[{"Name":"Introduction to Magnetic Dipole Moment","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Equations And Explanation","Duration":"12m 44s","ChapterTopicVideoID":21347,"CourseChapterTopicPlaylistID":99486,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/21347.jpeg","UploadDate":"2020-04-06T21:48:49.0370000","DurationForVideoObject":"PT12M44S","Description":null,"MetaTitle":"Equations And Explanation: Video + Workbook | Proprep","MetaDescription":"Magnetic Dipole Moment - Introduction to Magnetic Dipole Moment. 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/magnetic-dipole-moment/introduction-to-magnetic-dipole-moment/vid21427","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.235","Text":"Hello. In this lesson,"},{"Start":"00:02.235 ","End":"00:06.525","Text":"we\u0027re going to be speaking about the magnetic dipole moment."},{"Start":"00:06.525 ","End":"00:10.530","Text":"Now, a magnetic dipole is always going to be"},{"Start":"00:10.530 ","End":"00:16.275","Text":"some close loop that has a current I flowing through it."},{"Start":"00:16.275 ","End":"00:18.915","Text":"Now, usually when we\u0027re speaking about this,"},{"Start":"00:18.915 ","End":"00:25.125","Text":"we\u0027re speaking about either a circular loop where the current I flowing through"},{"Start":"00:25.125 ","End":"00:33.360","Text":"or some square loop with a current I flowing through it."},{"Start":"00:33.360 ","End":"00:36.980","Text":"That is what a magnetic dipole is."},{"Start":"00:36.980 ","End":"00:41.900","Text":"It some closed loop where the current I flowing through it."},{"Start":"00:41.900 ","End":"00:46.400","Text":"Now what we\u0027re going to define is the magnetic dipole moment,"},{"Start":"00:46.400 ","End":"00:48.350","Text":"and this is symbolized by Mu."},{"Start":"00:48.350 ","End":"00:56.015","Text":"It\u0027s also sometimes symbolized by m. That is equal to I,"},{"Start":"00:56.015 ","End":"00:58.175","Text":"the current flowing through the loop,"},{"Start":"00:58.175 ","End":"01:01.820","Text":"multiplied by the area of the loop."},{"Start":"01:01.820 ","End":"01:04.790","Text":"Now notice that the magnetic dipole moment, Mu,"},{"Start":"01:04.790 ","End":"01:09.335","Text":"is a vector and it\u0027s in the direction of the vector,"},{"Start":"01:09.335 ","End":"01:15.510","Text":"which is normal to the surface area of the loop."},{"Start":"01:17.210 ","End":"01:23.660","Text":"We can see that the direction of the magnetic dipole moment is always going to"},{"Start":"01:23.660 ","End":"01:29.875","Text":"be in the same direction as the vector which is normal to our surface area."},{"Start":"01:29.875 ","End":"01:35.615","Text":"Let\u0027s say that we\u0027re dealing with this square current loop."},{"Start":"01:35.615 ","End":"01:42.510","Text":"Our surface area of the current loop is going to be everything inside,"},{"Start":"01:42.510 ","End":"01:47.160","Text":"so the blue and that\u0027s our A and the vector normal to it."},{"Start":"01:47.160 ","End":"01:49.025","Text":"According to the right-hand rule,"},{"Start":"01:49.025 ","End":"01:54.265","Text":"we can see that because our current is flowing in this direction,"},{"Start":"01:54.265 ","End":"02:00.995","Text":"vector normal to it is going to be coming out towards us,"},{"Start":"02:00.995 ","End":"02:03.110","Text":"out of the page and towards us."},{"Start":"02:03.110 ","End":"02:06.500","Text":"That means that in this case over here,"},{"Start":"02:06.500 ","End":"02:13.200","Text":"our magnetic dipole moment is going to be also coming out towards us."},{"Start":"02:13.450 ","End":"02:16.655","Text":"Of course, as we said,"},{"Start":"02:16.655 ","End":"02:24.989","Text":"we can also write this instead of Mu as small m. If you see this, don\u0027t get confused."},{"Start":"02:24.989 ","End":"02:29.270","Text":"What can we do with the magnetic dipole moment?"},{"Start":"02:29.270 ","End":"02:33.965","Text":"Now, you\u0027ll remember from the chapter on electric dipoles,"},{"Start":"02:33.965 ","End":"02:35.300","Text":"when we have a dipole,"},{"Start":"02:35.300 ","End":"02:41.115","Text":"we can find our electric field when we\u0027re far away from our dipole."},{"Start":"02:41.115 ","End":"02:43.595","Text":"It\u0027s the same thing with the magnetic dipole."},{"Start":"02:43.595 ","End":"02:45.050","Text":"If we\u0027re located here,"},{"Start":"02:45.050 ","End":"02:50.209","Text":"which is far away from our magnetic dipole,"},{"Start":"02:50.209 ","End":"02:56.520","Text":"so we can find out what our field is at this point over here."},{"Start":"02:57.670 ","End":"03:03.870","Text":"What is the equation for finding the magnetic field at this point?"},{"Start":"03:04.850 ","End":"03:14.090","Text":"This is the equation for the magnetic field when we\u0027re far away from a magnetic dipole."},{"Start":"03:14.090 ","End":"03:15.815","Text":"When we\u0027re far away,"},{"Start":"03:15.815 ","End":"03:22.855","Text":"that means that our vector r or the size of our vector r,"},{"Start":"03:22.855 ","End":"03:29.945","Text":"is going to be much bigger than the square root of our area,"},{"Start":"03:29.945 ","End":"03:35.060","Text":"where the area is obviously the area of our current loop."},{"Start":"03:35.060 ","End":"03:39.425","Text":"When this occurs, then we can use"},{"Start":"03:39.425 ","End":"03:44.990","Text":"our equation for the magnetic field due to a magnetic dipole."},{"Start":"03:44.990 ","End":"03:48.815","Text":"The equation is equal to Mu_0 divided by 4Pi."},{"Start":"03:48.815 ","End":"03:56.040","Text":"Then that\u0027s multiplied by 3 our magnetic dipole moment dot product with"},{"Start":"03:56.040 ","End":"04:04.155","Text":"our r vector in the I vector direction minus our magnetic dipole moment divided by r^3,"},{"Start":"04:04.155 ","End":"04:06.600","Text":"which is the distance we"},{"Start":"04:06.600 ","End":"04:11.480","Text":"are from a magnetic dipole"},{"Start":"04:11.480 ","End":"04:16.045","Text":"to our point in space where we\u0027re measuring the magnetic field."},{"Start":"04:16.045 ","End":"04:19.690","Text":"How do we use this equation?"},{"Start":"04:19.940 ","End":"04:25.640","Text":"Now, of course, our magnetic dipole moment is in the direction of"},{"Start":"04:25.640 ","End":"04:31.350","Text":"the vector normal to the surface area of our current loop."},{"Start":"04:31.350 ","End":"04:39.300","Text":"Our I-hat is of course the unit vector pointing in our I vector direction."},{"Start":"04:41.510 ","End":"04:48.070","Text":"Now what we\u0027re going to do is we\u0027re going to see how this equation and how"},{"Start":"04:48.070 ","End":"04:55.475","Text":"the whole idea of a magnetic dipole is very similar to that of an electric dipole."},{"Start":"04:55.475 ","End":"04:58.015","Text":"Let\u0027s take a look at that here."},{"Start":"04:58.015 ","End":"05:04.660","Text":"Now, we know that an electric dipole has a positive charge and a negative charge,"},{"Start":"05:04.660 ","End":"05:05.710","Text":"here with the same size,"},{"Start":"05:05.710 ","End":"05:07.245","Text":"just a different sign."},{"Start":"05:07.245 ","End":"05:13.335","Text":"There\u0027s a distance d between the positive and negative charge."},{"Start":"05:13.335 ","End":"05:18.865","Text":"Then we know that the electric dipole moment is equal to q,"},{"Start":"05:18.865 ","End":"05:22.035","Text":"the charge multiplied by d vector,"},{"Start":"05:22.035 ","End":"05:29.045","Text":"which is the distance between the 2 charges in the direction that it is pointing."},{"Start":"05:29.045 ","End":"05:34.414","Text":"Then we know that if we\u0027re far away from this electric dipole,"},{"Start":"05:34.414 ","End":"05:38.060","Text":"then we can say that the electric field at some point far"},{"Start":"05:38.060 ","End":"05:42.050","Text":"away from the dipole is equal to k multiplied by"},{"Start":"05:42.050 ","End":"05:49.685","Text":"3p vector.r-hat in the r-hat direction"},{"Start":"05:49.685 ","End":"05:57.150","Text":"minus our p vector divided by r^3."},{"Start":"05:57.170 ","End":"06:02.650","Text":"Of course, this only works when we\u0027re far away from electric dipoles."},{"Start":"06:02.650 ","End":"06:09.750","Text":"That means that our r is a lot bigger than the distance between the 2 charges."},{"Start":"06:09.820 ","End":"06:14.030","Text":"Now notice how similar these 2 equations are."},{"Start":"06:14.030 ","End":"06:16.145","Text":"Here we have our k,"},{"Start":"06:16.145 ","End":"06:22.625","Text":"which is for our electrostatic equations and here we just have a different constant,"},{"Start":"06:22.625 ","End":"06:24.710","Text":"Mu_0 divided by 4Pi."},{"Start":"06:24.710 ","End":"06:27.530","Text":"Then here we have the exact same thing."},{"Start":"06:27.530 ","End":"06:32.055","Text":"Just instead of here we have the electric dipole moment."},{"Start":"06:32.055 ","End":"06:33.320","Text":"Here we have Mu,"},{"Start":"06:33.320 ","End":"06:35.570","Text":"which is the magnetic dipole moment,"},{"Start":"06:35.570 ","End":"06:38.790","Text":"and everything else is completely the same."},{"Start":"06:39.980 ","End":"06:47.105","Text":"Now what we\u0027re going to talk about is what happens when we have a magnetic dipole,"},{"Start":"06:47.105 ","End":"06:51.925","Text":"which is located in an external magnetic field."},{"Start":"06:51.925 ","End":"06:55.940","Text":"Let\u0027s say that we have our magnetic dipole,"},{"Start":"06:55.940 ","End":"06:58.040","Text":"which is this current loop in"},{"Start":"06:58.040 ","End":"07:02.345","Text":"this direction and the direction of the magnetic dipole moment,"},{"Start":"07:02.345 ","End":"07:09.065","Text":"Mu is perpendicular towards area of the current loop."},{"Start":"07:09.065 ","End":"07:14.060","Text":"Then let\u0027s say that we have our external magnetic field,"},{"Start":"07:14.060 ","End":"07:18.650","Text":"which is constant in the z-direction."},{"Start":"07:18.650 ","End":"07:22.630","Text":"We have a magnetic field like so."},{"Start":"07:22.630 ","End":"07:27.050","Text":"What happens in a case like this is that we have"},{"Start":"07:27.050 ","End":"07:31.445","Text":"a torque which will act on our magnetic dipole moment,"},{"Start":"07:31.445 ","End":"07:35.070","Text":"which will act to rotate it so that"},{"Start":"07:35.070 ","End":"07:40.775","Text":"our Mu vector is in the direction of our magnetic field."},{"Start":"07:40.775 ","End":"07:48.525","Text":"What we\u0027ll get is that our current loop will rotate to be like so"},{"Start":"07:48.525 ","End":"07:57.690","Text":"and our Mu vector will be pointing in the same direction parallel to our magnetic field."},{"Start":"07:58.700 ","End":"08:01.265","Text":"Now, why does this happen?"},{"Start":"08:01.265 ","End":"08:03.800","Text":"The magnetic field causes"},{"Start":"08:03.800 ","End":"08:08.150","Text":"forces which interact with the current flowing through the loop,"},{"Start":"08:08.150 ","End":"08:10.910","Text":"which causes a force over here going in"},{"Start":"08:10.910 ","End":"08:15.604","Text":"this direction and the force over here pointing in this direction."},{"Start":"08:15.604 ","End":"08:18.335","Text":"This will rotate our current loop,"},{"Start":"08:18.335 ","End":"08:24.865","Text":"or rotate our magnetic dipole to align with the external magnetic field."},{"Start":"08:24.865 ","End":"08:31.175","Text":"Now remember that we had a similar story with our electric dipole."},{"Start":"08:31.175 ","End":"08:36.950","Text":"If we had our electric dipole like so,"},{"Start":"08:36.950 ","End":"08:39.230","Text":"and this is our electric dipole moment,"},{"Start":"08:39.230 ","End":"08:46.415","Text":"and we placed it in some external E field, like so."},{"Start":"08:46.415 ","End":"08:49.940","Text":"Then we had forces which would pull"},{"Start":"08:49.940 ","End":"08:55.740","Text":"the positive charge upwards and the negative charge downwards"},{"Start":"08:55.950 ","End":"09:03.440","Text":"such that our electric dipole would rotate to"},{"Start":"09:03.440 ","End":"09:11.645","Text":"be pointing in the direction of the external E field."},{"Start":"09:11.645 ","End":"09:17.030","Text":"Now we said that the torque that is responsible for this rotation of"},{"Start":"09:17.030 ","End":"09:23.650","Text":"our electric dipole to align in the same direction as the external electric field."},{"Start":"09:23.650 ","End":"09:26.195","Text":"Our torque, which is also a vector,"},{"Start":"09:26.195 ","End":"09:35.400","Text":"is equal to our electric dipole moment cross-product with our external electric field."},{"Start":"09:35.650 ","End":"09:42.065","Text":"Then, of course, once our electric dipole is aligned with the external electric field,"},{"Start":"09:42.065 ","End":"09:44.480","Text":"our torque becomes zero."},{"Start":"09:44.480 ","End":"09:49.805","Text":"Similarly here with our magnetic field and magnetic dipole,"},{"Start":"09:49.805 ","End":"09:54.755","Text":"we have a torque which acts to rotate our magnetic dipole moment"},{"Start":"09:54.755 ","End":"10:01.010","Text":"into the direction of the external magnetic field."},{"Start":"10:01.010 ","End":"10:05.120","Text":"Our torque over here is"},{"Start":"10:05.120 ","End":"10:09.350","Text":"going to be a very similar story to what we see with the electric dipole."},{"Start":"10:09.350 ","End":"10:10.820","Text":"Our torque is equal to."},{"Start":"10:10.820 ","End":"10:13.145","Text":"Instead of electric dipole moment,"},{"Start":"10:13.145 ","End":"10:17.260","Text":"we\u0027re going to have our magnetic dipole moment cross-products"},{"Start":"10:17.260 ","End":"10:20.990","Text":"and then instead of our external electric field,"},{"Start":"10:20.990 ","End":"10:24.990","Text":"we have an external magnetic field."},{"Start":"10:26.170 ","End":"10:30.620","Text":"Now what about the energy of the system."},{"Start":"10:30.620 ","End":"10:34.085","Text":"With our electric dipole,"},{"Start":"10:34.085 ","End":"10:37.385","Text":"so we had that the energy of the system is equal to"},{"Start":"10:37.385 ","End":"10:44.890","Text":"our negative electric dipole moment dot product with our electric field."},{"Start":"10:44.890 ","End":"10:47.315","Text":"That means in the exact same way,"},{"Start":"10:47.315 ","End":"10:53.495","Text":"our energy for our magnetic dipole moment we\u0027ll have that u is equal to negative"},{"Start":"10:53.495 ","End":"11:00.435","Text":"our electric dipole moment dot product with our external B field."},{"Start":"11:00.435 ","End":"11:08.435","Text":"What does this mean? When our magnetic dipole moment"},{"Start":"11:08.435 ","End":"11:11.930","Text":"is not aligned with our external B field,"},{"Start":"11:11.930 ","End":"11:17.180","Text":"that means that we\u0027re going to have some kind of potential energy,"},{"Start":"11:17.180 ","End":"11:21.515","Text":"which means that we\u0027re going to want to move and rotate to a line."},{"Start":"11:21.515 ","End":"11:28.445","Text":"We\u0027re starting off with potential energy and as our magnetic dipole rotates around,"},{"Start":"11:28.445 ","End":"11:32.460","Text":"our u will become kinetic energy."},{"Start":"11:33.880 ","End":"11:41.180","Text":"The important equations to take from this lesson is the magnetic dipole moment,"},{"Start":"11:41.180 ","End":"11:44.990","Text":"Mu, is also sometimes denoted as m,"},{"Start":"11:44.990 ","End":"11:47.795","Text":"which is equal to the current flowing through the loop,"},{"Start":"11:47.795 ","End":"11:50.270","Text":"multiplied by the surface area of"},{"Start":"11:50.270 ","End":"11:55.210","Text":"the loop and in the direction normal to the surface area of the loop,"},{"Start":"11:55.210 ","End":"12:04.935","Text":"our equation for the magnetic field far away from our magnetic dipole, our torque,"},{"Start":"12:04.935 ","End":"12:07.685","Text":"which aims when our magnetic dipole,"},{"Start":"12:07.685 ","End":"12:09.710","Text":"is in an external magnetic field,"},{"Start":"12:09.710 ","End":"12:11.720","Text":"so our torque which aims to align"},{"Start":"12:11.720 ","End":"12:15.590","Text":"our magnetic dipole moment with our external electric field,"},{"Start":"12:15.590 ","End":"12:21.350","Text":"and the potential energy that we have in a system where"},{"Start":"12:21.350 ","End":"12:29.585","Text":"our magnetic dipole moment is not pointing in the same direction as our external B field,"},{"Start":"12:29.585 ","End":"12:35.150","Text":"which turns from potential energy to kinetic energy once we release"},{"Start":"12:35.150 ","End":"12:41.465","Text":"our magnetic dipole to rotate freely to align with our external field."},{"Start":"12:41.465 ","End":"12:44.520","Text":"That\u0027s the end of this lesson."}],"ID":21427},{"Watched":false,"Name":"Derivation Of Torque For Magnetic Dipole","Duration":"7m 58s","ChapterTopicVideoID":21348,"CourseChapterTopicPlaylistID":99486,"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":"00:09.870","Text":"we\u0027re going to speak about how we derive this equation for the magnetic dipole moment."},{"Start":"00:09.870 ","End":"00:14.250","Text":"First, in order to understand where this equation comes from,"},{"Start":"00:14.250 ","End":"00:16.491","Text":"let\u0027s look at some kind of"},{"Start":"00:16.491 ","End":"00:24.390","Text":"rectangular frame that is located inside an external magnetic field."},{"Start":"00:24.390 ","End":"00:29.925","Text":"Let\u0027s imagine that our external B field"},{"Start":"00:29.925 ","End":"00:35.880","Text":"is pointing in this direction and its value is B_0."},{"Start":"00:35.880 ","End":"00:41.050","Text":"That through this rectangular wire frame,"},{"Start":"00:41.050 ","End":"00:45.245","Text":"we have some current I flowing through."},{"Start":"00:45.245 ","End":"00:53.770","Text":"Now let\u0027s say that the dimensions of this wire frame is a by a,"},{"Start":"00:53.770 ","End":"00:57.210","Text":"so each side length is of length a."},{"Start":"00:57.340 ","End":"01:02.150","Text":"Now, if we draw this exact configuration but from a side view,"},{"Start":"01:02.150 ","End":"01:06.390","Text":"so we\u0027ll see this side of"},{"Start":"01:06.390 ","End":"01:12.740","Text":"our wire frame where here is the current is flowing in this anti-clockwise direction."},{"Start":"01:12.740 ","End":"01:15.475","Text":"The current at this edge,"},{"Start":"01:15.475 ","End":"01:18.890","Text":"we\u0027ll be coming out towards us."},{"Start":"01:18.890 ","End":"01:26.610","Text":"Then the current goes back at this side into the page."},{"Start":"01:27.730 ","End":"01:30.005","Text":"Now, as we know,"},{"Start":"01:30.005 ","End":"01:36.845","Text":"we have some normal vector over here,"},{"Start":"01:36.845 ","End":"01:38.900","Text":"which we called Mu,"},{"Start":"01:38.900 ","End":"01:43.280","Text":"which is normal to the surface area of our wire frame."},{"Start":"01:43.280 ","End":"01:45.620","Text":"It\u0027s at 90 degrees,"},{"Start":"01:45.620 ","End":"01:49.910","Text":"and this is the vector which describes the direction or"},{"Start":"01:49.910 ","End":"01:55.165","Text":"the orientation of our wire frame that has a current running through it."},{"Start":"01:55.165 ","End":"01:58.580","Text":"Let\u0027s again from the side view,"},{"Start":"01:58.580 ","End":"02:00.005","Text":"draw our B field."},{"Start":"02:00.005 ","End":"02:03.695","Text":"Our B field is going in this direction."},{"Start":"02:03.695 ","End":"02:08.675","Text":"Then we know that the angle between these 2 is Alpha."},{"Start":"02:08.675 ","End":"02:13.295","Text":"Now we know that the equation for force is equal to"},{"Start":"02:13.295 ","End":"02:19.880","Text":"BIL multiplied by sine of Alpha,"},{"Start":"02:19.880 ","End":"02:26.285","Text":"which is the angle between our Mu,"},{"Start":"02:26.285 ","End":"02:32.090","Text":"our normal vector to the surface area of our wire loop,"},{"Start":"02:32.090 ","End":"02:36.720","Text":"and our external magnetic field."},{"Start":"02:37.460 ","End":"02:40.250","Text":"Now, once we do that,"},{"Start":"02:40.250 ","End":"02:47.710","Text":"we can see that our force on this side is pointing in this direction."},{"Start":"02:47.710 ","End":"02:51.155","Text":"Also, if we did B cross I,"},{"Start":"02:51.155 ","End":"02:56.565","Text":"we would also get a vector for F pointing in this direction."},{"Start":"02:56.565 ","End":"03:01.550","Text":"Similarly, because here the current is coming in the direction out of the page,"},{"Start":"03:01.550 ","End":"03:04.220","Text":"so it\u0027s going to be in the opposite direction to here."},{"Start":"03:04.220 ","End":"03:09.800","Text":"We\u0027ll get that the force being applied on this side of the wire is equal,"},{"Start":"03:09.800 ","End":"03:13.080","Text":"but in this opposite direction over here."},{"Start":"03:13.580 ","End":"03:19.790","Text":"Now when we substitute that are L is equal to a,"},{"Start":"03:19.790 ","End":"03:22.715","Text":"the side of our wire frame is equal to a,"},{"Start":"03:22.715 ","End":"03:26.090","Text":"will get that our force is equal to"},{"Start":"03:26.090 ","End":"03:36.360","Text":"B_0 Ia sine of Alpha."},{"Start":"03:37.090 ","End":"03:42.890","Text":"Now, notice that because our forces are equal and opposite,"},{"Start":"03:42.890 ","End":"03:49.670","Text":"so the sum of all the forces acting on our wire frame is going to be equal to 0,"},{"Start":"03:49.670 ","End":"03:55.680","Text":"which means that the center"},{"Start":"03:55.680 ","End":"03:59.615","Text":"of mass of the wire frame which is located here, doesn\u0027t move."},{"Start":"03:59.615 ","End":"04:03.800","Text":"That\u0027s because the sum of the forces are equal to 0."},{"Start":"04:03.800 ","End":"04:13.520","Text":"Now what we want to do is we want to work out our torque or a magnetic dipole moment."},{"Start":"04:13.520 ","End":"04:18.965","Text":"Now, if we can see the sum of the forces are equal to 0."},{"Start":"04:18.965 ","End":"04:26.270","Text":"However, the sum of the moments or the sum of the torques is not equal to 0."},{"Start":"04:26.270 ","End":"04:33.800","Text":"My center of mass is located at the center of my wire frame,"},{"Start":"04:33.800 ","End":"04:35.960","Text":"so over here,"},{"Start":"04:35.960 ","End":"04:39.005","Text":"and it isn\u0027t moving,"},{"Start":"04:39.005 ","End":"04:44.960","Text":"and we can see that from my center of mass to here,"},{"Start":"04:44.960 ","End":"04:51.275","Text":"I have some vector r. Now I can work out the torque."},{"Start":"04:51.275 ","End":"04:55.025","Text":"My torque or my moment is equal to, as we know,"},{"Start":"04:55.025 ","End":"05:00.530","Text":"r cross F, which is equal to,"},{"Start":"05:00.530 ","End":"05:03.125","Text":"so the size of my r vector is,"},{"Start":"05:03.125 ","End":"05:05.435","Text":"as we can see, it\u0027s from the center of mass."},{"Start":"05:05.435 ","End":"05:08.450","Text":"This length is 1/2 the length of the wire frame,"},{"Start":"05:08.450 ","End":"05:18.020","Text":"which is simply a divided by 2 and then multiply it by my force,"},{"Start":"05:18.020 ","End":"05:27.205","Text":"so multiplied by B_0 Ia multiplied by sine of Alpha."},{"Start":"05:27.205 ","End":"05:28.705","Text":"This is, of course,"},{"Start":"05:28.705 ","End":"05:32.540","Text":"the size of the torque."},{"Start":"05:34.490 ","End":"05:41.185","Text":"That\u0027s the torque or the moment on this side of the wire frame."},{"Start":"05:41.185 ","End":"05:47.110","Text":"Now over here, we\u0027re going to have the exact same torque or the exact same size torque."},{"Start":"05:47.110 ","End":"05:51.810","Text":"We can say that the total torque acting on"},{"Start":"05:51.810 ","End":"05:54.040","Text":"this wire frame is going to be the torque"},{"Start":"05:54.040 ","End":"05:57.080","Text":"acting on this side and the torque on this side,"},{"Start":"05:57.080 ","End":"06:01.115","Text":"which is simply going to be 2 times what we have here,"},{"Start":"06:01.115 ","End":"06:08.890","Text":"a over 2 B_0 Ia sine of Alpha,"},{"Start":"06:10.130 ","End":"06:14.040","Text":"which is of course our 2 is going to cancel out."},{"Start":"06:14.040 ","End":"06:23.110","Text":"It\u0027s just a^2 B_0 I sine of Alpha."},{"Start":"06:23.530 ","End":"06:27.545","Text":"Now I can see that my a^2,"},{"Start":"06:27.545 ","End":"06:31.385","Text":"which is the area of my wire frame,"},{"Start":"06:31.385 ","End":"06:33.905","Text":"multiplied by I is"},{"Start":"06:33.905 ","End":"06:40.325","Text":"my magnetic dipole moment and"},{"Start":"06:40.325 ","End":"06:47.575","Text":"sine of Alpha is the angle between the dipole moment."},{"Start":"06:47.575 ","End":"06:50.920","Text":"So between a^2 I, the dipole moment,"},{"Start":"06:50.920 ","End":"06:55.930","Text":"and our external magnetic field B_0."},{"Start":"06:55.930 ","End":"07:00.700","Text":"Then therefore, I can write this as Mu,"},{"Start":"07:00.700 ","End":"07:03.130","Text":"which is my magnetic dipole moment,"},{"Start":"07:03.130 ","End":"07:09.610","Text":"which is a^2 I cross-product with my external B field,"},{"Start":"07:09.610 ","End":"07:12.940","Text":"which is my B_0 here specifically."},{"Start":"07:12.940 ","End":"07:18.700","Text":"We can see that if I wanted to find just the size of this and not the vector,"},{"Start":"07:18.700 ","End":"07:21.865","Text":"then I just take the size of my Mu,"},{"Start":"07:21.865 ","End":"07:23.950","Text":"which is a^2 I,"},{"Start":"07:23.950 ","End":"07:26.150","Text":"multiplied by the size of my B,"},{"Start":"07:26.150 ","End":"07:30.140","Text":"which is B_0 multiplied by sine of the angle between the,"},{"Start":"07:30.140 ","End":"07:32.255","Text":"which is sine of Alpha,"},{"Start":"07:32.255 ","End":"07:34.130","Text":"as we can see here."},{"Start":"07:34.130 ","End":"07:37.205","Text":"That\u0027s the end of this lesson."},{"Start":"07:37.205 ","End":"07:44.570","Text":"Now, obviously this derivation is also correct for every shape of a wire frame,"},{"Start":"07:44.570 ","End":"07:45.905","Text":"be it rectangular,"},{"Start":"07:45.905 ","End":"07:49.865","Text":"circular, triangular, and also of course,"},{"Start":"07:49.865 ","End":"07:57.690","Text":"for any orientation that it takes in space relative to the magnetic B field."}],"ID":21428},{"Watched":false,"Name":"Exercise 1","Duration":"9m 20s","ChapterTopicVideoID":21349,"CourseChapterTopicPlaylistID":99486,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.175","Text":"Hello, in this question we are given a magnetic dipole,"},{"Start":"00:05.175 ","End":"00:11.400","Text":"which is located at the origin with a magnetic dipole moment of Mu vector,"},{"Start":"00:11.400 ","End":"00:15.180","Text":"which is equal to Mu in the X-direction 0,0."},{"Start":"00:15.180 ","End":"00:20.500","Text":"That means that our magnetic dipole moment is like so."},{"Start":"00:20.570 ","End":"00:25.380","Text":"Then we\u0027re being asked to find the value of Mu"},{"Start":"00:25.380 ","End":"00:30.480","Text":"such that an electron positioned at 0 minus a,"},{"Start":"00:30.480 ","End":"00:38.230","Text":"0 with a velocity of 0,0 V will move in circular motion."},{"Start":"00:38.750 ","End":"00:41.195","Text":"Let\u0027s draw the problem."},{"Start":"00:41.195 ","End":"00:47.535","Text":"We have an electron positioned at negative a in the y direction."},{"Start":"00:47.535 ","End":"00:50.675","Text":"That\u0027s going to be something like over here,"},{"Start":"00:50.675 ","End":"00:56.240","Text":"where the magnitude of this distance is equal to a."},{"Start":"00:56.240 ","End":"01:00.390","Text":"Over here we have of course, our electron placed."},{"Start":"01:00.590 ","End":"01:09.500","Text":"We\u0027re being told that the velocity of the electron is v in the z direction."},{"Start":"01:09.500 ","End":"01:13.040","Text":"If this is our y-axis and this is our x-axis,"},{"Start":"01:13.040 ","End":"01:15.320","Text":"according to the right-hand rule,"},{"Start":"01:15.320 ","End":"01:21.790","Text":"our z-axis must be coming out of the page towards us."},{"Start":"01:21.790 ","End":"01:25.460","Text":"That means that the velocity of the electron,"},{"Start":"01:25.460 ","End":"01:28.535","Text":"which is in the z direction,"},{"Start":"01:28.535 ","End":"01:32.640","Text":"is towards us out of the page."},{"Start":"01:33.920 ","End":"01:37.640","Text":"In order for me to have circular motion,"},{"Start":"01:37.640 ","End":"01:40.160","Text":"that means I must have some kind of force"},{"Start":"01:40.160 ","End":"01:44.585","Text":"acting on my electron to make it move around in a circle."},{"Start":"01:44.585 ","End":"01:49.955","Text":"Now, this force is obviously going to come from my magnetic dipole,"},{"Start":"01:49.955 ","End":"01:52.860","Text":"which is located over here."},{"Start":"01:54.020 ","End":"02:00.170","Text":"In that case, if our force is coming from a magnetic dipole moment,"},{"Start":"02:00.170 ","End":"02:03.215","Text":"the first thing we have to do is we have to find"},{"Start":"02:03.215 ","End":"02:07.770","Text":"our magnetic field due to this dipole moment."},{"Start":"02:07.770 ","End":"02:09.230","Text":"Just to remind you,"},{"Start":"02:09.230 ","End":"02:14.100","Text":"the equation for a magnetic field is given by this equation."},{"Start":"02:15.710 ","End":"02:18.980","Text":"Before we start solving everything in,"},{"Start":"02:18.980 ","End":"02:24.005","Text":"we have our Mu vector dot r hat."},{"Start":"02:24.005 ","End":"02:26.570","Text":"Our Mu vector, as we know,"},{"Start":"02:26.570 ","End":"02:28.490","Text":"let\u0027s write this over here,"},{"Start":"02:28.490 ","End":"02:34.490","Text":"is equal to some kind of Mu in the x-direction."},{"Start":"02:34.490 ","End":"02:36.890","Text":"Now what is our r hat?"},{"Start":"02:36.890 ","End":"02:40.775","Text":"Our r hat is pointing from the origin,"},{"Start":"02:40.775 ","End":"02:46.025","Text":"from where our magnetic dipole is located, to our electron."},{"Start":"02:46.025 ","End":"02:50.090","Text":"The distance between the 2 is this distance a,"},{"Start":"02:50.090 ","End":"02:56.640","Text":"and it\u0027s a in the negative y direction."},{"Start":"02:56.640 ","End":"03:05.585","Text":"Given that, once we do Mu.r hat over here,"},{"Start":"03:05.585 ","End":"03:12.620","Text":"we\u0027ll get that this is equal to some value in the x direction dot,"},{"Start":"03:12.620 ","End":"03:15.325","Text":"some kind of value in the y direction."},{"Start":"03:15.325 ","End":"03:20.310","Text":"As we know, x.y is equal to 0."},{"Start":"03:20.310 ","End":"03:22.400","Text":"All of this crosses out,"},{"Start":"03:22.400 ","End":"03:26.060","Text":"which means that will simply get a minus from over here,"},{"Start":"03:26.060 ","End":"03:35.125","Text":"minus Mu 0 divided by 4 Pi multiplied by our Mu,"},{"Start":"03:35.125 ","End":"03:37.550","Text":"which is a Mu vector,"},{"Start":"03:37.550 ","End":"03:43.385","Text":"which is Mu 0,0 divided by r^3,"},{"Start":"03:43.385 ","End":"03:49.680","Text":"which is the distance between our electron and our dipole over here."},{"Start":"03:49.680 ","End":"03:54.825","Text":"That\u0027s a distance a, so a^3."},{"Start":"03:54.825 ","End":"03:57.365","Text":"Now what we can see is that our B field,"},{"Start":"03:57.365 ","End":"04:02.300","Text":"or magnetic field, is in the negative x-direction."},{"Start":"04:02.300 ","End":"04:05.315","Text":"I can draw it here like so."},{"Start":"04:05.315 ","End":"04:08.710","Text":"This is my B field."},{"Start":"04:09.200 ","End":"04:12.155","Text":"Now I found my B field,"},{"Start":"04:12.155 ","End":"04:19.410","Text":"and now I want to find my force due to my B field."},{"Start":"04:19.410 ","End":"04:23.000","Text":"As we know, our equation for magnetic force,"},{"Start":"04:23.000 ","End":"04:30.610","Text":"F is equal to qv cross B."},{"Start":"04:32.720 ","End":"04:35.130","Text":"Let\u0027s solve this."},{"Start":"04:35.130 ","End":"04:36.430","Text":"We have our q,"},{"Start":"04:36.430 ","End":"04:40.475","Text":"which is our electron charge,"},{"Start":"04:40.475 ","End":"04:43.160","Text":"and then we have a vector v,"},{"Start":"04:43.160 ","End":"04:45.170","Text":"so soon will plug this in,"},{"Start":"04:45.170 ","End":"04:48.005","Text":"cross our vector B."},{"Start":"04:48.005 ","End":"04:53.765","Text":"We\u0027ll have e v cross our B,"},{"Start":"04:53.765 ","End":"05:02.375","Text":"which is going to be negative Mu 0 divided by 4Pi a^3,"},{"Start":"05:02.375 ","End":"05:04.190","Text":"so all of our constants,"},{"Start":"05:04.190 ","End":"05:10.165","Text":"and now I\u0027m going to use the cross-products between our vector quantities."},{"Start":"05:10.165 ","End":"05:12.950","Text":"We saw that our v is in the z direction,"},{"Start":"05:12.950 ","End":"05:21.050","Text":"so that\u0027s our v 0,0,z or 1,"},{"Start":"05:21.050 ","End":"05:25.745","Text":"cross-product with our Mu vector over here."},{"Start":"05:25.745 ","End":"05:32.720","Text":"Which is our Mu in the x-direction 0,0."},{"Start":"05:32.720 ","End":"05:39.435","Text":"Now what we can do is we can do our cross multiplication."},{"Start":"05:39.435 ","End":"05:50.430","Text":"Let\u0027s do it here, so we have our ev multiplied by negative Mu 0 divided by 4Pi a^3."},{"Start":"05:51.170 ","End":"05:53.985","Text":"Now we\u0027re doing our first row."},{"Start":"05:53.985 ","End":"05:56.475","Text":"We cross out our first row,"},{"Start":"05:56.475 ","End":"06:00.885","Text":"and then we use 0 times 0 is 0,"},{"Start":"06:00.885 ","End":"06:03.060","Text":"minus 1 times 0,"},{"Start":"06:03.060 ","End":"06:07.210","Text":"which is 0, so here we can write 0."},{"Start":"06:07.490 ","End":"06:11.520","Text":"Now we\u0027re doing our second row,"},{"Start":"06:11.520 ","End":"06:13.935","Text":"so we can cross out our second row."},{"Start":"06:13.935 ","End":"06:17.745","Text":"Now we have 1 times Mu,"},{"Start":"06:17.745 ","End":"06:23.145","Text":"which is Mu, minus 0 times 0 which is 0."},{"Start":"06:23.145 ","End":"06:26.635","Text":"Now we\u0027re doing our third row."},{"Start":"06:26.635 ","End":"06:29.230","Text":"We have 0 times 0,"},{"Start":"06:29.230 ","End":"06:33.850","Text":"which is 0 minus 0 times Mu, which is 0."},{"Start":"06:33.850 ","End":"06:43.840","Text":"Now we can see that we have negative our charge of the electron v Mu"},{"Start":"06:43.840 ","End":"06:51.110","Text":"0 divided by 4Pi a^3 Mu"},{"Start":"06:51.110 ","End":"06:54.190","Text":"in the y direction."},{"Start":"06:54.190 ","End":"06:58.915","Text":"Now remember that our electron charge is also a minus."},{"Start":"06:58.915 ","End":"07:03.280","Text":"We\u0027ll have a minus and a minus which becomes a positive."},{"Start":"07:03.280 ","End":"07:08.020","Text":"Now we can see that our force is drawn in red,"},{"Start":"07:08.020 ","End":"07:14.420","Text":"is going to be pointing in the positive y direction."},{"Start":"07:14.420 ","End":"07:20.830","Text":"All right, so now we can see that our velocity of the electron is in the z direction,"},{"Start":"07:20.830 ","End":"07:26.430","Text":"and we have our F field pointing in the positive y-direction."},{"Start":"07:26.430 ","End":"07:35.300","Text":"We can see that we\u0027re going to have some kind of circular motion on the y, z plane."},{"Start":"07:35.620 ","End":"07:43.400","Text":"Now because of the symmetry of the problem our force is always going to be the same,"},{"Start":"07:43.400 ","End":"07:49.140","Text":"just pointing always inwards to this direction of circular motion."},{"Start":"07:49.850 ","End":"07:52.575","Text":"Now the last thing that we need,"},{"Start":"07:52.575 ","End":"07:54.490","Text":"let\u0027s just scroll down a little bit,"},{"Start":"07:54.490 ","End":"07:58.804","Text":"is that if I want this to be circular motion,"},{"Start":"07:58.804 ","End":"08:04.010","Text":"I have to say that my force is equal to the mass of the electron"},{"Start":"08:04.010 ","End":"08:10.405","Text":"multiplied by the electron\u0027s velocity squared divided by the radius."},{"Start":"08:10.405 ","End":"08:16.310","Text":"This is very important in order to define the circular motion."},{"Start":"08:16.310 ","End":"08:18.245","Text":"Now, of course,"},{"Start":"08:18.245 ","End":"08:21.720","Text":"over here my R,"},{"Start":"08:21.740 ","End":"08:24.645","Text":"the radius is equal to a,"},{"Start":"08:24.645 ","End":"08:27.495","Text":"because we\u0027re at distance a from the origin."},{"Start":"08:27.495 ","End":"08:32.670","Text":"Now I can say that my force, m_e,"},{"Start":"08:32.670 ","End":"08:35.825","Text":"v^2 squared divided by a,"},{"Start":"08:35.825 ","End":"08:41.215","Text":"is equal to the magnitude of my force due to the magnetic field."},{"Start":"08:41.215 ","End":"08:47.560","Text":"That\u0027s going to be the absolute value of my electron charge multiplied by"},{"Start":"08:47.560 ","End":"08:57.270","Text":"the velocity Mu 0 Mu divided by 4Pi a^3."},{"Start":"08:57.270 ","End":"09:02.780","Text":"Now we can see that some of these can cancel out."},{"Start":"09:02.780 ","End":"09:05.240","Text":"Now in my question,"},{"Start":"09:05.240 ","End":"09:07.355","Text":"I was asked to find Mu,"},{"Start":"09:07.355 ","End":"09:11.000","Text":"so that means that all I have to do is I have to rearrange"},{"Start":"09:11.000 ","End":"09:15.535","Text":"this equation in order to isolate out my Mu over here,"},{"Start":"09:15.535 ","End":"09:17.985","Text":"and that will be my answer."},{"Start":"09:17.985 ","End":"09:20.980","Text":"That\u0027s the end of this lesson."}],"ID":21429},{"Watched":false,"Name":"Exercise 2","Duration":"11m 36s","ChapterTopicVideoID":21350,"CourseChapterTopicPlaylistID":99486,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.905","Text":"Hello in this question we have half a spherical shell."},{"Start":"00:04.905 ","End":"00:09.180","Text":"Remember it\u0027s a shell which means that it\u0027s empty in the middle."},{"Start":"00:09.180 ","End":"00:14.425","Text":"We\u0027re being told that it has a charge density per unit area of Sigma."},{"Start":"00:14.425 ","End":"00:21.575","Text":"We\u0027re being told that the shell is rotating about the z-axis with angular velocity Omega."},{"Start":"00:21.575 ","End":"00:28.325","Text":"We\u0027re being asked what is the magnetic dipole moment of this shell?"},{"Start":"00:28.325 ","End":"00:33.155","Text":"Now the radius of the spherical shell is R, and of course,"},{"Start":"00:33.155 ","End":"00:39.030","Text":"R, Omega and Sigma are given to us in the question."},{"Start":"00:39.220 ","End":"00:43.400","Text":"Now, the first thing is that we know that"},{"Start":"00:43.400 ","End":"00:49.875","Text":"a magnetic dipole moment is given by our current"},{"Start":"00:49.875 ","End":"00:56.070","Text":"running through some close loop multiplied by"},{"Start":"00:56.070 ","End":"01:03.960","Text":"the surface area or the area inside that current loop and closed by the current loop."},{"Start":"01:03.960 ","End":"01:09.140","Text":"Of course the area is a vector because it\u0027s some size and"},{"Start":"01:09.140 ","End":"01:15.935","Text":"it\u0027s vector is in the direction perpendicular to its surface area."},{"Start":"01:15.935 ","End":"01:20.390","Text":"Now of course, we know that the spherical shell is"},{"Start":"01:20.390 ","End":"01:25.715","Text":"charged with this sigma and it\u0027s rotating about the z-axis."},{"Start":"01:25.715 ","End":"01:28.745","Text":"Because it\u0027s rotating and it has a charge,"},{"Start":"01:28.745 ","End":"01:34.700","Text":"that means that there are charges that are moving around in space."},{"Start":"01:34.700 ","End":"01:38.730","Text":"That means that we have a current."},{"Start":"01:40.420 ","End":"01:45.960","Text":"Our first mission is to find out what this current is."},{"Start":"01:45.960 ","End":"01:50.780","Text":"Whenever we have some charged object which is rotating,"},{"Start":"01:50.780 ","End":"01:54.365","Text":"we can find our current in 2 ways."},{"Start":"01:54.365 ","End":"01:57.320","Text":"Either we take our charge density and"},{"Start":"01:57.320 ","End":"02:01.505","Text":"multiply it by the velocity at which it is traveling,"},{"Start":"02:01.505 ","End":"02:10.300","Text":"and then we\u0027ll get some k. Our other way is to work out our dq by dt,"},{"Start":"02:10.300 ","End":"02:19.240","Text":"the amount of charge which passes some section that we make out per time interval."},{"Start":"02:19.240 ","End":"02:21.530","Text":"In a previous video,"},{"Start":"02:21.530 ","End":"02:25.100","Text":"we said that if we\u0027re given Sigma, our charged density,"},{"Start":"02:25.100 ","End":"02:30.590","Text":"it\u0027s easier in fact to use this dq by dt method rather"},{"Start":"02:30.590 ","End":"02:36.590","Text":"than Sigma multiplied by this vector for velocity over here."},{"Start":"02:36.590 ","End":"02:41.090","Text":"That\u0027s because it\u0027s a bit harder mathematically to work out in"},{"Start":"02:41.090 ","End":"02:46.115","Text":"which direction the vectors are pointing and to work with the units."},{"Start":"02:46.115 ","End":"02:48.530","Text":"If you ever see Sigma,"},{"Start":"02:48.530 ","End":"02:52.780","Text":"we\u0027re going to use this dq by dt method."},{"Start":"02:53.930 ","End":"03:00.455","Text":"As we said, we are going to use this dq by dt method."},{"Start":"03:00.455 ","End":"03:02.735","Text":"Let\u0027s do that."},{"Start":"03:02.735 ","End":"03:09.615","Text":"Our current I is going to be equal to dq by dt."},{"Start":"03:09.615 ","End":"03:15.275","Text":"This is equal to like when we were dealing with electrostatic."},{"Start":"03:15.275 ","End":"03:21.145","Text":"We know that our dq is going to be equal to our charge density Sigma."},{"Start":"03:21.145 ","End":"03:22.985","Text":"Because it\u0027s per unit area,"},{"Start":"03:22.985 ","End":"03:27.350","Text":"we\u0027re going to multiply it by unit area ds,"},{"Start":"03:27.350 ","End":"03:30.960","Text":"and of course divided by dt."},{"Start":"03:31.070 ","End":"03:33.990","Text":"Now what is our ds? Now,"},{"Start":"03:33.990 ","End":"03:37.010","Text":"because we\u0027re dealing with a spherical shell,"},{"Start":"03:37.010 ","End":"03:40.175","Text":"our ds is going to be equal to"},{"Start":"03:40.175 ","End":"03:48.270","Text":"R^2 sine of Phi d Theta d Phi."},{"Start":"03:48.270 ","End":"03:52.175","Text":"Of course, we\u0027re working in spherical coordinates."},{"Start":"03:52.175 ","End":"03:56.015","Text":"The clue was in that our shape is a half sphere,"},{"Start":"03:56.015 ","End":"03:59.130","Text":"then of course divided by dt."},{"Start":"03:59.960 ","End":"04:05.510","Text":"Of course, this is an element of area that we"},{"Start":"04:05.510 ","End":"04:11.120","Text":"use generally when we\u0027re dealing with spherical coordinates."},{"Start":"04:11.120 ","End":"04:17.715","Text":"This is going to be the element of area that we\u0027re going to use in our calculation."},{"Start":"04:17.715 ","End":"04:22.625","Text":"Now we\u0027re going to see that one of these values over here,"},{"Start":"04:22.625 ","End":"04:27.630","Text":"and the dt will be equal to some constant that we know."},{"Start":"04:28.490 ","End":"04:31.750","Text":"Let\u0027s take a look."},{"Start":"04:32.000 ","End":"04:37.160","Text":"We\u0027re looking for something that divided by dt is going to give us one of"},{"Start":"04:37.160 ","End":"04:41.585","Text":"our constants which is known in the question,"},{"Start":"04:41.585 ","End":"04:44.225","Text":"which has given to us in our question."},{"Start":"04:44.225 ","End":"04:51.835","Text":"We can see that d Theta by dt is simply equal to Omega over here."},{"Start":"04:51.835 ","End":"04:54.165","Text":"That works out really well."},{"Start":"04:54.165 ","End":"04:59.850","Text":"We have Sigma R^2 multiplied by Omega,"},{"Start":"04:59.850 ","End":"05:05.530","Text":"multiplied by sine Phi d Phi."},{"Start":"05:06.020 ","End":"05:09.510","Text":"Now, because I got a differential at the end,"},{"Start":"05:09.510 ","End":"05:11.550","Text":"so I still have this d Phi over here,"},{"Start":"05:11.550 ","End":"05:16.680","Text":"so that means that this I is in fact also dI."},{"Start":"05:16.680 ","End":"05:21.170","Text":"This is the amount of current running"},{"Start":"05:21.170 ","End":"05:26.980","Text":"through a tiny element of area in the spherical shell."},{"Start":"05:26.980 ","End":"05:29.940","Text":"That\u0027s the current running through here."},{"Start":"05:29.940 ","End":"05:32.150","Text":"Then of course we\u0027re going to integrate."},{"Start":"05:32.150 ","End":"05:35.270","Text":"We\u0027re going to sum up over everything here and then we\u0027ll get"},{"Start":"05:35.270 ","End":"05:40.920","Text":"the total current flowing through our half spherical shell."},{"Start":"05:42.260 ","End":"05:49.745","Text":"Now what I want to do is I want to work out the magnetic dipole moment of the shell."},{"Start":"05:49.745 ","End":"05:55.450","Text":"What I\u0027m going to do is I\u0027m going to split up the shell in to lots of"},{"Start":"05:55.450 ","End":"06:01.970","Text":"tiny little circular current loops."},{"Start":"06:01.970 ","End":"06:08.065","Text":"This is one strip along the shell and it\u0027s a very small width."},{"Start":"06:08.065 ","End":"06:12.940","Text":"As we know, I have some current flowing through in this direction."},{"Start":"06:12.940 ","End":"06:18.220","Text":"Now, my radius is up until this point over here and the current loop,"},{"Start":"06:18.220 ","End":"06:23.440","Text":"and I know that this angle over here is going to be some angle Phi."},{"Start":"06:23.440 ","End":"06:26.630","Text":"I\u0027m summing up along all the angles of Phi,"},{"Start":"06:26.630 ","End":"06:33.095","Text":"from Phi is equal to 0 and all the way until Phi is equal to Pi over 2,"},{"Start":"06:33.095 ","End":"06:36.175","Text":"because we\u0027re dealing with half a sphere."},{"Start":"06:36.175 ","End":"06:40.895","Text":"Now we\u0027re going to say that the distance between"},{"Start":"06:40.895 ","End":"06:47.030","Text":"our z-axis to the edge of our spherical shell is going to be"},{"Start":"06:47.030 ","End":"06:52.490","Text":"equal to R. As we can see r starts from"},{"Start":"06:52.490 ","End":"06:59.760","Text":"small at the top of the sphere and ends when r is equal to R over here."},{"Start":"07:01.510 ","End":"07:05.575","Text":"Now we know that our d Mu,"},{"Start":"07:05.575 ","End":"07:10.170","Text":"an infinitesimal magnetic dipole moments."},{"Start":"07:10.170 ","End":"07:14.405","Text":"That\u0027s the dipole moment which matches to this strip."},{"Start":"07:14.405 ","End":"07:23.614","Text":"This circular loop where we have current flowing is equal to our area."},{"Start":"07:23.614 ","End":"07:27.140","Text":"Our area enclosed in this circular loop is going to"},{"Start":"07:27.140 ","End":"07:32.365","Text":"be Pi multiplied by the radius^2 of the circular loop."},{"Start":"07:32.365 ","End":"07:37.535","Text":"Pi and the radius of this circular loop is"},{"Start":"07:37.535 ","End":"07:43.490","Text":"r^2 multiplied by our I."},{"Start":"07:43.490 ","End":"07:48.230","Text":"Here we\u0027re multiplying by our dI because it\u0027s"},{"Start":"07:48.230 ","End":"07:53.970","Text":"for a small circular loop and not yet the entire spherical shell."},{"Start":"07:55.250 ","End":"08:00.440","Text":"Now I can see that I\u0027m integrating along Phi and they have a sine Phi here."},{"Start":"08:00.440 ","End":"08:04.399","Text":"But what is this r exactly?"},{"Start":"08:04.399 ","End":"08:06.560","Text":"We can see here, as we said before,"},{"Start":"08:06.560 ","End":"08:09.935","Text":"that it\u0027s dependent on this angle Phi over here."},{"Start":"08:09.935 ","End":"08:12.335","Text":"When Phi is equal to 0,"},{"Start":"08:12.335 ","End":"08:14.525","Text":"our r is going to be really small,"},{"Start":"08:14.525 ","End":"08:16.295","Text":"and as our Phi grows,"},{"Start":"08:16.295 ","End":"08:24.765","Text":"our r is going to gradually increase in size until it is equal to our R over here."},{"Start":"08:24.765 ","End":"08:26.640","Text":"It\u0027s dependent on Phi,"},{"Start":"08:26.640 ","End":"08:31.550","Text":"and we can see that there\u0027s a 90 degree angle between it and the z-axis,"},{"Start":"08:31.550 ","End":"08:33.470","Text":"because that\u0027s how we defined it."},{"Start":"08:33.470 ","End":"08:40.400","Text":"We can say that our r is equal to from Pythagoras and our trig identities,"},{"Start":"08:40.400 ","End":"08:48.920","Text":"we can see that r is simply going to be equal to R multiplied by sine of Phi."},{"Start":"08:48.920 ","End":"08:53.705","Text":"Because it\u0027s the opposite side to our angle Phi."},{"Start":"08:53.705 ","End":"08:56.150","Text":"Now we can rewrite this as"},{"Start":"08:56.150 ","End":"09:05.895","Text":"Pi R^2 sine^2 Phi multiplied by dI,"},{"Start":"09:05.895 ","End":"09:15.850","Text":"and our dI is Sigma R^2 Omega sine Phi d Phi."},{"Start":"09:17.060 ","End":"09:20.210","Text":"Now it\u0027s time for me to sum up on"},{"Start":"09:20.210 ","End":"09:26.085","Text":"my entire spherical shell in order to find what my Mu is equal to."},{"Start":"09:26.085 ","End":"09:29.480","Text":"All I have to do is I have to integrate on my d Mu."},{"Start":"09:29.480 ","End":"09:32.750","Text":"Integrate over here, and of course integrate over here."},{"Start":"09:32.750 ","End":"09:37.220","Text":"Where my bounds for where my Phi is equal to 0"},{"Start":"09:37.220 ","End":"09:44.730","Text":"until my Phi is equal to Pi over 2."},{"Start":"09:45.850 ","End":"09:49.160","Text":"We won\u0027t solve this integral now,"},{"Start":"09:49.160 ","End":"09:51.035","Text":"but I\u0027ll just write the answer."},{"Start":"09:51.035 ","End":"09:53.780","Text":"If you want to solve this on your own and see"},{"Start":"09:53.780 ","End":"09:57.210","Text":"that you\u0027ve got the right answer that is advised."},{"Start":"09:57.210 ","End":"10:03.900","Text":"The integral is going to be equal to 2Pi R^4 divided by"},{"Start":"10:03.900 ","End":"10:11.385","Text":"3 multiplied by Sigma Omega and it\u0027s in the z-direction."},{"Start":"10:11.385 ","End":"10:14.000","Text":"First of all, we could have guessed that it would be in"},{"Start":"10:14.000 ","End":"10:16.820","Text":"the z-direction because each one of"},{"Start":"10:16.820 ","End":"10:25.005","Text":"these car interings is located on the xy-plane at some z value."},{"Start":"10:25.005 ","End":"10:31.220","Text":"But they\u0027re on the xy-plane or their projection is flat on the xy-plane,"},{"Start":"10:31.220 ","End":"10:38.160","Text":"which means that the vector normal to that is of course the z vector."},{"Start":"10:38.800 ","End":"10:42.095","Text":"We can see that that is going to be in the z-direction."},{"Start":"10:42.095 ","End":"10:44.435","Text":"Also from the right-hand rule,"},{"Start":"10:44.435 ","End":"10:48.110","Text":"we can see that if the current is flowing in this direction,"},{"Start":"10:48.110 ","End":"10:54.950","Text":"then our magnetic dipole moment is also going to be going in the z-direction."},{"Start":"10:54.950 ","End":"10:58.800","Text":"We can get that from 2 ways."},{"Start":"10:59.330 ","End":"11:03.425","Text":"Now that I know my magnetic dipole moment,"},{"Start":"11:03.425 ","End":"11:07.070","Text":"if I\u0027m asked later to find the torque."},{"Start":"11:07.070 ","End":"11:13.545","Text":"I know that that is equal to Mu cross B. I already have my Mu,"},{"Start":"11:13.545 ","End":"11:17.540","Text":"and I can find the magnetic field caused by this"},{"Start":"11:17.540 ","End":"11:22.395","Text":"current flowing through this half spherical shell."},{"Start":"11:22.395 ","End":"11:27.710","Text":"We can see that working out the magnetic dipole moment is extremely"},{"Start":"11:27.710 ","End":"11:33.310","Text":"useful and allows us to answer a wide range of questions."},{"Start":"11:33.310 ","End":"11:36.370","Text":"That\u0027s the end of this lesson."}],"ID":21430},{"Watched":false,"Name":"Exercise 3","Duration":"5m 36s","ChapterTopicVideoID":21351,"CourseChapterTopicPlaylistID":99486,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.085","Text":"Hello, in this lesson,"},{"Start":"00:02.085 ","End":"00:08.760","Text":"I\u0027m going to work out the magnetic dipole moment of this inductor or a coil."},{"Start":"00:08.760 ","End":"00:15.510","Text":"Here we have a coil which is of length L and has number of turns."},{"Start":"00:15.510 ","End":"00:24.810","Text":"How many times it\u0027s wound around N and it has a radius from the center of the coil,"},{"Start":"00:24.810 ","End":"00:26.370","Text":"a radius of a."},{"Start":"00:26.370 ","End":"00:33.525","Text":"Now let\u0027s imagine that we have a current I flowing through our coil"},{"Start":"00:33.525 ","End":"00:42.360","Text":"and what we want to do is we want to work out the magnetic dipole moment flowing through."},{"Start":"00:42.360 ","End":"00:48.530","Text":"We remember that the equation for the magnetic dipole moment is equal to the"},{"Start":"00:48.530 ","End":"00:55.705","Text":"current flowing through multiplied by our area enclosed by the current loop."},{"Start":"00:55.705 ","End":"00:57.905","Text":"When we\u0027re dealing with a coil,"},{"Start":"00:57.905 ","End":"01:03.800","Text":"we look at each turn as 1 current loop."},{"Start":"01:03.800 ","End":"01:08.220","Text":"In a coil, we have N current loops."},{"Start":"01:08.870 ","End":"01:16.535","Text":"For each coil we have that our current is equal to I multiplied by the area,"},{"Start":"01:16.535 ","End":"01:22.785","Text":"that is going to be Pi a^2 in this case over here,"},{"Start":"01:22.785 ","End":"01:29.700","Text":"and now if we use the right-hand rule to see which direction our vector is pointing in,"},{"Start":"01:29.700 ","End":"01:31.970","Text":"because of course our A is a vector."},{"Start":"01:31.970 ","End":"01:36.190","Text":"The vector is perpendicular to the surface area."},{"Start":"01:36.190 ","End":"01:41.389","Text":"Now we just have to see if it\u0027s going in the right direction or in the left direction."},{"Start":"01:41.389 ","End":"01:44.270","Text":"If we look at the right-hand rule,"},{"Start":"01:44.270 ","End":"01:49.190","Text":"our thumb is going in the direction of"},{"Start":"01:49.190 ","End":"01:54.650","Text":"our current and then our fingers curl around our thumb."},{"Start":"01:54.650 ","End":"02:01.280","Text":"We can see that our fingers are going to curl around and point in this direction."},{"Start":"02:01.280 ","End":"02:07.135","Text":"We get that our Mu is in this direction, going leftwards."},{"Start":"02:07.135 ","End":"02:12.230","Text":"If our current is flowing in this direction like so."},{"Start":"02:12.650 ","End":"02:17.620","Text":"If we say that this is the positive x-direction,"},{"Start":"02:17.620 ","End":"02:23.150","Text":"we\u0027ll have that this is pointing in the negative x-direction."},{"Start":"02:23.150 ","End":"02:30.445","Text":"Now if I want to know my total magnetic dipole moment for the entire coil,"},{"Start":"02:30.445 ","End":"02:34.010","Text":"I just multiply this dipole moment,"},{"Start":"02:34.010 ","End":"02:35.375","Text":"which is for 1 coil,"},{"Start":"02:35.375 ","End":"02:42.540","Text":"multiplied by the total number of turns,"},{"Start":"02:42.540 ","End":"02:48.185","Text":"which is N. I have that this is equal to N Mu over here,"},{"Start":"02:48.185 ","End":"02:53.315","Text":"which is equal to NI Pi a^2,"},{"Start":"02:53.315 ","End":"02:57.755","Text":"of course a minus over here in the x-direction,"},{"Start":"02:57.755 ","End":"03:01.625","Text":"and this is the magnetic dipole moment for"},{"Start":"03:01.625 ","End":"03:07.949","Text":"a coil with N turns where it has a current I flowing in this direction."},{"Start":"03:08.030 ","End":"03:13.204","Text":"Now let\u0027s see what I can do now that I have this calculation."},{"Start":"03:13.204 ","End":"03:19.400","Text":"Let\u0027s imagine that we have a magnetic field and it\u0027s"},{"Start":"03:19.400 ","End":"03:25.685","Text":"pointing in this direction and it\u0027s a constant."},{"Start":"03:25.685 ","End":"03:29.570","Text":"Now, we can see that the angle between"},{"Start":"03:29.570 ","End":"03:35.570","Text":"our coil and our B field is equal to some angle Alpha."},{"Start":"03:35.570 ","End":"03:41.285","Text":"Now, because we have a magnetic dipole moment and an external B field,"},{"Start":"03:41.285 ","End":"03:47.645","Text":"we know that some kind of torque is going to be acting on our coil and that that is"},{"Start":"03:47.645 ","End":"03:55.200","Text":"equal to the magnetic dipole moment cross the external B field."},{"Start":"03:56.600 ","End":"04:02.495","Text":"In which direction is my torque going to be acting?"},{"Start":"04:02.495 ","End":"04:09.245","Text":"We know from the cross-product and the right-hand rule that my torque"},{"Start":"04:09.245 ","End":"04:15.760","Text":"is always going to be acting in the direction such that my Mu,"},{"Start":"04:15.760 ","End":"04:18.170","Text":"my magnetic dipole moment,"},{"Start":"04:18.170 ","End":"04:22.845","Text":"is pointing in the direction of the magnetic field."},{"Start":"04:22.845 ","End":"04:27.395","Text":"My Tau, my torque is going to be trying to"},{"Start":"04:27.395 ","End":"04:31.895","Text":"rotate my coil so that it\u0027s in this direction."},{"Start":"04:31.895 ","End":"04:38.255","Text":"So that if my B field is pointing in this direction,"},{"Start":"04:38.255 ","End":"04:46.079","Text":"my torque is going to want that my coil will also be pointing in this direction aligned."},{"Start":"04:46.079 ","End":"04:49.795","Text":"That\u0027s going to be my direction of rotation."},{"Start":"04:49.795 ","End":"04:53.345","Text":"Let\u0027s write that this is my Tau and, of course,"},{"Start":"04:53.345 ","End":"04:58.789","Text":"if I wanted to work out the size of my vector,"},{"Start":"04:58.789 ","End":"05:04.385","Text":"that will simply be equal to the size of my magnetic dipole"},{"Start":"05:04.385 ","End":"05:10.670","Text":"moment Mu multiplied by the size of my magnetic field B,"},{"Start":"05:10.670 ","End":"05:16.325","Text":"multiplied by sine of the angle between the 2."},{"Start":"05:16.325 ","End":"05:18.485","Text":"That\u0027s the end of the lesson."},{"Start":"05:18.485 ","End":"05:25.669","Text":"This is the magnetic dipole moment of a coil and this will be the torque"},{"Start":"05:25.669 ","End":"05:33.820","Text":"acting to rotate our coil or our magnetic dipole moment given an external B field."},{"Start":"05:33.820 ","End":"05:36.670","Text":"That\u0027s the end of this lesson."}],"ID":21431},{"Watched":false,"Name":"Electric Engine","Duration":"12m 45s","ChapterTopicVideoID":21352,"CourseChapterTopicPlaylistID":99486,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.145","Text":"Hello. In this lesson,"},{"Start":"00:02.145 ","End":"00:06.525","Text":"we\u0027re going to be speaking about how an electric engine works."},{"Start":"00:06.525 ","End":"00:09.345","Text":"How does this work?"},{"Start":"00:09.345 ","End":"00:13.845","Text":"The electric engine is made up of 2 magnets,"},{"Start":"00:13.845 ","End":"00:18.630","Text":"1 with a North pole and the other one with a South pole."},{"Start":"00:18.630 ","End":"00:20.750","Text":"When we\u0027re dealing with magnets,"},{"Start":"00:20.750 ","End":"00:26.540","Text":"the magnetic field goes from North to South."},{"Start":"00:26.540 ","End":"00:29.180","Text":"Then in-between the 2 magnets,"},{"Start":"00:29.180 ","End":"00:34.790","Text":"we have this rectangular current loop with a current I flowing through."},{"Start":"00:34.790 ","End":"00:38.435","Text":"This current loop is attached to a ring,"},{"Start":"00:38.435 ","End":"00:42.050","Text":"or rather it\u0027s more cylinder, which is split."},{"Start":"00:42.050 ","End":"00:46.040","Text":"This is very important to note that there\u0027s a gap"},{"Start":"00:46.040 ","End":"00:50.865","Text":"here between the 2 halves of this ring or cylinder."},{"Start":"00:50.865 ","End":"00:54.485","Text":"That means that charges can pass through this gap,"},{"Start":"00:54.485 ","End":"00:59.020","Text":"which means that all the charges move along this current loop."},{"Start":"00:59.020 ","End":"01:05.720","Text":"Now, each half of this split ring is attached to some brush."},{"Start":"01:05.720 ","End":"01:14.030","Text":"Now the brush is made of some metal or a conducting material and this allows our ring to"},{"Start":"01:14.030 ","End":"01:17.825","Text":"rotate without causing too much friction"},{"Start":"01:17.825 ","End":"01:22.550","Text":"between itself and connecting to these wires over here."},{"Start":"01:22.550 ","End":"01:25.205","Text":"That\u0027s why these brushes are important."},{"Start":"01:25.205 ","End":"01:31.670","Text":"Then these wires are attached to a DC power supplies such as a battery."},{"Start":"01:31.670 ","End":"01:35.885","Text":"Now let\u0027s see how this entire system causes"},{"Start":"01:35.885 ","End":"01:41.970","Text":"a rotation in this current loop. What happens?"},{"Start":"01:41.970 ","End":"01:43.880","Text":"As we already discussed,"},{"Start":"01:43.880 ","End":"01:47.360","Text":"we have a magnetic field going from left to right over"},{"Start":"01:47.360 ","End":"01:51.935","Text":"here and because we have this DC power supply,"},{"Start":"01:51.935 ","End":"01:55.865","Text":"supplying some current around this loop over here."},{"Start":"01:55.865 ","End":"02:01.385","Text":"That means that we\u0027re going to have some magnetic dipole."},{"Start":"02:01.385 ","End":"02:05.825","Text":"In actual fact, this current loop is a magnetic dipole."},{"Start":"02:05.825 ","End":"02:09.500","Text":"Then, according to the right-hand rule,"},{"Start":"02:09.500 ","End":"02:18.505","Text":"the direction of this magnetic dipole is going to be in the downwards direction."},{"Start":"02:18.505 ","End":"02:26.480","Text":"Let\u0027s label this the direction of the magnetic dipole as the vector m. Now we can"},{"Start":"02:26.480 ","End":"02:34.430","Text":"see that our vector for our magnetic dipole is pointing in the downwards direction,"},{"Start":"02:34.430 ","End":"02:39.935","Text":"which is perpendicular to the direction of this external magnetic field."},{"Start":"02:39.935 ","End":"02:42.800","Text":"Then as we know from previous lessons,"},{"Start":"02:42.800 ","End":"02:47.644","Text":"there\u0027s going to be a torque acting on our magnetic dipole,"},{"Start":"02:47.644 ","End":"02:56.185","Text":"which is equal to the magnetic dipole vector cross-product with the external B field."},{"Start":"02:56.185 ","End":"03:01.700","Text":"As we know, what this means is that our magnetic dipole is going to want to"},{"Start":"03:01.700 ","End":"03:08.820","Text":"rotate such that it is aligned in the exact same direction as our external B field."},{"Start":"03:08.820 ","End":"03:11.840","Text":"It is going to rotate in this direction."},{"Start":"03:11.840 ","End":"03:15.800","Text":"As we can see, the anticlockwise direction,"},{"Start":"03:15.800 ","End":"03:18.050","Text":"which is the same as this direction over here,"},{"Start":"03:18.050 ","End":"03:21.140","Text":"which is the direction of the loop rotation."},{"Start":"03:21.140 ","End":"03:25.430","Text":"That means our current loop or our magnetic dipole is going to begin"},{"Start":"03:25.430 ","End":"03:31.400","Text":"rotating such that it is aligned to the external B field."},{"Start":"03:31.400 ","End":"03:36.985","Text":"Now another way of thinking about this is according to Lorentz\u0027s law."},{"Start":"03:36.985 ","End":"03:41.960","Text":"That is that if we have our current going into the page over here,"},{"Start":"03:41.960 ","End":"03:44.195","Text":"on a wire going into the page."},{"Start":"03:44.195 ","End":"03:49.205","Text":"The force due to the magnetic field will"},{"Start":"03:49.205 ","End":"03:54.965","Text":"induce a force downwards on the side of the current loop."},{"Start":"03:54.965 ","End":"03:58.490","Text":"Here we see that our current is traveling towards us,"},{"Start":"03:58.490 ","End":"03:59.765","Text":"out of the page."},{"Start":"03:59.765 ","End":"04:08.375","Text":"Which means that the magnet over here is going to exert a force in the upwards direction."},{"Start":"04:08.375 ","End":"04:11.990","Text":"Then we can see that, again,"},{"Start":"04:11.990 ","End":"04:17.350","Text":"the loop will rotate in this anticlockwise direction."},{"Start":"04:17.350 ","End":"04:22.070","Text":"Now we\u0027ve understood that our current loop is going to rotate."},{"Start":"04:22.070 ","End":"04:30.510","Text":"At some stage it\u0027s going to rotate such that it is perpendicular to this orientation."},{"Start":"04:30.510 ","End":"04:32.405","Text":"What does that mean?"},{"Start":"04:32.405 ","End":"04:36.890","Text":"That means that this side of the current loop is going to be at the top over"},{"Start":"04:36.890 ","End":"04:42.530","Text":"here and that this side of the current loop is going to be at the bottom."},{"Start":"04:42.530 ","End":"04:48.020","Text":"Then we\u0027ll see that our magnetic dipole moment is going to"},{"Start":"04:48.020 ","End":"04:54.150","Text":"be aligned with the external magnetic B field."},{"Start":"04:54.150 ","End":"04:58.145","Text":"If I draw this on the side over here,"},{"Start":"04:58.145 ","End":"05:08.790","Text":"we\u0027re going to have some situation like so where this is our magnetic moment."},{"Start":"05:08.790 ","End":"05:12.089","Text":"In the same way,"},{"Start":"05:12.089 ","End":"05:19.090","Text":"our B field is going to be exactly parallel."},{"Start":"05:19.160 ","End":"05:24.710","Text":"That means that we have no torque working at this exact moment,"},{"Start":"05:24.710 ","End":"05:26.905","Text":"excuse my poor drawing."},{"Start":"05:26.905 ","End":"05:33.755","Text":"How come our current loop is going to continue to rotate around?"},{"Start":"05:33.755 ","End":"05:39.275","Text":"First of all, from the rotation until we get to this 90 degrees orientation,"},{"Start":"05:39.275 ","End":"05:42.935","Text":"we\u0027re going to have some angular velocity,"},{"Start":"05:42.935 ","End":"05:45.095","Text":"which is going to help us overcome"},{"Start":"05:45.095 ","End":"05:49.895","Text":"this specific point where our current loop is located exactly"},{"Start":"05:49.895 ","End":"05:53.690","Text":"perpendicular to its current orientation and that"},{"Start":"05:53.690 ","End":"05:59.350","Text":"the dipole moment is exactly aligned in the direction of the external B field."},{"Start":"05:59.350 ","End":"06:02.615","Text":"Because we have this angular velocity,"},{"Start":"06:02.615 ","End":"06:09.850","Text":"we can overcome that point and that is how our current loop continues to rotate around."},{"Start":"06:09.850 ","End":"06:16.385","Text":"Once we\u0027ve gotten to the perpendicular position and we get past it,"},{"Start":"06:16.385 ","End":"06:20.960","Text":"we\u0027ll see that because of how this ring is built,"},{"Start":"06:20.960 ","End":"06:23.559","Text":"due to it being split,"},{"Start":"06:23.559 ","End":"06:26.595","Text":"this will also rotate."},{"Start":"06:26.595 ","End":"06:30.020","Text":"Then we\u0027ll have instead of it being in this formation we\u0027ll"},{"Start":"06:30.020 ","End":"06:35.330","Text":"have that this half is filled and this is not filled."},{"Start":"06:35.330 ","End":"06:38.190","Text":"Then it\u0027s at the perpendicular position."},{"Start":"06:38.190 ","End":"06:43.984","Text":"Then at some stage where this half is currently plus and this is minus,"},{"Start":"06:43.984 ","End":"06:51.585","Text":"our split ring will completely rotate such that this half from minus will become a plus."},{"Start":"06:51.585 ","End":"06:57.985","Text":"This half from being attached here and a plus will move here and become a minus."},{"Start":"06:57.985 ","End":"07:01.430","Text":"What\u0027s important to note is that due to"},{"Start":"07:01.430 ","End":"07:05.060","Text":"the splitting of the ring and the rotation of our current loop,"},{"Start":"07:05.060 ","End":"07:10.280","Text":"our split ring will also rotate and therefore the sign or"},{"Start":"07:10.280 ","End":"07:16.775","Text":"the charges on each half of the ring will switch signs as well."},{"Start":"07:16.775 ","End":"07:19.175","Text":"Therefore, if that happens,"},{"Start":"07:19.175 ","End":"07:23.430","Text":"then the direction of the current will also switch."},{"Start":"07:24.560 ","End":"07:30.140","Text":"Then the current will from here becoming towards us,"},{"Start":"07:30.140 ","End":"07:31.865","Text":"and here on this side,"},{"Start":"07:31.865 ","End":"07:34.175","Text":"will be traveling away from us."},{"Start":"07:34.175 ","End":"07:36.245","Text":"Then in that case,"},{"Start":"07:36.245 ","End":"07:44.090","Text":"our magnetic dipole moment will also be in the opposite direction as well."},{"Start":"07:44.090 ","End":"07:47.255","Text":"Instead of pointing in the right direction,"},{"Start":"07:47.255 ","End":"07:49.880","Text":"when it\u0027s located perpendicularly,"},{"Start":"07:49.880 ","End":"07:55.640","Text":"it will point in the left direction, like so."},{"Start":"07:55.640 ","End":"07:59.360","Text":"We\u0027re just switching our directions and then the magnetic dipole"},{"Start":"07:59.360 ","End":"08:03.755","Text":"is then opposite to the magnetic B field,"},{"Start":"08:03.755 ","End":"08:05.015","Text":"which means that, again,"},{"Start":"08:05.015 ","End":"08:10.970","Text":"we\u0027re going to have a torque acting on the current loop or acting on the magnetic dipole"},{"Start":"08:10.970 ","End":"08:14.060","Text":"in order to rotate the current loop such"},{"Start":"08:14.060 ","End":"08:17.615","Text":"that our new orientation for our magnetic dipole,"},{"Start":"08:17.615 ","End":"08:20.255","Text":"so in the opposite direction, will, again,"},{"Start":"08:20.255 ","End":"08:23.200","Text":"be pointing in the direction of the external B field,"},{"Start":"08:23.200 ","End":"08:25.485","Text":"which is in this direction."},{"Start":"08:25.485 ","End":"08:29.840","Text":"This is just going to carry on and carry on."},{"Start":"08:29.840 ","End":"08:32.135","Text":"Now let\u0027s try and draw"},{"Start":"08:32.135 ","End":"08:38.150","Text":"the magnetic dipole moment interacting with the magnetic field and the torque."},{"Start":"08:38.150 ","End":"08:43.460","Text":"At the beginning, we had our magnetic field going in this direction,"},{"Start":"08:43.460 ","End":"08:45.640","Text":"from North to South."},{"Start":"08:45.640 ","End":"08:51.890","Text":"Our magnetic dipole moment was pointing downwards like so."},{"Start":"08:51.890 ","End":"08:54.710","Text":"Then there was a torque acting on this,"},{"Start":"08:54.710 ","End":"08:59.240","Text":"such that our magnetic dipole moment wanted to rotate in"},{"Start":"08:59.240 ","End":"09:04.430","Text":"order to align with our external B field."},{"Start":"09:04.430 ","End":"09:06.545","Text":"Then at this point,"},{"Start":"09:06.545 ","End":"09:13.155","Text":"our magnetic dipole moment was at this position over here."},{"Start":"09:13.155 ","End":"09:15.750","Text":"It\u0027s at this position over here,"},{"Start":"09:15.750 ","End":"09:20.630","Text":"when our current loop rotates anticlockwise such that"},{"Start":"09:20.630 ","End":"09:25.795","Text":"it\u0027s perpendicular to its orientation that was drawn originally."},{"Start":"09:25.795 ","End":"09:28.475","Text":"That was this diagram over here."},{"Start":"09:28.475 ","End":"09:31.640","Text":"This was our first."},{"Start":"09:31.640 ","End":"09:36.170","Text":"Then after this, we move into the second stage."},{"Start":"09:36.170 ","End":"09:40.910","Text":"Our B field is still in the exact same position and now we\u0027re"},{"Start":"09:40.910 ","End":"09:47.480","Text":"located just below our orientation of the B field."},{"Start":"09:47.480 ","End":"09:50.764","Text":"Then at this exact moment,"},{"Start":"09:50.764 ","End":"09:59.930","Text":"our split ring rotates such that our positive side of the split ring becomes negative"},{"Start":"09:59.930 ","End":"10:03.635","Text":"because we\u0027re attached to a DC power supply and"},{"Start":"10:03.635 ","End":"10:08.945","Text":"our negative side of the split ring becomes positive."},{"Start":"10:08.945 ","End":"10:18.950","Text":"Then this switching of the signs means that our magnetic moment will change directions."},{"Start":"10:18.950 ","End":"10:26.584","Text":"Then due to how positives and negatives of our split ring changing,"},{"Start":"10:26.584 ","End":"10:30.680","Text":"so then will be in this position where"},{"Start":"10:30.680 ","End":"10:36.815","Text":"our magnetic dipole moment will be on this side over here."},{"Start":"10:36.815 ","End":"10:40.295","Text":"Now, if our positive and negative didn\u0027t change,"},{"Start":"10:40.295 ","End":"10:45.695","Text":"so when our magnetic dipole moment will move in this direction,"},{"Start":"10:45.695 ","End":"10:49.985","Text":"if our positive stayed positive and negative stayed negative,"},{"Start":"10:49.985 ","End":"10:58.470","Text":"then we would just be doing harmonic motion about our B field orientation."},{"Start":"10:58.470 ","End":"11:02.300","Text":"However, because of the split ring,"},{"Start":"11:02.300 ","End":"11:07.355","Text":"we\u0027re switching our charges on each side of the half cylinder,"},{"Start":"11:07.355 ","End":"11:11.045","Text":"which means that we\u0027re going to go from"},{"Start":"11:11.045 ","End":"11:18.930","Text":"this orientation to suddenly being in this orientation."},{"Start":"11:18.930 ","End":"11:22.745","Text":"It\u0027s going to be exactly equal and opposite."},{"Start":"11:22.745 ","End":"11:26.965","Text":"Imagine that the angle between here is 180 degrees."},{"Start":"11:26.965 ","End":"11:31.405","Text":"Then after it\u0027s in this position, again,"},{"Start":"11:31.405 ","End":"11:36.620","Text":"we\u0027re going to have this torque acting on now this oppositely oriented"},{"Start":"11:36.620 ","End":"11:42.710","Text":"magnetic moment in order to align it with the external B field."},{"Start":"11:42.710 ","End":"11:45.590","Text":"Then what we\u0027ll get is this torque such that"},{"Start":"11:45.590 ","End":"11:49.880","Text":"the magnetic moment rotates back to this orientation."},{"Start":"11:49.880 ","End":"11:53.173","Text":"Then we\u0027re back to Stage 1,"},{"Start":"11:53.173 ","End":"11:54.880","Text":"and 2, and 3."},{"Start":"11:54.880 ","End":"11:56.970","Text":"This was 2 and this is 3."},{"Start":"11:56.970 ","End":"12:04.270","Text":"That is how our current loop carries on rotating around in a circle."},{"Start":"12:04.270 ","End":"12:08.810","Text":"This is the basis of how an electric engine runs."},{"Start":"12:08.810 ","End":"12:16.785","Text":"Now, how will it work some electric component or some machine like a computer,"},{"Start":"12:16.785 ","End":"12:19.800","Text":"a car, an electric fan?"},{"Start":"12:19.800 ","End":"12:24.470","Text":"The current loop will have some screw attached over here,"},{"Start":"12:24.470 ","End":"12:28.310","Text":"which is attached to some nut or"},{"Start":"12:28.310 ","End":"12:32.870","Text":"bolt which will start rotating and then rotate whatever we want."},{"Start":"12:32.870 ","End":"12:38.015","Text":"Let\u0027s say it\u0027s a fan and then our fan will spin around."},{"Start":"12:38.015 ","End":"12:42.395","Text":"We can use this in lots of different machinery."},{"Start":"12:42.395 ","End":"12:45.510","Text":"That\u0027s the end of this lesson."}],"ID":21432}],"Thumbnail":null,"ID":99486}]
