all of this cancels out and we can move the minus to the other side."},{"Start":"13:32.080 ","End":"13:40.570","Text":"Then we\u0027re left with the Laplacian of this vector field A as being equal to negative,"},{"Start":"13:40.570 ","End":"13:45.220","Text":"from this negative over here, Mu naught J."},{"Start":"13:45.220 ","End":"13:49.480","Text":"Then we\u0027re left with a similar equation"},{"Start":"13:49.480 ","End":"13:55.915","Text":"to what we saw over here when dealing with the scalar potential."},{"Start":"13:55.915 ","End":"13:59.380","Text":"Here we have the Laplacian of scalar potential which"},{"Start":"13:59.380 ","End":"14:03.445","Text":"is equal to negative Rho over Epsilon naught."},{"Start":"14:03.445 ","End":"14:08.710","Text":"Similarly, we have the Laplacian of the vector potential,"},{"Start":"14:08.710 ","End":"14:12.415","Text":"which is equal to negative Mu naught J."},{"Start":"14:12.415 ","End":"14:16.225","Text":"We have similar type of equation."},{"Start":"14:16.225 ","End":"14:20.780","Text":"This is an equation to write on your equation sheets."},{"Start":"14:20.780 ","End":"14:22.250","Text":"As we can see,"},{"Start":"14:22.250 ","End":"14:26.840","Text":"we\u0027ve simplified this more complicated equation over here to"},{"Start":"14:26.840 ","End":"14:33.100","Text":"get Ampere\u0027s law when dealing with the vector potential."},{"Start":"14:33.100 ","End":"14:36.200","Text":"That is the end of this lesson."}],"ID":21452},{"Watched":false,"Name":"1.2 Exercise - Finding the Current Density","Duration":"8m 26s","ChapterTopicVideoID":21382,"CourseChapterTopicPlaylistID":99495,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.470","Text":"Hello. In this lesson we\u0027re going to be answering the following question."},{"Start":"00:04.470 ","End":"00:10.545","Text":"To find the current density J that created the potential A,"},{"Start":"00:10.545 ","End":"00:14.220","Text":"which is equal to C in the Phi direction,"},{"Start":"00:14.220 ","End":"00:16.590","Text":"which is given in cylindrical coordinates,"},{"Start":"00:16.590 ","End":"00:19.230","Text":"where C is a constant."},{"Start":"00:19.230 ","End":"00:23.790","Text":"We\u0027re currently trying to find J."},{"Start":"00:23.790 ","End":"00:32.550","Text":"There is 2 ways either we can calculate the magnetic field by taking the rotor of A."},{"Start":"00:32.550 ","End":"00:38.170","Text":"Then when we take the rotor of B,"},{"Start":"00:39.710 ","End":"00:47.090","Text":"we will get Mu_naught J and then we can isolate out the J."},{"Start":"00:47.090 ","End":"00:49.765","Text":"This is method number 1."},{"Start":"00:49.765 ","End":"00:55.400","Text":"Method number 2 is what we saw in the previous lesson,"},{"Start":"00:55.400 ","End":"00:59.665","Text":"where we can take the Laplacian of A,"},{"Start":"00:59.665 ","End":"01:03.000","Text":"which as we know, is equal to Mu_naught J,"},{"Start":"01:03.000 ","End":"01:05.355","Text":"and then we can isolate out J."},{"Start":"01:05.355 ","End":"01:07.460","Text":"We\u0027re going to solve this both ways,"},{"Start":"01:07.460 ","End":"01:09.890","Text":"so we can just see how to solve it."},{"Start":"01:09.890 ","End":"01:13.680","Text":"Let\u0027s first solve it via method 1."},{"Start":"01:14.060 ","End":"01:18.675","Text":"First, we want to find the magnetic field,"},{"Start":"01:18.675 ","End":"01:25.266","Text":"as we saw by taking the rotor of A,"},{"Start":"01:25.266 ","End":"01:31.730","Text":"this of course, we know that A is only in the Phi direction,"},{"Start":"01:31.730 ","End":"01:33.750","Text":"so it only has a Phi component."},{"Start":"01:33.750 ","End":"01:40.310","Text":"We can say that A_r is equal to A_z because we\u0027re in cylindrical coordinates,"},{"Start":"01:40.310 ","End":"01:43.180","Text":"which is equal to 0."},{"Start":"01:43.180 ","End":"01:50.100","Text":"Therefore, if I just plug in the equation for the rotor."},{"Start":"01:50.100 ","End":"01:54.030","Text":"Remember we have Nabla cross multiplied with A."},{"Start":"01:54.030 ","End":"02:04.920","Text":"What we\u0027ll have is negative dAPhi by dz in the r-direction"},{"Start":"02:04.920 ","End":"02:11.370","Text":"plus 1 divided by r of d by dr"},{"Start":"02:11.370 ","End":"02:18.855","Text":"of rA in the Phi component of A in the z-direction."},{"Start":"02:18.855 ","End":"02:23.555","Text":"This is what we\u0027ll be left with because all our other components for A is 0."},{"Start":"02:23.555 ","End":"02:26.090","Text":"Now this, as we can see,"},{"Start":"02:26.090 ","End":"02:30.830","Text":"A of Phi component is just a constant in the Phi direction."},{"Start":"02:30.830 ","End":"02:33.455","Text":"When we take its derivative with respect to z,"},{"Start":"02:33.455 ","End":"02:37.500","Text":"this will be equal to 0 because there\u0027s no z variable."},{"Start":"02:37.500 ","End":"02:41.970","Text":"We\u0027re just going to be left with 1 divided by r of"},{"Start":"02:41.970 ","End":"02:47.225","Text":"d by dr of r multiplied by A in the Phi direction,"},{"Start":"02:47.225 ","End":"02:52.290","Text":"which is just the constant C in the z-direction."},{"Start":"02:52.290 ","End":"02:54.560","Text":"When we take the derivative of rC,"},{"Start":"02:54.560 ","End":"03:03.480","Text":"we\u0027re just going to be left with C. What we have is C divided by r in the z-direction."},{"Start":"03:04.760 ","End":"03:08.205","Text":"This is our magnetic field and now,"},{"Start":"03:08.205 ","End":"03:14.310","Text":"in order to get J is equal to 1"},{"Start":"03:14.310 ","End":"03:21.480","Text":"divided by Mu_naught of the rotor of B."},{"Start":"03:21.480 ","End":"03:24.935","Text":"The same thing, I\u0027m not going to write out the whole equation,"},{"Start":"03:24.935 ","End":"03:26.495","Text":"because as we know,"},{"Start":"03:26.495 ","End":"03:32.465","Text":"we have, our B field has only a z component."},{"Start":"03:32.465 ","End":"03:41.835","Text":"That means that our Br is equal to our BPhi component which is equal to 0."},{"Start":"03:41.835 ","End":"03:50.730","Text":"Therefore, what we are left with is 1 divided by Mu_naught."},{"Start":"03:50.730 ","End":"03:53.630","Text":"Then for the rotor of B,"},{"Start":"03:53.630 ","End":"03:58.910","Text":"we\u0027re left with 1 divided by r of dBz,"},{"Start":"03:58.910 ","End":"04:08.475","Text":"the z component divided by dPhi in the r-direction"},{"Start":"04:08.475 ","End":"04:18.930","Text":"minus dBz by dr in the Phi direction."},{"Start":"04:18.930 ","End":"04:24.305","Text":"Here we can look at our magnetic field,"},{"Start":"04:24.305 ","End":"04:29.400","Text":"and we can see that there\u0027s no variable Phi."},{"Start":"04:29.400 ","End":"04:35.180","Text":"We only have the variable r. If we take the derivative of the z component,"},{"Start":"04:35.180 ","End":"04:38.505","Text":"which is C divided by r by Phi it will be equal to 0."},{"Start":"04:38.505 ","End":"04:40.430","Text":"All of this is equal to 0."},{"Start":"04:40.430 ","End":"04:45.525","Text":"Then we\u0027re left with 1 divided by Mu_naught of"},{"Start":"04:45.525 ","End":"04:55.748","Text":"negative d by dr of Bz which is C divided by r,"},{"Start":"04:55.748 ","End":"05:01.020","Text":"and this is all going to be in the Phi direction."},{"Start":"05:01.020 ","End":"05:11.710","Text":"C divided by r is going to be equal to negative C divided by r^2."},{"Start":"05:11.710 ","End":"05:15.410","Text":"Then the negative will cancel out with this negative over here."},{"Start":"05:15.410 ","End":"05:24.425","Text":"What we\u0027ll get is that J is equal to 1 divided by Mu_naught."},{"Start":"05:24.425 ","End":"05:28.310","Text":"The negative and the negative cancel out of C divided by"},{"Start":"05:28.310 ","End":"05:36.330","Text":"r^2 in the Phi direction."},{"Start":"05:37.660 ","End":"05:43.100","Text":"This was the first method finding first"},{"Start":"05:43.100 ","End":"05:48.740","Text":"the magnetic field from our vector potential and then through our magnetic field,"},{"Start":"05:48.740 ","End":"05:50.810","Text":"finding our current density."},{"Start":"05:50.810 ","End":"05:52.640","Text":"Now we\u0027ll do method 2,"},{"Start":"05:52.640 ","End":"05:57.750","Text":"which is by using the Laplacian of a vector potential."},{"Start":"05:57.950 ","End":"06:00.080","Text":"If we rearrange this,"},{"Start":"06:00.080 ","End":"06:09.420","Text":"we get that J is equal to 1 divided by Mu_naught multiplied by the Laplacian of A,"},{"Start":"06:09.420 ","End":"06:11.250","Text":"of a vector potential."},{"Start":"06:11.250 ","End":"06:14.150","Text":"This is 1 divided by Mu_naught."},{"Start":"06:14.150 ","End":"06:17.465","Text":"If we look in our equation sheets for"},{"Start":"06:17.465 ","End":"06:21.740","Text":"the Laplacian when we\u0027re dealing with cylindrical coordinates."},{"Start":"06:21.740 ","End":"06:24.140","Text":"If you don\u0027t have this on your equation sheet,"},{"Start":"06:24.140 ","End":"06:27.095","Text":"please look it up online."},{"Start":"06:27.095 ","End":"06:29.720","Text":"For the Laplacian in all coordinates,"},{"Start":"06:29.720 ","End":"06:33.620","Text":"cartesian, cylindrical, and spherical."},{"Start":"06:33.620 ","End":"06:36.995","Text":"What will be left with here, again,"},{"Start":"06:36.995 ","End":"06:43.665","Text":"where A_z is equal to A_r which is equal to 0."},{"Start":"06:43.665 ","End":"06:49.665","Text":"Because we only have the Phi coordinates for our vector potential."},{"Start":"06:49.665 ","End":"06:53.720","Text":"What this is equal to over here,now I\u0027m doing the Laplacian,"},{"Start":"06:53.720 ","End":"07:04.175","Text":"is negative 2 divided by r^2 multiplied by dAPhi"},{"Start":"07:04.175 ","End":"07:13.895","Text":"by dPhi minus APhi"},{"Start":"07:13.895 ","End":"07:20.250","Text":"divided by r^2 in the Phi direction."},{"Start":"07:21.320 ","End":"07:26.550","Text":"Our Phi component of the A vector is a constant."},{"Start":"07:26.550 ","End":"07:29.530","Text":"If I take its derivative with respect to Phi,"},{"Start":"07:29.530 ","End":"07:31.545","Text":"it will be equal to 0."},{"Start":"07:31.545 ","End":"07:33.435","Text":"This here is 0,"},{"Start":"07:33.435 ","End":"07:37.560","Text":"and then what we\u0027re left with is A Phi."},{"Start":"07:37.560 ","End":"07:44.210","Text":"We have just C divided by r^2 of course,"},{"Start":"07:44.210 ","End":"07:47.450","Text":"we have to take into account the one divided by Mu_naught."},{"Start":"07:47.450 ","End":"07:52.610","Text":"We have one divided by Mu_naught and then that is multiplied by A Phi,"},{"Start":"07:52.610 ","End":"08:02.415","Text":"which is C divided by r^2 and again in the Phi direction."},{"Start":"08:02.415 ","End":"08:07.985","Text":"This is the answer and we can see that we got the same answer using both methods."},{"Start":"08:07.985 ","End":"08:12.035","Text":"Of course, these 2 methods are correct so long as"},{"Start":"08:12.035 ","End":"08:17.990","Text":"the divergence of A over the vector potential is equal to 0."},{"Start":"08:17.990 ","End":"08:20.540","Text":"Which if we plug this into this equation,"},{"Start":"08:20.540 ","End":"08:22.955","Text":"we see that it\u0027s equal to 0."},{"Start":"08:22.955 ","End":"08:27.030","Text":"That is the end of this lesson."}],"ID":21453},{"Watched":false,"Name":"1.3 Poisson_s Equation for Vector Potential","Duration":"10m 38s","ChapterTopicVideoID":21383,"CourseChapterTopicPlaylistID":99495,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.115","Text":"Hello. In this lesson,"},{"Start":"00:02.115 ","End":"00:06.425","Text":"we\u0027re going to be speaking about Poisson\u0027s equation for vector potential."},{"Start":"00:06.425 ","End":"00:13.020","Text":"Up until now, we\u0027ve discussed how we can calculate the magnetic field"},{"Start":"00:13.020 ","End":"00:20.060","Text":"by taking the rotor of the vector potential A,"},{"Start":"00:20.060 ","End":"00:26.415","Text":"where the vector potential is some vector,"},{"Start":"00:26.415 ","End":"00:30.335","Text":"and it\u0027s some function that through it,"},{"Start":"00:30.335 ","End":"00:34.535","Text":"we can calculate the magnetic field that is present."},{"Start":"00:34.535 ","End":"00:37.235","Text":"Then from this equation,"},{"Start":"00:37.235 ","End":"00:43.970","Text":"we got via using Ampere\u0027s Law, the Laplacian equation,"},{"Start":"00:43.970 ","End":"00:47.000","Text":"where we got that the Laplacian of"},{"Start":"00:47.000 ","End":"00:53.440","Text":"this vector potential A is equal to negative Mu_naught J,"},{"Start":"00:53.440 ","End":"00:57.420","Text":"where J is of course, the current density."},{"Start":"00:57.790 ","End":"01:00.530","Text":"In a previous lesson,"},{"Start":"01:00.530 ","End":"01:03.805","Text":"we discussed what the Laplacian means,"},{"Start":"01:03.805 ","End":"01:06.695","Text":"and we saw that the Laplacian of"},{"Start":"01:06.695 ","End":"01:12.110","Text":"a vector potential is different to the Laplacian of a scalar potential,"},{"Start":"01:12.110 ","End":"01:17.525","Text":"isn\u0027t just the divergence of the gradient of it."},{"Start":"01:17.525 ","End":"01:21.350","Text":"What we do is we do the Laplacian,"},{"Start":"01:21.350 ","End":"01:28.860","Text":"we take this derivative on each component of this vector field A."},{"Start":"01:29.890 ","End":"01:40.005","Text":"In other words, what this means is we can take the Laplacian of the x component A_x,"},{"Start":"01:40.005 ","End":"01:44.770","Text":"and we\u0027ll get negative Mu_naught of J_x,"},{"Start":"01:45.470 ","End":"01:48.465","Text":"the x component of J,"},{"Start":"01:48.465 ","End":"01:53.840","Text":"and of course we can do the same thing for the next component of A_y,"},{"Start":"01:53.840 ","End":"01:58.060","Text":"and we\u0027ll get that this is equal to negative Mu_naught J_y,"},{"Start":"01:58.060 ","End":"02:00.105","Text":"and of course for z,"},{"Start":"02:00.105 ","End":"02:07.270","Text":"the Laplacian of A_z is equal to negative Mu_naught J_z."},{"Start":"02:07.310 ","End":"02:12.390","Text":"All of this we\u0027ve seen in the previous lessons."},{"Start":"02:13.010 ","End":"02:23.190","Text":"Each one of these is Poisson\u0027s equation for a scalar potential."},{"Start":"02:23.530 ","End":"02:26.945","Text":"Because if we look at this component,"},{"Start":"02:26.945 ","End":"02:32.525","Text":"so this is A_x, the x component of our vector field A."},{"Start":"02:32.525 ","End":"02:35.555","Text":"However, even though it\u0027s the x-component,"},{"Start":"02:35.555 ","End":"02:37.215","Text":"it isn\u0027t a vector,"},{"Start":"02:37.215 ","End":"02:39.795","Text":"it itself is a scalar."},{"Start":"02:39.795 ","End":"02:42.561","Text":"Similarly for A_y and A_z,"},{"Start":"02:42.561 ","End":"02:45.515","Text":"they\u0027re just the different components of this vector."},{"Start":"02:45.515 ","End":"02:49.760","Text":"However, they themselves are scalar values as seen."},{"Start":"02:49.760 ","End":"02:54.530","Text":"We don\u0027t have the arrow on top showing that it\u0027s a vector."},{"Start":"02:54.530 ","End":"03:01.020","Text":"What we have are various Poisson equations."},{"Start":"03:01.640 ","End":"03:08.390","Text":"We\u0027ve seen this before when we had Poisson\u0027s equation for a scalar potential,"},{"Start":"03:08.390 ","End":"03:12.160","Text":"so let\u0027s just write Poisson\u0027s equation for scalar potential."},{"Start":"03:12.160 ","End":"03:19.250","Text":"We had the Laplacian of this scalar potential in electricity,"},{"Start":"03:19.250 ","End":"03:25.970","Text":"which was equal to negative the charge density divided by Epsilon_naught."},{"Start":"03:25.970 ","End":"03:31.100","Text":"Then we saw that in order to find this scalar potential,"},{"Start":"03:31.100 ","End":"03:36.255","Text":"we had 1 divided by 4Pi Epsilon naught"},{"Start":"03:36.255 ","End":"03:43.245","Text":"multiplied by the integral of Rho dv tag,"},{"Start":"03:43.245 ","End":"03:52.350","Text":"divided by the magnitude of the vectors r minus r tag."},{"Start":"03:53.960 ","End":"03:57.950","Text":"In other words, this over here,"},{"Start":"03:57.950 ","End":"04:01.725","Text":"the numerator is our dq,"},{"Start":"04:01.725 ","End":"04:04.125","Text":"our element of charge,"},{"Start":"04:04.125 ","End":"04:08.025","Text":"and this is the radius, the distance."},{"Start":"04:08.025 ","End":"04:15.350","Text":"This is how we can get the potential for the electric field from,"},{"Start":"04:15.350 ","End":"04:19.265","Text":"let\u0027s say, a point charge or some shape."},{"Start":"04:19.265 ","End":"04:24.745","Text":"Similarly here, this is the exact same equation."},{"Start":"04:24.745 ","End":"04:28.450","Text":"Just here, instead of Phi, we have A_x."},{"Start":"04:28.450 ","End":"04:36.210","Text":"In other words, we can calculate A_x via Poisson\u0027s equation over here."},{"Start":"04:36.210 ","End":"04:46.680","Text":"The units is just Mu_ divided by 4 Pi,"},{"Start":"04:46.680 ","End":"04:52.305","Text":"and this is just multiplied by the integral of"},{"Start":"04:52.305 ","End":"04:58.680","Text":"J_x divided by the magnitude of r minus r tag,"},{"Start":"04:58.680 ","End":"05:02.470","Text":"and all of this dv tag."},{"Start":"05:03.080 ","End":"05:07.495","Text":"Then I can write the same thing for all the other components,"},{"Start":"05:07.495 ","End":"05:11.026","Text":"so let\u0027s say A_y would be the exact same thing;"},{"Start":"05:11.026 ","End":"05:16.690","Text":"Mu_naught divided by 4 Pi multiplied by the integral"},{"Start":"05:16.690 ","End":"05:22.525","Text":"of J_y divided by r minus r tag,"},{"Start":"05:22.525 ","End":"05:25.745","Text":"the magnitude dv tag."},{"Start":"05:25.745 ","End":"05:30.490","Text":"Of course, I\u0027ll have the exact same thing for A_z."},{"Start":"05:30.490 ","End":"05:35.950","Text":"What we can do is we can put all of these together and write out"},{"Start":"05:35.950 ","End":"05:41.920","Text":"the equation for the vector potential A at some radius r,"},{"Start":"05:41.920 ","End":"05:48.155","Text":"so it will be equal to Mu_naught divided by 4 Pi multiplied by the"},{"Start":"05:48.155 ","End":"05:55.485","Text":"integral of J vector at point r tag,"},{"Start":"05:55.485 ","End":"05:56.870","Text":"so where the current is,"},{"Start":"05:56.870 ","End":"06:03.870","Text":"the current is at r tag divided by r minus r tag,"},{"Start":"06:03.870 ","End":"06:09.360","Text":"and the magnitude of this distance, dV tag."},{"Start":"06:09.740 ","End":"06:12.425","Text":"This distance over here,"},{"Start":"06:12.425 ","End":"06:16.880","Text":"r minus r tag is the distance between the current,"},{"Start":"06:16.880 ","End":"06:19.400","Text":"the location of all the currents,"},{"Start":"06:19.400 ","End":"06:26.760","Text":"and the point r where I am calculating the vector potential."},{"Start":"06:28.820 ","End":"06:32.525","Text":"It\u0027s just the distance between where the current is"},{"Start":"06:32.525 ","End":"06:37.480","Text":"and the point in space that I\u0027m calculating this whole equation for."},{"Start":"06:37.480 ","End":"06:42.395","Text":"In other words, if here I have my origin,"},{"Start":"06:42.395 ","End":"06:48.000","Text":"and here I have an area with some current density J,"},{"Start":"06:48.000 ","End":"06:54.800","Text":"and I want to check my vector potential at this point over here,"},{"Start":"06:54.800 ","End":"07:00.875","Text":"so what we\u0027ll have is that this vector to this point over here is our r vector."},{"Start":"07:00.875 ","End":"07:10.155","Text":"The vector that will sum over every single small volume of this J,"},{"Start":"07:10.155 ","End":"07:11.790","Text":"is our r tag,"},{"Start":"07:11.790 ","End":"07:13.879","Text":"so r tag is changing,"},{"Start":"07:13.879 ","End":"07:17.390","Text":"and it\u0027s constantly summing up the current."},{"Start":"07:17.390 ","End":"07:21.335","Text":"Then if we have r minus r tag,"},{"Start":"07:21.335 ","End":"07:27.245","Text":"so that is going to give us this vector over here,"},{"Start":"07:27.245 ","End":"07:31.890","Text":"so this is r minus r tag."},{"Start":"07:32.690 ","End":"07:34.905","Text":"Or we can write it rather,"},{"Start":"07:34.905 ","End":"07:36.260","Text":"where here we go,"},{"Start":"07:36.260 ","End":"07:38.135","Text":"minus r tag,"},{"Start":"07:38.135 ","End":"07:41.645","Text":"so that will take us to the origin plus r,"},{"Start":"07:41.645 ","End":"07:44.810","Text":"so we get to this point over here."},{"Start":"07:44.810 ","End":"07:47.515","Text":"Instead of going like so,"},{"Start":"07:47.515 ","End":"07:49.740","Text":"we can just go straight like this,"},{"Start":"07:49.740 ","End":"07:53.090","Text":"and that gives us this vector over here."},{"Start":"07:53.090 ","End":"07:57.245","Text":"So the distance between where our current is"},{"Start":"07:57.245 ","End":"08:01.640","Text":"and the point in space where I\u0027m calculating my vector potential."},{"Start":"08:01.640 ","End":"08:05.400","Text":"That\u0027s what this r minus r tag is."},{"Start":"08:06.590 ","End":"08:10.865","Text":"This is the vector potential given J,"},{"Start":"08:10.865 ","End":"08:16.680","Text":"which is current density per unit area,"},{"Start":"08:16.680 ","End":"08:19.640","Text":"we spoke about this in one of the previous chapters."},{"Start":"08:19.640 ","End":"08:22.100","Text":"If you don\u0027t remember, please go back."},{"Start":"08:22.100 ","End":"08:25.010","Text":"But what happens if we have k,"},{"Start":"08:25.010 ","End":"08:28.580","Text":"which is current density per unit length?"},{"Start":"08:28.580 ","End":"08:33.700","Text":"Our A at some point r,"},{"Start":"08:33.700 ","End":"08:38.020","Text":"is going to be equal to Mu_naught divided by"},{"Start":"08:38.020 ","End":"08:45.890","Text":"4 Pi multiplied by the integral of k at the certain point."},{"Start":"08:45.890 ","End":"08:55.350","Text":"Again, r tag divided by r minus r tag magnitude."},{"Start":"08:55.350 ","End":"08:58.545","Text":"But this time it\u0027s ds tag,"},{"Start":"08:58.545 ","End":"09:01.685","Text":"because this is per unit length,"},{"Start":"09:01.685 ","End":"09:04.265","Text":"we spoke about this."},{"Start":"09:04.265 ","End":"09:10.190","Text":"If we have the vector potential and we\u0027re given I,"},{"Start":"09:10.190 ","End":"09:17.179","Text":"which is the current density for a current traveling along a wire,"},{"Start":"09:17.179 ","End":"09:24.860","Text":"so this is going to be equal to Mu_naught divided by 4 Pi multiplied by the"},{"Start":"09:24.860 ","End":"09:31.100","Text":"integral of I dl tag"},{"Start":"09:31.100 ","End":"09:41.400","Text":"divided by r minus r tag vector."},{"Start":"09:42.610 ","End":"09:47.945","Text":"These are 3 equations for your equation sheets."},{"Start":"09:47.945 ","End":"09:49.310","Text":"Please write them down."},{"Start":"09:49.310 ","End":"09:50.465","Text":"Now a little note,"},{"Start":"09:50.465 ","End":"09:52.850","Text":"the current density is over here,"},{"Start":"09:52.850 ","End":"09:56.810","Text":"have to be some finite density."},{"Start":"09:56.810 ","End":"10:01.505","Text":"We can\u0027t have an infinite current density be at J, k,"},{"Start":"10:01.505 ","End":"10:06.500","Text":"or I, and if we do have an infinite current density,"},{"Start":"10:06.500 ","End":"10:10.230","Text":"then we cannot use these equations."},{"Start":"10:11.240 ","End":"10:16.910","Text":"The current density must be finite to use these equations,"},{"Start":"10:16.910 ","End":"10:20.460","Text":"so be at J, k, or I."},{"Start":"10:20.630 ","End":"10:25.790","Text":"That is the end of this lesson explaining how"},{"Start":"10:25.790 ","End":"10:30.600","Text":"to calculate the vector potential from current density,"},{"Start":"10:30.600 ","End":"10:32.075","Text":"and in the next lesson,"},{"Start":"10:32.075 ","End":"10:35.660","Text":"we\u0027re going to do a worked example."},{"Start":"10:35.660 ","End":"10:38.460","Text":"That\u0027s the end of this lesson."}],"ID":21454},{"Watched":false,"Name":"1.4 Exercise - Vector Potential of Finite Wire","Duration":"25m 9s","ChapterTopicVideoID":21384,"CourseChapterTopicPlaylistID":99495,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.355","Text":"Hello. In this lesson,"},{"Start":"00:02.355 ","End":"00:04.710","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.710 ","End":"00:11.235","Text":"We have a finite wire of length L with a current I flowing through it."},{"Start":"00:11.235 ","End":"00:14.400","Text":"It\u0027s placed along the z-axis."},{"Start":"00:14.400 ","End":"00:19.465","Text":"Question Number 1 is to calculate the vector potential."},{"Start":"00:19.465 ","End":"00:26.060","Text":"Let\u0027s write out our equation for vector potential that we saw in the previous lesson."},{"Start":"00:26.060 ","End":"00:29.585","Text":"We saw that the vector potential at some kind of vector I,"},{"Start":"00:29.585 ","End":"00:35.195","Text":"we\u0027ll write what that is in a minute is equal to Mu naught"},{"Start":"00:35.195 ","End":"00:40.960","Text":"divided by 4Pi multiplied by the integral of the current."},{"Start":"00:40.960 ","End":"00:42.650","Text":"So here it\u0027s I."},{"Start":"00:42.650 ","End":"00:47.390","Text":"Then if we have I, we remember that we have dl vector and of course,"},{"Start":"00:47.390 ","End":"00:49.385","Text":"the l is a tag,"},{"Start":"00:49.385 ","End":"00:57.775","Text":"divided by the magnitude of this I vector minus our I tag vector."},{"Start":"00:57.775 ","End":"01:04.745","Text":"What we want to do is we want to calculate the potential at some kind of point."},{"Start":"01:04.745 ","End":"01:09.365","Text":"This is where we want to calculate this at some kind of point,"},{"Start":"01:09.365 ","End":"01:11.630","Text":"which is in space,"},{"Start":"01:11.630 ","End":"01:13.550","Text":"it has some x kind of value,"},{"Start":"01:13.550 ","End":"01:14.870","Text":"some kind of y value,"},{"Start":"01:14.870 ","End":"01:17.400","Text":"and some kind of z value."},{"Start":"01:18.040 ","End":"01:22.265","Text":"If we take this to be at the origin,"},{"Start":"01:22.265 ","End":"01:25.703","Text":"we can say that this is our I vector."},{"Start":"01:25.703 ","End":"01:29.045","Text":"It\u0027s the vector from the origin pointing to"},{"Start":"01:29.045 ","End":"01:34.615","Text":"the place that we want to calculate the vector potential."},{"Start":"01:34.615 ","End":"01:42.480","Text":"We can write out that r is simply equal to the vector,"},{"Start":"01:42.480 ","End":"01:46.630","Text":"so x, y, z."},{"Start":"01:46.760 ","End":"01:51.120","Text":"Now we want to know what the r tag vector is."},{"Start":"01:51.120 ","End":"01:55.940","Text":"If we remember, the r tag vector is the vector that"},{"Start":"01:55.940 ","End":"02:00.879","Text":"sums up across the wire all of the currents."},{"Start":"02:00.879 ","End":"02:08.120","Text":"So here it\u0027s I. It\u0027s a vector that is summing up all of the current on the wire,"},{"Start":"02:08.120 ","End":"02:11.700","Text":"and it points, of course,"},{"Start":"02:11.700 ","End":"02:14.630","Text":"in the direction of travel of the current."},{"Start":"02:14.630 ","End":"02:18.815","Text":"We can say that our r vector if we want to sum up,"},{"Start":"02:18.815 ","End":"02:23.450","Text":"let\u0027s say the current density over here."},{"Start":"02:23.450 ","End":"02:29.810","Text":"Our r tag vector is going to be this vector here from"},{"Start":"02:29.810 ","End":"02:36.815","Text":"the origin until this point over here where we\u0027re summing up the current over here."},{"Start":"02:36.815 ","End":"02:42.860","Text":"Now let\u0027s write out what our r tag vector is equal to."},{"Start":"02:42.860 ","End":"02:49.490","Text":"We can see that it\u0027s going to be some z value because we can see we\u0027re along the z-axis."},{"Start":"02:49.490 ","End":"02:52.520","Text":"But of course, the z value is constantly changing."},{"Start":"02:52.520 ","End":"02:53.975","Text":"Right now it\u0027s up to here."},{"Start":"02:53.975 ","End":"02:55.505","Text":"It can be up to here,"},{"Start":"02:55.505 ","End":"02:58.160","Text":"can be up to this area or this,"},{"Start":"02:58.160 ","End":"03:00.455","Text":"so we\u0027ll write it out as z tag."},{"Start":"03:00.455 ","End":"03:02.574","Text":"This is what\u0027s changing."},{"Start":"03:02.574 ","End":"03:06.980","Text":"Of course, this is a vector so what direction is it in?"},{"Start":"03:06.980 ","End":"03:10.900","Text":"It\u0027s in the z-hat direction."},{"Start":"03:10.900 ","End":"03:13.580","Text":"This is our I tag vector."},{"Start":"03:13.580 ","End":"03:15.635","Text":"We can notice that"},{"Start":"03:15.635 ","End":"03:18.539","Text":"our dl tag vector"},{"Start":"03:21.520 ","End":"03:26.103","Text":"is also equal to this over here."},{"Start":"03:26.103 ","End":"03:28.415","Text":"We have this dl,"},{"Start":"03:28.415 ","End":"03:31.581","Text":"this is some length."},{"Start":"03:31.581 ","End":"03:34.605","Text":"So dl is this change in z."},{"Start":"03:34.605 ","End":"03:41.475","Text":"It\u0027s equal simply dz or rather dz tag."},{"Start":"03:41.475 ","End":"03:46.790","Text":"We can see that our dl vector is dz tag,"},{"Start":"03:46.790 ","End":"03:49.160","Text":"and of course, it\u0027s a vector."},{"Start":"03:49.160 ","End":"03:56.925","Text":"This also has to be a vector and we can see it\u0027s in the z-hat direction."},{"Start":"03:56.925 ","End":"04:05.510","Text":"Now we want to calculate the magnitude of r minus our r tag vector."},{"Start":"04:05.510 ","End":"04:09.840","Text":"As we know, that\u0027s going to be the square root of x."},{"Start":"04:09.840 ","End":"04:11.390","Text":"Here we don\u0027t have an x-component,"},{"Start":"04:11.390 ","End":"04:13.595","Text":"so it\u0027s just x^2, y,"},{"Start":"04:13.595 ","End":"04:15.140","Text":"again, we don\u0027t have a y component,"},{"Start":"04:15.140 ","End":"04:17.855","Text":"so it\u0027s y^2 plus."},{"Start":"04:17.855 ","End":"04:20.540","Text":"Then here we have z components in both."},{"Start":"04:20.540 ","End":"04:23.110","Text":"We have r minus r tag."},{"Start":"04:23.110 ","End":"04:30.920","Text":"We have (z minus z tag)^2,"},{"Start":"04:30.920 ","End":"04:33.335","Text":"and we take the square root of this."},{"Start":"04:33.335 ","End":"04:36.725","Text":"What I could do is I could convert this into"},{"Start":"04:36.725 ","End":"04:43.055","Text":"cylindrical coordinates and then my x^2 plus y^2 would just be equal to I^2."},{"Start":"04:43.055 ","End":"04:44.540","Text":"We could do that,"},{"Start":"04:44.540 ","End":"04:48.245","Text":"which could be useful over here because we are anyway dealing with this I."},{"Start":"04:48.245 ","End":"04:50.570","Text":"But in the meantime, I\u0027ll leave it like this."},{"Start":"04:50.570 ","End":"04:55.800","Text":"I\u0027ll just write a note that this could also be written as r^2."},{"Start":"04:56.700 ","End":"05:00.130","Text":"Now a little note with this."},{"Start":"05:00.130 ","End":"05:02.775","Text":"This z is constant."},{"Start":"05:02.775 ","End":"05:06.340","Text":"This z is speaking about the z of"},{"Start":"05:06.340 ","End":"05:11.425","Text":"the location where we\u0027re calculating the vector potential."},{"Start":"05:11.425 ","End":"05:14.780","Text":"This z is a constant."},{"Start":"05:14.780 ","End":"05:19.810","Text":"However, this z tag is a variable,"},{"Start":"05:19.810 ","End":"05:24.760","Text":"and this is what we\u0027re integrating according to z tag."},{"Start":"05:24.760 ","End":"05:28.150","Text":"That\u0027s why it\u0027s important that here you don\u0027t forget"},{"Start":"05:28.150 ","End":"05:31.450","Text":"the tag so that you don\u0027t integrate according to z."},{"Start":"05:31.450 ","End":"05:37.625","Text":"You want to integrate specifically according to the tag because that is the variable."},{"Start":"05:37.625 ","End":"05:44.880","Text":"Remember, we\u0027re summing all of the currents and this point is static."},{"Start":"05:45.110 ","End":"05:52.655","Text":"Now let\u0027s plug in the values and let\u0027s calculate the vector potential."},{"Start":"05:52.655 ","End":"06:02.450","Text":"We have that A at our point r is equal to Mu naught divided by"},{"Start":"06:02.450 ","End":"06:09.530","Text":"4Pi of the integral where our bounds are from"},{"Start":"06:09.530 ","End":"06:18.195","Text":"negative l divided by 2 until positive l divided by 2(I)."},{"Start":"06:18.195 ","End":"06:21.540","Text":"Then we have our dl tag,"},{"Start":"06:21.540 ","End":"06:30.890","Text":"which we saw was equal to dz tag in the z direction divided by our minus I tag,"},{"Start":"06:30.890 ","End":"06:35.210","Text":"which is the square root of x^2 plus y^2"},{"Start":"06:35.210 ","End":"06:42.510","Text":"plus (z minus z tag)^2."},{"Start":"06:43.240 ","End":"06:49.205","Text":"Now, what we want to do is we want to use different variables."},{"Start":"06:49.205 ","End":"06:51.080","Text":"Let\u0027s just call this over here."},{"Start":"06:51.080 ","End":"06:53.675","Text":"This is a constant, of course, x and y."},{"Start":"06:53.675 ","End":"06:56.240","Text":"Let\u0027s just label this as b."},{"Start":"06:56.240 ","End":"07:04.240","Text":"Then over here, let\u0027s say that t is equal to this over here."},{"Start":"07:04.240 ","End":"07:09.420","Text":"This is t, which is z minus z tag."},{"Start":"07:09.420 ","End":"07:12.090","Text":"Then, if we take dt,"},{"Start":"07:12.090 ","End":"07:14.610","Text":"then what will we left with,"},{"Start":"07:14.610 ","End":"07:16.350","Text":"z as a constant,"},{"Start":"07:16.350 ","End":"07:20.980","Text":"so we will be left with negative dz tag."},{"Start":"07:20.980 ","End":"07:26.045","Text":"Now let\u0027s continue like so,"},{"Start":"07:26.045 ","End":"07:28.400","Text":"where we\u0027re substituting this in."},{"Start":"07:28.400 ","End":"07:36.560","Text":"We have Mu naught divided by 4Pi,"},{"Start":"07:36.560 ","End":"07:41.060","Text":"and then we have the integral of I. I is a constant,"},{"Start":"07:41.060 ","End":"07:43.130","Text":"so we can take it outside as well."},{"Start":"07:43.130 ","End":"07:46.100","Text":"Instead of dz tag,"},{"Start":"07:46.100 ","End":"07:50.110","Text":"we are using dt."},{"Start":"07:50.110 ","End":"07:53.330","Text":"We just have dt and of course,"},{"Start":"07:53.330 ","End":"08:01.530","Text":"we have this minus because here we have dz tag and here we have dt is negative dz tag,"},{"Start":"08:01.530 ","End":"08:03.100","Text":"so this must be negative dt."},{"Start":"08:03.100 ","End":"08:05.780","Text":"We can take the negative over here."},{"Start":"08:05.780 ","End":"08:10.280","Text":"Then, we have, of course,"},{"Start":"08:10.280 ","End":"08:12.170","Text":"this is in the z-direction,"},{"Start":"08:12.170 ","End":"08:14.839","Text":"which is also constant, so we place it here."},{"Start":"08:14.839 ","End":"08:24.120","Text":"This is divided by the square root of this constant b plus t^2."},{"Start":"08:25.920 ","End":"08:32.365","Text":"Now we want to plug in the bounds for when we\u0027re using this new variable,"},{"Start":"08:32.365 ","End":"08:37.930","Text":"t. What we saw is that we have negative l divided by 2"},{"Start":"08:37.930 ","End":"08:48.170","Text":"for z tag so we have the bound from z."},{"Start":"08:50.400 ","End":"08:54.475","Text":"Then we have negative,"},{"Start":"08:54.475 ","End":"08:56.245","Text":"negative l divided by 2."},{"Start":"08:56.245 ","End":"09:04.940","Text":"That\u0027s plus l divided by 2 until z minus l divided by 2."},{"Start":"09:07.680 ","End":"09:12.700","Text":"All we\u0027ve done is we\u0027ve taken this equation over here and we\u0027ve kept z,"},{"Start":"09:12.700 ","End":"09:16.795","Text":"of course, is a constant so we leave it like so."},{"Start":"09:16.795 ","End":"09:21.220","Text":"Just in place of z tag we\u0027ve plugged in our balance."},{"Start":"09:21.220 ","End":"09:24.070","Text":"Because that tag is the variable and that\u0027s how we\u0027ve"},{"Start":"09:24.070 ","End":"09:26.830","Text":"changed our bounds for this variable"},{"Start":"09:26.830 ","End":"09:31.735","Text":"t. Now what we need to remember"},{"Start":"09:31.735 ","End":"09:38.140","Text":"is that we\u0027re integrating now this which is an immediate integral."},{"Start":"09:38.140 ","End":"09:48.610","Text":"The integral of dt divided by the square root of some constant"},{"Start":"09:48.610 ","End":"09:55.675","Text":"plus t^2 is equal to"},{"Start":"09:55.675 ","End":"10:05.150","Text":"ln of t plus the square root of t^2 plus b."},{"Start":"10:06.030 ","End":"10:17.440","Text":"This is important to remember and to even write in your equation sheets."},{"Start":"10:17.440 ","End":"10:22.990","Text":"This isn\u0027t the answer but this is a very important identity to remember."},{"Start":"10:22.990 ","End":"10:27.080","Text":"Now we can just write out the answer."},{"Start":"10:27.090 ","End":"10:34.915","Text":"We have negative I Mu naught in the z-direction"},{"Start":"10:34.915 ","End":"10:42.730","Text":"divided by 4Pi and multiplied by the solution to this integral."},{"Start":"10:42.730 ","End":"10:52.197","Text":"Which is ln of t plus the square root of t squared plus b"},{"Start":"10:52.197 ","End":"10:57.675","Text":"in the bounds of z plus"},{"Start":"10:57.675 ","End":"11:04.270","Text":"l divided by 2 to z minus l divided by 2."},{"Start":"11:05.790 ","End":"11:10.570","Text":"Now when we plug this in according to law of ln\u0027s we get"},{"Start":"11:10.570 ","End":"11:16.060","Text":"negative I Mu naught divided by 4Pi."},{"Start":"11:16.060 ","End":"11:21.205","Text":"I\u0027ll add in the z hat later and then we have a ln of,"},{"Start":"11:21.205 ","End":"11:23.305","Text":"and according to the law of ln\u0027s we have"},{"Start":"11:23.305 ","End":"11:32.740","Text":"z"},{"Start":"11:32.740 ","End":"11:34.090","Text":"minus"},{"Start":"11:34.090 ","End":"11:35.170","Text":"l divided"},{"Start":"11:35.170 ","End":"11:39.580","Text":"by 2 plus the square root of t^2,"},{"Start":"11:39.580 ","End":"11:45.910","Text":"so z minus l divided by 2^2 plus instead of b we"},{"Start":"11:45.910 ","End":"11:54.355","Text":"have x^2 plus y^2 divided by,"},{"Start":"11:54.355 ","End":"12:04.960","Text":"so t which is z plus l divided by 2 plus the square root of t^2 so"},{"Start":"12:04.960 ","End":"12:09.880","Text":"z plus l over 2^2"},{"Start":"12:09.880 ","End":"12:16.700","Text":"plus b which we said was x^2 plus y^2."},{"Start":"12:19.500 ","End":"12:23.454","Text":"Of course, this is in the z-hat direction."},{"Start":"12:23.454 ","End":"12:26.680","Text":"Now please note because there\u0027s a minus here we could"},{"Start":"12:26.680 ","End":"12:30.580","Text":"have just turn this into a positive and then just"},{"Start":"12:30.580 ","End":"12:37.090","Text":"flipped over the values over here so then here we would have had z plus l"},{"Start":"12:37.090 ","End":"12:44.635","Text":"divided by 2 and in the denominator over here we would have had z minus l over 2 in both."},{"Start":"12:44.635 ","End":"12:52.590","Text":"That\u0027s if we changed this minus into a positive but we decided to leave it like so,"},{"Start":"12:52.590 ","End":"12:55.085","Text":"so this is the answer."},{"Start":"12:55.085 ","End":"12:59.305","Text":"This is the final answer to question number 1."},{"Start":"12:59.305 ","End":"13:01.495","Text":"Now let\u0027s answer question number 2."},{"Start":"13:01.495 ","End":"13:07.550","Text":"Calculate the magnetic field at a point above the center of the wire."},{"Start":"13:07.710 ","End":"13:10.360","Text":"To answer question 2,"},{"Start":"13:10.360 ","End":"13:16.030","Text":"I have simplified our previous answer and I took away the minus sign."},{"Start":"13:16.030 ","End":"13:22.555","Text":"I flipped these values around the balance because the minus sign I took away."},{"Start":"13:22.555 ","End":"13:26.980","Text":"We\u0027re left with this answer over here with pluses in"},{"Start":"13:26.980 ","End":"13:31.840","Text":"the numerator and the minus in the denominator."},{"Start":"13:31.840 ","End":"13:33.120","Text":"That\u0027s what I did."},{"Start":"13:33.120 ","End":"13:37.845","Text":"We\u0027re calculating the magnetic field at a point above the wire."},{"Start":"13:37.845 ","End":"13:40.950","Text":"Let\u0027s just erase all of this,"},{"Start":"13:40.950 ","End":"13:42.570","Text":"we don\u0027t need this right now."},{"Start":"13:42.570 ","End":"13:46.810","Text":"What we\u0027re doing is we\u0027re calculating the magnetic field at"},{"Start":"13:46.810 ","End":"13:51.210","Text":"this point directly above the center of the wire,"},{"Start":"13:51.210 ","End":"13:56.660","Text":"so let\u0027s say that this is the x-axis."},{"Start":"13:57.840 ","End":"14:01.300","Text":"If we\u0027re directly above the wire on"},{"Start":"14:01.300 ","End":"14:06.520","Text":"the x-axis and we\u0027re right at the center of the wire,"},{"Start":"14:06.520 ","End":"14:13.180","Text":"then that means that over here at this point over here the center of the wire z is"},{"Start":"14:13.180 ","End":"14:16.300","Text":"equal to 0 and because we\u0027re not going"},{"Start":"14:16.300 ","End":"14:20.830","Text":"in or out of the page that means y is also equal to 0."},{"Start":"14:20.830 ","End":"14:25.150","Text":"We have this wire which is located only"},{"Start":"14:25.150 ","End":"14:29.268","Text":"on the z-axis and we\u0027re directly above its center."},{"Start":"14:29.268 ","End":"14:33.085","Text":"That means that it\u0027s at the origin over here,"},{"Start":"14:33.085 ","End":"14:40.105","Text":"with respect to the z-axis if z is 0 and we\u0027re not going in or out of the page,"},{"Start":"14:40.105 ","End":"14:45.260","Text":"we\u0027re just going straight up so that means y is also equal to 0."},{"Start":"14:45.690 ","End":"14:53.020","Text":"We just have this variable over here x."},{"Start":"14:53.020 ","End":"15:01.480","Text":"As we\u0027ve seen the magnetic field is given by the rotor,"},{"Start":"15:01.480 ","End":"15:07.179","Text":"taking the rotor of our A vector or vector potential."},{"Start":"15:07.179 ","End":"15:12.775","Text":"What we want to do is we want to take the rotor of"},{"Start":"15:12.775 ","End":"15:18.670","Text":"B at a certain point so we\u0027ve seen at this point x,"},{"Start":"15:18.670 ","End":"15:27.700","Text":"somewhere above where y is equal to 0 as we just said and so is z."},{"Start":"15:27.700 ","End":"15:33.175","Text":"This is equal to the rotor of A where"},{"Start":"15:33.175 ","End":"15:38.905","Text":"of course we said that y is equal to z is equal to 0."},{"Start":"15:38.905 ","End":"15:47.450","Text":"We plug in these values for y and z is equal to 0 into this equation over here."},{"Start":"15:49.290 ","End":"15:55.900","Text":"We just do the determinant so we can calculate"},{"Start":"15:55.900 ","End":"15:59.125","Text":"the determinant or the rotor by writing it out"},{"Start":"15:59.125 ","End":"16:03.295","Text":"also like this as opposed to the other way that I\u0027ve done."},{"Start":"16:03.295 ","End":"16:06.835","Text":"Then we have the x-direction, y-direction,"},{"Start":"16:06.835 ","End":"16:10.690","Text":"and z-direction dx dy"},{"Start":"16:10.690 ","End":"16:17.890","Text":"dz and then we have the components of our vector potential."},{"Start":"16:17.890 ","End":"16:22.300","Text":"We can see that we only have potential in the z-direction so 0,"},{"Start":"16:22.300 ","End":"16:26.545","Text":"0 and then we have A_z,"},{"Start":"16:26.545 ","End":"16:36.355","Text":"and then what we\u0027re left with is d of A_z according to y."},{"Start":"16:36.355 ","End":"16:40.690","Text":"We take the derivative of all of this with respect to y in"},{"Start":"16:40.690 ","End":"16:51.010","Text":"the x-direction minus dA_z by dx in the y-direction."},{"Start":"16:51.010 ","End":"16:58.050","Text":"Then minus the derivative of all of this with respect to x and then in the y-direction."},{"Start":"16:59.190 ","End":"17:02.238","Text":"Let\u0027s write all of this out."},{"Start":"17:02.238 ","End":"17:03.950","Text":"We\u0027ll write it over here."},{"Start":"17:03.950 ","End":"17:11.090","Text":"We have Mu naught I divided by"},{"Start":"17:11.090 ","End":"17:16.180","Text":"4Pi and then I\u0027m going to split up the ln so I\u0027m just going to"},{"Start":"17:16.180 ","End":"17:21.800","Text":"write ln of the numerator minus ln of the denominator."},{"Start":"17:21.800 ","End":"17:28.673","Text":"According to law of ln\u0027s and take the derivative so let\u0027s do first the numerator."},{"Start":"17:28.673 ","End":"17:31.940","Text":"We have the derivative of ln"},{"Start":"17:31.940 ","End":"17:35.500","Text":"of the numerators of course just 1 divided by the numerator,"},{"Start":"17:35.500 ","End":"17:40.810","Text":"so 1 divided by z plus l divided by"},{"Start":"17:40.810 ","End":"17:45.640","Text":"2 plus the square root of z plus"},{"Start":"17:45.640 ","End":"17:52.420","Text":"l divided by 2^2 plus x^2 plus y^2."},{"Start":"17:54.510 ","End":"18:05.470","Text":"Then, we want to multiply this by the first inner derivative with respect to y."},{"Start":"18:05.470 ","End":"18:09.325","Text":"Here we have this square root."},{"Start":"18:09.325 ","End":"18:19.870","Text":"Then we\u0027re multiplying all of this by 1 divided by 2 multiplied by"},{"Start":"18:19.870 ","End":"18:28.420","Text":"the square root of z plus l divided"},{"Start":"18:28.420 ","End":"18:38.050","Text":"by 2^2 plus x^2 plus y^2."},{"Start":"18:38.050 ","End":"18:44.740","Text":"All of this is multiplied by the inner derivative, so y^2."},{"Start":"18:44.740 ","End":"18:49.370","Text":"All of this is multiplied by 2y."},{"Start":"18:49.770 ","End":"18:53.380","Text":"Then we\u0027re going to add to this,"},{"Start":"18:53.380 ","End":"18:58.435","Text":"the derivative of the next ln,"},{"Start":"18:58.435 ","End":"19:00.520","Text":"ln of the denominator."},{"Start":"19:00.520 ","End":"19:05.410","Text":"We have plus, but then we have a minus because it\u0027s ln of all"},{"Start":"19:05.410 ","End":"19:11.125","Text":"of this minus ln of all of this, so negative."},{"Start":"19:11.125 ","End":"19:14.950","Text":"Then we have the exact same thing that we have over here,"},{"Start":"19:14.950 ","End":"19:20.150","Text":"just with z minus l divided by 2."},{"Start":"19:23.310 ","End":"19:27.190","Text":"That is the exact same thing that we did over"},{"Start":"19:27.190 ","End":"19:34.330","Text":"here but the only difference is that we have these minuses and this is of course,"},{"Start":"19:34.330 ","End":"19:38.275","Text":"as we saw in the x-direction."},{"Start":"19:38.275 ","End":"19:48.055","Text":"Then to this, we subtract so minus dA_z by dx in the y-direction."},{"Start":"19:48.055 ","End":"19:53.500","Text":"We did the exact same calculation just our derivatives are with respect to x"},{"Start":"19:53.500 ","End":"19:59.890","Text":"so instead of having 2y here and here, we have 2x."},{"Start":"19:59.890 ","End":"20:01.660","Text":"It\u0027s the exact same thing,"},{"Start":"20:01.660 ","End":"20:08.820","Text":"just instead of with respect to y,"},{"Start":"20:08.820 ","End":"20:13.680","Text":"where here we had y^2 so now we\u0027re taking the derivative with respect to x,"},{"Start":"20:13.680 ","End":"20:15.165","Text":"where we have x^2."},{"Start":"20:15.165 ","End":"20:20.525","Text":"It\u0027s the exact same thing just here we\u0027re going to be multiplying by 2x."},{"Start":"20:20.525 ","End":"20:24.115","Text":"This is the answer in the meantime."},{"Start":"20:24.115 ","End":"20:29.050","Text":"Now, what we want to do is we want to substitute"},{"Start":"20:29.050 ","End":"20:35.245","Text":"in what we have for our values for y and z."},{"Start":"20:35.245 ","End":"20:39.805","Text":"We know that y is equal to z is equal to 0."},{"Start":"20:39.805 ","End":"20:41.395","Text":"If we do that,"},{"Start":"20:41.395 ","End":"20:47.440","Text":"then this is equal to 0 so this entire term cancels out."},{"Start":"20:47.440 ","End":"20:52.465","Text":"This is equal to 0 so this entire term cancels out."},{"Start":"20:52.465 ","End":"20:56.740","Text":"Then we can see our x component is equal to 0."},{"Start":"20:56.740 ","End":"20:59.050","Text":"Then we\u0027re left over here with"},{"Start":"20:59.050 ","End":"21:07.765","Text":"negative Mu naught I divided by 4 Pi multiplied by."},{"Start":"21:07.765 ","End":"21:14.050","Text":"We know that z is also equal to 0 so we have 1 divided by this is 0,"},{"Start":"21:14.050 ","End":"21:20.110","Text":"l divided by 2 plus the square root of 0,"},{"Start":"21:20.110 ","End":"21:27.250","Text":"l divided by 2^2 plus x^2,"},{"Start":"21:27.250 ","End":"21:33.775","Text":"and then y^2 is equal to 0 so we just leave it at this."},{"Start":"21:33.775 ","End":"21:39.220","Text":"This is multiplied by 1 divided by 2,"},{"Start":"21:39.220 ","End":"21:42.010","Text":"the square root of l divided by"},{"Start":"21:42.010 ","End":"21:48.100","Text":"2^2 plus x^2 again"},{"Start":"21:48.100 ","End":"21:52.990","Text":"because z and y is equal to 0."},{"Start":"21:52.990 ","End":"21:54.940","Text":"Then multiply by 2x,"},{"Start":"21:54.940 ","End":"21:57.265","Text":"so let\u0027s just write out our 2x here,"},{"Start":"21:57.265 ","End":"22:05.240","Text":"and then we add plus negative and then the exact same thing."},{"Start":"22:06.630 ","End":"22:09.940","Text":"Then we\u0027re left with this."},{"Start":"22:09.940 ","End":"22:16.040","Text":"What we can see is that these two are like terms."},{"Start":"22:17.520 ","End":"22:26.455","Text":"Then what we can do is we can just simplify out our like terms to be left with"},{"Start":"22:26.455 ","End":"22:35.170","Text":"negative Mu naught I divided by 4 Pi and then multiply this."},{"Start":"22:35.170 ","End":"22:38.560","Text":"We have in both,"},{"Start":"22:38.560 ","End":"22:47.665","Text":"we have x divided by the square root of l over"},{"Start":"22:47.665 ","End":"22:57.350","Text":"2^2 plus x^2 and then all of this is multiplied by"},{"Start":"22:58.260 ","End":"23:09.134","Text":"negative l divided by negative l over 2 plus"},{"Start":"23:09.134 ","End":"23:10.980","Text":"the square root of l"},{"Start":"23:10.980 ","End":"23:11.430","Text":"over"},{"Start":"23:11.430 ","End":"23:15.810","Text":"2^2"},{"Start":"23:16.650 ","End":"23:21.580","Text":"plus x^2."},{"Start":"23:21.580 ","End":"23:29.800","Text":"This is one bracket and then the other bracket is l over 2 plus the square root of l"},{"Start":"23:29.800 ","End":"23:36.850","Text":"over 2^2 plus"},{"Start":"23:36.850 ","End":"23:41.410","Text":"x^2."},{"Start":"23:41.410 ","End":"23:44.230","Text":"Of course, this is in the y-direction."},{"Start":"23:44.230 ","End":"23:51.200","Text":"Then we can finally simplify this out even more by"},{"Start":"23:51.510 ","End":"23:56.290","Text":"expanding these brackets because then we see that we"},{"Start":"23:56.290 ","End":"24:02.275","Text":"have this term and this term which will cancel out."},{"Start":"24:02.275 ","End":"24:06.100","Text":"If you expand the brackets and play around with this a bit more,"},{"Start":"24:06.100 ","End":"24:07.600","Text":"we\u0027ll get the final answer,"},{"Start":"24:07.600 ","End":"24:17.290","Text":"which is Mu_0Il divided by 4 Pi x multiplied by"},{"Start":"24:17.290 ","End":"24:20.200","Text":"the square root of l divided by"},{"Start":"24:20.200 ","End":"24:28.760","Text":"2^2 plus x^2 in the y-direction."},{"Start":"24:28.950 ","End":"24:31.300","Text":"This is the final answer."},{"Start":"24:31.300 ","End":"24:34.620","Text":"This is our magnetic field at the point x (0,"},{"Start":"24:34.620 ","End":"24:38.635","Text":"0), right in the center of our wire."},{"Start":"24:38.635 ","End":"24:41.500","Text":"You can check this answer."},{"Start":"24:41.500 ","End":"24:43.900","Text":"If you remember in the chapter where we dealt with"},{"Start":"24:43.900 ","End":"24:47.650","Text":"Biot-Savart law so you can check what we got for"},{"Start":"24:47.650 ","End":"24:55.510","Text":"the magnetic field of a wire and you\u0027ll see that it is the same answer as over here."},{"Start":"24:55.510 ","End":"24:57.520","Text":"To solve these types of questions,"},{"Start":"24:57.520 ","End":"25:01.585","Text":"you can either use Biot-Savart or in certain cases,"},{"Start":"25:01.585 ","End":"25:06.355","Text":"use the vector potential and you\u0027ll get the exact same onset."},{"Start":"25:06.355 ","End":"25:09.410","Text":"That\u0027s the end of this lesson."}],"ID":21455},{"Watched":false,"Name":"1.5 Calculating the Vector Potential of a Given Magnetic Field","Duration":"5m 26s","ChapterTopicVideoID":21385,"CourseChapterTopicPlaylistID":99495,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.510","Text":"Hello. In this lesson we\u0027re going to learn how to calculate"},{"Start":"00:03.510 ","End":"00:07.920","Text":"the vector potential given a magnetic field."},{"Start":"00:07.920 ","End":"00:12.435","Text":"In the previous lessons we"},{"Start":"00:12.435 ","End":"00:19.635","Text":"saw that to calculate the vector potential A at some point in space r,"},{"Start":"00:19.635 ","End":"00:25.320","Text":"we would have to take Mu naught divided by 4 Pi,"},{"Start":"00:25.320 ","End":"00:31.895","Text":"and multiply it by the integral of our current density J,"},{"Start":"00:31.895 ","End":"00:35.735","Text":"with at point r vector tag,"},{"Start":"00:35.735 ","End":"00:41.795","Text":"dv tag, divided by the distance,"},{"Start":"00:41.795 ","End":"00:48.750","Text":"which was r vector minus r tag vector and the magnitude of that."},{"Start":"00:49.100 ","End":"00:52.549","Text":"This was using the current density."},{"Start":"00:52.549 ","End":"00:55.670","Text":"In this lesson, we\u0027re going to see how we can get"},{"Start":"00:55.670 ","End":"01:00.160","Text":"our vector potential from a magnetic field."},{"Start":"01:00.160 ","End":"01:07.470","Text":"We\u0027ve seen that the rotor of A is equal to B."},{"Start":"01:07.470 ","End":"01:11.600","Text":"This is what we\u0027re going to play around with."},{"Start":"01:12.440 ","End":"01:15.635","Text":"This method that we\u0027re going to show now,"},{"Start":"01:15.635 ","End":"01:19.210","Text":"and we\u0027ve rearranged this equation for A."},{"Start":"01:19.210 ","End":"01:25.180","Text":"We will use this version when the B fields can be found using Ampere\u0027s Law."},{"Start":"01:25.180 ","End":"01:30.835","Text":"If we are given a question where the magnetic field can be calculated using Ampere\u0027s law,"},{"Start":"01:30.835 ","End":"01:36.025","Text":"then we will use this version of the equation to find the vector potential."},{"Start":"01:36.025 ","End":"01:41.448","Text":"Just a reminder in which cases we can use Ampere\u0027s law to find the magnetic field,"},{"Start":"01:41.448 ","End":"01:43.615","Text":"it\u0027s when we have infinite wire,"},{"Start":"01:43.615 ","End":"01:44.710","Text":"or an infinite plane,"},{"Start":"01:44.710 ","End":"01:46.780","Text":"or an infinite cylinder,"},{"Start":"01:46.780 ","End":"01:49.070","Text":"or an infinite spring,"},{"Start":"01:49.070 ","End":"01:51.970","Text":"and when we have a toroid."},{"Start":"01:53.900 ","End":"01:58.400","Text":"We\u0027ve already seen similar mathematical processes when we"},{"Start":"01:58.400 ","End":"02:01.970","Text":"were looking at the original Ampere\u0027s law that"},{"Start":"02:01.970 ","End":"02:10.120","Text":"said that the rotor of the magnetic field is equal to Mu naught J."},{"Start":"02:10.130 ","End":"02:13.355","Text":"We saw that this is the differential form,"},{"Start":"02:13.355 ","End":"02:17.760","Text":"and that we can convert it to the integral form"},{"Start":"02:18.520 ","End":"02:27.440","Text":"by calculating the closed loop integral on B.dl."},{"Start":"02:27.440 ","End":"02:37.880","Text":"Where this was equal to the integral of Mu naught J.ds."},{"Start":"02:38.200 ","End":"02:44.480","Text":"Where of course this integral is on the closed surface area that"},{"Start":"02:44.480 ","End":"02:51.000","Text":"is confined within this closed loop."},{"Start":"02:52.400 ","End":"02:55.445","Text":"Let\u0027s remember how we did this."},{"Start":"02:55.445 ","End":"02:56.900","Text":"We\u0027ve already seen this calculation,"},{"Start":"02:56.900 ","End":"02:59.135","Text":"but let\u0027s just remember."},{"Start":"02:59.135 ","End":"03:05.415","Text":"What we can write out is our rotor of a magnetic field."},{"Start":"03:05.415 ","End":"03:07.790","Text":"This was our original equation,"},{"Start":"03:07.790 ","End":"03:12.150","Text":"which is equal to Mu naught J."},{"Start":"03:12.150 ","End":"03:17.540","Text":"Then we\u0027ll integrate along both sides."},{"Start":"03:17.540 ","End":"03:21.995","Text":"Then we\u0027re integrating along surface areas."},{"Start":"03:21.995 ","End":"03:25.070","Text":"Here we have the rotor of B ds,"},{"Start":"03:25.070 ","End":"03:29.495","Text":"and here we have the integral of Mu J."},{"Start":"03:29.495 ","End":"03:31.715","Text":"Again, because of this s over here,"},{"Start":"03:31.715 ","End":"03:36.480","Text":"we have ds over here as well."},{"Start":"03:36.480 ","End":"03:44.045","Text":"Then this side, through Stokes\u0027 law,"},{"Start":"03:44.045 ","End":"03:48.550","Text":"what we get is Stokes\u0027 law says that the integral of"},{"Start":"03:48.550 ","End":"03:53.950","Text":"the rotor ds of some kind of vector is equal to"},{"Start":"03:53.950 ","End":"03:57.880","Text":"the closed loop integral and some kind of"},{"Start":"03:57.880 ","End":"04:05.970","Text":"line integral of B.dl."},{"Start":"04:05.970 ","End":"04:07.650","Text":"This side is Stokes\u0027 law."},{"Start":"04:07.650 ","End":"04:10.885","Text":"Then this is just equal to the same thing."},{"Start":"04:10.885 ","End":"04:15.710","Text":"Mu naught of J.ds."},{"Start":"04:18.020 ","End":"04:23.710","Text":"Now we\u0027ll do the exact same thing for A, our vector potential."},{"Start":"04:23.710 ","End":"04:31.630","Text":"What we have is the closed loop integral of A.dl."},{"Start":"04:31.630 ","End":"04:35.290","Text":"Instead of the rotor of B.ds,"},{"Start":"04:35.290 ","End":"04:38.150","Text":"we have the rotor of A.ds."},{"Start":"04:38.150 ","End":"04:43.240","Text":"This is equal to the integral along the closed surface area"},{"Start":"04:43.240 ","End":"04:49.520","Text":"enclosed in this loop of B.ds."},{"Start":"04:49.520 ","End":"04:55.765","Text":"Every time we have an integral of some kind of vector through some kind of surface area,"},{"Start":"04:55.765 ","End":"04:57.520","Text":"that\u0027s what ds is."},{"Start":"04:57.520 ","End":"05:01.514","Text":"We can just call this the flux."},{"Start":"05:01.514 ","End":"05:04.875","Text":"In this case we have a magnetic field."},{"Start":"05:04.875 ","End":"05:08.590","Text":"This is the magnetic flux."},{"Start":"05:09.860 ","End":"05:14.705","Text":"This is the equation that is"},{"Start":"05:14.705 ","End":"05:19.100","Text":"relevant and that you should definitely write on your equation sheets."},{"Start":"05:19.100 ","End":"05:24.060","Text":"In the next lesson we\u0027re going to do a worked example using this equation."},{"Start":"05:24.060 ","End":"05:26.720","Text":"That\u0027s the end of this lesson."}],"ID":21456},{"Watched":false,"Name":"1.6 Exercise - Infinite Spring","Duration":"9m 11s","ChapterTopicVideoID":21386,"CourseChapterTopicPlaylistID":99495,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.463","Text":"Hello. In this lesson we\u0027re going to be answering the following question"},{"Start":"00:04.463 ","End":"00:09.120","Text":": an infinite spring has turn density per unit length n,"},{"Start":"00:09.120 ","End":"00:12.285","Text":"radius a, and current I flowing through it."},{"Start":"00:12.285 ","End":"00:15.465","Text":"Calculate the vector potential."},{"Start":"00:15.465 ","End":"00:20.970","Text":"Here we have a spring of infinite length."},{"Start":"00:20.970 ","End":"00:24.225","Text":"Let\u0027s just draw this."},{"Start":"00:24.225 ","End":"00:25.770","Text":"It\u0027s of infinite length."},{"Start":"00:25.770 ","End":"00:34.665","Text":"It has a radius over here of a and the number of turns per unit length is"},{"Start":"00:34.665 ","End":"00:40.730","Text":"n. We\u0027ve already spoken"},{"Start":"00:40.730 ","End":"00:46.205","Text":"about how to calculate the magnetic field of a spring using Ampere\u0027s law,"},{"Start":"00:46.205 ","End":"00:51.215","Text":"it is equal to Mu naught multiplied by the current flowing through it,"},{"Start":"00:51.215 ","End":"00:53.060","Text":"multiplied by n,"},{"Start":"00:53.060 ","End":"00:54.710","Text":"the turn density,"},{"Start":"00:54.710 ","End":"00:58.830","Text":"and it is in the z direction."},{"Start":"00:58.940 ","End":"01:05.570","Text":"Of course, this is the magnetic field when the radius is smaller than a,"},{"Start":"01:05.570 ","End":"01:08.375","Text":"as in we\u0027re located within the spring."},{"Start":"01:08.375 ","End":"01:14.045","Text":"Of course, as we saw in the chapter dealing with Ampere\u0027s law,"},{"Start":"01:14.045 ","End":"01:18.920","Text":"the magnetic field is 0 when r is greater than a."},{"Start":"01:18.920 ","End":"01:21.395","Text":"When we\u0027re located outside of the spring,"},{"Start":"01:21.395 ","End":"01:24.690","Text":"there is no magnetic field."},{"Start":"01:25.430 ","End":"01:33.375","Text":"We know that the magnetic field is constant inside the spring."},{"Start":"01:33.375 ","End":"01:37.175","Text":"Now what we want to do is we want to calculate the vector potential."},{"Start":"01:37.175 ","End":"01:41.900","Text":"In the previous lesson, we saw that the vector potential is equal to"},{"Start":"01:41.900 ","End":"01:48.745","Text":"the closed loop integral of A.dl."},{"Start":"01:48.745 ","End":"01:57.870","Text":"All of this was equal to the integral along the surface area of B.ds."},{"Start":"01:58.700 ","End":"02:02.135","Text":"If our magnetic field is constant,"},{"Start":"02:02.135 ","End":"02:08.020","Text":"that means that our vector potential is also going to be constant."},{"Start":"02:08.020 ","End":"02:17.385","Text":"What we do is we take a little loop inside the spring."},{"Start":"02:17.385 ","End":"02:23.330","Text":"We can see through the right-hand rule that magnetic field lines are in this z direction,"},{"Start":"02:23.330 ","End":"02:33.035","Text":"then our vector potential is going to be from the right-hand rule in this direction,"},{"Start":"02:33.035 ","End":"02:35.910","Text":"like so in the Phi direction."},{"Start":"02:35.910 ","End":"02:38.790","Text":"That we can see,"},{"Start":"02:38.790 ","End":"02:40.125","Text":"and as we said,"},{"Start":"02:40.125 ","End":"02:42.760","Text":"our A is constant,"},{"Start":"02:42.760 ","End":"02:49.940","Text":"because we saw from Ampere\u0027s law that the magnetic field inside a spring is constant."},{"Start":"02:49.940 ","End":"02:55.430","Text":"Our A is this integral along this loop over here."},{"Start":"02:55.430 ","End":"03:00.500","Text":"We have A multiplied by the circumference of the circle,"},{"Start":"03:00.500 ","End":"03:02.195","Text":"which is just 2Pi."},{"Start":"03:02.195 ","End":"03:05.540","Text":"We\u0027ll say that the radius of it is r,"},{"Start":"03:05.540 ","End":"03:11.330","Text":"so 2Pi r. This is equal to this integral of B.ds,"},{"Start":"03:11.330 ","End":"03:13.940","Text":"where, of course, we said that B is constant."},{"Start":"03:13.940 ","End":"03:19.595","Text":"It\u0027s just the magnetic field multiplied by the surface area of this loop,"},{"Start":"03:19.595 ","End":"03:25.680","Text":"which is simply equal to Pi r^2."},{"Start":"03:26.180 ","End":"03:28.655","Text":"You may have noticed that this is"},{"Start":"03:28.655 ","End":"03:33.605","Text":"identical to what we did in Ampere\u0027s law to calculate B."},{"Start":"03:33.605 ","End":"03:37.415","Text":"Here the vector potential is,"},{"Start":"03:37.415 ","End":"03:39.330","Text":"if we go back to Ampere\u0027s law,"},{"Start":"03:39.330 ","End":"03:40.760","Text":"in the place of B,"},{"Start":"03:40.760 ","End":"03:46.030","Text":"and B is in the place of the current in Ampere\u0027s law."},{"Start":"03:46.030 ","End":"03:50.960","Text":"We\u0027re just doing the exact same thing as Ampere\u0027s law to find the magnetic field."},{"Start":"03:50.960 ","End":"03:53.345","Text":"Just instead of B,"},{"Start":"03:53.345 ","End":"03:55.130","Text":"we place A,"},{"Start":"03:55.130 ","End":"03:59.790","Text":"and instead of the current, we place B."},{"Start":"04:01.370 ","End":"04:05.945","Text":"We\u0027re just doing Ampere\u0027s law to find the magnetic field,"},{"Start":"04:05.945 ","End":"04:08.360","Text":"but in order to calculate the vector potential,"},{"Start":"04:08.360 ","End":"04:15.840","Text":"we\u0027re just changing the variables and also the units."},{"Start":"04:15.840 ","End":"04:20.460","Text":"If we have, I\u0027ll just write it here, Ampere\u0027s law."},{"Start":"04:20.460 ","End":"04:25.020","Text":"So Ampere\u0027s law to find the B field."},{"Start":"04:25.020 ","End":"04:31.575","Text":"We have our variable B and our Variable I for the current."},{"Start":"04:31.575 ","End":"04:36.560","Text":"In this case to calculate the vector potential,"},{"Start":"04:36.560 ","End":"04:40.340","Text":"we do the exact same thing as here."},{"Start":"04:40.340 ","End":"04:42.740","Text":"However, instead of B,"},{"Start":"04:42.740 ","End":"04:47.180","Text":"we switch it out to our vector potential,"},{"Start":"04:47.180 ","End":"04:51.875","Text":"and instead of I, we switch it out with the magnetic field."},{"Start":"04:51.875 ","End":"04:56.030","Text":"Everything else, the mathematics is exactly the same."},{"Start":"04:56.030 ","End":"04:58.460","Text":"Then we can cancel out the Pi\u0027s,"},{"Start":"04:58.460 ","End":"05:00.020","Text":"we can cancel out the r,"},{"Start":"05:00.020 ","End":"05:09.105","Text":"and then what we get is that A=Br divided by 2."},{"Start":"05:09.105 ","End":"05:12.730","Text":"Of course we said it\u0027s in the Phi direction."},{"Start":"05:12.830 ","End":"05:17.070","Text":"We can put it in the Phi direction."},{"Start":"05:17.070 ","End":"05:22.610","Text":"If we want to add in over here now our magnetic field,"},{"Start":"05:22.610 ","End":"05:28.500","Text":"so we can just say that A=B,"},{"Start":"05:28.500 ","End":"05:34.800","Text":"which we saw was equal to Mu naught I n multiplied by"},{"Start":"05:34.800 ","End":"05:41.580","Text":"r divided by 2 in the Phi direction."},{"Start":"05:41.580 ","End":"05:43.200","Text":"This is, of course,"},{"Start":"05:43.200 ","End":"05:46.930","Text":"when r is smaller than A."},{"Start":"05:48.020 ","End":"05:54.610","Text":"What will be the vector potential outside of the spring?"},{"Start":"05:54.610 ","End":"05:59.200","Text":"Again, we take this loop like so."},{"Start":"05:59.200 ","End":"06:04.585","Text":"This time, this is r. Of course,"},{"Start":"06:04.585 ","End":"06:07.690","Text":"the vector potential is the exact same."},{"Start":"06:07.690 ","End":"06:10.150","Text":"Again, for A.dl,"},{"Start":"06:10.150 ","End":"06:14.505","Text":"we\u0027re going to have a multiplied by this loop length."},{"Start":"06:14.505 ","End":"06:16.845","Text":"We have, again,"},{"Start":"06:16.845 ","End":"06:23.510","Text":"A multiplied by 2Pi r, is A.dl."},{"Start":"06:23.510 ","End":"06:28.150","Text":"Again, the magnetic field inside over here is constant."},{"Start":"06:28.150 ","End":"06:30.730","Text":"That means that our vector potential is constant,"},{"Start":"06:30.730 ","End":"06:32.660","Text":"and that\u0027s why we don\u0027t have to integrate."},{"Start":"06:32.660 ","End":"06:38.020","Text":"We can just add this up and this is equal to B.ds."},{"Start":"06:38.020 ","End":"06:46.035","Text":"Now, notice the magnetic field is only flowing in this loop over here."},{"Start":"06:46.035 ","End":"06:51.200","Text":"There\u0027s only magnetic flux through this loop of radius A."},{"Start":"06:51.200 ","End":"06:52.640","Text":"Because as we saw,"},{"Start":"06:52.640 ","End":"06:58.870","Text":"the magnetic field only flows inside the spring, not outside."},{"Start":"06:58.870 ","End":"07:03.290","Text":"When we\u0027re looking at this large loop of radius r,"},{"Start":"07:03.290 ","End":"07:08.075","Text":"we can see that only in this section in blue,"},{"Start":"07:08.075 ","End":"07:14.465","Text":"where we have magnetic flux and the rest of this area will have no magnetic flux,"},{"Start":"07:14.465 ","End":"07:19.430","Text":"because we saw that there\u0027s no magnetic field in this region where r"},{"Start":"07:19.430 ","End":"07:25.900","Text":"is greater than a."},{"Start":"07:26.240 ","End":"07:30.815","Text":"In that case, just because the magnetic field in this region is equal to 0,"},{"Start":"07:30.815 ","End":"07:35.300","Text":"we\u0027re calculating the magnetic flux in this whole loop."},{"Start":"07:35.300 ","End":"07:39.930","Text":"There is an area in this loop where there is magnetic flux,"},{"Start":"07:39.930 ","End":"07:43.125","Text":"so we multiply B.ds."},{"Start":"07:43.125 ","End":"07:47.110","Text":"It\u0027s the total surface area over here in blue,"},{"Start":"07:47.110 ","End":"07:55.225","Text":"which is just equal to Pi multiplied by the radius, which is a^2."},{"Start":"07:55.225 ","End":"07:58.020","Text":"Again, we can cancel this out."},{"Start":"07:58.020 ","End":"08:03.770","Text":"Then what we get is that A is equal to the magnetic field,"},{"Start":"08:03.770 ","End":"08:12.060","Text":"which is equal to Mu naught I n multiplied by a^2,"},{"Start":"08:12.060 ","End":"08:16.815","Text":"and then divided by 2r."},{"Start":"08:16.815 ","End":"08:20.460","Text":"Again, it\u0027s in the Phi direction,"},{"Start":"08:20.460 ","End":"08:24.190","Text":"and this is when r is greater than a."},{"Start":"08:24.740 ","End":"08:28.070","Text":"This is the final answer."},{"Start":"08:28.070 ","End":"08:34.160","Text":"This is the vector potential inside and outside of the spring."},{"Start":"08:34.160 ","End":"08:38.765","Text":"Just remember that to calculate the vector potential when you have the magnetic field,"},{"Start":"08:38.765 ","End":"08:44.690","Text":"you\u0027re doing the same thing as calculating the magnetic field using Ampere\u0027s law."},{"Start":"08:44.690 ","End":"08:47.885","Text":"But instead of doing it like so,"},{"Start":"08:47.885 ","End":"08:50.630","Text":"you switch out the magnetic field and the equation for"},{"Start":"08:50.630 ","End":"08:53.860","Text":"Ampere\u0027s law for the vector potential,"},{"Start":"08:53.860 ","End":"08:57.320","Text":"and you switch out the current when using Ampere\u0027s law to"},{"Start":"08:57.320 ","End":"09:01.565","Text":"calculate the B field for the B field itself."},{"Start":"09:01.565 ","End":"09:03.464","Text":"Then you can, through this,"},{"Start":"09:03.464 ","End":"09:05.350","Text":"calculate the vector potential."},{"Start":"09:05.350 ","End":"09:09.040","Text":"It\u0027s the exact same situation."},{"Start":"09:09.040 ","End":"09:11.930","Text":"That\u0027s the end of this lesson."}],"ID":21457},{"Watched":false,"Name":"1.7 Exercise - Infinite Cylinder","Duration":"25m 46s","ChapterTopicVideoID":21387,"CourseChapterTopicPlaylistID":99495,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.025","Text":"Hello. In this lesson,"},{"Start":"00:02.025 ","End":"00:05.040","Text":"we\u0027re going to be answering the following question."},{"Start":"00:05.040 ","End":"00:08.670","Text":"Calculate the vector potential of an infinite cylinder of"},{"Start":"00:08.670 ","End":"00:12.810","Text":"radius a that has current I flowing through it."},{"Start":"00:12.810 ","End":"00:17.080","Text":"The current density is uniform throughout the cylinder."},{"Start":"00:17.180 ","End":"00:23.985","Text":"Let\u0027s just draw a section of the cylinder."},{"Start":"00:23.985 ","End":"00:33.710","Text":"Let\u0027s say that this is its axis of symmetry and this is the cylinder and of"},{"Start":"00:33.710 ","End":"00:43.795","Text":"course we know that it is of radius a and we have this current I flowing through it."},{"Start":"00:43.795 ","End":"00:46.905","Text":"Let\u0027s write out what J is."},{"Start":"00:46.905 ","End":"00:50.740","Text":"J because I is a uniform,"},{"Start":"00:50.810 ","End":"00:55.010","Text":"J is equal to I divided by S,"},{"Start":"00:55.010 ","End":"00:57.980","Text":"where S is of course the surface area of"},{"Start":"00:57.980 ","End":"01:01.880","Text":"the cross-section or the cross-sectional surface area."},{"Start":"01:01.880 ","End":"01:08.280","Text":"We can say that this is equal to I divided by Pi a^2."},{"Start":"01:10.220 ","End":"01:17.550","Text":"In other words, let\u0027s call this J Naught and then we can say that our J vector is equal"},{"Start":"01:17.550 ","End":"01:24.270","Text":"to this J Naught and let\u0027s define this as the z axis,"},{"Start":"01:24.270 ","End":"01:28.210","Text":"J Naught in the z direction."},{"Start":"01:28.580 ","End":"01:32.350","Text":"Now in order to calculate the vector potential,"},{"Start":"01:32.350 ","End":"01:36.565","Text":"we know that we can use the magnetic field to do that."},{"Start":"01:36.565 ","End":"01:41.090","Text":"Let\u0027s calculate the magnetic field."},{"Start":"01:41.930 ","End":"01:51.860","Text":"First of all, we know that we can take the closed loop integral of B.dl and"},{"Start":"01:51.860 ","End":"02:01.630","Text":"that this is going to be equal to Mu Naught J.ds."},{"Start":"02:01.630 ","End":"02:05.715","Text":"We\u0027re calculating the magnetic field."},{"Start":"02:05.715 ","End":"02:07.980","Text":"Let\u0027s do this. First of all,"},{"Start":"02:07.980 ","End":"02:10.670","Text":"we saw that I is uniform."},{"Start":"02:10.670 ","End":"02:14.480","Text":"We weren\u0027t told that I is changing and we can"},{"Start":"02:14.480 ","End":"02:19.320","Text":"assume that it\u0027s uniform in this infinite cylinder."},{"Start":"02:19.330 ","End":"02:25.520","Text":"What we can do is we can take some loop like"},{"Start":"02:25.520 ","End":"02:34.870","Text":"so in this direction with radius r. Then,"},{"Start":"02:34.870 ","End":"02:36.290","Text":"because I is uniform,"},{"Start":"02:36.290 ","End":"02:43.175","Text":"we can just write over here that B multiplied by the circumference of this loop,"},{"Start":"02:43.175 ","End":"02:47.585","Text":"which is just 2 Pi multiplied by its radius r,"},{"Start":"02:47.585 ","End":"02:53.780","Text":"is equal to Mu Naught multiplied by the integral of J.ds,"},{"Start":"02:53.780 ","End":"02:57.950","Text":"where of course J is a constant, it\u0027s uniform."},{"Start":"02:57.950 ","End":"03:03.439","Text":"Again, we can just take Mu Naught and then we have our J,"},{"Start":"03:03.439 ","End":"03:06.770","Text":"which is equal to J Naught without"},{"Start":"03:06.770 ","End":"03:11.285","Text":"the vector multiplied by the surface area of this loop,"},{"Start":"03:11.285 ","End":"03:18.760","Text":"which is just Pi r^2."},{"Start":"03:18.760 ","End":"03:20.620","Text":"Now, of course, this loop,"},{"Start":"03:20.620 ","End":"03:23.785","Text":"this is the magnetic field."},{"Start":"03:23.785 ","End":"03:28.375","Text":"Then we can cancel out Pi from both sides and"},{"Start":"03:28.375 ","End":"03:34.720","Text":"r. Then we get that the magnetic field is equal to Mu Naught,"},{"Start":"03:34.720 ","End":"03:39.705","Text":"J Naught, r divided by 2."},{"Start":"03:39.705 ","End":"03:41.800","Text":"Using the right-hand rule,"},{"Start":"03:41.800 ","End":"03:45.025","Text":"we can see that the current is pointing in the z direction."},{"Start":"03:45.025 ","End":"03:46.180","Text":"Our thumb points in"},{"Start":"03:46.180 ","End":"03:53.980","Text":"the z direction and our fingers curl in the direction of the magnetic field."},{"Start":"03:53.980 ","End":"03:59.690","Text":"Our fingers curl in the Phi direction."},{"Start":"04:00.380 ","End":"04:05.040","Text":"This is our B vector, of course,"},{"Start":"04:05.040 ","End":"04:07.935","Text":"when r is less than a,"},{"Start":"04:07.935 ","End":"04:10.555","Text":"when we\u0027re located inside the cylinder."},{"Start":"04:10.555 ","End":"04:14.425","Text":"Now what happens when we\u0027re located outside the cylinder?"},{"Start":"04:14.425 ","End":"04:18.070","Text":"The direction of the magnetic field is"},{"Start":"04:18.070 ","End":"04:22.120","Text":"the same and now we have r where r is greater than a."},{"Start":"04:22.120 ","End":"04:24.055","Text":"Again, we\u0027re doing this."},{"Start":"04:24.055 ","End":"04:29.300","Text":"We have B multiplied by 2 Pi r,"},{"Start":"04:29.300 ","End":"04:32.515","Text":"which is equal to this side,"},{"Start":"04:32.515 ","End":"04:36.385","Text":"Mu Naught multiplied by J ds."},{"Start":"04:36.385 ","End":"04:39.470","Text":"We have J Naught,"},{"Start":"04:39.470 ","End":"04:41.970","Text":"and then the surface area of the loop."},{"Start":"04:41.970 ","End":"04:48.155","Text":"Now notice that our J is only flowing through the cylinder."},{"Start":"04:48.155 ","End":"04:52.825","Text":"There\u0027s no J or there\u0027s no current outside of the cylinder."},{"Start":"04:52.825 ","End":"05:02.340","Text":"The current is only flowing in this loop over here of radius a."},{"Start":"05:03.750 ","End":"05:06.740","Text":"In the space outside of the cylinder,"},{"Start":"05:06.740 ","End":"05:08.420","Text":"we have no current or no J,"},{"Start":"05:08.420 ","End":"05:10.160","Text":"so we can\u0027t sum it up."},{"Start":"05:10.160 ","End":"05:12.950","Text":"What we do is we multiply J Naught by"},{"Start":"05:12.950 ","End":"05:17.570","Text":"the surface area or the cross-sectional area of the cylinder,"},{"Start":"05:17.570 ","End":"05:20.850","Text":"which is just Pi a^2."},{"Start":"05:20.990 ","End":"05:26.420","Text":"Because all of this space over here, there\u0027s no J."},{"Start":"05:26.420 ","End":"05:35.400","Text":"Here J is equal to zero so we\u0027re only summing up in this area over here."},{"Start":"05:35.400 ","End":"05:44.270","Text":"Then we can again cancel out the Pi\u0027s and then we get that B is equal to Mu Naught,"},{"Start":"05:44.270 ","End":"05:50.725","Text":"J Naught, a^2 divided by 2r."},{"Start":"05:50.725 ","End":"05:58.120","Text":"Again, we know that this is in the Phi direction from the right-hand rule."},{"Start":"05:58.790 ","End":"06:04.900","Text":"Of course, this is when r is greater than a."},{"Start":"06:04.900 ","End":"06:12.530","Text":"We\u0027re using these 2 values over here."},{"Start":"06:13.640 ","End":"06:16.255","Text":"Saying as we have the magnetic fields,"},{"Start":"06:16.255 ","End":"06:20.555","Text":"we want to calculate now the vector potential."},{"Start":"06:20.555 ","End":"06:24.130","Text":"We\u0027ve seen that the equation for calculating"},{"Start":"06:24.130 ","End":"06:28.750","Text":"the vector potential is the closed loop integral of a,"},{"Start":"06:28.750 ","End":"06:31.735","Text":"the vector potential dot dl,"},{"Start":"06:31.735 ","End":"06:41.150","Text":"which is equal to the surface integral of B.ds."},{"Start":"06:41.630 ","End":"06:46.735","Text":"Just before we do the calculation and notes on the directions,"},{"Start":"06:46.735 ","End":"06:49.240","Text":"we see that the B field,"},{"Start":"06:49.240 ","End":"06:52.570","Text":"the magnetic field is in the Phi direction."},{"Start":"06:52.570 ","End":"06:54.175","Text":"From the right-hand rule,"},{"Start":"06:54.175 ","End":"06:58.690","Text":"we can say that the vector potential is in the z direction."},{"Start":"06:58.690 ","End":"07:02.395","Text":"If B curls around in the Phi direction,"},{"Start":"07:02.395 ","End":"07:06.980","Text":"then our thumb points up in the direction of the vector potential,"},{"Start":"07:06.980 ","End":"07:09.955","Text":"so the vector potentials in the z direction."},{"Start":"07:09.955 ","End":"07:16.870","Text":"Now, another way of looking at this is to remember that we\u0027ve"},{"Start":"07:16.870 ","End":"07:24.545","Text":"previously seen that the rotor of the vector potential is equal to our magnetic field."},{"Start":"07:24.545 ","End":"07:26.915","Text":"As we\u0027ve seen over here,"},{"Start":"07:26.915 ","End":"07:30.760","Text":"our magnetic field is only in the phi direction."},{"Start":"07:30.760 ","End":"07:35.600","Text":"Because our magnetic field is only in this Phi direction,"},{"Start":"07:35.600 ","End":"07:45.720","Text":"that means that the rotor of the vector potential must also only have a Phi component,"},{"Start":"07:46.310 ","End":"07:49.235","Text":"because the components as the direction,"},{"Start":"07:49.235 ","End":"07:53.885","Text":"so if the magnetic field is in the Phi direction only has a Phi components,"},{"Start":"07:53.885 ","End":"07:58.685","Text":"then the rotor of a also has only a Phi component."},{"Start":"07:58.685 ","End":"08:04.550","Text":"If we look at our equation sheet and we look for the equation for"},{"Start":"08:04.550 ","End":"08:10.640","Text":"the rotor of a vector field and we look only at its Phi component."},{"Start":"08:10.640 ","End":"08:13.115","Text":"Then we\u0027ll see that the Phi component,"},{"Start":"08:13.115 ","End":"08:18.580","Text":"the 1 that has Phi hat next to it is just d A_r"},{"Start":"08:18.580 ","End":"08:26.515","Text":"by dz minus d A_z by dr."},{"Start":"08:26.515 ","End":"08:34.364","Text":"This is the Phi component of the rotor of a vector field."},{"Start":"08:34.364 ","End":"08:40.105","Text":"Here we can see 2 components."},{"Start":"08:40.105 ","End":"08:43.090","Text":"We have the partial differential with respect to"},{"Start":"08:43.090 ","End":"08:47.455","Text":"z and the partial differential with respect to r. Now,"},{"Start":"08:47.455 ","End":"08:49.375","Text":"let\u0027s take a look at z."},{"Start":"08:49.375 ","End":"08:53.680","Text":"We were told in the question that we have an infinite cylinder."},{"Start":"08:53.680 ","End":"08:56.545","Text":"What does that mean if we have an infinite cylinder?"},{"Start":"08:56.545 ","End":"08:59.305","Text":"Now we\u0027re just looking at this section of the cylinder."},{"Start":"08:59.305 ","End":"09:04.930","Text":"But if we would move down to this section or move up to a section above it."},{"Start":"09:04.930 ","End":"09:10.644","Text":"If we would go just up infinity or down infinity along the cylinder,"},{"Start":"09:10.644 ","End":"09:14.245","Text":"we would have the exact same question in front of us."},{"Start":"09:14.245 ","End":"09:18.055","Text":"Nothing would change because this is an infinite cylinder."},{"Start":"09:18.055 ","End":"09:22.090","Text":"We could look up 1 meter down 1 meter and"},{"Start":"09:22.090 ","End":"09:26.140","Text":"our diagram and how we would solve this question is exactly the same."},{"Start":"09:26.140 ","End":"09:31.255","Text":"In other words, we have symmetry in the z-direction."},{"Start":"09:31.255 ","End":"09:36.835","Text":"If we have z-symmetry or symmetry in the z direction,"},{"Start":"09:36.835 ","End":"09:40.580","Text":"this means that d A,"},{"Start":"09:40.580 ","End":"09:47.340","Text":"any components of A by dz is equal to 0."},{"Start":"09:47.340 ","End":"09:53.050","Text":"Because our problem isn\u0027t changing as we go up and down the z-axis,"},{"Start":"09:53.050 ","End":"09:59.620","Text":"which means that the component with respect to z is constant."},{"Start":"09:59.620 ","End":"10:04.870","Text":"Of course, if we take the derivative of a constant, we get 0."},{"Start":"10:04.870 ","End":"10:08.635","Text":"Therefore, we can say that this is equal to 0."},{"Start":"10:08.635 ","End":"10:13.615","Text":"All we\u0027re left with for the phi component of the rotor of"},{"Start":"10:13.615 ","End":"10:20.000","Text":"A is negative d A _z by dr."},{"Start":"10:22.980 ","End":"10:29.875","Text":"In that case, we can look at this over here, a_z."},{"Start":"10:29.875 ","End":"10:35.320","Text":"We can see from this that we\u0027re left with only a_z,"},{"Start":"10:35.320 ","End":"10:39.370","Text":"the z component of our vector potential."},{"Start":"10:39.370 ","End":"10:42.775","Text":"If we have the z component of a vector potential,"},{"Start":"10:42.775 ","End":"10:47.020","Text":"that means that our vector potential is only in"},{"Start":"10:47.020 ","End":"10:53.540","Text":"the z-direction because all the other components cancel out given this problem set."},{"Start":"10:53.700 ","End":"11:00.910","Text":"Given a_z, that means that A is going"},{"Start":"11:00.910 ","End":"11:07.390","Text":"to be equal to some A value only in the z-direction."},{"Start":"11:07.390 ","End":"11:11.000","Text":"We have a z component for A."},{"Start":"11:11.130 ","End":"11:15.640","Text":"That means that our vector potential is in the z-direction."},{"Start":"11:15.640 ","End":"11:18.610","Text":"This is a more complicated explanation"},{"Start":"11:18.610 ","End":"11:21.670","Text":"for why our vector potential is in the z direction."},{"Start":"11:21.670 ","End":"11:25.510","Text":"Alternatively, you could have just used the right-hand rule,"},{"Start":"11:25.510 ","End":"11:32.890","Text":"the magnetic field curls in the phi direction so our thumb points up in the z direction."},{"Start":"11:32.890 ","End":"11:36.880","Text":"That will tell us already that our vector potential is in the z direction."},{"Start":"11:36.880 ","End":"11:41.530","Text":"But if you want the mathematical explanation here it is."},{"Start":"11:41.530 ","End":"11:48.290","Text":"This z over here tells us that it\u0027s in the z-direction."},{"Start":"11:49.020 ","End":"11:52.810","Text":"Let\u0027s calculate A."},{"Start":"11:52.810 ","End":"11:56.605","Text":"What I\u0027m going to do is I\u0027m going to take,"},{"Start":"11:56.605 ","End":"12:04.210","Text":"so I see that my A is in this direction so this is my A."},{"Start":"12:04.210 ","End":"12:11.575","Text":"Then what I\u0027m going to do is I\u0027m going to form a rectangular loop like so."},{"Start":"12:11.575 ","End":"12:15.700","Text":"The clue is given in this where we see a loop over here."},{"Start":"12:15.700 ","End":"12:21.290","Text":"A loop going through my magnetic field loop."},{"Start":"12:21.630 ","End":"12:24.910","Text":"Then I can define,"},{"Start":"12:24.910 ","End":"12:35.560","Text":"so what I\u0027m going to do is I\u0027m going to define that the A located at r,"},{"Start":"12:35.560 ","End":"12:38.935","Text":"is equal to 0, is equal to 0."},{"Start":"12:38.935 ","End":"12:42.820","Text":"Because if I take that r is equal to 0,"},{"Start":"12:42.820 ","End":"12:51.790","Text":"I\u0027m looking at how much magnetic field passes in side this loop."},{"Start":"12:51.790 ","End":"12:53.365","Text":"That means that here,"},{"Start":"12:53.365 ","End":"12:55.000","Text":"my A over here,"},{"Start":"12:55.000 ","End":"12:59.395","Text":"this line over here will be over here at the origin,"},{"Start":"12:59.395 ","End":"13:02.245","Text":"so at this point in the center."},{"Start":"13:02.245 ","End":"13:07.060","Text":"Then I can say that the magnetic field passing through here is just going to"},{"Start":"13:07.060 ","End":"13:12.710","Text":"be equal to 0 because there\u0027s not even a loop to go through."},{"Start":"13:13.220 ","End":"13:16.065","Text":"A at r is equal to 0,"},{"Start":"13:16.065 ","End":"13:18.705","Text":"is equal to 0."},{"Start":"13:18.705 ","End":"13:28.645","Text":"Then I\u0027m going to say that this length over here is equal to L. Now,"},{"Start":"13:28.645 ","End":"13:31.330","Text":"let\u0027s do this equation."},{"Start":"13:31.330 ","End":"13:36.830","Text":"We have the closed loop integral of A.dl."},{"Start":"13:39.030 ","End":"13:41.455","Text":"If we go up here,"},{"Start":"13:41.455 ","End":"13:45.260","Text":"we already said that A over here is equal to 0."},{"Start":"13:46.410 ","End":"13:51.160","Text":"A over here is equal to 0."},{"Start":"13:51.160 ","End":"13:53.650","Text":"We go up here or down here,"},{"Start":"13:53.650 ","End":"13:56.680","Text":"A is equal to 0, so we don\u0027t add anything."},{"Start":"13:56.680 ","End":"13:58.525","Text":"Then we go across here,"},{"Start":"13:58.525 ","End":"14:03.720","Text":"where here we have no addition to the vector potential,"},{"Start":"14:03.720 ","End":"14:09.855","Text":"because this line over here is perpendicular to the direction of vector potential."},{"Start":"14:09.855 ","End":"14:12.860","Text":"The vector potential is in the z direction,"},{"Start":"14:12.860 ","End":"14:14.905","Text":"not the I direction."},{"Start":"14:14.905 ","End":"14:20.995","Text":"Here we have 0, and then here as we go up this length of L,"},{"Start":"14:20.995 ","End":"14:25.225","Text":"we have vector potential A as we go up."},{"Start":"14:25.225 ","End":"14:28.330","Text":"Then we return to our starting position where again,"},{"Start":"14:28.330 ","End":"14:31.060","Text":"this is perpendicular to the direction of A."},{"Start":"14:31.060 ","End":"14:33.670","Text":"We don\u0027t add anything, and here,"},{"Start":"14:33.670 ","End":"14:39.040","Text":"we return to this where of course we defined that A over here is"},{"Start":"14:39.040 ","End":"14:45.620","Text":"equal to 0 because no magnetic field passes through a loop at r is equal to 0."},{"Start":"14:46.200 ","End":"14:55.465","Text":"In actual fact, this integral of A.dl is just going to be equal to this length over here,"},{"Start":"14:55.465 ","End":"15:01.855","Text":"A multiplied by the length L. Here we have A multiplied by L,"},{"Start":"15:01.855 ","End":"15:08.930","Text":"which is equal to the integral of B.ds."},{"Start":"15:10.470 ","End":"15:13.195","Text":"As we can see B,"},{"Start":"15:13.195 ","End":"15:21.740","Text":"we\u0027re trying to see how much magnetic field passes through this surface area over here."},{"Start":"15:21.810 ","End":"15:27.800","Text":"We\u0027ve seen that our magnetic field is in the phi direction."},{"Start":"15:28.560 ","End":"15:34.840","Text":"We want to see how much B is coming through this surface area,"},{"Start":"15:34.840 ","End":"15:38.095","Text":"which is perpendicular to its direction of travel."},{"Start":"15:38.095 ","End":"15:43.060","Text":"If the magnetic field is in the phi direction,"},{"Start":"15:43.060 ","End":"15:45.880","Text":"that means that this plane,"},{"Start":"15:45.880 ","End":"15:48.790","Text":"the surface area over here,"},{"Start":"15:48.790 ","End":"15:52.975","Text":"which is in the plane which is perpendicular to the magnetic field,"},{"Start":"15:52.975 ","End":"16:00.680","Text":"so that means that our integral is going to be according to r and z."},{"Start":"16:01.260 ","End":"16:06.610","Text":"In other words, we have this surface area like so,"},{"Start":"16:06.610 ","End":"16:10.570","Text":"and our B is traveling into it like this."},{"Start":"16:10.570 ","End":"16:17.110","Text":"Our B is in the phi direction and we know that it is perpendicular to this plane."},{"Start":"16:17.110 ","End":"16:22.105","Text":"That means that the plane is going to be in the z direction,"},{"Start":"16:22.105 ","End":"16:25.750","Text":"which we\u0027ve already seen and in the I direction,"},{"Start":"16:25.750 ","End":"16:27.775","Text":"which is also clear from this."},{"Start":"16:27.775 ","End":"16:31.210","Text":"This is dr direction,"},{"Start":"16:31.210 ","End":"16:36.470","Text":"and of course this is dz direction."},{"Start":"16:36.630 ","End":"16:42.490","Text":"Let\u0027s see, for r is less than a."},{"Start":"16:42.490 ","End":"16:44.995","Text":"We\u0027re located in the cylinder."},{"Start":"16:44.995 ","End":"16:50.200","Text":"We\u0027re looking at this integral over here of Mu Naught, J Naught,"},{"Start":"16:50.200 ","End":"16:55.360","Text":"r divided by 2, dr,"},{"Start":"16:55.360 ","End":"17:00.730","Text":"dz, where z were going from 0 to L,"},{"Start":"17:00.730 ","End":"17:08.065","Text":"and I were going from 0 to r. We can do r tags over here to separate this,"},{"Start":"17:08.065 ","End":"17:14.620","Text":"and so we can see that there\u0027s no z variable."},{"Start":"17:14.620 ","End":"17:17.530","Text":"We just multiply all of this by L,"},{"Start":"17:17.530 ","End":"17:21.790","Text":"so this L will cancel out with this L over here."},{"Start":"17:21.790 ","End":"17:29.260","Text":"What we\u0027re left with is A is equal to the integral from 0 to r of Mu Naught,"},{"Start":"17:29.260 ","End":"17:35.830","Text":"J Naught, r tag divided by 2 dr tag."},{"Start":"17:36.170 ","End":"17:45.375","Text":"Then this is going to integrate to r tags squared divided by 2,"},{"Start":"17:45.375 ","End":"17:55.080","Text":"and then we just substitute in r. This is going to be equal to Mu Naught, J Naught,"},{"Start":"17:55.510 ","End":"17:59.290","Text":"r^2 divided by 2 multiplied by 2,"},{"Start":"17:59.290 ","End":"18:00.730","Text":"so divided by 4,"},{"Start":"18:00.730 ","End":"18:03.055","Text":"and this is of course,"},{"Start":"18:03.055 ","End":"18:07.795","Text":"in the z direction as previously discussed."},{"Start":"18:07.795 ","End":"18:15.343","Text":"This is when we\u0027re located inside the cylinder."},{"Start":"18:15.343 ","End":"18:21.645","Text":"This is the vector potential inside the cylinder and now,"},{"Start":"18:21.645 ","End":"18:25.250","Text":"let\u0027s calculate when r is greater than a,"},{"Start":"18:25.250 ","End":"18:28.970","Text":"so we\u0027re located outside of the cylinder."},{"Start":"18:28.970 ","End":"18:33.365","Text":"What does that mean? We\u0027ve taken our a,"},{"Start":"18:33.365 ","End":"18:39.047","Text":"this over here, and this is our loop now."},{"Start":"18:39.047 ","End":"18:42.765","Text":"We can erase this."},{"Start":"18:42.765 ","End":"18:45.450","Text":"This is our loop,"},{"Start":"18:45.450 ","End":"18:49.500","Text":"so that means that we\u0027re going to have 2 integrals."},{"Start":"18:49.500 ","End":"18:52.365","Text":"We have the integral."},{"Start":"18:52.365 ","End":"18:57.855","Text":"Again, a multiplied by L for the exact same reasoning as before,"},{"Start":"18:57.855 ","End":"19:01.830","Text":"here we defined that a is equal to 0,"},{"Start":"19:01.830 ","End":"19:03.975","Text":"the vector potential is equal to 0."},{"Start":"19:03.975 ","End":"19:08.084","Text":"Here, is perpendicular to the direction of the vector potential,"},{"Start":"19:08.084 ","End":"19:10.320","Text":"as is this over here,"},{"Start":"19:10.320 ","End":"19:13.560","Text":"it has a 0 effect."},{"Start":"19:13.560 ","End":"19:17.100","Text":"We just take this side of the loop over here,"},{"Start":"19:17.100 ","End":"19:22.990","Text":"A multiplied by L, so that\u0027s here."},{"Start":"19:23.600 ","End":"19:27.030","Text":"Now we split the integral into 2,"},{"Start":"19:27.030 ","End":"19:31.140","Text":"we have the integral of the magnetic field that"},{"Start":"19:31.140 ","End":"19:35.505","Text":"is traveling up until this region over here,"},{"Start":"19:35.505 ","End":"19:43.425","Text":"up until here, at this radius a. Whilst we\u0027re located,"},{"Start":"19:43.425 ","End":"19:46.545","Text":"still inside the cylinder."},{"Start":"19:46.545 ","End":"19:49.740","Text":"The magnetic field inside the cylinder and then we"},{"Start":"19:49.740 ","End":"19:54.270","Text":"have the magnetic fields when we\u0027re located outside the cylinder,"},{"Start":"19:54.270 ","End":"19:58.780","Text":"which is this region over here."},{"Start":"19:59.300 ","End":"20:06.390","Text":"We have the integral again for the surface area of B over here,"},{"Start":"20:06.390 ","End":"20:08.280","Text":"so we have Mu Naught,"},{"Start":"20:08.280 ","End":"20:17.250","Text":"J Naught, r tag divided by 2, dr tag dz."},{"Start":"20:17.250 ","End":"20:22.215","Text":"Here, r is from 0 up until a."},{"Start":"20:22.215 ","End":"20:28.350","Text":"From the origin up until the edge of our cylinder and z is from"},{"Start":"20:28.350 ","End":"20:35.730","Text":"0 until L and then to this we add on."},{"Start":"20:35.730 ","End":"20:38.680","Text":"I\u0027ll scroll down here."},{"Start":"20:38.930 ","End":"20:44.655","Text":"The integral in this region over here where we\u0027re located outside of the cylinder."},{"Start":"20:44.655 ","End":"20:49.440","Text":"Again, we\u0027re integrating according to ds,"},{"Start":"20:49.440 ","End":"20:56.190","Text":"so it\u0027s still a double integral and it\u0027s still according to dr tag dz,"},{"Start":"20:56.190 ","End":"20:59.250","Text":"but now we\u0027re located outside of the cylinder."},{"Start":"20:59.250 ","End":"21:08.070","Text":"We\u0027re looking at the magnetic field where r is greater than a."},{"Start":"21:08.070 ","End":"21:10.815","Text":"This was a mistake here, greater than a."},{"Start":"21:10.815 ","End":"21:16.000","Text":"That is equal to Mu Naught, J Naught,"},{"Start":"21:16.250 ","End":"21:26.025","Text":"a^2 divided by 2r and then we have dr tag by dz."},{"Start":"21:26.025 ","End":"21:28.170","Text":"Where now the bounds are,"},{"Start":"21:28.170 ","End":"21:32.535","Text":"so for r, we\u0027re going from the radius a."},{"Start":"21:32.535 ","End":"21:36.075","Text":"From when we exit the realm of the cylinder,"},{"Start":"21:36.075 ","End":"21:45.824","Text":"so we\u0027re looking at this area over here from a until r. Where, of course,"},{"Start":"21:45.824 ","End":"21:53.870","Text":"this is r or this and z is still from 0"},{"Start":"21:53.870 ","End":"22:02.655","Text":"to L. Now let\u0027s see what this is equal to."},{"Start":"22:02.655 ","End":"22:06.765","Text":"Here, we have the exact same thing,"},{"Start":"22:06.765 ","End":"22:12.240","Text":"so let\u0027s do this integral first."},{"Start":"22:12.240 ","End":"22:16.200","Text":"We see that there is no z variables,"},{"Start":"22:16.200 ","End":"22:18.840","Text":"so we can just multiply everything by L,"},{"Start":"22:18.840 ","End":"22:20.940","Text":"and also over here, similarly,"},{"Start":"22:20.940 ","End":"22:23.700","Text":"we can see that there\u0027s no z variables,"},{"Start":"22:23.700 ","End":"22:27.060","Text":"so we can multiply everything by L and then we divide both sides by"},{"Start":"22:27.060 ","End":"22:31.470","Text":"L. This L will cancel out the L from this integral will cancel out,"},{"Start":"22:31.470 ","End":"22:33.825","Text":"and the L from this integral will cancel out,"},{"Start":"22:33.825 ","End":"22:36.135","Text":"we can just leave it like so."},{"Start":"22:36.135 ","End":"22:42.660","Text":"Then we\u0027re left with the integral from 0-a of Mu Naught,"},{"Start":"22:42.660 ","End":"22:49.210","Text":"J Naught by tag divided by 2 dr tag."},{"Start":"22:49.370 ","End":"22:51.780","Text":"Let\u0027s do that integral."},{"Start":"22:51.780 ","End":"22:54.465","Text":"It\u0027s just going to be the exact same thing as here,"},{"Start":"22:54.465 ","End":"22:57.720","Text":"but just with a^2 instead."},{"Start":"22:57.720 ","End":"23:00.690","Text":"Let\u0027s just erase all of this."},{"Start":"23:00.690 ","End":"23:02.220","Text":"We have the same integral,"},{"Start":"23:02.220 ","End":"23:06.100","Text":"Mu Naught, J Naught,"},{"Start":"23:06.110 ","End":"23:14.430","Text":"a^2 divided by 4 in the z direction,"},{"Start":"23:14.430 ","End":"23:17.490","Text":"but now we have plus."},{"Start":"23:17.490 ","End":"23:19.830","Text":"Again, we\u0027ve already done this L integral,"},{"Start":"23:19.830 ","End":"23:27.525","Text":"so we\u0027re left with plus the integral from a until r of Mu Naught,"},{"Start":"23:27.525 ","End":"23:34.395","Text":"J Naught, a^2 divided by 2r, dr tag."},{"Start":"23:34.395 ","End":"23:39.490","Text":"Here, our r is in the denominator."},{"Start":"23:39.590 ","End":"23:43.980","Text":"In other words, we\u0027re integrating 1 divided by rdr,"},{"Start":"23:43.980 ","End":"23:47.020","Text":"which as we know, gives us ln."},{"Start":"23:48.110 ","End":"23:52.095","Text":"Let\u0027s just write out this in the meantime."},{"Start":"23:52.095 ","End":"23:57.345","Text":"What we\u0027re going to have is Mu Naught, J Naught,"},{"Start":"23:57.345 ","End":"24:03.180","Text":"a^2 divided by 2 multiplied by ln of r,"},{"Start":"24:03.180 ","End":"24:05.760","Text":"minus ln of a, or in other words,"},{"Start":"24:05.760 ","End":"24:10.750","Text":"ln of r divided by a from the law of ln\u0027s."},{"Start":"24:11.930 ","End":"24:15.570","Text":"Now we can just plug all of this in,"},{"Start":"24:15.570 ","End":"24:24.880","Text":"so we can get that a is equal to this over here,"},{"Start":"24:25.250 ","End":"24:27.810","Text":"when r is less than a."},{"Start":"24:27.810 ","End":"24:32.685","Text":"It\u0027s equal to Mu Naught, J Naught,"},{"Start":"24:32.685 ","End":"24:40.500","Text":"r^2 divided by 4 in the z direction when r is less than A."},{"Start":"24:40.500 ","End":"24:45.840","Text":"A is equal to, so Mu Naught,"},{"Start":"24:45.840 ","End":"24:54.705","Text":"J Naught, a^2."},{"Start":"24:54.705 ","End":"25:00.600","Text":"This is also a common for both of these divided by 2"},{"Start":"25:00.600 ","End":"25:07.650","Text":"and then multiplied by 1 divided by 2 for here,"},{"Start":"25:07.650 ","End":"25:15.510","Text":"plus ln of r divided by a and this is,"},{"Start":"25:15.510 ","End":"25:20.820","Text":"of course, also in the z direction for r is greater than a."},{"Start":"25:20.820 ","End":"25:26.890","Text":"This is the vector potential when we\u0027re located outside of the cylinder."},{"Start":"25:28.400 ","End":"25:32.220","Text":"This is the final answer to this question."},{"Start":"25:32.220 ","End":"25:36.960","Text":"This is the vector potential when we\u0027re inside the cylinder and"},{"Start":"25:36.960 ","End":"25:42.975","Text":"this is the vector potential when we\u0027re located outside of the cylinder."},{"Start":"25:42.975 ","End":"25:46.480","Text":"That is the end of this lesson."}],"ID":21458},{"Watched":false,"Name":"1.8 Boundary Conditions for Vector Potential","Duration":"14m 27s","ChapterTopicVideoID":21388,"CourseChapterTopicPlaylistID":99495,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.040","Text":"Hello. In this lesson,"},{"Start":"00:02.040 ","End":"00:07.515","Text":"we\u0027re going to be learning about the boundary conditions for the vector potential."},{"Start":"00:07.515 ","End":"00:13.635","Text":"In the cases where we can\u0027t use Ampere\u0027s law in order to calculate the magnetic field,"},{"Start":"00:13.635 ","End":"00:17.565","Text":"what we will do is we will find the vector potential"},{"Start":"00:17.565 ","End":"00:22.650","Text":"and we will find it in the different regions."},{"Start":"00:22.650 ","End":"00:30.840","Text":"Then we will connect those potentials. What do I mean?"},{"Start":"00:30.840 ","End":"00:37.060","Text":"Let\u0027s say I have some plane that has current density,"},{"Start":"00:37.060 ","End":"00:40.505","Text":"K traveling along it."},{"Start":"00:40.505 ","End":"00:47.700","Text":"Here we have K or linear current density."},{"Start":"00:48.310 ","End":"00:54.695","Text":"In this case, I\u0027ll have vector potential a above the plane."},{"Start":"00:54.695 ","End":"01:01.320","Text":"I\u0027ll also have vector potential b below the plane."},{"Start":"01:02.420 ","End":"01:06.800","Text":"What we want to do is we want to find the components"},{"Start":"01:06.800 ","End":"01:12.233","Text":"of the a\u0027s that are perpendicular to the plane,"},{"Start":"01:12.233 ","End":"01:19.265","Text":"and the component of a of the vector potential that is parallel to the plane."},{"Start":"01:19.265 ","End":"01:28.440","Text":"Let\u0027s begin with looking at the components of a which are perpendicular."},{"Start":"01:29.240 ","End":"01:39.670","Text":"The first thing that we do is we calibrate this system such that div A,"},{"Start":"01:39.670 ","End":"01:45.470","Text":"the divergence of the vector potential is equal to 0."},{"Start":"01:45.470 ","End":"01:55.295","Text":"Then what we do is we build a small Gaussian surface to"},{"Start":"01:55.295 ","End":"01:57.840","Text":"envelope"},{"Start":"02:03.650 ","End":"02:06.930","Text":"the surface like so."},{"Start":"02:06.930 ","End":"02:09.375","Text":"Our surface over here."},{"Start":"02:09.375 ","End":"02:14.880","Text":"Then what we want to do is we want to calculate the vector potential over here."},{"Start":"02:14.880 ","End":"02:22.065","Text":"We\u0027re very close to our plane with K flowing through it."},{"Start":"02:22.065 ","End":"02:27.315","Text":"Here we have a perpendicular above."},{"Start":"02:27.315 ","End":"02:31.720","Text":"We also want to calculate the vector potential over"},{"Start":"02:31.720 ","End":"02:37.190","Text":"here where we have A perpendicular below."},{"Start":"02:38.900 ","End":"02:43.720","Text":"Now what I want to do is I want to look at the flux."},{"Start":"02:44.600 ","End":"02:50.510","Text":"Now calculating the flux of vector potential."},{"Start":"02:50.510 ","End":"02:55.070","Text":"If I think that this cube,"},{"Start":"02:55.070 ","End":"03:00.155","Text":"this Gaussian surface that is enveloping my plane,"},{"Start":"03:00.155 ","End":"03:03.515","Text":"is very close to the plane."},{"Start":"03:03.515 ","End":"03:06.440","Text":"This top is almost parallel to"},{"Start":"03:06.440 ","End":"03:15.365","Text":"the plane and right on top of it as is the bottom side."},{"Start":"03:15.365 ","End":"03:23.599","Text":"Then I can see that there won\u0027t be any flux of vector potential in this direction,"},{"Start":"03:23.599 ","End":"03:27.755","Text":"in the direction that is parallel to the plane,"},{"Start":"03:27.755 ","End":"03:29.900","Text":"or in this direction."},{"Start":"03:29.900 ","End":"03:38.955","Text":"We can say that the flux that is parallel will be equal to 0."},{"Start":"03:38.955 ","End":"03:47.125","Text":"Now what we\u0027re trying to find is the flux of A that is perpendicular to the plane."},{"Start":"03:47.125 ","End":"03:51.490","Text":"It\u0027s these black arrows over here."},{"Start":"03:52.700 ","End":"04:01.550","Text":"The flux for the perpendicular components will be just like we saw in Gauss\u0027s law."},{"Start":"04:01.550 ","End":"04:08.455","Text":"It will be equal to the vector potential that is perpendicular above."},{"Start":"04:08.455 ","End":"04:14.645","Text":"This over here multiplied by the surface area of"},{"Start":"04:14.645 ","End":"04:22.592","Text":"this side minus A perpendicular below,"},{"Start":"04:22.592 ","End":"04:27.575","Text":"this arrow over here multiplied by S,"},{"Start":"04:27.575 ","End":"04:31.370","Text":"the surface area of the bottom-most side."},{"Start":"04:32.720 ","End":"04:40.695","Text":"From this, we can use Gauss\u0027s theorem to get to the flux."},{"Start":"04:40.695 ","End":"04:44.015","Text":"Just before I do this, I\u0027m reminding you that"},{"Start":"04:44.015 ","End":"04:48.574","Text":"the parallel components are equal to 0 because the cube,"},{"Start":"04:48.574 ","End":"04:54.915","Text":"the top and bottom sides are very close to the plane."},{"Start":"04:54.915 ","End":"04:59.740","Text":"There\u0027s no flux parallel to the plane."},{"Start":"05:01.010 ","End":"05:03.605","Text":"If we use Gauss\u0027s theorem,"},{"Start":"05:03.605 ","End":"05:11.415","Text":"we can take the divergence of A which is equal to 0."},{"Start":"05:11.415 ","End":"05:15.650","Text":"From Gauss\u0027s law, we can say that if we integrate according to"},{"Start":"05:15.650 ","End":"05:20.780","Text":"dv both sides according"},{"Start":"05:20.780 ","End":"05:26.450","Text":"to the volume of the cube also here,"},{"Start":"05:26.450 ","End":"05:28.880","Text":"the volume of the cube."},{"Start":"05:28.880 ","End":"05:38.430","Text":"Then what we get over here on this side is going to be the flux of the vector potential."},{"Start":"05:38.430 ","End":"05:44.000","Text":"This comes from Gauss\u0027s theorem where of course the flux of"},{"Start":"05:44.000 ","End":"05:51.005","Text":"the vector potential is the flux through this Gaussian envelope in blue over here."},{"Start":"05:51.005 ","End":"05:54.890","Text":"This is the amount of flux in the envelope,"},{"Start":"05:54.890 ","End":"06:00.870","Text":"or rather the flux through the Gaussian envelope."},{"Start":"06:01.010 ","End":"06:03.229","Text":"Of course from this side,"},{"Start":"06:03.229 ","End":"06:08.075","Text":"the integral on 0 dv is equal to 0."},{"Start":"06:08.075 ","End":"06:13.790","Text":"From this, I get therefore that"},{"Start":"06:13.790 ","End":"06:20.000","Text":"my vector potential flux has to be"},{"Start":"06:20.000 ","End":"06:27.750","Text":"equal to 0 because the divergence of the vector potential is 0."},{"Start":"06:28.940 ","End":"06:37.595","Text":"Therefore, the flux of the vector potential is equal to 0."},{"Start":"06:37.595 ","End":"06:44.700","Text":"That means that this is equal to 0."},{"Start":"06:44.700 ","End":"06:48.725","Text":"Then if we rearrange this from here,"},{"Start":"06:48.725 ","End":"06:57.950","Text":"we divide everything by S. What we get is that the perpendicular component of a above"},{"Start":"06:57.950 ","End":"07:01.775","Text":"the plane over here is equal to"},{"Start":"07:01.775 ","End":"07:07.890","Text":"the perpendicular component of a below with a plane over here."},{"Start":"07:08.880 ","End":"07:14.545","Text":"Therefore, if the perpendicular component just above the plane,"},{"Start":"07:14.545 ","End":"07:19.044","Text":"and the perpendicular component just below the plane are equal."},{"Start":"07:19.044 ","End":"07:23.635","Text":"Then that means, that the perpendicular component of A,"},{"Start":"07:23.635 ","End":"07:29.455","Text":"as a whole of the vector potential to the plane is continuous."},{"Start":"07:29.455 ","End":"07:36.320","Text":"The perpendicular component of the vector potential must be continuous."},{"Start":"07:37.830 ","End":"07:43.885","Text":"Now let\u0027s look at the parallel component to the plane."},{"Start":"07:43.885 ","End":"07:51.525","Text":"Now we\u0027re looking at a parallel to plane."},{"Start":"07:51.525 ","End":"07:58.065","Text":"Festival, we\u0027ve seen that the rotor of A,"},{"Start":"07:58.065 ","End":"08:02.290","Text":"is equal to the magnetic field."},{"Start":"08:03.390 ","End":"08:07.210","Text":"This was the differential form of the equation."},{"Start":"08:07.210 ","End":"08:13.555","Text":"We saw that the integral form of the equation is the closed loop integral,"},{"Start":"08:13.555 ","End":"08:18.745","Text":"and me loop of A.dl,"},{"Start":"08:18.745 ","End":"08:26.710","Text":"which we saw was equal to the integral of B.ds,"},{"Start":"08:26.710 ","End":"08:32.785","Text":"on the surface, which is confined to this closed loop."},{"Start":"08:32.785 ","End":"08:38.870","Text":"We saw that this was equal to the magnetic flux."},{"Start":"08:40.260 ","End":"08:44.980","Text":"Now again, I\u0027m going to draw plane,"},{"Start":"08:44.980 ","End":"08:47.920","Text":"like using Ampere\u0027s law."},{"Start":"08:47.920 ","End":"08:50.290","Text":"The plane, it\u0027s not a plane,"},{"Start":"08:50.290 ","End":"08:54.115","Text":"it\u0027s rather a frame,"},{"Start":"08:54.115 ","End":"08:58.880","Text":"that goes above and below the plane."},{"Start":"08:58.980 ","End":"09:05.260","Text":"When dealing with Ampere\u0027s law and current density, those chapters."},{"Start":"09:05.260 ","End":"09:10.764","Text":"We saw that if the current density is flowing in some direction,"},{"Start":"09:10.764 ","End":"09:15.085","Text":"the magnetic field is going to be perpendicular to it."},{"Start":"09:15.085 ","End":"09:20.545","Text":"Here, the k is flowing in this rightwards direction."},{"Start":"09:20.545 ","End":"09:25.360","Text":"That means that the magnetic field will be above the plane,"},{"Start":"09:25.360 ","End":"09:28.465","Text":"flowing in this direction,"},{"Start":"09:28.465 ","End":"09:35.750","Text":"and below the plane will be flowing in the opposite but parallel direction."},{"Start":"09:37.170 ","End":"09:41.140","Text":"If k is in the x direction,"},{"Start":"09:41.140 ","End":"09:48.505","Text":"then our magnetic field is first of all have to be parallel to the plane,"},{"Start":"09:48.505 ","End":"09:54.745","Text":"that means it can be only in the y direction because the z direction is going up."},{"Start":"09:54.745 ","End":"09:57.850","Text":"It\u0027s in the positive or negative y direction,"},{"Start":"09:57.850 ","End":"10:02.050","Text":"depending on which side of the plane we\u0027re looking at."},{"Start":"10:02.050 ","End":"10:05.140","Text":"Of course, we can see that it\u0027s going to be in"},{"Start":"10:05.140 ","End":"10:08.605","Text":"this direction above the plane and this direction below."},{"Start":"10:08.605 ","End":"10:10.655","Text":"Because of the right-hand rule,"},{"Start":"10:10.655 ","End":"10:18.385","Text":"a thumb points in the k direction, in this x-direction."},{"Start":"10:18.385 ","End":"10:23.005","Text":"Then we can see that below the plane our magnetic field is going in,"},{"Start":"10:23.005 ","End":"10:27.320","Text":"and then above it loops around the plane like so,"},{"Start":"10:27.840 ","End":"10:33.440","Text":"and it travels in this direction."},{"Start":"10:35.310 ","End":"10:42.820","Text":"As before where I took my Gaussian surface to be very close to the plane."},{"Start":"10:42.820 ","End":"10:46.870","Text":"To here, this frame or this loop,"},{"Start":"10:46.870 ","End":"10:49.750","Text":"is very close to the surface."},{"Start":"10:49.750 ","End":"10:51.955","Text":"So this upper bar,"},{"Start":"10:51.955 ","End":"10:56.605","Text":"and this lower bar, are right next to my plane."},{"Start":"10:56.605 ","End":"11:04.580","Text":"In that case, I can say that my magnetic flux is equal to 0."},{"Start":"11:06.000 ","End":"11:08.245","Text":"This is equal to 0."},{"Start":"11:08.245 ","End":"11:10.750","Text":"Now let\u0027s calculate out our A.dl,"},{"Start":"11:10.750 ","End":"11:14.320","Text":"which will of course be equal to 0."},{"Start":"11:14.320 ","End":"11:16.690","Text":"We\u0027re looking at the, of course,"},{"Start":"11:16.690 ","End":"11:18.925","Text":"the perpendicular components of A,"},{"Start":"11:18.925 ","End":"11:24.640","Text":"so we can say that this is A parallel,"},{"Start":"11:24.640 ","End":"11:27.370","Text":"this is, A, parallel above,"},{"Start":"11:27.370 ","End":"11:29.785","Text":"and then we can draw over here."},{"Start":"11:29.785 ","End":"11:32.770","Text":"We have, A parallel below."},{"Start":"11:32.770 ","End":"11:35.495","Text":"I\u0027ve drawn them in the same direction,"},{"Start":"11:35.495 ","End":"11:38.685","Text":"it doesn\u0027t really make a difference because,"},{"Start":"11:38.685 ","End":"11:44.610","Text":"if they were in opposite directions or both of them were in the rightwards direction,"},{"Start":"11:44.610 ","End":"11:47.505","Text":"then we\u0027ll just get a minus in our calculation."},{"Start":"11:47.505 ","End":"11:51.390","Text":"It\u0027s just easier in this case to work with it like this, and of course,"},{"Start":"11:51.390 ","End":"11:54.460","Text":"if they\u0027re in different directions,"},{"Start":"11:54.460 ","End":"11:58.120","Text":"one of them, then we\u0027ll just get a minus when we solve this."},{"Start":"11:58.120 ","End":"12:02.635","Text":"What we\u0027ll have then for A.dl,"},{"Start":"12:02.635 ","End":"12:05.230","Text":"we go along here."},{"Start":"12:05.230 ","End":"12:10.315","Text":"We\u0027ll have a parallel above,"},{"Start":"12:10.315 ","End":"12:13.810","Text":"multiplied by the length of this side,"},{"Start":"12:13.810 ","End":"12:16.165","Text":"let\u0027s say that the length is L,"},{"Start":"12:16.165 ","End":"12:18.460","Text":"and then we go down,"},{"Start":"12:18.460 ","End":"12:19.735","Text":"and here of course,"},{"Start":"12:19.735 ","End":"12:21.400","Text":"we have perpendicular components,"},{"Start":"12:21.400 ","End":"12:24.250","Text":"which we\u0027re not looking at so it\u0027s 0."},{"Start":"12:24.250 ","End":"12:26.590","Text":"Then we go over here,"},{"Start":"12:26.590 ","End":"12:28.045","Text":"so we have minus,"},{"Start":"12:28.045 ","End":"12:35.485","Text":"and then we have A parallel below again multiplied by the length of this side,"},{"Start":"12:35.485 ","End":"12:41.470","Text":"which is L, and this is of course equal to the magnetic flux."},{"Start":"12:41.470 ","End":"12:44.245","Text":"This side is of course equal to this,"},{"Start":"12:44.245 ","End":"12:46.345","Text":"which we said that was the magnetic flux,"},{"Start":"12:46.345 ","End":"12:48.655","Text":"which we said in this case over here,"},{"Start":"12:48.655 ","End":"12:50.665","Text":"was equal to 0."},{"Start":"12:50.665 ","End":"12:55.210","Text":"In that case we can divide both sides by L,"},{"Start":"12:55.210 ","End":"12:57.475","Text":"and rearrange, and what we get,"},{"Start":"12:57.475 ","End":"13:00.625","Text":"is that, A parallel above,"},{"Start":"13:00.625 ","End":"13:05.395","Text":"is equal to A parallel below."},{"Start":"13:05.395 ","End":"13:14.570","Text":"In other words, the parallel component of A to the plane is also continuous."},{"Start":"13:16.680 ","End":"13:20.275","Text":"I could have said that at the beginning of the lesson,"},{"Start":"13:20.275 ","End":"13:25.315","Text":"but I wanted to just show you the logic of how we get to that."},{"Start":"13:25.315 ","End":"13:31.285","Text":"Both the parallel and the perpendicular components of the vector potential,"},{"Start":"13:31.285 ","End":"13:35.570","Text":"are going to be continuous above and below the plane."},{"Start":"13:36.270 ","End":"13:40.030","Text":"The vector potential is continuous."},{"Start":"13:40.030 ","End":"13:45.505","Text":"However, I want to remind you that the magnetic field is not continuous."},{"Start":"13:45.505 ","End":"13:52.960","Text":"B above, does not equal B below."},{"Start":"13:52.960 ","End":"13:55.299","Text":"The magnetic field isn\u0027t continuous,"},{"Start":"13:55.299 ","End":"13:57.040","Text":"but its potential is."},{"Start":"13:57.040 ","End":"14:02.200","Text":"We saw a similar thing when dealing with electric fields and potentials,"},{"Start":"14:02.200 ","End":"14:06.430","Text":"where we saw that the electric potential was also continuous."},{"Start":"14:06.430 ","End":"14:08.845","Text":"However, its derivative,"},{"Start":"14:08.845 ","End":"14:11.785","Text":"or its field was not continuous."},{"Start":"14:11.785 ","End":"14:15.865","Text":"It\u0027s same here, our vector potential, is continuous,"},{"Start":"14:15.865 ","End":"14:20.165","Text":"however, the derivative,"},{"Start":"14:20.165 ","End":"14:24.000","Text":"its field is not continuous."},{"Start":"14:24.000 ","End":"14:27.890","Text":"That is the end of this lesson."}],"ID":21459},{"Watched":false,"Name":"1.9 Multipole Expansion and Equation for Magnetic Dipole","Duration":"12m 21s","ChapterTopicVideoID":21389,"CourseChapterTopicPlaylistID":99495,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.025","Text":"Hello. In this lesson,"},{"Start":"00:02.025 ","End":"00:03.525","Text":"we\u0027re going to do"},{"Start":"00:03.525 ","End":"00:09.645","Text":"the multiple expansion and we\u0027re going to find the equation for a magnetic dipole."},{"Start":"00:09.645 ","End":"00:14.775","Text":"Let\u0027s imagine that we have some current loop like"},{"Start":"00:14.775 ","End":"00:20.595","Text":"so that has a current I flowing through it."},{"Start":"00:20.595 ","End":"00:24.255","Text":"Then let\u0027s say that we have this point in space,"},{"Start":"00:24.255 ","End":"00:33.120","Text":"and we want to find the vector potential at this point r in space."},{"Start":"00:34.130 ","End":"00:36.915","Text":"This vector is of course,"},{"Start":"00:36.915 ","End":"00:40.370","Text":"our r vector and we\u0027ve already seen in"},{"Start":"00:40.370 ","End":"00:45.730","Text":"previous lessons that we can calculate the vector potential at r,"},{"Start":"00:45.730 ","End":"00:47.835","Text":"by using this equation,"},{"Start":"00:47.835 ","End":"00:51.945","Text":"Mu naught I divided by"},{"Start":"00:51.945 ","End":"00:59.065","Text":"4 Pi multiplied by the integral of 1,"},{"Start":"00:59.065 ","End":"01:09.780","Text":"divided by the magnitude of r minus r tag dl."},{"Start":"01:10.160 ","End":"01:18.290","Text":"But of course this is dl tag because it relates to the r tag over here."},{"Start":"01:19.790 ","End":"01:22.545","Text":"This is our I vector,"},{"Start":"01:22.545 ","End":"01:24.690","Text":"and then what is our r tag vector?"},{"Start":"01:24.690 ","End":"01:27.285","Text":"It\u0027s this vector over here,"},{"Start":"01:27.285 ","End":"01:36.920","Text":"r tag and it just sums up along the length or dl of the loop."},{"Start":"01:37.910 ","End":"01:44.335","Text":"It will move to here and then to here and so on and so forth,"},{"Start":"01:44.335 ","End":"01:48.680","Text":"all the way until it closes the circle."},{"Start":"01:48.980 ","End":"01:53.830","Text":"The multiple expansion,"},{"Start":"01:54.110 ","End":"01:59.565","Text":"is to expand this equation."},{"Start":"01:59.565 ","End":"02:02.030","Text":"This is an exact equation,"},{"Start":"02:02.030 ","End":"02:04.190","Text":"but it\u0027s to expand this equation,"},{"Start":"02:04.190 ","End":"02:12.405","Text":"opening up all the different powers of r. In other words,"},{"Start":"02:12.405 ","End":"02:20.890","Text":"it means to take this equation and expand it such that we have something,"},{"Start":"02:20.890 ","End":"02:27.770","Text":"all sorts of constants divided by r plus again the same something,"},{"Start":"02:27.770 ","End":"02:29.650","Text":"or actually a different something,"},{"Start":"02:29.650 ","End":"02:38.615","Text":"divided by r squared plus another something divided by r cubed plus,"},{"Start":"02:38.615 ","End":"02:41.070","Text":"and so on and so forth."},{"Start":"02:41.510 ","End":"02:47.210","Text":"In case where we\u0027re calculating our vector potential at a point very,"},{"Start":"02:47.210 ","End":"02:53.660","Text":"very far away from the loop we can express that by saying that r vector is much,"},{"Start":"02:53.660 ","End":"02:57.890","Text":"much larger than the dimensions of the loop, so a much,"},{"Start":"02:57.890 ","End":"03:00.980","Text":"much larger than I tag vector,"},{"Start":"03:00.980 ","End":"03:04.170","Text":"which gives us the dimension of the loop."},{"Start":"03:04.460 ","End":"03:08.450","Text":"When r vector is really large,"},{"Start":"03:08.450 ","End":"03:12.245","Text":"then what we can see in this multiple expansion is that"},{"Start":"03:12.245 ","End":"03:16.160","Text":"each next term in this series"},{"Start":"03:16.160 ","End":"03:21.370","Text":"is going to be much smaller than the term that comes before it."},{"Start":"03:21.370 ","End":"03:27.185","Text":"So this will be the largest term then this divided by r squared,"},{"Start":"03:27.185 ","End":"03:32.120","Text":"and then this over here divided by r cubed and so on and so forth."},{"Start":"03:32.120 ","End":"03:37.910","Text":"Sometimes when these terms are so small and this term exists,"},{"Start":"03:37.910 ","End":"03:44.070","Text":"then we can just cancel out these terms because they won\u0027t change much."},{"Start":"03:44.480 ","End":"03:49.610","Text":"As an example, if this term is equal to, let\u0027s say,"},{"Start":"03:49.610 ","End":"03:54.005","Text":"100 and this term is equal to 0.1,"},{"Start":"03:54.005 ","End":"03:57.230","Text":"and this term is equal to 0.01."},{"Start":"03:57.230 ","End":"04:05.240","Text":"Then of course, we can just take this answer alone and cross out these because adding 0.1"},{"Start":"04:05.240 ","End":"04:13.879","Text":"plus 0.001 isn\u0027t going to change our result in a way that\u0027s meaningful."},{"Start":"04:13.879 ","End":"04:16.520","Text":"However, when is this useful?"},{"Start":"04:16.520 ","End":"04:20.015","Text":"When this term doesn\u0027t exist."},{"Start":"04:20.015 ","End":"04:23.570","Text":"What we\u0027re going to see is that sometimes we can calculate"},{"Start":"04:23.570 ","End":"04:29.180","Text":"the vector potential and the term divided by r is equal to 0."},{"Start":"04:29.180 ","End":"04:32.270","Text":"We know that because there\u0027s a current loop over here,"},{"Start":"04:32.270 ","End":"04:35.780","Text":"the vector potential can actually be equal to 0."},{"Start":"04:35.780 ","End":"04:39.550","Text":"It is going to be equal to something, but its value is very,"},{"Start":"04:39.550 ","End":"04:43.040","Text":"very small, so the 0 value doesn\u0027t help us."},{"Start":"04:43.040 ","End":"04:49.318","Text":"Then we\u0027ll look to this value which will be small,"},{"Start":"04:49.318 ","End":"04:53.270","Text":"and will give us some larger degree of accuracy."},{"Start":"04:53.270 ","End":"04:58.565","Text":"Then if this exists, this term,"},{"Start":"04:58.565 ","End":"05:02.390","Text":"then we\u0027ll write that our vector potential is equal to this term,"},{"Start":"05:02.390 ","End":"05:05.485","Text":"and if this term is also equal to 0,"},{"Start":"05:05.485 ","End":"05:08.355","Text":"then we\u0027ll move on to the next term."},{"Start":"05:08.355 ","End":"05:10.835","Text":"What we\u0027ll see is eventually as we go down,"},{"Start":"05:10.835 ","End":"05:16.460","Text":"we\u0027ll get some term over here that is a non-zero value."},{"Start":"05:16.460 ","End":"05:20.740","Text":"That is when multipole expansion is useful."},{"Start":"05:20.740 ","End":"05:26.970","Text":"Of course we\u0027re going to see an example if this explanation wasn\u0027t clear enough."},{"Start":"05:26.970 ","End":"05:30.450","Text":"Now let\u0027s look at the first component,"},{"Start":"05:30.450 ","End":"05:37.835","Text":"this divided by r. This is the monopole component,"},{"Start":"05:37.835 ","End":"05:43.940","Text":"so the vector potential of the monopole is equal to"},{"Start":"05:43.940 ","End":"05:52.160","Text":"Mu naught divided by 4 Pi multiplied by the current I,"},{"Start":"05:52.160 ","End":"05:57.500","Text":"multiplied by the closed loop or closed circuit integral of"},{"Start":"05:57.500 ","End":"06:05.520","Text":"dl tag divided by r. Let\u0027s break this down."},{"Start":"06:05.520 ","End":"06:08.900","Text":"What we can see is that here we have"},{"Start":"06:08.900 ","End":"06:12.530","Text":"the closed loop integral of dl tag like we had here,"},{"Start":"06:12.530 ","End":"06:17.840","Text":"so we\u0027re integrating along this loop, like so,"},{"Start":"06:17.840 ","End":"06:23.280","Text":"taking all the dls and adding them together, like so,"},{"Start":"06:23.280 ","End":"06:32.430","Text":"until we get to the end where we reach our starting point over here."},{"Start":"06:32.440 ","End":"06:37.055","Text":"When we do we do this closed loop integral on dl,"},{"Start":"06:37.055 ","End":"06:40.670","Text":"then it\u0027s equivalent to integrating."},{"Start":"06:40.670 ","End":"06:44.240","Text":"Because we\u0027re of course summing up the magnitude and"},{"Start":"06:44.240 ","End":"06:48.450","Text":"the direction of each of these dl vectors."},{"Start":"06:48.450 ","End":"06:52.050","Text":"If we arrive at the same point,"},{"Start":"06:52.050 ","End":"06:55.475","Text":"it\u0027s like taking an integral of the displacement."},{"Start":"06:55.475 ","End":"06:58.175","Text":"Even though we\u0027ve gone all around here,"},{"Start":"06:58.175 ","End":"07:01.775","Text":"we start and end at the exact same point,"},{"Start":"07:01.775 ","End":"07:04.790","Text":"which means that our displacement is 0,"},{"Start":"07:04.790 ","End":"07:08.630","Text":"and if our displacement is 0,"},{"Start":"07:08.630 ","End":"07:14.190","Text":"then this integral over here is equal to 0."},{"Start":"07:14.810 ","End":"07:18.610","Text":"Then we get that all of this is equal to 0."},{"Start":"07:18.610 ","End":"07:25.187","Text":"Just to note, by the way, this r is the magnitude of this r vector."},{"Start":"07:25.187 ","End":"07:27.160","Text":"We get that A mono,"},{"Start":"07:27.160 ","End":"07:29.620","Text":"this over here is equal to 0,"},{"Start":"07:29.620 ","End":"07:32.860","Text":"which makes sense because as we know,"},{"Start":"07:32.860 ","End":"07:37.260","Text":"this will be a magnetic field."},{"Start":"07:37.260 ","End":"07:41.200","Text":"The monopole over here is a magnetic monopole and we"},{"Start":"07:41.200 ","End":"07:45.830","Text":"know that there is no such thing as a magnetic monopole."},{"Start":"07:46.700 ","End":"07:51.235","Text":"In this example, specifically when we have a closed loop,"},{"Start":"07:51.235 ","End":"07:57.100","Text":"so the first component is always going to be equal to 0."},{"Start":"07:57.100 ","End":"08:00.680","Text":"Now we can see that this crosses out it\u0027s equal to 0,"},{"Start":"08:00.680 ","End":"08:05.465","Text":"which means that we have to move to the next component,"},{"Start":"08:05.465 ","End":"08:08.980","Text":"to the 1 divided by r squared."},{"Start":"08:09.320 ","End":"08:11.870","Text":"This is the second order,"},{"Start":"08:11.870 ","End":"08:16.880","Text":"and this is what will give me the vector potential over here,"},{"Start":"08:16.880 ","End":"08:19.745","Text":"because this component, the first order,"},{"Start":"08:19.745 ","End":"08:23.490","Text":"will always be equal to 0."},{"Start":"08:24.860 ","End":"08:31.490","Text":"Now let\u0027s move over to our second component or our second-order."},{"Start":"08:31.490 ","End":"08:38.565","Text":"This is the vector potential of the dipole at point"},{"Start":"08:38.565 ","End":"08:41.330","Text":"r. Here we could also say that this is a point"},{"Start":"08:41.330 ","End":"08:47.315","Text":"r. This is equal to Mu naught divided by 4 Pi,"},{"Start":"08:47.315 ","End":"08:57.230","Text":"multiplied by m vector cross product with r hat divided by r squared."},{"Start":"08:57.230 ","End":"09:07.455","Text":"Where of course, r again is the magnitude of this vector r. This lowercase m,"},{"Start":"09:07.455 ","End":"09:15.930","Text":"so m vector, is equal to the magnetic dipole moment."},{"Start":"09:16.940 ","End":"09:21.150","Text":"We\u0027ve already seen what this is equal to."},{"Start":"09:21.150 ","End":"09:28.220","Text":"M is equal to the current"},{"Start":"09:28.220 ","End":"09:35.360","Text":"that flows through the loop multiplied by the surface area of the loop."},{"Start":"09:35.360 ","End":"09:38.255","Text":"We\u0027ve seen this equation before and over here,"},{"Start":"09:38.255 ","End":"09:46.565","Text":"because here we\u0027ve written that our vector potential is denoted by this A."},{"Start":"09:46.565 ","End":"09:48.050","Text":"Here I\u0027m just going to erase it,"},{"Start":"09:48.050 ","End":"09:51.940","Text":"but we\u0027ve seen it with the A in previous chapters."},{"Start":"09:51.940 ","End":"09:57.830","Text":"Here just not to confuse the surface area of the loop with the vector potential,"},{"Start":"09:57.830 ","End":"10:04.415","Text":"so we\u0027ll denote it with S. This is the current flowing through"},{"Start":"10:04.415 ","End":"10:14.010","Text":"the loop and this is the surface area of the loop."},{"Start":"10:14.010 ","End":"10:19.745","Text":"Of course, the direction of this vector is perpendicular to the plane."},{"Start":"10:19.745 ","End":"10:24.730","Text":"This sign over here means perpendicular."},{"Start":"10:24.730 ","End":"10:31.985","Text":"Again, this r hat and this r over here refer to this r vector."},{"Start":"10:31.985 ","End":"10:37.115","Text":"So r hat is of course the unit vector of this and"},{"Start":"10:37.115 ","End":"10:43.680","Text":"r over here is just the magnitude of the vector squared."},{"Start":"10:43.880 ","End":"10:48.380","Text":"We can say that this vector over here in green,"},{"Start":"10:48.380 ","End":"10:49.715","Text":"if you can see it,"},{"Start":"10:49.715 ","End":"10:53.000","Text":"is our r hat."},{"Start":"10:54.080 ","End":"10:58.790","Text":"This over here is an equation for the equation sheets,"},{"Start":"10:58.790 ","End":"11:04.805","Text":"and this gives us the vector potential of a magnetic dipole,"},{"Start":"11:04.805 ","End":"11:08.225","Text":"where m over here is the magnetic dipole moment."},{"Start":"11:08.225 ","End":"11:10.980","Text":"Remember this equation over here."},{"Start":"11:11.780 ","End":"11:20.824","Text":"This is the vector potential of the magnetic dipole, and of course,"},{"Start":"11:20.824 ","End":"11:27.680","Text":"it is also going to give us the largest order for"},{"Start":"11:27.680 ","End":"11:32.870","Text":"any magnetic multipole because this"},{"Start":"11:32.870 ","End":"11:38.600","Text":"over here we saw the first order is always going to be equal to 0 as we saw."},{"Start":"11:38.600 ","End":"11:46.235","Text":"This is the largest order in the multipole expansion,"},{"Start":"11:46.235 ","End":"11:51.385","Text":"which will give us the vector potentials in general."},{"Start":"11:51.385 ","End":"11:55.640","Text":"Then the next element is not really relevant."},{"Start":"11:55.640 ","End":"11:57.320","Text":"We won\u0027t look at it because it\u0027s"},{"Start":"11:57.320 ","End":"12:04.040","Text":"quite a complicated equation and also it very rarely comes up,"},{"Start":"12:04.040 ","End":"12:06.620","Text":"so we\u0027re not going to take a look at it."},{"Start":"12:06.620 ","End":"12:10.460","Text":"This is the equation to find the vector potential,"},{"Start":"12:10.460 ","End":"12:15.770","Text":"especially when we\u0027re located very very far away from a magnetic dipole,"},{"Start":"12:15.770 ","End":"12:18.830","Text":"where this is of course, the dipole."},{"Start":"12:18.830 ","End":"12:22.140","Text":"That is the end of this lesson."}],"ID":21460},{"Watched":false,"Name":"1.10 Exercise – Thick Plane with Uniform Current Density","Duration":"38m 29s","ChapterTopicVideoID":21390,"CourseChapterTopicPlaylistID":99495,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.890","Text":"Hello. In this lesson,"},{"Start":"00:01.890 ","End":"00:04.375","Text":"we\u0027re going to be answering the following questions."},{"Start":"00:04.375 ","End":"00:07.485","Text":"An infinite plane of width d,"},{"Start":"00:07.485 ","End":"00:10.710","Text":"so here is its width d,"},{"Start":"00:10.710 ","End":"00:13.259","Text":"is parallel to the x-y plane,"},{"Start":"00:13.259 ","End":"00:15.780","Text":"where the x-y plane is at its center."},{"Start":"00:15.780 ","End":"00:19.829","Text":"There is a uniform current density of J is equal to"},{"Start":"00:19.829 ","End":"00:25.365","Text":"J Naught flowing in the x-direction through the plane."},{"Start":"00:25.365 ","End":"00:30.570","Text":"Question number 1 is to find the direction of the vector potential,"},{"Start":"00:30.570 ","End":"00:33.240","Text":"so that\u0027s what we\u0027re going to start with."},{"Start":"00:33.240 ","End":"00:38.459","Text":"We already know that the equation for the vector potential is"},{"Start":"00:38.459 ","End":"00:46.005","Text":"the rotor of A is equal to B."},{"Start":"00:46.005 ","End":"00:50.540","Text":"The direction of the magnetic field we can see here is,"},{"Start":"00:50.540 ","End":"00:55.850","Text":"if we use the right-hand rule and we put our thumb in the x-direction,"},{"Start":"00:55.850 ","End":"01:02.090","Text":"so here, the denotation here is that the x-axis is out of the page."},{"Start":"01:02.090 ","End":"01:05.498","Text":"We point our thumb in the direction out of the page."},{"Start":"01:05.498 ","End":"01:07.070","Text":"Then the fingers of"},{"Start":"01:07.070 ","End":"01:10.630","Text":"our right hand are going to point in the direction of the magnetic field."},{"Start":"01:10.630 ","End":"01:15.004","Text":"What we can see is that above the y-plane,"},{"Start":"01:15.004 ","End":"01:20.145","Text":"the magnetic field is going to be in this direction,"},{"Start":"01:20.145 ","End":"01:24.420","Text":"and below the y-plane or the y-axis,"},{"Start":"01:24.420 ","End":"01:27.980","Text":"sorry, are fingers on your right hand are"},{"Start":"01:27.980 ","End":"01:32.740","Text":"going to be pointing or curling in this direction."},{"Start":"01:32.740 ","End":"01:39.380","Text":"Another way to do this is to say that we know that the rotor of"},{"Start":"01:39.380 ","End":"01:47.055","Text":"the magnetic field is going to be equal to Mu Naught multiplied by the J vector,"},{"Start":"01:47.055 ","End":"01:49.565","Text":"and we know that over here,"},{"Start":"01:49.565 ","End":"01:55.554","Text":"we\u0027re given in the question that our J vector is in the x-direction."},{"Start":"01:55.554 ","End":"01:59.955","Text":"Therefore, if our J vectors in the x- direction,"},{"Start":"01:59.955 ","End":"02:01.170","Text":"that means that the rotor of"},{"Start":"02:01.170 ","End":"02:10.490","Text":"the magnetic field must also be in the x-direction."},{"Start":"02:10.490 ","End":"02:15.295","Text":"All the other components y and z will cancel out."},{"Start":"02:15.295 ","End":"02:22.004","Text":"Let\u0027s try and calculate what the rotor of B in the x-direction will be."},{"Start":"02:22.004 ","End":"02:30.680","Text":"We have the rotor of B in the x-direction is going to be equal to,"},{"Start":"02:30.680 ","End":"02:35.824","Text":"so we have D over B_z by"},{"Start":"02:35.824 ","End":"02:44.655","Text":"D_y minus D over B_y, by D_z."},{"Start":"02:44.655 ","End":"02:50.943","Text":"This, as we saw, is equal to Mu Naught multiplied by J Naught,"},{"Start":"02:50.943 ","End":"02:56.770","Text":"the x-component, because J_x is equal to J Naught as we saw."},{"Start":"02:57.110 ","End":"03:03.484","Text":"Let\u0027s look at this. Our first integral over differential over here is"},{"Start":"03:03.484 ","End":"03:10.279","Text":"taking the differential of the z component of the B field with respect to y."},{"Start":"03:10.279 ","End":"03:17.119","Text":"However, what we can see is that we have symmetry in the y-direction. Why is this?"},{"Start":"03:17.119 ","End":"03:22.715","Text":"Because we have this infinite plane of width d. D,"},{"Start":"03:22.715 ","End":"03:26.670","Text":"this is clearly in the z-direction,"},{"Start":"03:27.590 ","End":"03:31.315","Text":"the width because we\u0027re given in the diagram,"},{"Start":"03:31.315 ","End":"03:33.205","Text":"the x and y directions."},{"Start":"03:33.205 ","End":"03:38.439","Text":"The width is in the z-direction and in the z-direction we have this finite width."},{"Start":"03:38.439 ","End":"03:42.700","Text":"However, we have an infinite plane with respect to x and y."},{"Start":"03:42.700 ","End":"03:45.085","Text":"If y is infinite,"},{"Start":"03:45.085 ","End":"03:50.210","Text":"then that means we have symmetry in the y-direction."},{"Start":"03:50.420 ","End":"03:57.609","Text":"I wrote here the plane is infinite in the y-direction as given in the question."},{"Start":"03:57.609 ","End":"03:59.545","Text":"Therefore, we have symmetry in y,"},{"Start":"03:59.545 ","End":"04:01.670","Text":"and if we have symmetry in y,"},{"Start":"04:01.670 ","End":"04:06.760","Text":"any partial integral that will take with respect to y,"},{"Start":"04:06.760 ","End":"04:08.560","Text":"d by d_y of anything,"},{"Start":"04:08.560 ","End":"04:14.685","Text":"so you can sleuth anything over here is going to be equal to 0."},{"Start":"04:14.685 ","End":"04:22.180","Text":"Therefore, what we get is that our magnetic field is equal to the integral of this,"},{"Start":"04:22.180 ","End":"04:30.975","Text":"so what we\u0027ll have is some magnetic field with respect to z,"},{"Start":"04:30.975 ","End":"04:35.710","Text":"with z as some variable."},{"Start":"04:38.380 ","End":"04:42.454","Text":"As we can see over here,"},{"Start":"04:42.454 ","End":"04:47.795","Text":"it\u0027s going to be in the y-direction."},{"Start":"04:47.795 ","End":"04:50.990","Text":"I\u0027m not going to do the integral now because"},{"Start":"04:50.990 ","End":"04:53.989","Text":"I don\u0027t want to calculate the field in this way."},{"Start":"04:53.989 ","End":"04:58.205","Text":"But just so you can see that also using this method,"},{"Start":"04:58.205 ","End":"05:04.459","Text":"we get that the magnetic field is going to be in the y-direction plus or minus."},{"Start":"05:04.459 ","End":"05:09.379","Text":"We can see, but we can see over here that using the right-hand rule,"},{"Start":"05:09.379 ","End":"05:13.990","Text":"we also got that the magnetic field is in the y-direction."},{"Start":"05:13.990 ","End":"05:18.140","Text":"Now that we\u0027ve understood this using the magnetic field,"},{"Start":"05:18.140 ","End":"05:21.409","Text":"the question was to find the direction of the vector potential,"},{"Start":"05:21.409 ","End":"05:25.580","Text":"where of course the vector potential is this vector A."},{"Start":"05:25.580 ","End":"05:32.865","Text":"We\u0027re going to use the exact same method to find the direction of A."},{"Start":"05:32.865 ","End":"05:38.059","Text":"What we just calculated was that the B field is in the y-direction,"},{"Start":"05:38.059 ","End":"05:41.405","Text":"so it only has a y component."},{"Start":"05:41.405 ","End":"05:45.705","Text":"Therefore, the rotor of"},{"Start":"05:45.705 ","End":"05:54.270","Text":"A is going to be equal to B in the y-direction,"},{"Start":"05:54.270 ","End":"06:00.020","Text":"so that means that the rotor of A just like we did over here if B is in the y-direction,"},{"Start":"06:00.020 ","End":"06:02.225","Text":"what we saw if J is in the x-direction,"},{"Start":"06:02.225 ","End":"06:04.909","Text":"then the rotor of B has to be in the x-direction."},{"Start":"06:04.909 ","End":"06:08.239","Text":"Therefore, if here our B is in the y-direction,"},{"Start":"06:08.239 ","End":"06:12.220","Text":"the rotor of A has to be in the y-direction."},{"Start":"06:12.260 ","End":"06:20.449","Text":"Then what we can do is we can write out the rotor of A in"},{"Start":"06:20.449 ","End":"06:28.785","Text":"the y-direction as being equal to d A_x by d_z,"},{"Start":"06:28.785 ","End":"06:31.759","Text":"so the partial differential of the x-component"},{"Start":"06:31.759 ","End":"06:35.644","Text":"of the vector potential with respect to z,"},{"Start":"06:35.644 ","End":"06:39.500","Text":"minus the partial differential of"},{"Start":"06:39.500 ","End":"06:46.980","Text":"the vector potential in the z-direction with respect to x."},{"Start":"06:46.980 ","End":"06:50.430","Text":"This is going to be equal to B_y."},{"Start":"06:50.430 ","End":"06:53.269","Text":"Just as we saw in the question,"},{"Start":"06:53.269 ","End":"06:58.159","Text":"we\u0027re being told that we have an infinite plane with respect to x and y,"},{"Start":"06:58.159 ","End":"07:01.775","Text":"and as we saw just a few moments ago,"},{"Start":"07:01.775 ","End":"07:03.710","Text":"the plane is infinite,"},{"Start":"07:03.710 ","End":"07:07.775","Text":"so if it\u0027s infinite here in the y-direction in the cemetery y."},{"Start":"07:07.775 ","End":"07:14.580","Text":"What we have over here is also that it\u0027s infinite in the x-direction."},{"Start":"07:14.580 ","End":"07:17.555","Text":"If it\u0027s infinite in the x-direction,"},{"Start":"07:17.555 ","End":"07:20.375","Text":"then there\u0027s symmetry in the x-direction."},{"Start":"07:20.375 ","End":"07:23.344","Text":"The exact same logic that we used over here."},{"Start":"07:23.344 ","End":"07:27.110","Text":"Because it\u0027s infinite in the x-direction,"},{"Start":"07:27.110 ","End":"07:31.189","Text":"so we can say that any partial differential with respect"},{"Start":"07:31.189 ","End":"07:35.839","Text":"to x of anything is going to be equal to 0."},{"Start":"07:35.839 ","End":"07:39.200","Text":"Just like we had over here for the exact same reasons."},{"Start":"07:39.200 ","End":"07:42.559","Text":"In that case, this over here,"},{"Start":"07:42.559 ","End":"07:45.215","Text":"we have a partial differential with respect to x,"},{"Start":"07:45.215 ","End":"07:48.005","Text":"so this is going to be equal to 0."},{"Start":"07:48.005 ","End":"07:58.450","Text":"Therefore, what we\u0027re left with is d A_x by d_z is equal to B_y."},{"Start":"07:59.300 ","End":"08:05.539","Text":"Therefore, what we get is that our vector potential"},{"Start":"08:05.539 ","End":"08:12.125","Text":"is equal to A with z as a variable."},{"Start":"08:12.125 ","End":"08:20.455","Text":"Again over here in the x-direction."},{"Start":"08:20.455 ","End":"08:23.705","Text":"This is the answer to question number 1."},{"Start":"08:23.705 ","End":"08:30.235","Text":"What we see is that the direction of the vector potential is in the x-direction."},{"Start":"08:30.235 ","End":"08:34.770","Text":"Now let\u0027s answer question number 2."},{"Start":"08:34.770 ","End":"08:41.350","Text":"Question number 2 is asking us to find the equation for the vector potential."},{"Start":"08:41.350 ","End":"08:45.625","Text":"In other words, what we\u0027re being asked is to calculate A."},{"Start":"08:45.625 ","End":"08:47.635","Text":"In order to calculate A,"},{"Start":"08:47.635 ","End":"08:51.670","Text":"what we\u0027re going to do is we\u0027re going to calculate B."},{"Start":"08:51.670 ","End":"08:53.589","Text":"We already have the direction,"},{"Start":"08:53.589 ","End":"08:57.440","Text":"so we\u0027re calculating now the magnitude of B."},{"Start":"08:57.860 ","End":"09:02.785","Text":"In order to do this, we\u0027re going to use Ampere\u0027s law."},{"Start":"09:02.785 ","End":"09:04.795","Text":"Ampere\u0027s law, as we remember,"},{"Start":"09:04.795 ","End":"09:10.050","Text":"is the closed loop integral of B.dl,"},{"Start":"09:10.050 ","End":"09:19.648","Text":"which is equal to Mu Naught multiplied by the integral of J.ds."},{"Start":"09:19.648 ","End":"09:23.369","Text":"Now, what we want to do is we want to"},{"Start":"09:23.369 ","End":"09:28.124","Text":"find the magnitude of the magnetic field in the whole area."},{"Start":"09:28.124 ","End":"09:35.415","Text":"We\u0027re going to be looking at the regions above and below the plane and inside the plane."},{"Start":"09:35.415 ","End":"09:38.324","Text":"Let\u0027s build Ampere\u0027s loop."},{"Start":"09:38.324 ","End":"09:41.055","Text":"We\u0027re going to have this length over here,"},{"Start":"09:41.055 ","End":"09:43.440","Text":"which is of length L,"},{"Start":"09:43.440 ","End":"09:49.090","Text":"and it goes through the plane lake."},{"Start":"09:49.090 ","End":"09:53.114","Text":"Then it connects over here at the bottom."},{"Start":"09:53.114 ","End":"10:02.135","Text":"Of course, the length of the bottom is also L. This side of the equation,"},{"Start":"10:02.135 ","End":"10:07.444","Text":"so now we\u0027re going to be looking outside of the plane."},{"Start":"10:07.444 ","End":"10:12.735","Text":"We have our magnetic field multiplied by dl."},{"Start":"10:12.735 ","End":"10:14.370","Text":"What is our dl?"},{"Start":"10:14.370 ","End":"10:16.110","Text":"As we saw over here,"},{"Start":"10:16.110 ","End":"10:19.469","Text":"our magnetic field is going in this direction and over here,"},{"Start":"10:19.469 ","End":"10:21.850","Text":"it\u0027s going in this direction."},{"Start":"10:22.340 ","End":"10:26.880","Text":"This is in the anticlockwise direction,"},{"Start":"10:26.880 ","End":"10:29.070","Text":"but it\u0027s fine, they don\u0027t cancel out."},{"Start":"10:29.070 ","End":"10:33.750","Text":"What we have is B multiplied by 2L, and of course,"},{"Start":"10:33.750 ","End":"10:43.289","Text":"rather because our B field"},{"Start":"10:43.289 ","End":"10:44.970","Text":"is traveling in the y-direction,"},{"Start":"10:44.970 ","End":"10:48.480","Text":"it\u0027s perpendicular to these sides over here,"},{"Start":"10:48.480 ","End":"10:50.820","Text":"which are in the z direction,"},{"Start":"10:50.820 ","End":"10:52.560","Text":"so they cancel out."},{"Start":"10:52.560 ","End":"10:59.670","Text":"Then this is equal to Mu Naught multiplied by J.ds. What is?"},{"Start":"10:59.670 ","End":"11:03.449","Text":"Our J is a uniform and it\u0027s equal to J Naught."},{"Start":"11:03.449 ","End":"11:10.305","Text":"Then ds, we\u0027re looking at this region over here where there\u0027s current density,"},{"Start":"11:10.305 ","End":"11:13.334","Text":"so it\u0027s only this area over here."},{"Start":"11:13.334 ","End":"11:17.085","Text":"It doesn\u0027t include these white sections."},{"Start":"11:17.085 ","End":"11:22.035","Text":"We can see that the length is d and the width is L,"},{"Start":"11:22.035 ","End":"11:25.049","Text":"so multiplied by d,"},{"Start":"11:25.049 ","End":"11:29.219","Text":"multiplied by L. Then what we can do is we can"},{"Start":"11:29.219 ","End":"11:33.240","Text":"cancel out the L\u0027s from both sides and divide both sides by 2."},{"Start":"11:33.240 ","End":"11:40.620","Text":"What we get is that B is equal to 1/2 of Mu Naught, J Naught,"},{"Start":"11:40.620 ","End":"11:45.329","Text":"d. As we saw before,"},{"Start":"11:45.329 ","End":"11:51.585","Text":"this is going to be with regards to the directions."},{"Start":"11:51.585 ","End":"12:00.149","Text":"This is going to be in the negative y-direction when we\u0027re above the plane."},{"Start":"12:00.149 ","End":"12:05.410","Text":"When z, because this is in the z axis,"},{"Start":"12:09.560 ","End":"12:12.570","Text":"so this is going to be d divided by 2,"},{"Start":"12:12.570 ","End":"12:15.180","Text":"so greater than d divided by 2."},{"Start":"12:15.180 ","End":"12:20.100","Text":"This is going to be equal to 1/2 Mu Naught,"},{"Start":"12:20.100 ","End":"12:25.530","Text":"J Naught, d in the positive y-direction."},{"Start":"12:25.530 ","End":"12:33.420","Text":"When z, over here we have negative d divided by 2,"},{"Start":"12:33.420 ","End":"12:38.470","Text":"so less than negative d divided by 2."},{"Start":"12:40.130 ","End":"12:46.484","Text":"Now, let\u0027s find out what the magnetic field is inside."},{"Start":"12:46.484 ","End":"12:50.684","Text":"Again, I\u0027m drawing my Ampere\u0027s loop."},{"Start":"12:50.684 ","End":"12:57.599","Text":"Again, it\u0027s of length L. What we have"},{"Start":"12:57.599 ","End":"13:05.490","Text":"here is that our magnetic field is going to be equal to, again this."},{"Start":"13:05.490 ","End":"13:08.400","Text":"B multiplied by the same length,"},{"Start":"13:08.400 ","End":"13:10.905","Text":"so 2L, again,"},{"Start":"13:10.905 ","End":"13:15.990","Text":"because the B is parallel to the top and bottom sides,"},{"Start":"13:15.990 ","End":"13:19.500","Text":"but it\u0027s perpendicular to the left and right side,"},{"Start":"13:19.500 ","End":"13:22.245","Text":"so it cancels out, so it\u0027s just 2L."},{"Start":"13:22.245 ","End":"13:27.405","Text":"Then this is equal to Mu Naught, J Naught,"},{"Start":"13:27.405 ","End":"13:29.580","Text":"and then our ds this time,"},{"Start":"13:29.580 ","End":"13:33.644","Text":"so it\u0027s all of this area enclosed in the loop."},{"Start":"13:33.644 ","End":"13:38.700","Text":"As we can see, it\u0027s going to be if this is height z,"},{"Start":"13:38.700 ","End":"13:46.470","Text":"so it\u0027s going to be multiplied by 2z because we want the total length over here."},{"Start":"13:46.470 ","End":"13:50.767","Text":"2z multiplied by L,"},{"Start":"13:50.767 ","End":"13:54.645","Text":"so we have L times z plus z."},{"Start":"13:54.645 ","End":"14:00.090","Text":"Then we can cancel out the L\u0027s from both sides and the 2 from both sides."},{"Start":"14:00.090 ","End":"14:08.035","Text":"Then what we get is that B is equal to Mu Naught, J Naught z."},{"Start":"14:08.035 ","End":"14:11.809","Text":"As we can see,"},{"Start":"14:11.809 ","End":"14:14.049","Text":"so we can look over here."},{"Start":"14:14.049 ","End":"14:19.919","Text":"Over here, it\u0027s going to be in the negative y-direction above the center of the plane."},{"Start":"14:19.919 ","End":"14:24.360","Text":"This is going to be in the negative y-direction."},{"Start":"14:24.360 ","End":"14:26.744","Text":"However, when z is negative,"},{"Start":"14:26.744 ","End":"14:28.320","Text":"when z is down here,"},{"Start":"14:28.320 ","End":"14:30.524","Text":"we\u0027re going to have a negative value for z,"},{"Start":"14:30.524 ","End":"14:34.094","Text":"which when we multiply it by this negative over here,"},{"Start":"14:34.094 ","End":"14:35.624","Text":"it\u0027s going to get a positive."},{"Start":"14:35.624 ","End":"14:39.509","Text":"Then we get positive y-direction, which makes sense."},{"Start":"14:39.509 ","End":"14:41.130","Text":"We can just leave this."},{"Start":"14:41.130 ","End":"14:44.969","Text":"This is for when z is"},{"Start":"14:44.969 ","End":"14:53.610","Text":"between negative d divided"},{"Start":"14:53.610 ","End":"14:57.310","Text":"by 2 and d divided by 2."},{"Start":"14:58.130 ","End":"15:02.159","Text":"Now, we have the magnetic field in all the regions."},{"Start":"15:02.159 ","End":"15:05.939","Text":"What we\u0027re going to do is we want to find the vector potential."},{"Start":"15:05.939 ","End":"15:10.769","Text":"The same relationship that occurs between the magnetic field and"},{"Start":"15:10.769 ","End":"15:16.965","Text":"the current density occurs between the vector potential and the magnetic field."},{"Start":"15:16.965 ","End":"15:21.975","Text":"What we have is an equation for the closed loop integral of"},{"Start":"15:21.975 ","End":"15:32.949","Text":"A.dl is equal to the surface integral of B.ds."},{"Start":"15:35.120 ","End":"15:39.699","Text":"Now, our loop is going to be."},{"Start":"15:40.790 ","End":"15:47.715","Text":"This side of the equation is calculating the magnetic flux."},{"Start":"15:47.715 ","End":"15:50.490","Text":"In order to calculate the magnetic flux,"},{"Start":"15:50.490 ","End":"15:56.445","Text":"we want an area that is perpendicular to the direction of the magnetic field"},{"Start":"15:56.445 ","End":"16:02.985","Text":"so that we can calculate how much magnetic field lines are passing through that area."},{"Start":"16:02.985 ","End":"16:08.550","Text":"As we saw, the magnetic field is always in the y-direction, positive or negative."},{"Start":"16:08.550 ","End":"16:17.200","Text":"That means that our loop is going to have to be perpendicular to the y-direction."},{"Start":"16:19.130 ","End":"16:24.960","Text":"Because this side is pretty much calculating the flux of the magnetic field."},{"Start":"16:24.960 ","End":"16:26.594","Text":"In order to do that,"},{"Start":"16:26.594 ","End":"16:29.955","Text":"we have to see what direction the magnetic field is in."},{"Start":"16:29.955 ","End":"16:33.090","Text":"We saw that it\u0027s in the y-direction."},{"Start":"16:33.090 ","End":"16:34.290","Text":"If it\u0027s in the y-direction,"},{"Start":"16:34.290 ","End":"16:36.015","Text":"to calculate the magnetic flux,"},{"Start":"16:36.015 ","End":"16:39.839","Text":"the loop, similar to what we did for the Ampere\u0027s loop,"},{"Start":"16:39.839 ","End":"16:41.850","Text":"has to be perpendicular."},{"Start":"16:41.850 ","End":"16:46.410","Text":"This sign means perpendicular to the y-axis."},{"Start":"16:46.410 ","End":"16:48.870","Text":"That\u0027s from this side."},{"Start":"16:48.870 ","End":"16:50.640","Text":"From this side over here,"},{"Start":"16:50.640 ","End":"16:58.770","Text":"because we saw in the previous question that the vector potential is in the x-direction,"},{"Start":"16:58.770 ","End":"17:01.485","Text":"that means that our loop,"},{"Start":"17:01.485 ","End":"17:03.525","Text":"at least some of the sides of the loop,"},{"Start":"17:03.525 ","End":"17:09.950","Text":"are going to have to be parallel to the x-axis."},{"Start":"17:09.950 ","End":"17:15.969","Text":"Because of this, because we saw that it\u0027s in the x-direction,"},{"Start":"17:15.969 ","End":"17:18.190","Text":"our loop has to be parallel."},{"Start":"17:18.190 ","End":"17:23.390","Text":"This sign has to be parallel to the x-axis."},{"Start":"17:23.690 ","End":"17:28.285","Text":"Let\u0027s draw it in pink."},{"Start":"17:28.285 ","End":"17:31.300","Text":"What we have is we\u0027re just going to have a loop that is"},{"Start":"17:31.300 ","End":"17:35.185","Text":"perpendicular to this loop that we had."},{"Start":"17:35.185 ","End":"17:38.470","Text":"It\u0027s going to go like this,"},{"Start":"17:38.470 ","End":"17:41.890","Text":"then down like so,"},{"Start":"17:41.890 ","End":"17:46.194","Text":"then like this and like that."},{"Start":"17:46.194 ","End":"17:52.120","Text":"Excuse my drawing. Where this is the x-direction."},{"Start":"17:52.120 ","End":"17:58.645","Text":"It\u0027s coming out of the page and this is the z direction."},{"Start":"17:58.645 ","End":"18:01.790","Text":"It\u0027s going up."},{"Start":"18:02.070 ","End":"18:07.810","Text":"In black, I\u0027m going to draw the direction of the current density."},{"Start":"18:07.810 ","End":"18:14.815","Text":"We saw that the current density was going along the x-axis always it\u0027s uniform."},{"Start":"18:14.815 ","End":"18:19.195","Text":"Also here, it\u0027s going along the x-axis."},{"Start":"18:19.195 ","End":"18:23.425","Text":"What is our problem over here?"},{"Start":"18:23.425 ","End":"18:27.264","Text":"When we looked at this loop, for Ampere\u0027s Law,"},{"Start":"18:27.264 ","End":"18:31.750","Text":"we saw that we decided to go in this anticlockwise direction."},{"Start":"18:31.750 ","End":"18:35.829","Text":"What we had was B times this length L over"},{"Start":"18:35.829 ","End":"18:40.330","Text":"here nothing because the magnetic field is perpendicular to here."},{"Start":"18:40.330 ","End":"18:42.025","Text":"Then we go here."},{"Start":"18:42.025 ","End":"18:48.730","Text":"Our magnetic field was in the same direction as our direction of travel."},{"Start":"18:48.730 ","End":"18:52.060","Text":"Then we added again BL,"},{"Start":"18:52.060 ","End":"18:54.280","Text":"and then we went up here."},{"Start":"18:54.280 ","End":"18:58.434","Text":"We saw that again we have this 0 contribution."},{"Start":"18:58.434 ","End":"19:00.309","Text":"What we had was because"},{"Start":"19:00.309 ","End":"19:06.039","Text":"our magnetic field was always in our direction of travel throughout this loop,"},{"Start":"19:06.039 ","End":"19:09.010","Text":"we got this 2L over here."},{"Start":"19:09.010 ","End":"19:10.945","Text":"What\u0027s our problem here?"},{"Start":"19:10.945 ","End":"19:13.659","Text":"If we choose the same thing,"},{"Start":"19:13.659 ","End":"19:15.609","Text":"so our direction of travel,"},{"Start":"19:15.609 ","End":"19:17.664","Text":"let\u0027s say is in this direction."},{"Start":"19:17.664 ","End":"19:20.439","Text":"Then here and then here and here."},{"Start":"19:20.439 ","End":"19:24.400","Text":"Here we\u0027ll have J is 0,"},{"Start":"19:24.400 ","End":"19:28.749","Text":"L. Then we go down, again, it\u0027s perpendicular."},{"Start":"19:28.749 ","End":"19:31.945","Text":"0 contribution just like on this side."},{"Start":"19:31.945 ","End":"19:37.540","Text":"Then here, we can see that our j or"},{"Start":"19:37.540 ","End":"19:44.169","Text":"our A is opposing our direction of travel."},{"Start":"19:44.169 ","End":"19:49.100","Text":"What we would have Is AL over here."},{"Start":"19:49.200 ","End":"19:54.790","Text":"AL plus, so we go down plus 0,"},{"Start":"19:54.790 ","End":"19:58.900","Text":"and then we have minus AL."},{"Start":"19:58.900 ","End":"20:02.799","Text":"Because it\u0027s in the opposite direction of travel,"},{"Start":"20:02.799 ","End":"20:04.149","Text":"and then plus 0 again."},{"Start":"20:04.149 ","End":"20:07.819","Text":"What we have is that this is equal to 0."},{"Start":"20:08.610 ","End":"20:11.319","Text":"Then if this is equal to 0,"},{"Start":"20:11.319 ","End":"20:17.979","Text":"then that means that the magnetic flux here is also equal to 0, which it isn\u0027t."},{"Start":"20:17.979 ","End":"20:22.539","Text":"We can\u0027t have that R vector potential over here is equal to 0."},{"Start":"20:22.539 ","End":"20:29.390","Text":"We know that that is not true because we\u0027ve seen that we clearly have vector potential."},{"Start":"20:29.910 ","End":"20:33.279","Text":"We\u0027ve dealt with in previous chapters,"},{"Start":"20:33.279 ","End":"20:35.860","Text":"this issue with Ampere\u0027s loop,"},{"Start":"20:35.860 ","End":"20:39.114","Text":"where something can cancel out."},{"Start":"20:39.114 ","End":"20:42.130","Text":"But we know that it isn\u0027t actually canceling out."},{"Start":"20:42.130 ","End":"20:46.700","Text":"It\u0027s just the loop itself isn\u0027t good."},{"Start":"20:47.310 ","End":"20:57.460","Text":"This also can\u0027t be equal to 0 because we know that the rotor of A is equal to B."},{"Start":"20:57.460 ","End":"21:07.059","Text":"We know that B is not equal to 0 over here because we got values for what B is."},{"Start":"21:07.059 ","End":"21:09.564","Text":"If B is not equal to 0,"},{"Start":"21:09.564 ","End":"21:12.220","Text":"then the rotor of A can be equal to 0."},{"Start":"21:12.220 ","End":"21:14.979","Text":"If the rotor of A can\u0027t be equal to 0,"},{"Start":"21:14.979 ","End":"21:19.225","Text":"then this equation doesn\u0027t make sense."},{"Start":"21:19.225 ","End":"21:23.270","Text":"This loop just doesn\u0027t work for this question."},{"Start":"21:23.940 ","End":"21:27.250","Text":"What do we do?"},{"Start":"21:27.250 ","End":"21:30.115","Text":"I\u0027ve crossed all of this out."},{"Start":"21:30.115 ","End":"21:33.309","Text":"In order to find a better loop, first of all,"},{"Start":"21:33.309 ","End":"21:37.345","Text":"what I\u0027m going to have to do is I\u0027m going to have to calibrate the A."},{"Start":"21:37.345 ","End":"21:45.565","Text":"Or in other words, find a point that I just define that at that point A is equal to 0."},{"Start":"21:45.565 ","End":"21:51.055","Text":"Now the reason that I can do that is because A is a vector potential,"},{"Start":"21:51.055 ","End":"21:56.685","Text":"which means that it is correct up until some constant."},{"Start":"21:56.685 ","End":"22:02.199","Text":"I can have this Az in the x direction plus some constant,"},{"Start":"22:02.670 ","End":"22:09.295","Text":"which we\u0027ve seen when integrating without having the initial conditions."},{"Start":"22:09.295 ","End":"22:12.770","Text":"Because it\u0027s going to be something plus a constant,"},{"Start":"22:12.770 ","End":"22:18.795","Text":"I can define that constant because here I\u0027m not giving initial conditions."},{"Start":"22:18.795 ","End":"22:29.630","Text":"I can define that constant to be anything such that at some point A will be equal to 0."},{"Start":"22:30.870 ","End":"22:34.675","Text":"What we\u0027re going to do is we\u0027re going to say that"},{"Start":"22:34.675 ","End":"22:38.320","Text":"A point we want it to be at z is equal to 0."},{"Start":"22:38.320 ","End":"22:41.964","Text":"A at the point z is equal to 0."},{"Start":"22:41.964 ","End":"22:46.250","Text":"We\u0027re going to define it as being equal to 0."},{"Start":"22:49.590 ","End":"22:53.215","Text":"If z is equal to 0,"},{"Start":"22:53.215 ","End":"22:55.810","Text":"A is equal to 0."},{"Start":"22:55.810 ","End":"22:59.245","Text":"That means that kz is equal to 0 over here."},{"Start":"22:59.245 ","End":"23:01.795","Text":"On the whole x-y plane."},{"Start":"23:01.795 ","End":"23:06.220","Text":"The x-y plane that lies at z is equal to 0."},{"Start":"23:06.220 ","End":"23:11.269","Text":"Our vector potential A is equal to 0."},{"Start":"23:12.390 ","End":"23:19.960","Text":"Because we were in the question told that this is an infinite plane in the x-y direction."},{"Start":"23:19.960 ","End":"23:25.039","Text":"That means that there\u0027s symmetry in the x-y direction."},{"Start":"23:25.530 ","End":"23:28.330","Text":"There is symmetry in the x-y plane,"},{"Start":"23:28.330 ","End":"23:33.440","Text":"and therefore A is uniform and the x-y plane."},{"Start":"23:34.350 ","End":"23:40.420","Text":"Now that we know that on the x-y plane,"},{"Start":"23:40.420 ","End":"23:43.269","Text":"A is equal to 0,"},{"Start":"23:43.269 ","End":"23:45.730","Text":"now we can redraw the loop."},{"Start":"23:45.730 ","End":"23:50.560","Text":"Similarly to what we did previously."},{"Start":"23:50.560 ","End":"23:56.500","Text":"However, this time it\u0027s along z is equal to 0."},{"Start":"23:56.500 ","End":"24:00.444","Text":"This is the x-direction."},{"Start":"24:00.444 ","End":"24:04.374","Text":"This is the z-direction."},{"Start":"24:04.374 ","End":"24:13.609","Text":"This line over here is on z is equal to 0."},{"Start":"24:14.340 ","End":"24:18.655","Text":"Then we can see that our,"},{"Start":"24:18.655 ","End":"24:25.134","Text":"sorry, our vector potential as we saw is traveling in the x direction."},{"Start":"24:25.134 ","End":"24:29.005","Text":"Then here, as we defined it,"},{"Start":"24:29.005 ","End":"24:31.705","Text":"A is equal to 0 along here."},{"Start":"24:31.705 ","End":"24:38.725","Text":"Now, when we do the closed loop integral of A.dL."},{"Start":"24:38.725 ","End":"24:43.060","Text":"What we have is A multiplied by this length over here,"},{"Start":"24:43.060 ","End":"24:48.789","Text":"which is L. Then we go down the sides,"},{"Start":"24:48.789 ","End":"24:57.579","Text":"which is equal to 0 because it is perpendicular to our vector field up over here,"},{"Start":"24:57.579 ","End":"25:01.989","Text":"which is equal to 0 as we defined it and up again,"},{"Start":"25:01.989 ","End":"25:07.479","Text":"the side which is again equal to 0 because it\u0027s perpendicular to the vector field."},{"Start":"25:07.479 ","End":"25:10.870","Text":"We just have A multiplied by L,"},{"Start":"25:10.870 ","End":"25:16.270","Text":"which is equal to the integral of B.ds."},{"Start":"25:16.270 ","End":"25:20.215","Text":"We\u0027re integrating this area over here."},{"Start":"25:20.215 ","End":"25:25.060","Text":"Ds over here, because we can see that B isn\u0027t uniform,"},{"Start":"25:25.060 ","End":"25:28.690","Text":"it changes with respect to z."},{"Start":"25:28.690 ","End":"25:30.699","Text":"Here it is uniform,"},{"Start":"25:30.699 ","End":"25:33.819","Text":"but inside it\u0027s with respect to z."},{"Start":"25:33.819 ","End":"25:38.784","Text":"What we have is a double integral for ds, for area."},{"Start":"25:38.784 ","End":"25:44.059","Text":"Then we have the magnetic field."},{"Start":"25:44.660 ","End":"25:50.160","Text":"Of course, we know that the magnetic field only has a y component,"},{"Start":"25:50.160 ","End":"25:53.240","Text":"it\u0027s only in the y direction as we saw."},{"Start":"25:53.240 ","End":"25:58.180","Text":"We can just write B_y and also we know it\u0027s going to be perpendicular to"},{"Start":"25:58.180 ","End":"26:02.890","Text":"this plane and y is perpendicular to this loop,"},{"Start":"26:02.890 ","End":"26:05.454","Text":"which is in the x-z plane,"},{"Start":"26:05.454 ","End":"26:08.720","Text":"so y is perpendicular to both x and z."},{"Start":"26:09.240 ","End":"26:18.085","Text":"Then this is going to be so ds is dx by d_z, as we said."},{"Start":"26:18.085 ","End":"26:21.520","Text":"It\u0027s along the z and x axes."},{"Start":"26:21.520 ","End":"26:25.345","Text":"When we\u0027re integrating according to x,"},{"Start":"26:25.345 ","End":"26:29.080","Text":"so we\u0027re going to be going from 0"},{"Start":"26:29.080 ","End":"26:33.970","Text":"until allocate the total length over here is L as we saw."},{"Start":"26:33.970 ","End":"26:38.110","Text":"From z, we\u0027re going to be going from 0,"},{"Start":"26:38.110 ","End":"26:42.590","Text":"z is equal to 0 until z."},{"Start":"26:42.840 ","End":"26:49.700","Text":"We can put a tag over here so as not to confuse the 2."},{"Start":"26:51.630 ","End":"26:54.550","Text":"As we can see, all of our B\u0027s,"},{"Start":"26:54.550 ","End":"26:55.929","Text":"let\u0027s scroll down,"},{"Start":"26:55.929 ","End":"27:03.079","Text":"all of our values for the magnetic field in the different regions are independent of x."},{"Start":"27:03.260 ","End":"27:06.030","Text":"It\u0027s constant in the x area."},{"Start":"27:06.030 ","End":"27:09.749","Text":"We also saw that a is uniform in the x-y plane."},{"Start":"27:09.749 ","End":"27:11.774","Text":"Also in the x-axis."},{"Start":"27:11.774 ","End":"27:14.840","Text":"When we integrate with respect to x,"},{"Start":"27:14.840 ","End":"27:18.559","Text":"it\u0027s just going to be equal to x."},{"Start":"27:18.559 ","End":"27:25.822","Text":"Then we substitute in L minus Naught which is just going to be L. Then multiply"},{"Start":"27:25.822 ","End":"27:35.529","Text":"it by the integral of 0 to z of B_yd_z tag."},{"Start":"27:35.529 ","End":"27:37.239","Text":"Now we can cross off,"},{"Start":"27:37.239 ","End":"27:44.379","Text":"we can divide both sides by L. What we\u0027re going to have is that A is equal"},{"Start":"27:44.379 ","End":"27:53.260","Text":"to the integral of 0 to z of B_yd_z tag."},{"Start":"27:53.260 ","End":"27:56.770","Text":"What we can see here is we have 2 regions."},{"Start":"27:56.770 ","End":"28:00.849","Text":"Now, this is meant to go above the plane."},{"Start":"28:00.849 ","End":"28:04.840","Text":"We have the region that goes up until here,"},{"Start":"28:04.840 ","End":"28:06.263","Text":"the top of the plane,"},{"Start":"28:06.263 ","End":"28:08.589","Text":"so this region over here,"},{"Start":"28:08.589 ","End":"28:14.770","Text":"and then we have the region above which is above the plane."},{"Start":"28:14.770 ","End":"28:17.960","Text":"It\u0027s this area over here."},{"Start":"28:18.300 ","End":"28:20.769","Text":"As we saw in those regions,"},{"Start":"28:20.769 ","End":"28:22.510","Text":"we have a different magnetic field,"},{"Start":"28:22.510 ","End":"28:23.904","Text":"so we\u0027re going to add up."},{"Start":"28:23.904 ","End":"28:32.510","Text":"We\u0027re going to integrate from 0 until the top of the plane which is d divided by 2."},{"Start":"28:32.820 ","End":"28:35.439","Text":"If this total height is d,"},{"Start":"28:35.439 ","End":"28:43.329","Text":"then from z is equal to 0 until the top it\u0027s half of d. From 0 until here,"},{"Start":"28:43.329 ","End":"28:45.535","Text":"this is d divided by 2."},{"Start":"28:45.535 ","End":"28:49.760","Text":"Then we\u0027re also going to integrate on this region over here,"},{"Start":"28:49.760 ","End":"28:52.585","Text":"but, of course, in this loop."},{"Start":"28:52.585 ","End":"28:54.880","Text":"All of this is going on in here."},{"Start":"28:54.880 ","End":"28:57.020","Text":"It\u0027s just not very clear."},{"Start":"28:57.420 ","End":"29:02.210","Text":"0 until d divided by 2."},{"Start":"29:02.910 ","End":"29:08.745","Text":"In that region, so what we\u0027re looking at is when z is smaller than d divided by 2,"},{"Start":"29:08.745 ","End":"29:11.364","Text":"which is this region over here."},{"Start":"29:11.364 ","End":"29:15.819","Text":"We\u0027re integrating along Mu Naught, J Naught,"},{"Start":"29:15.819 ","End":"29:20.440","Text":"z, d_z tag,"},{"Start":"29:20.440 ","End":"29:22.918","Text":"and this will also be tag."},{"Start":"29:22.918 ","End":"29:26.289","Text":"That\u0027s this integral over here,"},{"Start":"29:26.289 ","End":"29:29.634","Text":"and then we\u0027re adding on the integral in this region over here."},{"Start":"29:29.634 ","End":"29:37.530","Text":"It\u0027s an integral from d divided by 2 until the top of our initial region which is z."},{"Start":"29:37.530 ","End":"29:41.980","Text":"In total, we\u0027re going from 0 until z."},{"Start":"29:43.350 ","End":"29:46.855","Text":"Then now we\u0027re looking at the region in the magnetic field"},{"Start":"29:46.855 ","End":"29:50.320","Text":"that is greater than d divided by 2."},{"Start":"29:50.320 ","End":"29:52.592","Text":"That\u0027s this region over here,"},{"Start":"29:52.592 ","End":"29:57.340","Text":"so we\u0027re integrating along half of Mu Naught,"},{"Start":"29:57.340 ","End":"30:00.340","Text":"J Naught d,"},{"Start":"30:00.340 ","End":"30:03.949","Text":"and then d_z tag."},{"Start":"30:05.760 ","End":"30:09.520","Text":"Then if we do these integrals,"},{"Start":"30:09.520 ","End":"30:14.620","Text":"so what we\u0027ll have over here is Mu Naught,"},{"Start":"30:14.620 ","End":"30:19.090","Text":"J Naught, and then we have the integral on z."},{"Start":"30:19.090 ","End":"30:23.380","Text":"We\u0027ll have z tag squared divided by 2,"},{"Start":"30:23.380 ","End":"30:25.240","Text":"and then when we substitute in,"},{"Start":"30:25.240 ","End":"30:32.690","Text":"so we\u0027ll have d divided by 2 squared and then all of this divided by 2."},{"Start":"30:32.940 ","End":"30:37.120","Text":"Then the 0 will just make minus 0."},{"Start":"30:37.120 ","End":"30:41.500","Text":"Then plus, I forgot"},{"Start":"30:41.500 ","End":"30:43.659","Text":"a minus over here because we\u0027re looking at"},{"Start":"30:43.659 ","End":"30:46.030","Text":"this region over here in the negative y direction,"},{"Start":"30:46.030 ","End":"30:49.240","Text":"so we can just write here that it\u0027s in the plus y direction,"},{"Start":"30:49.240 ","End":"30:50.965","Text":"add a minus over here."},{"Start":"30:50.965 ","End":"30:52.959","Text":"I just forgot the minus."},{"Start":"30:52.959 ","End":"30:55.330","Text":"This will then become minus."},{"Start":"30:55.330 ","End":"30:57.679","Text":"Then here, we don\u0027t have a z,"},{"Start":"30:57.679 ","End":"31:01.150","Text":"so it\u0027s just going to be equal to plus minus,"},{"Start":"31:01.150 ","End":"31:05.649","Text":"so 1/2 of Mu Naught,"},{"Start":"31:05.649 ","End":"31:08.500","Text":"J Naught, d,"},{"Start":"31:08.500 ","End":"31:16.945","Text":"and then multiplied by z tag where we substitute in z minus d divided by 2."},{"Start":"31:16.945 ","End":"31:19.659","Text":"Then this is equal to,"},{"Start":"31:19.659 ","End":"31:22.000","Text":"if I take out common factors,"},{"Start":"31:22.000 ","End":"31:25.869","Text":"it\u0027s going to be equal to negative Mu Naught,"},{"Start":"31:25.869 ","End":"31:28.907","Text":"J Naught d,"},{"Start":"31:28.907 ","End":"31:37.310","Text":"divided by 2 multiplied by z minus d divided by 4."},{"Start":"31:37.440 ","End":"31:48.535","Text":"This is going to be the vector potential for this region over here above z is equal to 0."},{"Start":"31:48.535 ","End":"31:52.149","Text":"Of course, because our vector potential is symmetrical,"},{"Start":"31:52.149 ","End":"31:58.189","Text":"so we\u0027ll get the exact same result below, z is equal to 0."},{"Start":"31:59.880 ","End":"32:07.929","Text":"This is all given that z is bigger than d divided by 2,"},{"Start":"32:07.929 ","End":"32:14.665","Text":"but what if z is smaller than d divided by 2?"},{"Start":"32:14.665 ","End":"32:20.689","Text":"In other words, we\u0027re located inside over here."},{"Start":"32:20.760 ","End":"32:24.115","Text":"In that case, the integral is even easier."},{"Start":"32:24.115 ","End":"32:33.320","Text":"We have that A is equal to the integral from 0 until z of this region."},{"Start":"32:34.470 ","End":"32:40.825","Text":"This is going to be this over here."},{"Start":"32:40.825 ","End":"32:44.200","Text":"It\u0027s Mu Naught,"},{"Start":"32:44.200 ","End":"32:46.570","Text":"the negative because of this, Mu Naught,"},{"Start":"32:46.570 ","End":"32:54.729","Text":"J Naught, z, d_z tag."},{"Start":"32:54.729 ","End":"32:57.159","Text":"Then this is simply going to be equal to,"},{"Start":"32:57.159 ","End":"32:59.185","Text":"once I substitute in all the values,"},{"Start":"32:59.185 ","End":"33:02.620","Text":"negative Mu Naught,"},{"Start":"33:02.620 ","End":"33:08.260","Text":"J Naught, z^2 divided by 2."},{"Start":"33:08.260 ","End":"33:12.129","Text":"In other words, I\u0027ll write it over here,"},{"Start":"33:12.129 ","End":"33:17.360","Text":"A, as a function of z,"},{"Start":"33:17.360 ","End":"33:27.550","Text":"is equal to negative Mu Naught, J Naught,"},{"Start":"33:27.550 ","End":"33:32.529","Text":"z^2 divided by 2 when the absolute value"},{"Start":"33:32.529 ","End":"33:37.540","Text":"of z is smaller than d divided by 2 because it\u0027s symmetrical,"},{"Start":"33:37.540 ","End":"33:43.070","Text":"so z can be positive or negative as we said."},{"Start":"33:44.550 ","End":"33:53.859","Text":"It is equal to negative Mu Naught, J Naught,"},{"Start":"33:53.859 ","End":"34:01.075","Text":"d divided by 2 of z minus d divided by 4,"},{"Start":"34:01.075 ","End":"34:08.530","Text":"again, when the absolute value of z is bigger or greater than d divided by 2."},{"Start":"34:08.530 ","End":"34:17.739","Text":"Again, because z can be larger than Naught or smaller."},{"Start":"34:17.739 ","End":"34:19.659","Text":"That takes care of that and, of course,"},{"Start":"34:19.659 ","End":"34:21.940","Text":"as we said before, this is, of course,"},{"Start":"34:21.940 ","End":"34:26.634","Text":"in the x direction as we calculated."},{"Start":"34:26.634 ","End":"34:33.460","Text":"This is the answer to question number 2 and this is the end of the lesson."},{"Start":"34:33.460 ","End":"34:37.555","Text":"However, I\u0027m going to show you now a second way to"},{"Start":"34:37.555 ","End":"34:42.680","Text":"calculate the vector potential using the router equation."},{"Start":"34:43.980 ","End":"34:48.969","Text":"Here, I\u0027ve just rewritten what we got for our magnetic field."},{"Start":"34:48.969 ","End":"34:52.420","Text":"Let\u0027s work out the second way."},{"Start":"34:52.420 ","End":"34:54.309","Text":"Let\u0027s see how we do it."},{"Start":"34:54.309 ","End":"34:57.925","Text":"What we can do is we can take the router of"},{"Start":"34:57.925 ","End":"35:03.880","Text":"the vector potential which we know is equal to the magnetic field."},{"Start":"35:03.880 ","End":"35:06.835","Text":"We\u0027ve already seen this equation before."},{"Start":"35:06.835 ","End":"35:11.905","Text":"Then because we know that the magnetic field is only in the y direction,"},{"Start":"35:11.905 ","End":"35:14.989","Text":"it only has a y component,"},{"Start":"35:18.780 ","End":"35:22.850","Text":"we have to take the y component of the router,"},{"Start":"35:22.850 ","End":"35:30.550","Text":"and then what we\u0027re left with is d A_x by d_z."},{"Start":"35:30.550 ","End":"35:35.754","Text":"We also had the partial differential with respect to y."},{"Start":"35:35.754 ","End":"35:40.258","Text":"However, that canceled out because we know,"},{"Start":"35:40.258 ","End":"35:43.936","Text":"because the plane is infinite in the x-y direction,"},{"Start":"35:43.936 ","End":"35:49.929","Text":"that taking the derivative with respect to y will equal 0."},{"Start":"35:49.929 ","End":"35:53.509","Text":"This is the equation that we\u0027re left with."},{"Start":"35:54.150 ","End":"35:58.939","Text":"Then in order to find A_x,"},{"Start":"35:59.940 ","End":"36:03.594","Text":"which is as a function of z,"},{"Start":"36:03.594 ","End":"36:10.129","Text":"we just integrate B_y with respect to z."},{"Start":"36:11.040 ","End":"36:18.824","Text":"Then what we get is Mu Naught, J Naught,"},{"Start":"36:18.824 ","End":"36:27.595","Text":"d divided by 2 and because this is an integral without limits,"},{"Start":"36:27.595 ","End":"36:29.920","Text":"so plus some constant C_1,"},{"Start":"36:29.920 ","End":"36:35.440","Text":"so this is when z is smaller than negative d divided by 2."},{"Start":"36:35.440 ","End":"36:39.685","Text":"Then we have negative Mu Naught,"},{"Start":"36:39.685 ","End":"36:46.030","Text":"J Naught, z^2 divided by 2 plus another constant C_2."},{"Start":"36:46.030 ","End":"36:50.785","Text":"This is when z is between negative d divided by 2"},{"Start":"36:50.785 ","End":"36:57.480","Text":"and d divided by 2 and negative Mu Naught,"},{"Start":"36:57.480 ","End":"37:01.455","Text":"J Naught, d divided by 2 plus,"},{"Start":"37:01.455 ","End":"37:03.359","Text":"again, another constant,"},{"Start":"37:03.359 ","End":"37:10.285","Text":"C_3, and this is when z is bigger than d divided by 2."},{"Start":"37:10.285 ","End":"37:15.130","Text":"Then in order to calculate what the constants are,"},{"Start":"37:15.130 ","End":"37:19.899","Text":"we apply our initial condition where we said that"},{"Start":"37:19.899 ","End":"37:26.710","Text":"A_x at z is equal 0 is equal to 0."},{"Start":"37:26.710 ","End":"37:30.564","Text":"Remember we did this calibration."},{"Start":"37:30.564 ","End":"37:35.230","Text":"Then we just substitute in over here the like values."},{"Start":"37:35.230 ","End":"37:41.289","Text":"Here, we\u0027ll substitute in negative d divided by 2 into z and"},{"Start":"37:41.289 ","End":"37:48.495","Text":"equate it to this given this and that\u0027s it."},{"Start":"37:48.495 ","End":"37:56.800","Text":"What we\u0027ll find is that C_2 is equal to 0 from this and then"},{"Start":"37:56.800 ","End":"38:05.275","Text":"C_1 and C_2 you can just calculate from subbing in these values that are on the border."},{"Start":"38:05.275 ","End":"38:07.525","Text":"These are on the border with 1 another,"},{"Start":"38:07.525 ","End":"38:10.690","Text":"so making that at this point negative d divided by 2,"},{"Start":"38:10.690 ","End":"38:12.549","Text":"these equations are equal."},{"Start":"38:12.549 ","End":"38:15.174","Text":"Then at this point,"},{"Start":"38:15.174 ","End":"38:17.125","Text":"positive d divided by 2,"},{"Start":"38:17.125 ","End":"38:22.202","Text":"these equations are equal given that C_2 is equal to 0,"},{"Start":"38:22.202 ","End":"38:24.774","Text":"because of our initial condition."},{"Start":"38:24.774 ","End":"38:28.910","Text":"That is the end of this lesson."}],"ID":21461}],"Thumbnail":null,"ID":99495}]

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