[{"Name":"Introduction to Biot Savart Law","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"The Equation","Duration":"24m 6s","ChapterTopicVideoID":21319,"CourseChapterTopicPlaylistID":99469,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/21319.jpeg","UploadDate":"2020-04-06T15:10:56.7370000","DurationForVideoObject":"PT24M6S","Description":null,"MetaTitle":"1 The Equation: Video + Workbook | Proprep","MetaDescription":"Biot Savart Law - Introduction to Biot Savart Law. 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/biot-savart-law/introduction-to-biot-savart-law/vid21395","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.755","Text":"Hello. In this lesson,"},{"Start":"00:01.755 ","End":"00:04.740","Text":"we\u0027re going to be learning about Biot Savart Law."},{"Start":"00:04.740 ","End":"00:06.030","Text":"But before we do that,"},{"Start":"00:06.030 ","End":"00:09.465","Text":"let\u0027s review what we already know about magnetism."},{"Start":"00:09.465 ","End":"00:12.960","Text":"We\u0027ve already learned about Lawrence\u0027s law,"},{"Start":"00:12.960 ","End":"00:21.495","Text":"where we saw that the force due to a magnetic field on 1 charge is equal to q,"},{"Start":"00:21.495 ","End":"00:27.930","Text":"the charge multiplied by its velocity cross product with the magnetic field."},{"Start":"00:27.930 ","End":"00:30.585","Text":"This was the force on 1 charge,"},{"Start":"00:30.585 ","End":"00:36.510","Text":"and the magnetic force on a current carrying wire was equal to the"},{"Start":"00:36.510 ","End":"00:39.860","Text":"current multiplied by the length of the"},{"Start":"00:39.860 ","End":"00:44.430","Text":"current and cross product with the magnetic field."},{"Start":"00:44.430 ","End":"00:47.600","Text":"This is the force due to the magnetic field."},{"Start":"00:47.600 ","End":"00:53.905","Text":"Biot Savart Law speaks about how the magnetic field comes about in the first place."},{"Start":"00:53.905 ","End":"00:56.475","Text":"How is the magnetic field formed?"},{"Start":"00:56.475 ","End":"01:01.370","Text":"What we\u0027re going to learn in this lesson is that it\u0027s formed by current."},{"Start":"01:01.370 ","End":"01:08.890","Text":"However, this is not to be confused between this magnetic field acting on this current."},{"Start":"01:08.890 ","End":"01:14.600","Text":"In these equations we have some magnetic field which was formed somewhere else."},{"Start":"01:14.600 ","End":"01:19.010","Text":"The magnetic fields is present in a region where"},{"Start":"01:19.010 ","End":"01:26.050","Text":"a different object or body has current flowing through it."},{"Start":"01:26.050 ","End":"01:33.559","Text":"Then we see how this magnetic field is acting on this current."},{"Start":"01:33.559 ","End":"01:38.050","Text":"This is what you have to remember from these equations over here."},{"Start":"01:38.050 ","End":"01:45.560","Text":"What we\u0027re going to do now is we\u0027re going to see how current forms this magnetic field."},{"Start":"01:46.070 ","End":"01:52.000","Text":"Here I wrote a note that when we\u0027re speaking about the B field in these equations,"},{"Start":"01:52.000 ","End":"01:54.970","Text":"we\u0027re speaking about some external magnetic field"},{"Start":"01:54.970 ","End":"01:58.390","Text":"which has present and it is acting in a region."},{"Start":"01:58.390 ","End":"02:05.060","Text":"Then that region, a charge or a current passes through and we see the effect."},{"Start":"02:05.120 ","End":"02:08.315","Text":"Now what we\u0027re going to do is we\u0027re going to see"},{"Start":"02:08.315 ","End":"02:13.919","Text":"how this external magnetic field is formed."},{"Start":"02:14.020 ","End":"02:17.240","Text":"The external magnetic field, as we just said,"},{"Start":"02:17.240 ","End":"02:22.400","Text":"is formed by a different current or charge passing through."},{"Start":"02:22.400 ","End":"02:29.435","Text":"Let\u0027s imagine that we have this piece of wire over here and it has"},{"Start":"02:29.435 ","End":"02:33.110","Text":"a length dl and through"},{"Start":"02:33.110 ","End":"02:38.479","Text":"the wire we have a current I passing through and we know this current."},{"Start":"02:38.479 ","End":"02:42.905","Text":"This current is of course different to this current."},{"Start":"02:42.905 ","End":"02:47.000","Text":"This is a current which is in a magnetic field and"},{"Start":"02:47.000 ","End":"02:52.060","Text":"this current now that we\u0027re speaking about is what forms the magnetic field."},{"Start":"02:52.060 ","End":"02:57.445","Text":"Now what I want to do is I want to calculate the magnetic field in this region."},{"Start":"02:57.445 ","End":"03:01.175","Text":"What I\u0027m going to do is I\u0027m going to choose a random point,"},{"Start":"03:01.175 ","End":"03:02.960","Text":"let\u0027s say over here."},{"Start":"03:02.960 ","End":"03:09.240","Text":"I\u0027m going to try and work out what my magnetic field is at this point."},{"Start":"03:09.740 ","End":"03:13.310","Text":"In order to calculate this magnetic field,"},{"Start":"03:13.310 ","End":"03:17.490","Text":"I\u0027m going to use Biot Savart Law."},{"Start":"03:17.750 ","End":"03:23.685","Text":"This is the equation for Biot Savart\u0027s Law."},{"Start":"03:23.685 ","End":"03:28.370","Text":"It is equal to Mu naught divided by 4 Pi."},{"Start":"03:28.370 ","End":"03:29.405","Text":"These are of course,"},{"Start":"03:29.405 ","End":"03:32.210","Text":"constants that we\u0027ve already seen."},{"Start":"03:32.210 ","End":"03:38.855","Text":"Then that is multiplied by the current I multiplied by dl."},{"Start":"03:38.855 ","End":"03:42.950","Text":"Notice that the vector is above dl so dl is"},{"Start":"03:42.950 ","End":"03:50.405","Text":"this short length of wire in the direction of dl is in the direction of the current."},{"Start":"03:50.405 ","End":"03:57.295","Text":"I can just redraw this like so and then this will be my dl vector."},{"Start":"03:57.295 ","End":"04:02.265","Text":"This length cross product with my I vector."},{"Start":"04:02.265 ","End":"04:04.155","Text":"What is my r vector?"},{"Start":"04:04.155 ","End":"04:12.395","Text":"My r vector is the vector from my small piece of wire until my magnetic field."},{"Start":"04:12.395 ","End":"04:16.595","Text":"This is my r vector and then all of this"},{"Start":"04:16.595 ","End":"04:21.720","Text":"is divided by the magnitude of my r vector cubed."},{"Start":"04:22.130 ","End":"04:27.080","Text":"What\u0027s important to note here is that this r rarely is from"},{"Start":"04:27.080 ","End":"04:32.495","Text":"the piece of wire dl until the point where we\u0027re measuring the magnetic field."},{"Start":"04:32.495 ","End":"04:35.120","Text":"Why am I highlighting this and repeating it?"},{"Start":"04:35.120 ","End":"04:41.480","Text":"Because this r vector isn\u0027t necessarily from where the origin is."},{"Start":"04:41.480 ","End":"04:47.405","Text":"Sometimes what I will have is that my origin will be over here."},{"Start":"04:47.405 ","End":"04:53.400","Text":"In which case I\u0027ll have to find a vector to this point over here and this"},{"Start":"04:53.400 ","End":"05:00.145","Text":"will be I tag vector plus the r vector and that will give me this coordinate."},{"Start":"05:00.145 ","End":"05:03.650","Text":"Sometimes we\u0027ll have to play around with the coordinates a little bit."},{"Start":"05:03.650 ","End":"05:08.250","Text":"What\u0027s important to remember is that this r is from"},{"Start":"05:08.250 ","End":"05:13.640","Text":"the piece of wire dl until our point we\u0027re measuring the magnetic field."},{"Start":"05:13.640 ","End":"05:18.740","Text":"Of course, this is equal to dB because we\u0027re"},{"Start":"05:18.740 ","End":"05:24.780","Text":"calculating the magnetic field over here due to the small length of wire dl."},{"Start":"05:24.780 ","End":"05:29.610","Text":"If we have some wire of length l,"},{"Start":"05:29.610 ","End":"05:35.015","Text":"in order to find the total magnetic field at this point due to the entire wire,"},{"Start":"05:35.015 ","End":"05:38.100","Text":"of course, we\u0027re going to have to integrate."},{"Start":"05:39.320 ","End":"05:45.455","Text":"An alternative way of writing Biot Savart Law is the exact same thing."},{"Start":"05:45.455 ","End":"05:50.445","Text":"However, instead of cross multiplying with r vector,"},{"Start":"05:50.445 ","End":"05:55.385","Text":"we have r hat and that means that instead of dividing by r cubed,"},{"Start":"05:55.385 ","End":"06:04.410","Text":"we divide by r squared and that\u0027s because I vector is equal to r in the r hat direction."},{"Start":"06:04.410 ","End":"06:09.810","Text":"That\u0027s the only difference to notice."},{"Start":"06:11.030 ","End":"06:17.690","Text":"There are a few ways to work out the direction of the magnetic field."},{"Start":"06:17.690 ","End":"06:21.835","Text":"The first way is just to solve this cross product."},{"Start":"06:21.835 ","End":"06:24.020","Text":"Now, in order to solve the cross products,"},{"Start":"06:24.020 ","End":"06:29.800","Text":"you use the rules or you can work it out using the determinant. It\u0027s the same thing."},{"Start":"06:29.800 ","End":"06:31.620","Text":"You write dl as a vector,"},{"Start":"06:31.620 ","End":"06:34.380","Text":"you write dl as a vector and then you do"},{"Start":"06:34.380 ","End":"06:38.030","Text":"the cross products and then you\u0027ll get it in terms of your coordinates."},{"Start":"06:38.030 ","End":"06:42.050","Text":"Let\u0027s say in Cartesian you\u0027ll get it in terms of x, y, and z."},{"Start":"06:42.050 ","End":"06:43.955","Text":"That\u0027s the first way,"},{"Start":"06:43.955 ","End":"06:48.050","Text":"just doing it straightforward how it\u0027s written in the equation."},{"Start":"06:48.050 ","End":"06:53.460","Text":"The other way is to first calculate the magnitude of the magnetic field."},{"Start":"06:53.780 ","End":"06:57.335","Text":"We just multiply everything together,"},{"Start":"06:57.335 ","End":"07:00.050","Text":"and then in order to find the direction,"},{"Start":"07:00.050 ","End":"07:03.275","Text":"we then use the right hand rule."},{"Start":"07:03.275 ","End":"07:05.690","Text":"In the previous chapters,"},{"Start":"07:05.690 ","End":"07:08.885","Text":"we\u0027ve spoken extensively about the right hand rule,"},{"Start":"07:08.885 ","End":"07:13.160","Text":"but let\u0027s go over it really quickly again now."},{"Start":"07:13.160 ","End":"07:18.555","Text":"As we know, if we have an equation, whether cross products."},{"Start":"07:18.555 ","End":"07:22.445","Text":"Let\u0027s say we have that a is equal to"},{"Start":"07:22.445 ","End":"07:29.935","Text":"b cross c. The first thing that we can do, there\u0027s 2 methods."},{"Start":"07:29.935 ","End":"07:37.610","Text":"The first method is we point our thumb in the direction of b."},{"Start":"07:37.920 ","End":"07:43.205","Text":"Our thumb would be in the direction of b and let\u0027s say it\u0027s like so."},{"Start":"07:43.205 ","End":"07:47.200","Text":"This would represent something in the direction of b."},{"Start":"07:47.200 ","End":"07:51.385","Text":"Our fore finger represents the value at"},{"Start":"07:51.385 ","End":"07:58.100","Text":"c so we\u0027d point out fore finger in the direction of c."},{"Start":"07:59.000 ","End":"08:03.285","Text":"Then that means that we\u0027re left with our middle finger,"},{"Start":"08:03.285 ","End":"08:07.710","Text":"which in this case would be coming out of the page."},{"Start":"08:07.710 ","End":"08:10.155","Text":"Let\u0027s draw it like so."},{"Start":"08:10.155 ","End":"08:17.235","Text":"Coming out of the page and then the rest of our hand is irrelevant."},{"Start":"08:17.235 ","End":"08:24.825","Text":"This is our middle finger and this will represent out of the page the direction of a."},{"Start":"08:24.825 ","End":"08:29.962","Text":"Thumb is the first value in the cross-product,"},{"Start":"08:29.962 ","End":"08:32.805","Text":"forefinger is the second value and the cross product,"},{"Start":"08:32.805 ","End":"08:36.660","Text":"and the middle finger is the answer."},{"Start":"08:36.660 ","End":"08:39.660","Text":"This is the first method for the right-hand rule."},{"Start":"08:39.660 ","End":"08:45.480","Text":"The next method that we have is to"},{"Start":"08:45.480 ","End":"08:52.395","Text":"hold our 4 fingers out in the direction of b."},{"Start":"08:52.395 ","End":"08:56.820","Text":"Let\u0027s say that it\u0027s something like this."},{"Start":"08:56.820 ","End":"09:00.510","Text":"Then what we do, so this represents the direction of b,"},{"Start":"09:00.510 ","End":"09:02.610","Text":"the direction of the 4 fingers,"},{"Start":"09:02.610 ","End":"09:06.150","Text":"and then we curl them in,"},{"Start":"09:06.150 ","End":"09:16.290","Text":"till they meet c. This curl to meet c. I\u0027ll say that c is over here,"},{"Start":"09:16.290 ","End":"09:21.405","Text":"so we curl them in to meet c. Then in this scenario,"},{"Start":"09:21.405 ","End":"09:27.340","Text":"our thumb represents the direction of a."},{"Start":"09:28.190 ","End":"09:31.395","Text":"Let\u0027s look at this example for instance."},{"Start":"09:31.395 ","End":"09:34.920","Text":"Here, if we use this version first."},{"Start":"09:34.920 ","End":"09:39.135","Text":"Our thumb will be pointing rightwards,"},{"Start":"09:39.135 ","End":"09:46.110","Text":"then our thumb is in the direction of dl,"},{"Start":"09:46.110 ","End":"09:47.970","Text":"which is rightwards,"},{"Start":"09:47.970 ","End":"09:51.810","Text":"and then our pointing finger is in the direction of c,"},{"Start":"09:51.810 ","End":"09:57.300","Text":"which in our case is r. Then if you do that,"},{"Start":"09:57.300 ","End":"10:04.620","Text":"what you\u0027ll get is that the magnetic field has to be coming out of the page."},{"Start":"10:04.620 ","End":"10:09.180","Text":"The magnetic field is like so coming out of the page."},{"Start":"10:09.180 ","End":"10:11.520","Text":"This is the sign for out of the page,"},{"Start":"10:11.520 ","End":"10:14.955","Text":"and this would be the sign for inside the page."},{"Start":"10:14.955 ","End":"10:17.985","Text":"The magnetic field is coming out of the page,"},{"Start":"10:17.985 ","End":"10:19.755","Text":"and this is with this method."},{"Start":"10:19.755 ","End":"10:23.835","Text":"Now let\u0027s look at the second method for the right-hand rule."},{"Start":"10:23.835 ","End":"10:26.115","Text":"In the second method,"},{"Start":"10:26.115 ","End":"10:31.415","Text":"we put our 4 fingers in the direction of dl vector."},{"Start":"10:31.415 ","End":"10:35.180","Text":"Our 4 fingers are pointing right,"},{"Start":"10:35.180 ","End":"10:37.700","Text":"and then we want to curl them in,"},{"Start":"10:37.700 ","End":"10:43.925","Text":"in the direction of our r. If you can imagine you place your hand on the screen,"},{"Start":"10:43.925 ","End":"10:46.980","Text":"where your little finger,"},{"Start":"10:46.980 ","End":"10:51.300","Text":"your pinky finger is resting on the dl vector,"},{"Start":"10:51.300 ","End":"10:58.725","Text":"for instance, and all the other fingers are resting on the pinky finger stacked up."},{"Start":"10:58.725 ","End":"11:05.100","Text":"Then you curl your fingers in to meet this i vector over here."},{"Start":"11:05.100 ","End":"11:09.375","Text":"Then you\u0027ll see that your thumb is pointing upwards,"},{"Start":"11:09.375 ","End":"11:12.435","Text":"which represents the direction of the b field."},{"Start":"11:12.435 ","End":"11:14.700","Text":"That means that the b field, again,"},{"Start":"11:14.700 ","End":"11:18.780","Text":"we see this method is coming out of the page."},{"Start":"11:18.780 ","End":"11:26.295","Text":"Now, another thing to notice with the b field is that the b field,"},{"Start":"11:26.295 ","End":"11:28.665","Text":"if we have a wire carrying a current,"},{"Start":"11:28.665 ","End":"11:34.980","Text":"the magnetic field is always rotating around it, like so."},{"Start":"11:34.980 ","End":"11:39.315","Text":"It forms a circle around it."},{"Start":"11:39.315 ","End":"11:45.465","Text":"What we\u0027re going to have is these circles going like so,"},{"Start":"11:45.465 ","End":"11:48.375","Text":"and what our right hand rule shows us,"},{"Start":"11:48.375 ","End":"11:53.860","Text":"going around the back of the wire,"},{"Start":"11:54.290 ","End":"11:58.050","Text":"what the right-hand rule shows us about the direction of"},{"Start":"11:58.050 ","End":"12:02.400","Text":"the b field is that at this point over here,"},{"Start":"12:02.400 ","End":"12:04.679","Text":"it\u0027s coming out of the page,"},{"Start":"12:04.679 ","End":"12:14.145","Text":"which means that it\u0027s traveling in this direction around the wire like so,"},{"Start":"12:14.145 ","End":"12:19.600","Text":"and similarly here like this."},{"Start":"12:19.850 ","End":"12:24.780","Text":"We can see that at the bottom of the circle, over here,"},{"Start":"12:24.780 ","End":"12:28.440","Text":"the magnetic field is going into the page,"},{"Start":"12:28.440 ","End":"12:29.940","Text":"and at the top over here,"},{"Start":"12:29.940 ","End":"12:35.055","Text":"it\u0027s coming out of the page and it goes around the wire like so."},{"Start":"12:35.055 ","End":"12:41.145","Text":"I just want to show you that it works or so if we have the same setup,"},{"Start":"12:41.145 ","End":"12:42.990","Text":"let\u0027s say over here,"},{"Start":"12:42.990 ","End":"12:49.604","Text":"so we can see that our dl is in this direction and our current is in this direction."},{"Start":"12:49.604 ","End":"12:52.200","Text":"We have the exact same setup like so."},{"Start":"12:52.200 ","End":"12:55.380","Text":"However, this time what we want to do is,"},{"Start":"12:55.380 ","End":"12:59.805","Text":"we want to measure the magnetic field over here."},{"Start":"12:59.805 ","End":"13:03.165","Text":"We draw our i vector,"},{"Start":"13:03.165 ","End":"13:07.810","Text":"sorry, our i vector like so."},{"Start":"13:08.090 ","End":"13:13.439","Text":"Now we can see that our dl vector is going in the rightwards direction,"},{"Start":"13:13.439 ","End":"13:15.855","Text":"but we\u0027re measuring this point down here."},{"Start":"13:15.855 ","End":"13:17.925","Text":"Let\u0027s use this."},{"Start":"13:17.925 ","End":"13:22.170","Text":"Here we have our 4 fingers pointing in"},{"Start":"13:22.170 ","End":"13:27.270","Text":"the direction of the current or in the direction of our dl,"},{"Start":"13:27.270 ","End":"13:32.370","Text":"and then we curl them in to meet our i vector."},{"Start":"13:32.370 ","End":"13:37.275","Text":"In actual fact, what we have is we have over here,"},{"Start":"13:37.275 ","End":"13:40.110","Text":"dl and over here,"},{"Start":"13:40.110 ","End":"13:45.840","Text":"r. We\u0027re curling our fingers so that they go from dl and meet r,"},{"Start":"13:45.840 ","End":"13:52.139","Text":"and then we\u0027ll see that our thumb is going to be pointing inside the page."},{"Start":"13:52.139 ","End":"13:54.450","Text":"You can try that now if you want."},{"Start":"13:54.450 ","End":"13:58.060","Text":"That means that at this point over here,"},{"Start":"13:58.820 ","End":"14:04.335","Text":"the magnetic field is going in to the page."},{"Start":"14:04.335 ","End":"14:09.465","Text":"That corresponds to the exact same setup that we had over here."},{"Start":"14:09.465 ","End":"14:13.380","Text":"Just now we\u0027re checking the magnetic field at this point over here,"},{"Start":"14:13.380 ","End":"14:15.570","Text":"and at the bottom,"},{"Start":"14:15.570 ","End":"14:20.115","Text":"because our i vector is pointing downwards,"},{"Start":"14:20.115 ","End":"14:24.525","Text":"just like here, and we get that the b field is going into the page."},{"Start":"14:24.525 ","End":"14:28.545","Text":"We just get the direction of the b field at that point,"},{"Start":"14:28.545 ","End":"14:31.740","Text":"and if it\u0027s coming out at this point over here,"},{"Start":"14:31.740 ","End":"14:36.735","Text":"then that means that this is the direction that the b field is going in."},{"Start":"14:36.735 ","End":"14:41.430","Text":"If we\u0027re looking from this side,"},{"Start":"14:41.430 ","End":"14:47.860","Text":"we can see that the magnetic field is anticlockwise."},{"Start":"14:49.430 ","End":"14:55.500","Text":"One last method that we can use with our right hand."},{"Start":"14:55.500 ","End":"15:01.350","Text":"This is the third right-hand rule if you will for magnetic fields,"},{"Start":"15:01.350 ","End":"15:09.015","Text":"is if you draw your thumb or you place your thumb in the direction of dl,"},{"Start":"15:09.015 ","End":"15:16.674","Text":"so here it would be in the direction of b and this is the thumb."},{"Start":"15:16.674 ","End":"15:22.870","Text":"Then what you do is you wrap your fingers"},{"Start":"15:22.870 ","End":"15:30.715","Text":"around in the direction of the magnetic field."},{"Start":"15:30.715 ","End":"15:36.775","Text":"All of this will give you a. What does that mean?"},{"Start":"15:36.775 ","End":"15:41.230","Text":"If you pointed your thumb in the direction of the current,"},{"Start":"15:41.230 ","End":"15:45.160","Text":"so here it represents the direction of the current"},{"Start":"15:45.160 ","End":"15:48.796","Text":"because dl represents also the direction of the current,"},{"Start":"15:48.796 ","End":"15:54.710","Text":"you would see that your fingers would wrap around like so in this direction,"},{"Start":"15:54.710 ","End":"16:01.330","Text":"which is exactly what we expected from what we calculated before."},{"Start":"16:02.340 ","End":"16:04.870","Text":"The fingers will just wrap around and"},{"Start":"16:04.870 ","End":"16:08.560","Text":"this method can actually also help you to remember that"},{"Start":"16:08.560 ","End":"16:17.420","Text":"the magnetic field just forms concentric field lines around the current carrying wire."},{"Start":"16:18.270 ","End":"16:24.325","Text":"Let\u0027s scroll down and let\u0027s give this method a little practice."},{"Start":"16:24.325 ","End":"16:31.760","Text":"Let\u0027s imagine that my wire carrying current is going into the page,"},{"Start":"16:31.760 ","End":"16:34.950","Text":"I\u0027ll draw it in black so as not to confuse,"},{"Start":"16:34.950 ","End":"16:40.690","Text":"so this is my wire carrying current and it\u0027s going into the page,"},{"Start":"16:40.690 ","End":"16:46.315","Text":"which means that my dl vector is also going into the page."},{"Start":"16:46.315 ","End":"16:51.535","Text":"What I want to do is I want to find the direction of the magnetic field."},{"Start":"16:51.535 ","End":"16:53.620","Text":"The first thing that I know,"},{"Start":"16:53.620 ","End":"16:56.350","Text":"the first thing that you always have to remember is"},{"Start":"16:56.350 ","End":"17:01.360","Text":"the magnetic field forms concentric circles of"},{"Start":"17:01.360 ","End":"17:10.045","Text":"field lines around the current carrying wire so this is of course a perfect circle."},{"Start":"17:10.045 ","End":"17:13.960","Text":"Now what I want to do is I want to find the direction of the field lines."},{"Start":"17:13.960 ","End":"17:17.770","Text":"Is it going clockwise or anticlockwise?"},{"Start":"17:17.770 ","End":"17:24.410","Text":"Let\u0027s try using each different right-hand rule that we have."},{"Start":"17:24.480 ","End":"17:27.670","Text":"Let\u0027s use this method first."},{"Start":"17:27.670 ","End":"17:29.275","Text":"First of all,"},{"Start":"17:29.275 ","End":"17:33.175","Text":"our thumb is in the direction of our dl vector,"},{"Start":"17:33.175 ","End":"17:36.745","Text":"so our thumb is pointing inwards."},{"Start":"17:36.745 ","End":"17:42.325","Text":"Then we have to put our pointing finger in the direction of r vector,"},{"Start":"17:42.325 ","End":"17:47.420","Text":"so let\u0027s look at this point over here."},{"Start":"17:47.810 ","End":"17:50.870","Text":"Let\u0027s draw our r vector,"},{"Start":"17:50.870 ","End":"17:53.814","Text":"so this is our r vector."},{"Start":"17:53.814 ","End":"17:57.550","Text":"If we put our thumb into the page,"},{"Start":"17:57.550 ","End":"18:00.580","Text":"our pointing finger pointing in this direction,"},{"Start":"18:00.580 ","End":"18:06.055","Text":"we\u0027ll get that our middle finger is pointing like so out here."},{"Start":"18:06.055 ","End":"18:09.790","Text":"I\u0027m going to draw my middle finger in red to represent the answer."},{"Start":"18:09.790 ","End":"18:13.720","Text":"My middle finger is going to be pointing in this direction,"},{"Start":"18:13.720 ","End":"18:18.115","Text":"feel free to try this out."},{"Start":"18:18.115 ","End":"18:20.754","Text":"This is the direction of my dB,"},{"Start":"18:20.754 ","End":"18:23.110","Text":"which means that over here,"},{"Start":"18:23.110 ","End":"18:29.080","Text":"my magnetic field is going to be going in this direction."},{"Start":"18:29.080 ","End":"18:36.175","Text":"Then just make sure I can draw my r vector over here now,"},{"Start":"18:36.175 ","End":"18:41.560","Text":"so I\u0027ll move my r vector to this point."},{"Start":"18:41.560 ","End":"18:45.910","Text":"Now again, I\u0027ll put my thumb into the page,"},{"Start":"18:45.910 ","End":"18:53.439","Text":"my pointing finger will point in the direction of my r vector and then my middle finger,"},{"Start":"18:53.439 ","End":"18:57.895","Text":"in this case will be pointing in this direction."},{"Start":"18:57.895 ","End":"19:00.775","Text":"That means that over here,"},{"Start":"19:00.775 ","End":"19:04.345","Text":"when I\u0027m looking at some point over here,"},{"Start":"19:04.345 ","End":"19:07.765","Text":"the magnetic field is traveling in this direction."},{"Start":"19:07.765 ","End":"19:11.620","Text":"What we can see is that my magnetic field is traveling in"},{"Start":"19:11.620 ","End":"19:17.965","Text":"the clockwise direction and that is from this method."},{"Start":"19:17.965 ","End":"19:21.190","Text":"Now let\u0027s look at the second method."},{"Start":"19:21.190 ","End":"19:26.590","Text":"In this method, I align all 4 of my fingers up in"},{"Start":"19:26.590 ","End":"19:32.688","Text":"the direction of my dl vector or in the direction of my current."},{"Start":"19:32.688 ","End":"19:36.220","Text":"All 4 fingers should be pointing inside"},{"Start":"19:36.220 ","End":"19:42.430","Text":"the page and then I curl them in in the direction of my r vector."},{"Start":"19:42.430 ","End":"19:46.570","Text":"Let\u0027s choose some arbitrary point over here."},{"Start":"19:46.570 ","End":"19:50.605","Text":"This is my r vector,"},{"Start":"19:50.605 ","End":"19:59.780","Text":"so my 4 fingers are pointing in like so and they curl in the direction of my r."},{"Start":"20:00.260 ","End":"20:04.725","Text":"What we\u0027re going to do is our 4 fingers are pointing down"},{"Start":"20:04.725 ","End":"20:08.505","Text":"in the direction of the current and then what we do is we"},{"Start":"20:08.505 ","End":"20:14.920","Text":"curl them upwards in this direction like"},{"Start":"20:14.920 ","End":"20:18.280","Text":"so to meet here and then you\u0027ll see that"},{"Start":"20:18.280 ","End":"20:22.640","Text":"your thumb is pointing in this rightwards direction,"},{"Start":"20:22.640 ","End":"20:25.465","Text":"which means that at this point"},{"Start":"20:25.465 ","End":"20:31.765","Text":"the magnetic field is going in this rightwards direction like so."},{"Start":"20:31.765 ","End":"20:38.218","Text":"Again, we get this clockwise direction and you can also try it at a different point,"},{"Start":"20:38.218 ","End":"20:41.380","Text":"let\u0027s say over here."},{"Start":"20:41.380 ","End":"20:44.919","Text":"This is now the r vector."},{"Start":"20:44.919 ","End":"20:48.370","Text":"Again, 4 fingers in the direction of the current,"},{"Start":"20:48.370 ","End":"20:53.030","Text":"and then we curl them in to the r vector."},{"Start":"20:53.490 ","End":"21:00.790","Text":"Now, our thumb should be pointing in something like this direction."},{"Start":"21:00.790 ","End":"21:02.695","Text":"This is the direction of the thumb."},{"Start":"21:02.695 ","End":"21:04.870","Text":"At this point over here,"},{"Start":"21:04.870 ","End":"21:08.755","Text":"we can see that the current is trying to move upwards,"},{"Start":"21:08.755 ","End":"21:13.790","Text":"which is in the exact direction of this clockwise motion."},{"Start":"21:15.210 ","End":"21:23.140","Text":"This method, in this case is maybe slightly more complicated or more difficult to see,"},{"Start":"21:23.140 ","End":"21:27.010","Text":"but you see we can still get the same answer and then this,"},{"Start":"21:27.010 ","End":"21:29.575","Text":"which is by far the easiest method,"},{"Start":"21:29.575 ","End":"21:34.104","Text":"you just point your thumb in the direction of the current,"},{"Start":"21:34.104 ","End":"21:37.810","Text":"so your thumb is pointing down the page and then you\u0027ll see that"},{"Start":"21:37.810 ","End":"21:45.400","Text":"your fingers are curved in this direction automatically."},{"Start":"21:45.400 ","End":"21:47.500","Text":"Your thumb is pointing down,"},{"Start":"21:47.500 ","End":"21:49.435","Text":"your palm is somewhere over here,"},{"Start":"21:49.435 ","End":"21:53.215","Text":"and your fingers are curled in this clockwise direction,"},{"Start":"21:53.215 ","End":"21:57.160","Text":"so this method is definitely the easiest when"},{"Start":"21:57.160 ","End":"22:02.545","Text":"working on calculating the direction of the magnetic field,"},{"Start":"22:02.545 ","End":"22:05.800","Text":"and all that is to remember in these 3 methods"},{"Start":"22:05.800 ","End":"22:09.535","Text":"is that they\u0027re all using only the right hand,"},{"Start":"22:09.535 ","End":"22:12.385","Text":"they\u0027re just different versions of the right-hand rule."},{"Start":"22:12.385 ","End":"22:15.685","Text":"This method is quite difficult"},{"Start":"22:15.685 ","End":"22:19.720","Text":"to figure out relatively if you\u0027re dealing with this configuration,"},{"Start":"22:19.720 ","End":"22:22.090","Text":"but of course you can still use it but"},{"Start":"22:22.090 ","End":"22:26.710","Text":"just really notice what you\u0027re calculating and what you\u0027re working out."},{"Start":"22:26.710 ","End":"22:31.420","Text":"One thing to note when using this method is"},{"Start":"22:31.420 ","End":"22:34.150","Text":"that your thumb will point in the direction of"},{"Start":"22:34.150 ","End":"22:37.435","Text":"the field at that point that you\u0027re measuring."},{"Start":"22:37.435 ","End":"22:42.100","Text":"Let\u0027s say we\u0027re looking at this point over here,"},{"Start":"22:42.100 ","End":"22:46.825","Text":"so this is our r vector."},{"Start":"22:46.825 ","End":"22:51.310","Text":"Again, if we point all 4 fingers in the direction of the current and curl"},{"Start":"22:51.310 ","End":"22:56.185","Text":"them to the direction of our vector,"},{"Start":"22:56.185 ","End":"22:59.150","Text":"so curl them to this point,"},{"Start":"23:01.560 ","End":"23:04.180","Text":"let\u0027s draw the thumb in black,"},{"Start":"23:04.180 ","End":"23:07.880","Text":"your thumb will be pointing in this direction."},{"Start":"23:08.190 ","End":"23:13.345","Text":"That means that at this point over here,"},{"Start":"23:13.345 ","End":"23:18.490","Text":"the direction of the magnetic field is in this direction."},{"Start":"23:18.490 ","End":"23:22.360","Text":"Similarly, if you look at this point, let\u0027s say,"},{"Start":"23:22.360 ","End":"23:26.590","Text":"so you\u0027ll get really that your thumb is pointing in this direction,"},{"Start":"23:26.590 ","End":"23:30.460","Text":"this will be the direction of the thumb so that means that at"},{"Start":"23:30.460 ","End":"23:34.630","Text":"this point the magnetic field is pointing in the same direction,"},{"Start":"23:34.630 ","End":"23:38.575","Text":"so that might make this method easier to use."},{"Start":"23:38.575 ","End":"23:43.180","Text":"This is the equation to remember for Biot Savart law,"},{"Start":"23:43.180 ","End":"23:50.290","Text":"which describes the magnetic field caused by a current carrying wire and to remember"},{"Start":"23:50.290 ","End":"23:53.470","Text":"the 3 methods for the right-hand rule to"},{"Start":"23:53.470 ","End":"23:57.760","Text":"calculate the direction of the magnetic field and of course,"},{"Start":"23:57.760 ","End":"24:04.050","Text":"we can also just calculate it by using this equation for the cross product."},{"Start":"24:04.050 ","End":"24:06.910","Text":"That\u0027s the end of the lesson."}],"ID":21395},{"Watched":false,"Name":"Exercise 1","Duration":"28m 29s","ChapterTopicVideoID":21320,"CourseChapterTopicPlaylistID":99469,"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":"00:04.725","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.725 ","End":"00:08.955","Text":"We\u0027re being asked to show that the magnitude of"},{"Start":"00:08.955 ","End":"00:13.695","Text":"the magnetic field caused by this current-carrying wire,"},{"Start":"00:13.695 ","End":"00:21.090","Text":"a distance a from the wire is equal to this equation over here."},{"Start":"00:21.090 ","End":"00:26.820","Text":"Here we have a finite wire with a current I flowing through it."},{"Start":"00:26.820 ","End":"00:29.205","Text":"We\u0027re looking at this point over here."},{"Start":"00:29.205 ","End":"00:32.050","Text":"It descends a above the wire."},{"Start":"00:32.050 ","End":"00:37.010","Text":"We want to calculate the magnitude of the magnetic field over here."},{"Start":"00:37.010 ","End":"00:43.295","Text":"If we draw a line connecting the edge of the wire to this point,"},{"Start":"00:43.295 ","End":"00:45.495","Text":"there is an angle of Alpha 1,"},{"Start":"00:45.495 ","End":"00:53.435","Text":"and if we connect a line from this edge of the wire over here to this point over here,"},{"Start":"00:53.435 ","End":"00:56.815","Text":"the angle formed is Alpha 2."},{"Start":"00:56.815 ","End":"01:01.775","Text":"The first thing that we\u0027re going to do in order to calculate the magnitude over here,"},{"Start":"01:01.775 ","End":"01:05.315","Text":"we\u0027re going to write out Biot-Savart\u0027s Law."},{"Start":"01:05.315 ","End":"01:08.505","Text":"We know that db."},{"Start":"01:08.505 ","End":"01:11.930","Text":"The magnetic field at this point formed from"},{"Start":"01:11.930 ","End":"01:18.170","Text":"a small section of the wire is equal to Mu naught divided by"},{"Start":"01:18.170 ","End":"01:21.769","Text":"4 Phi multiplied by I"},{"Start":"01:21.769 ","End":"01:25.685","Text":"dl vector cross-product with"},{"Start":"01:25.685 ","End":"01:31.025","Text":"the I vector divided by the magnitude of the I vector cubed,"},{"Start":"01:31.025 ","End":"01:36.265","Text":"or alternatively, it\u0027s also equal to Mu naught divided by 4 Phi"},{"Start":"01:36.265 ","End":"01:42.995","Text":"I dl cross product with the r hat vector,"},{"Start":"01:42.995 ","End":"01:46.820","Text":"and then it\u0027s divided by the magnitude of r^2."},{"Start":"01:46.820 ","End":"01:49.740","Text":"Here we can use either equation."},{"Start":"01:49.740 ","End":"01:52.725","Text":"First of all, let\u0027s mark out our dl."},{"Start":"01:52.725 ","End":"01:56.055","Text":"Let\u0027s choose this length over here."},{"Start":"01:56.055 ","End":"02:00.950","Text":"This is a small portion of the wire and it\u0027s of length dl."},{"Start":"02:00.950 ","End":"02:05.045","Text":"Then we know that our r vector, our radius,"},{"Start":"02:05.045 ","End":"02:08.930","Text":"is the vector from our piece of wire"},{"Start":"02:08.930 ","End":"02:13.395","Text":"until the point where we\u0027re measuring the magnetic field."},{"Start":"02:13.395 ","End":"02:16.300","Text":"This is our r vector."},{"Start":"02:17.270 ","End":"02:21.475","Text":"First of all, let\u0027s define where origin is."},{"Start":"02:21.475 ","End":"02:25.115","Text":"The easiest that we can do is to define our origin at"},{"Start":"02:25.115 ","End":"02:29.555","Text":"this point where we\u0027re measuring the magnetic field."},{"Start":"02:29.555 ","End":"02:33.215","Text":"Let\u0027s say that this is the x-direction,"},{"Start":"02:33.215 ","End":"02:35.830","Text":"and this is the y-direction."},{"Start":"02:35.830 ","End":"02:38.850","Text":"Of course, it doesn\u0027t matter,"},{"Start":"02:38.850 ","End":"02:40.970","Text":"we could define it at any other place,"},{"Start":"02:40.970 ","End":"02:45.305","Text":"but here it\u0027s slightly easier to work with."},{"Start":"02:45.305 ","End":"02:48.965","Text":"Now let\u0027s define our dl vector."},{"Start":"02:48.965 ","End":"02:57.240","Text":"We can say dl vector."},{"Start":"02:57.240 ","End":"03:03.290","Text":"We know that our dl vector is always in the direction of our current."},{"Start":"03:03.290 ","End":"03:10.395","Text":"We can say that is equal to dx in the x direction."},{"Start":"03:10.395 ","End":"03:16.860","Text":"It\u0027s traveling in the x-direction and then there\u0027s a small change in x each time."},{"Start":"03:18.680 ","End":"03:21.695","Text":"Now what I want to do is first,"},{"Start":"03:21.695 ","End":"03:26.460","Text":"I\u0027m going to work with this equation over here."},{"Start":"03:26.630 ","End":"03:29.120","Text":"Then, later in the lesson,"},{"Start":"03:29.120 ","End":"03:32.360","Text":"I\u0027ll work with this equation and show that we get the same answer."},{"Start":"03:32.360 ","End":"03:36.330","Text":"Now, because I\u0027m first working out the magnitude,"},{"Start":"03:36.330 ","End":"03:40.015","Text":"we\u0027re not being asked to work out the direction, but I\u0027ll do it anyway."},{"Start":"03:40.015 ","End":"03:47.115","Text":"Let\u0027s work out the magnitude of dl cross r hat."},{"Start":"03:47.115 ","End":"03:50.825","Text":"As we know, this is going to be equal to the magnitude"},{"Start":"03:50.825 ","End":"03:56.150","Text":"of dl multiplied by the magnitude of r hat,"},{"Start":"03:56.150 ","End":"04:00.650","Text":"multiplied by sine of the angle between the 2."},{"Start":"04:00.650 ","End":"04:08.760","Text":"The magnitude of dl is of course this over here, dx."},{"Start":"04:08.760 ","End":"04:12.170","Text":"This is the size and this is the direction."},{"Start":"04:12.170 ","End":"04:13.925","Text":"So the size is the magnitude."},{"Start":"04:13.925 ","End":"04:15.815","Text":"The magnitude of r hat,"},{"Start":"04:15.815 ","End":"04:19.400","Text":"every vector that has a hat on top means that it\u0027s a unit vector,"},{"Start":"04:19.400 ","End":"04:21.485","Text":"so it\u0027s multiplied by 1."},{"Start":"04:21.485 ","End":"04:24.610","Text":"Then let\u0027s see what the angle is."},{"Start":"04:24.610 ","End":"04:29.720","Text":"As we know, this is the direction of our dl."},{"Start":"04:29.720 ","End":"04:33.770","Text":"This is the direction of our r vector,"},{"Start":"04:33.770 ","End":"04:37.040","Text":"which means it\u0027s the angle between the 2,"},{"Start":"04:37.040 ","End":"04:40.160","Text":"which means that this is our angle Alpha."},{"Start":"04:40.160 ","End":"04:43.490","Text":"We have dx multiplied by 1,"},{"Start":"04:43.490 ","End":"04:46.820","Text":"multiplied by sine of Alpha,"},{"Start":"04:46.820 ","End":"04:50.245","Text":"where this as Alpha over here."},{"Start":"04:50.245 ","End":"04:53.585","Text":"Because we have defined our origin over here,"},{"Start":"04:53.585 ","End":"04:57.905","Text":"we can say that the distance between the origin to"},{"Start":"04:57.905 ","End":"05:03.390","Text":"our piece of wire is a distance of x away."},{"Start":"05:05.810 ","End":"05:12.450","Text":"Now I\u0027m going to use this in order to find the magnitude of my r vector."},{"Start":"05:12.450 ","End":"05:14.465","Text":"Then I want to square it."},{"Start":"05:14.465 ","End":"05:19.360","Text":"I take my r vector squared and this is equal to,"},{"Start":"05:19.360 ","End":"05:20.800","Text":"so it\u0027s the hypotenuse,"},{"Start":"05:20.800 ","End":"05:22.870","Text":"my I vector is the hypotenuse."},{"Start":"05:22.870 ","End":"05:30.220","Text":"From Pythagoras, we get that r^2 is equal to a^2+x^2."},{"Start":"05:30.350 ","End":"05:35.215","Text":"Now what I can do is I can just plug this all into my equation."},{"Start":"05:35.215 ","End":"05:45.180","Text":"I\u0027ll get that db vector is equal to Mu naught divided by 4 Phi multiplied by I."},{"Start":"05:45.180 ","End":"05:46.875","Text":"All of these are constants."},{"Start":"05:46.875 ","End":"05:49.850","Text":"Then I have dl cross r hat,"},{"Start":"05:49.850 ","End":"05:58.145","Text":"which I got was equal to dx sine of Alpha divided by r^2,"},{"Start":"05:58.145 ","End":"05:59.600","Text":"which as we saw,"},{"Start":"05:59.600 ","End":"06:03.610","Text":"is equal to a^2+x^2."},{"Start":"06:03.610 ","End":"06:06.230","Text":"Now what I want to do is I want to find"},{"Start":"06:06.230 ","End":"06:10.850","Text":"the total magnetic field at this point from the entire wire."},{"Start":"06:10.850 ","End":"06:16.440","Text":"This will just give me the magnetic field from this section of wire."},{"Start":"06:16.670 ","End":"06:21.260","Text":"Of course, I have to add in because I\u0027ve written this as a vector."},{"Start":"06:21.260 ","End":"06:23.030","Text":"From using the right-hand rule,"},{"Start":"06:23.030 ","End":"06:27.545","Text":"if I point my thumb in this rightwards direction down the wire,"},{"Start":"06:27.545 ","End":"06:33.170","Text":"I\u0027ll get my fingers curl around it in a way"},{"Start":"06:33.170 ","End":"06:39.530","Text":"that I see that my magnetic field at this point over here is coming out of the page."},{"Start":"06:39.530 ","End":"06:46.715","Text":"My fingers are curling in the direction that they\u0027re coming towards us,"},{"Start":"06:46.715 ","End":"06:49.140","Text":"out of the screen."},{"Start":"06:49.700 ","End":"06:54.005","Text":"Therefore, at this point over here,"},{"Start":"06:54.005 ","End":"06:57.410","Text":"if this is how I define my x and y directions,"},{"Start":"06:57.410 ","End":"07:00.380","Text":"that means that this point where I\u0027m measuring the magnetic field,"},{"Start":"07:00.380 ","End":"07:03.060","Text":"it\u0027s in the z direction."},{"Start":"07:03.380 ","End":"07:07.025","Text":"In order to find the total magnetic field over here,"},{"Start":"07:07.025 ","End":"07:10.715","Text":"I of course, have to integrate on all of this."},{"Start":"07:10.715 ","End":"07:14.855","Text":"Now, notice that my variable over here is x."},{"Start":"07:14.855 ","End":"07:19.710","Text":"However, my angle Alpha is dependent on x."},{"Start":"07:19.710 ","End":"07:22.260","Text":"We see that as x grows,"},{"Start":"07:22.260 ","End":"07:23.640","Text":"my Alpha grows,"},{"Start":"07:23.640 ","End":"07:26.115","Text":"and as x becomes smaller,"},{"Start":"07:26.115 ","End":"07:31.000","Text":"my Alpha approaches a 90-degree angle."},{"Start":"07:31.000 ","End":"07:34.970","Text":"I can see that my Alpha is also a variable."},{"Start":"07:34.970 ","End":"07:41.700","Text":"What I want to do is I want to convert this equation to just 1 variable."},{"Start":"07:41.700 ","End":"07:49.745","Text":"What I\u0027m going to do is I\u0027m going to use my angle Alpha as my only variable."},{"Start":"07:49.745 ","End":"07:54.110","Text":"I\u0027m going to convert my x in terms of Alpha."},{"Start":"07:54.110 ","End":"08:00.245","Text":"Then I\u0027ll have 1 easy variable that I can integrate across."},{"Start":"08:00.245 ","End":"08:06.870","Text":"The first thing that I notice is that here I have a right-angled triangle."},{"Start":"08:06.870 ","End":"08:10.745","Text":"What I want to do is I want to figure out this angle,"},{"Start":"08:10.745 ","End":"08:15.155","Text":"which is relevant to my right-angle triangle."},{"Start":"08:15.155 ","End":"08:19.550","Text":"As then I can use this triangle and the trig identities"},{"Start":"08:19.550 ","End":"08:24.365","Text":"associated with it in order to do this conversion from x to Alpha."},{"Start":"08:24.365 ","End":"08:27.260","Text":"Of course, I\u0027m going to choose this angle over"},{"Start":"08:27.260 ","End":"08:30.885","Text":"here because I know what my Alpha is over here,"},{"Start":"08:30.885 ","End":"08:33.150","Text":"so it\u0027s easy to find this angle."},{"Start":"08:33.150 ","End":"08:37.890","Text":"This angle over here is Phi minus Alpha,"},{"Start":"08:37.890 ","End":"08:41.590","Text":"or in other words, a 180 minus Alpha."},{"Start":"08:43.550 ","End":"08:47.800","Text":"In order to find what this angle is,"},{"Start":"08:47.800 ","End":"08:54.560","Text":"from SOHCAHTOA, we know that tan is opposite over adjacent."},{"Start":"08:54.560 ","End":"08:57.210","Text":"Tan of this angle,"},{"Start":"08:57.210 ","End":"09:03.125","Text":"so tan of Phi minus Alpha is equal to opposite,"},{"Start":"09:03.125 ","End":"09:06.980","Text":"which is a divided by adjacent,"},{"Start":"09:06.980 ","End":"09:13.220","Text":"divided by x. I want to isolate out my x,"},{"Start":"09:13.220 ","End":"09:19.255","Text":"so I\u0027ll multiply both sides by x and divide both sides by tan of this angle."},{"Start":"09:19.255 ","End":"09:25.435","Text":"What I\u0027ll be left with is that x is equal to a divided by"},{"Start":"09:25.435 ","End":"09:33.175","Text":"tan of Phi minus Alpha or what is 1 divided by tan?"},{"Start":"09:33.175 ","End":"09:36.010","Text":"It\u0027s cotangent or cot."},{"Start":"09:36.010 ","End":"09:45.814","Text":"So what I\u0027m left with is a multiplied by cot of Phi minus Alpha."},{"Start":"09:45.814 ","End":"09:48.870","Text":"What is cost? Of course,"},{"Start":"09:48.870 ","End":"09:50.610","Text":"it\u0027s 1 divided by 10."},{"Start":"09:50.610 ","End":"09:54.915","Text":"I can write this as a multiplied by,"},{"Start":"09:54.915 ","End":"10:00.375","Text":"so tangent is sine divided by cosine."},{"Start":"10:00.375 ","End":"10:11.290","Text":"Cotangent is going to be cosine of the angle divided by sine of the angle."},{"Start":"10:11.780 ","End":"10:17.100","Text":"Cotangent is just the inverse of the tangent."},{"Start":"10:17.100 ","End":"10:19.770","Text":"Now we are carrying on."},{"Start":"10:19.770 ","End":"10:25.085","Text":"I have a and then cosine of Pi minus Alpha is"},{"Start":"10:25.085 ","End":"10:31.650","Text":"the same as negative cosine of Alpha,"},{"Start":"10:31.650 ","End":"10:39.660","Text":"and sine of Pi minus Alpha is the same as sine of Alpha."},{"Start":"10:39.660 ","End":"10:45.600","Text":"These are things that are useful to remember, useful identities."},{"Start":"10:45.600 ","End":"10:52.080","Text":"Therefore, what we have is that x is equal to negative"},{"Start":"10:52.080 ","End":"10:58.515","Text":"a multiplied by cosine divided by sine,"},{"Start":"10:58.515 ","End":"11:04.330","Text":"which is just cotangent of Alpha."},{"Start":"11:05.360 ","End":"11:11.235","Text":"Now I have my x and what I want to find is what my dx is,"},{"Start":"11:11.235 ","End":"11:18.102","Text":"so I\u0027m going to find my change in x with respect to my change in the angle Alpha."},{"Start":"11:18.102 ","End":"11:26.100","Text":"That is simply going to be equal to negative a which is a constant and the derivative of"},{"Start":"11:26.100 ","End":"11:35.895","Text":"cotangent of Alpha is simply equal to negative 1 divided by sine squared of Alpha."},{"Start":"11:35.895 ","End":"11:38.955","Text":"If you don\u0027t know this off by heart,"},{"Start":"11:38.955 ","End":"11:45.885","Text":"you can just take the derivative of cosine divided by sine,"},{"Start":"11:45.885 ","End":"11:49.740","Text":"where you use the rule when taking"},{"Start":"11:49.740 ","End":"11:54.510","Text":"the derivative when dealing with fractions and you\u0027ll get the same answer."},{"Start":"11:54.510 ","End":"12:01.290","Text":"Therefore, we get that our dx if we multiply both sides by d Alpha,"},{"Start":"12:01.290 ","End":"12:05.805","Text":"so the minus and the minus cancel out and what we\u0027re left with is"},{"Start":"12:05.805 ","End":"12:13.390","Text":"ad Alpha divided by sine squared Alpha."},{"Start":"12:13.390 ","End":"12:17.970","Text":"Now we have our dx and x so let\u0027s plug this into our equation."},{"Start":"12:17.970 ","End":"12:26.985","Text":"We get that db is equal to Mu_0I divided by 4Pi and then we have"},{"Start":"12:26.985 ","End":"12:36.375","Text":"dx so we have ad Alpha divided by sine squared Alpha."},{"Start":"12:36.375 ","End":"12:42.210","Text":"This is dx and this is multiplied by sine of Alpha."},{"Start":"12:42.210 ","End":"12:47.535","Text":"Then all of this is divided by a squared plus x squared."},{"Start":"12:47.535 ","End":"12:49.530","Text":"What is x squared?"},{"Start":"12:49.530 ","End":"12:52.620","Text":"It\u0027s negative a cotangent of Alpha squared,"},{"Start":"12:52.620 ","End":"13:00.300","Text":"so it\u0027s just positive a squared cotangent of Alpha squared."},{"Start":"13:00.300 ","End":"13:05.910","Text":"Then our sine Alpha over here can cancel out with 1 of these sines of Alpha."},{"Start":"13:05.910 ","End":"13:10.950","Text":"Now we have Mu_0I divided by"},{"Start":"13:10.950 ","End":"13:18.420","Text":"4Pi multiplied by so then we have ad Alpha divided by,"},{"Start":"13:18.420 ","End":"13:26.730","Text":"so our sine Alpha goes down over here and we can also take out the a squared."},{"Start":"13:26.730 ","End":"13:32.130","Text":"Multiplied by a squared and then inside the brackets we have 1"},{"Start":"13:32.130 ","End":"13:38.970","Text":"plus cotangent squared of Alpha."},{"Start":"13:38.970 ","End":"13:42.240","Text":"Then 1 of this a over here in"},{"Start":"13:42.240 ","End":"13:45.945","Text":"the numerator can cancel out with 1 of the as in the denominator."},{"Start":"13:45.945 ","End":"13:56.040","Text":"Then what we\u0027ll get is that this is equal to Mu_0I divided by 4Pi."},{"Start":"13:56.040 ","End":"14:02.849","Text":"We\u0027ll move this a over here and then we\u0027re left with d Alpha divided"},{"Start":"14:02.849 ","End":"14:13.030","Text":"by sine of Alpha multiplied by 1 plus cot squared Alpha."},{"Start":"14:14.960 ","End":"14:20.400","Text":"The next thing that we can simplify over here is this cot squared Alpha."},{"Start":"14:20.400 ","End":"14:25.200","Text":"As we said, cotangent is the inverse of tangent."},{"Start":"14:25.200 ","End":"14:30.405","Text":"What is cot squared Alpha? Let\u0027s scroll down."},{"Start":"14:30.405 ","End":"14:40.755","Text":"That is equal to cosine squared of Alpha divided by sine squared of Alpha."},{"Start":"14:40.755 ","End":"14:44.890","Text":"This is of course, cot squared Alpha."},{"Start":"14:45.470 ","End":"14:50.355","Text":"Whenever you see sine squared or cosine squared,"},{"Start":"14:50.355 ","End":"14:56.415","Text":"what you\u0027re meant to remember is that this is the basic and very important identity,"},{"Start":"14:56.415 ","End":"15:06.015","Text":"1 is equal to cosine squared Alpha plus sine squared Alpha, basic identity."},{"Start":"15:06.015 ","End":"15:10.500","Text":"Now what we can do is we can isolate out our cosine squared."},{"Start":"15:10.500 ","End":"15:20.490","Text":"What we get is that our cosine squared Alpha is equal to 1 minus sine squared Alpha."},{"Start":"15:20.490 ","End":"15:23.400","Text":"Now we can plug this in."},{"Start":"15:23.400 ","End":"15:28.590","Text":"What we get is that cot squared Alpha is equal to cosine squared Alpha,"},{"Start":"15:28.590 ","End":"15:33.675","Text":"which is 1 minus sine squared Alpha and"},{"Start":"15:33.675 ","End":"15:39.510","Text":"all of this is divided by sine squared Alpha."},{"Start":"15:39.510 ","End":"15:41.970","Text":"Now if we split this up,"},{"Start":"15:41.970 ","End":"15:46.530","Text":"so what we get is 1 divided by sine squared"},{"Start":"15:46.530 ","End":"15:52.620","Text":"Alpha and then we have minus sine squared Alpha divided by sine squared Alpha,"},{"Start":"15:52.620 ","End":"15:55.360","Text":"which is just minus 1."},{"Start":"15:56.030 ","End":"16:01.140","Text":"Using this identity, let\u0027s plug this into this equation."},{"Start":"16:01.140 ","End":"16:09.120","Text":"What we get is that db is equal to Mu_0I divided by"},{"Start":"16:09.120 ","End":"16:18.690","Text":"4Pi a multiplied by d Alpha divided by sine of Alpha."},{"Start":"16:18.690 ","End":"16:22.770","Text":"Then we have 1 plus cot squared Alpha,"},{"Start":"16:22.770 ","End":"16:24.765","Text":"so 1 plus,"},{"Start":"16:24.765 ","End":"16:32.220","Text":"and then cot squared Alpha is 1 divided by sine squared alpha minus 1."},{"Start":"16:32.220 ","End":"16:34.680","Text":"These cancel out."},{"Start":"16:34.680 ","End":"16:40.260","Text":"Therefore we get that db is simply equal to Mu_0I divided by"},{"Start":"16:40.260 ","End":"16:46.770","Text":"4Pi a and then we have d Alpha divided by,"},{"Start":"16:46.770 ","End":"16:52.350","Text":"so then we have sine of Alpha multiplied"},{"Start":"16:52.350 ","End":"16:57.704","Text":"by 1 divided by sine squared of Alpha."},{"Start":"16:57.704 ","End":"17:01.560","Text":"This sine of Alpha will cancel with 1 of these signs of Alpha"},{"Start":"17:01.560 ","End":"17:05.910","Text":"and then we have Mu_0I divided by"},{"Start":"17:05.910 ","End":"17:14.496","Text":"4Pi a multiplied by d Alpha divided by 1 divided by sine Alpha."},{"Start":"17:14.496 ","End":"17:17.085","Text":"What we have in fact in the end,"},{"Start":"17:17.085 ","End":"17:20.069","Text":"because here we have a denominator and the denominator,"},{"Start":"17:20.069 ","End":"17:25.770","Text":"so it goes up to the numerator so I have sine of Alpha d Alpha."},{"Start":"17:25.770 ","End":"17:32.685","Text":"Of course, all of this is in the z direction so I just added z hat everywhere."},{"Start":"17:32.685 ","End":"17:36.990","Text":"If I want to find the total magnetic field,"},{"Start":"17:36.990 ","End":"17:42.390","Text":"then of course what I\u0027m going to have to do is integrate along this,"},{"Start":"17:42.390 ","End":"17:45.780","Text":"so I have to integrate along this as well."},{"Start":"17:45.780 ","End":"17:50.590","Text":"Let\u0027s just scroll up to see what our bounds are going to be."},{"Start":"17:51.380 ","End":"17:57.075","Text":"What I can see is that we\u0027re always looking at this angle over here Alpha,"},{"Start":"17:57.075 ","End":"18:02.205","Text":"between the direction of our current and our eigenvector."},{"Start":"18:02.205 ","End":"18:06.000","Text":"When we start at this end over here,"},{"Start":"18:06.000 ","End":"18:10.860","Text":"so I vector is going to be like so."},{"Start":"18:10.860 ","End":"18:14.745","Text":"This will be I vector in this case, in which case,"},{"Start":"18:14.745 ","End":"18:19.215","Text":"this is the angle between the current and our I vector."},{"Start":"18:19.215 ","End":"18:22.620","Text":"This will be representing our angle Alpha,"},{"Start":"18:22.620 ","End":"18:26.685","Text":"so our lower bound is Alpha 1."},{"Start":"18:26.685 ","End":"18:34.020","Text":"Then what happens when we get to our final point along this wire,"},{"Start":"18:34.020 ","End":"18:37.485","Text":"so this is our upper bound so in this case,"},{"Start":"18:37.485 ","End":"18:42.690","Text":"this is going to be our I vector and what we can see is that"},{"Start":"18:42.690 ","End":"18:45.360","Text":"we\u0027re going from I current to our I vector so"},{"Start":"18:45.360 ","End":"18:48.030","Text":"we can imagine that this is the direction of our current,"},{"Start":"18:48.030 ","End":"18:49.920","Text":"although of course it\u0027s not flowing here."},{"Start":"18:49.920 ","End":"18:53.130","Text":"We\u0027re looking at this angle over here."},{"Start":"18:53.130 ","End":"18:56.850","Text":"This angle over here is Alpha,"},{"Start":"18:56.850 ","End":"19:00.570","Text":"so what is Alpha?"},{"Start":"19:00.570 ","End":"19:02.895","Text":"Alpha is equal to,"},{"Start":"19:02.895 ","End":"19:08.340","Text":"its angles on a straight line is equal to 180."},{"Start":"19:08.340 ","End":"19:14.685","Text":"Alpha is equal to Pi minus Alpha 2 minus this angle over here."},{"Start":"19:14.685 ","End":"19:21.000","Text":"Our upper bound is just going to be Pi minus Alpha 2."},{"Start":"19:21.000 ","End":"19:24.930","Text":"Now when we do this integration,"},{"Start":"19:24.930 ","End":"19:34.445","Text":"what we\u0027re going to have is Mu_0I divided by 4Pi a and then the"},{"Start":"19:34.445 ","End":"19:40.505","Text":"integral of sine Alpha is equal to negative cosine of Alpha"},{"Start":"19:40.505 ","End":"19:47.450","Text":"and then our bounds are from Alpha 1 until Pi minus Alpha 2."},{"Start":"19:47.450 ","End":"19:49.970","Text":"Now let\u0027s plug that in."},{"Start":"19:49.970 ","End":"19:59.175","Text":"What we\u0027ll get is that this is equal to Mu_0I divided by 4Pi a,"},{"Start":"19:59.175 ","End":"20:04.310","Text":"and then what we have is that this is multiplied by negative."},{"Start":"20:04.310 ","End":"20:08.390","Text":"Then we have cosine of Pi minus Alpha 2,"},{"Start":"20:08.390 ","End":"20:18.625","Text":"which is the same as just negative cosine of Alpha 2."},{"Start":"20:18.625 ","End":"20:24.535","Text":"Then we have cosine minus cosine of Alpha 1,"},{"Start":"20:24.535 ","End":"20:28.940","Text":"so minus cosine of Alpha 1."},{"Start":"20:28.940 ","End":"20:34.760","Text":"Then the negative over here all cancels out."},{"Start":"20:34.760 ","End":"20:43.340","Text":"What we simply get is that the B field is equal to Mu_0I divided by 4Pi a"},{"Start":"20:43.340 ","End":"20:48.230","Text":"multiplied by cosine of"},{"Start":"20:48.230 ","End":"20:55.430","Text":"Alpha 2 plus cosine of Alpha 1 and in the question,"},{"Start":"20:55.430 ","End":"20:58.748","Text":"we were asked to prove that Mu_0I divided by 4Pi"},{"Start":"20:58.748 ","End":"21:04.930","Text":"a cosine of Alpha 1 plus cosine of Alpha 2."},{"Start":"21:04.970 ","End":"21:08.185","Text":"There we approved it."},{"Start":"21:08.185 ","End":"21:13.460","Text":"What I\u0027m going to do now is I\u0027m going to wrap everything out and I want to"},{"Start":"21:13.460 ","End":"21:19.910","Text":"look at a case where this point over here,"},{"Start":"21:19.910 ","End":"21:23.870","Text":"where we\u0027re measuring the magnetic field is located right at"},{"Start":"21:23.870 ","End":"21:28.875","Text":"the center of the wire or above the center of the wire."},{"Start":"21:28.875 ","End":"21:35.785","Text":"Let\u0027s imagine that this wire is of length l,"},{"Start":"21:35.785 ","End":"21:42.730","Text":"and if this point over here is right in the center of the wire,"},{"Start":"21:42.730 ","End":"21:48.910","Text":"so that would mean that Alpha 1 would be equal to Alpha 2."},{"Start":"21:48.910 ","End":"21:54.460","Text":"In that case, we will get that our magnetic field is simply"},{"Start":"21:54.460 ","End":"22:02.430","Text":"equal to 2 cosine of Alpha 1."},{"Start":"22:02.430 ","End":"22:04.620","Text":"Let\u0027s just say that this is all equal to Alpha,"},{"Start":"22:04.620 ","End":"22:07.920","Text":"so 2 cosine of Alpha divided by the 4,"},{"Start":"22:07.920 ","End":"22:13.590","Text":"so what we\u0027d have is Mu_0 I multiplied by 2 divided by 4,"},{"Start":"22:13.590 ","End":"22:20.420","Text":"so divided by 2 Pi a cosine of Alpha."},{"Start":"22:21.390 ","End":"22:28.105","Text":"Now what I want to do is I want to write the magnetic fields with respect"},{"Start":"22:28.105 ","End":"22:31.480","Text":"to my length a and my length l that I\u0027m"},{"Start":"22:31.480 ","End":"22:35.990","Text":"given rather than with respect to the angle Alpha."},{"Start":"22:36.540 ","End":"22:42.730","Text":"Let\u0027s take a look. If my magnetic field is directly above the center of the wire,"},{"Start":"22:42.730 ","End":"22:50.545","Text":"that means that this length over here will be l divided by 2."},{"Start":"22:50.545 ","End":"22:54.700","Text":"Then I can see that I again have a right angle triangle,"},{"Start":"22:54.700 ","End":"22:59.890","Text":"so my hypotenuse over here is going to be equal to"},{"Start":"22:59.890 ","End":"23:07.639","Text":"the square root of l/2 squared plus a squared."},{"Start":"23:08.310 ","End":"23:12.070","Text":"Now I want to know what cosine of Alpha is,"},{"Start":"23:12.070 ","End":"23:15.760","Text":"where this angle over here we said is Alpha."},{"Start":"23:15.760 ","End":"23:19.570","Text":"Cosine of Alpha is equal to,"},{"Start":"23:19.570 ","End":"23:25.045","Text":"from SOHCAHTOA, cosine is adjacent over hypotenuse."},{"Start":"23:25.045 ","End":"23:29.440","Text":"My adjacent side to the angle is l/2,"},{"Start":"23:29.440 ","End":"23:38.000","Text":"and my hypotenuse is the square root of l divided by 2 squared plus a squared."},{"Start":"23:39.330 ","End":"23:46.225","Text":"Therefore, if we plug this in to my equation for the b field,"},{"Start":"23:46.225 ","End":"23:53.125","Text":"I\u0027ll have Mu_0 I multiplied by l divided by 2."},{"Start":"23:53.125 ","End":"24:02.525","Text":"Then the 2 from here and the 2 from here will multiply together to give me 4 Pi a."},{"Start":"24:02.525 ","End":"24:07.155","Text":"Then all of this will be multiplied by 1 divided by"},{"Start":"24:07.155 ","End":"24:13.930","Text":"the square root of l/2 squared plus a squared."},{"Start":"24:15.750 ","End":"24:19.705","Text":"This is how we calculate the magnetic field"},{"Start":"24:19.705 ","End":"24:23.530","Text":"when we\u0027re looking at a point right in the center."},{"Start":"24:23.530 ","End":"24:32.485","Text":"Let\u0027s say we have a square loop wire with a current going like so."},{"Start":"24:32.485 ","End":"24:38.095","Text":"If we wanted to measure the magnetic field at some point over here in the middle,"},{"Start":"24:38.095 ","End":"24:45.295","Text":"we could calculate the magnetic field from 1 of the wires."},{"Start":"24:45.295 ","End":"24:51.595","Text":"Let\u0027s say that this length is a and this length is also a,"},{"Start":"24:51.595 ","End":"24:54.565","Text":"so this length over here will be a/2."},{"Start":"24:54.565 ","End":"24:57.580","Text":"Then we just calculate it in the same way that we just did,"},{"Start":"24:57.580 ","End":"25:03.025","Text":"and then all we have to do is we multiply it by 4 because there\u0027s 4 sides."},{"Start":"25:03.025 ","End":"25:06.205","Text":"Similarly, if we had any other shape,"},{"Start":"25:06.205 ","End":"25:10.540","Text":"such as a regular hexagon,"},{"Start":"25:10.540 ","End":"25:13.810","Text":"and we\u0027re calculating the magnetic field over here."},{"Start":"25:13.810 ","End":"25:20.260","Text":"Again, we would calculate the magnetic fields due to 1 of the wires,"},{"Start":"25:20.260 ","End":"25:28.254","Text":"and then this time we would multiply it by 6 because there\u0027s 6 sides to a hexagon."},{"Start":"25:28.254 ","End":"25:35.170","Text":"One last example that I want to speak about is what if a wire is infinitely long?"},{"Start":"25:35.170 ","End":"25:37.885","Text":"If our wire is infinitely long,"},{"Start":"25:37.885 ","End":"25:42.100","Text":"that means that the length of wire l is much"},{"Start":"25:42.100 ","End":"25:47.140","Text":"greater than the distance at which we\u0027re measuring the magnetic field,"},{"Start":"25:47.140 ","End":"25:49.300","Text":"so the distance away from the wire."},{"Start":"25:49.300 ","End":"25:51.225","Text":"Here it was a."},{"Start":"25:51.225 ","End":"25:54.060","Text":"So l is much greater than a,"},{"Start":"25:54.060 ","End":"25:58.170","Text":"and that corresponds to an infinite wire."},{"Start":"25:58.170 ","End":"26:01.880","Text":"Then what we would have is that,"},{"Start":"26:01.880 ","End":"26:06.595","Text":"if we have infinity plus a small number,"},{"Start":"26:06.595 ","End":"26:09.280","Text":"then it\u0027s the same as just having infinity."},{"Start":"26:09.280 ","End":"26:14.920","Text":"What we would be left with here is that our B field would be equal"},{"Start":"26:14.920 ","End":"26:22.900","Text":"to Mu_0 IL divided by 4 Pi a."},{"Start":"26:22.900 ","End":"26:26.740","Text":"Then here we would have 1 divided by,"},{"Start":"26:26.740 ","End":"26:29.680","Text":"so infinity plus a is just the same as infinity,"},{"Start":"26:29.680 ","End":"26:32.980","Text":"so we would have infinity divided by 2 squared,"},{"Start":"26:32.980 ","End":"26:35.020","Text":"and then we take the square root of that."},{"Start":"26:35.020 ","End":"26:39.565","Text":"That\u0027s like just having infinity divided by 2."},{"Start":"26:39.565 ","End":"26:43.165","Text":"This L and this L over here will cancel out,"},{"Start":"26:43.165 ","End":"26:47.470","Text":"and our 2 will go up to the numerator,"},{"Start":"26:47.470 ","End":"26:50.860","Text":"so we can just rub that out and write that over here."},{"Start":"26:50.860 ","End":"26:54.310","Text":"Then this 2 and the 4 will cancel out."},{"Start":"26:54.310 ","End":"26:59.410","Text":"So what will be left with is that the B field for an infinitely long wire is"},{"Start":"26:59.410 ","End":"27:05.720","Text":"equal to Mu_0 I divided by 2Pi a."},{"Start":"27:07.980 ","End":"27:11.094","Text":"If we want to write this in an equation,"},{"Start":"27:11.094 ","End":"27:14.125","Text":"a is just the distance away from the wire."},{"Start":"27:14.125 ","End":"27:18.730","Text":"In general, when we\u0027re dealing with equations,"},{"Start":"27:18.730 ","End":"27:22.030","Text":"we call that r. R is the radius,"},{"Start":"27:22.030 ","End":"27:24.070","Text":"which is the distance away from the wire."},{"Start":"27:24.070 ","End":"27:32.390","Text":"Let\u0027s just erase this and replace it with r. Then what we\u0027re left with is an equation."},{"Start":"27:32.430 ","End":"27:36.070","Text":"This is an equation to write in your equation sheets,"},{"Start":"27:36.070 ","End":"27:40.540","Text":"and this is the B field for an infinitely long wire."},{"Start":"27:40.540 ","End":"27:42.550","Text":"This is, of course, the magnitude,"},{"Start":"27:42.550 ","End":"27:44.605","Text":"and if you want to look at the direction,"},{"Start":"27:44.605 ","End":"27:50.395","Text":"so let\u0027s say that this is our infinitely long wire,"},{"Start":"27:50.395 ","End":"27:53.470","Text":"of course, we\u0027re just going to use the right-hand rule"},{"Start":"27:53.470 ","End":"27:56.410","Text":"depending on what direction the current is traveling in."},{"Start":"27:56.410 ","End":"27:58.870","Text":"Let\u0027s say it\u0027s traveling like so,"},{"Start":"27:58.870 ","End":"28:02.545","Text":"so what we can do is we can point our thumb in the direction of the current,"},{"Start":"28:02.545 ","End":"28:05.605","Text":"and then our 4 fingers will curl like this."},{"Start":"28:05.605 ","End":"28:09.340","Text":"If we\u0027re looking at a point above the wire,"},{"Start":"28:09.340 ","End":"28:11.980","Text":"the magnetic field is coming out towards us."},{"Start":"28:11.980 ","End":"28:17.065","Text":"Of course, the magnetic field forms concentric circles around the wire,"},{"Start":"28:17.065 ","End":"28:19.270","Text":"so over here below the wire,"},{"Start":"28:19.270 ","End":"28:22.765","Text":"the magnetic field will be going in,"},{"Start":"28:22.765 ","End":"28:26.050","Text":"and this is a distance r away."},{"Start":"28:26.050 ","End":"28:29.060","Text":"That\u0027s the end of this lesson."}],"ID":21396},{"Watched":false,"Name":"Exercise 2","Duration":"14m 33s","ChapterTopicVideoID":21321,"CourseChapterTopicPlaylistID":99469,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.950","Text":"Hello. In this lesson,"},{"Start":"00:01.950 ","End":"00:04.500","Text":"we\u0027re going to be answering the following question:"},{"Start":"00:04.500 ","End":"00:08.580","Text":"Calculate the magnetic field at point y below"},{"Start":"00:08.580 ","End":"00:11.625","Text":"the middle of a current carrying wire of length"},{"Start":"00:11.625 ","End":"00:15.960","Text":"L. The current is flowing in the x direction."},{"Start":"00:15.960 ","End":"00:26.535","Text":"First of all, let\u0027s mark that this is the x direction and that our wire"},{"Start":"00:26.535 ","End":"00:30.060","Text":"over here is of length"},{"Start":"00:30.060 ","End":"00:38.735","Text":"L. We\u0027re being told that our current is flowing in the x direction,"},{"Start":"00:38.735 ","End":"00:41.995","Text":"like so, draw it in green."},{"Start":"00:41.995 ","End":"00:45.590","Text":"This is the direction that our current is flowing."},{"Start":"00:45.590 ","End":"00:50.270","Text":"We\u0027re being asked to calculate the magnetic field at"},{"Start":"00:50.270 ","End":"00:55.150","Text":"distance y from the center of the wire."},{"Start":"00:55.150 ","End":"00:58.200","Text":"Let\u0027s say that the center is around about here."},{"Start":"00:58.200 ","End":"01:02.140","Text":"So this is the origin."},{"Start":"01:03.440 ","End":"01:10.730","Text":"Let\u0027s say that this is the point where we\u0027re trying to calculate our B field,"},{"Start":"01:10.730 ","End":"01:16.309","Text":"and this distance is a distance y from the center."},{"Start":"01:16.309 ","End":"01:21.550","Text":"What we might as well do to make this easier without any minus signs,"},{"Start":"01:21.550 ","End":"01:29.200","Text":"so we can say that the downwards direction is the y direction."},{"Start":"01:30.200 ","End":"01:33.725","Text":"In order to calculate this,"},{"Start":"01:33.725 ","End":"01:37.130","Text":"what I\u0027m going to do is I\u0027m going to use Biot-Savart\u0027s law."},{"Start":"01:37.130 ","End":"01:40.820","Text":"So let\u0027s just write out the equation for Biot-Savart\u0027s law."},{"Start":"01:40.820 ","End":"01:44.750","Text":"It says that dB is equal to,"},{"Start":"01:44.750 ","End":"01:49.820","Text":"so this is the magnetic field at this point due to a small slice of this wire,"},{"Start":"01:49.820 ","End":"01:57.910","Text":"is equal to Mu_0 divided by 4 Pi multiplied by"},{"Start":"01:58.010 ","End":"02:03.965","Text":"Idl cross multiplied by the I vector"},{"Start":"02:03.965 ","End":"02:10.990","Text":"divided by the magnitude of the r vector cubed."},{"Start":"02:10.990 ","End":"02:18.360","Text":"What we\u0027re going to do is we\u0027re going to choose some point over here."},{"Start":"02:18.610 ","End":"02:27.840","Text":"This is going to be some small slice of wire of infinitely small length dl."},{"Start":"02:29.830 ","End":"02:33.440","Text":"If this we said was the origin,"},{"Start":"02:33.440 ","End":"02:35.645","Text":"so it\u0027s going to be,"},{"Start":"02:35.645 ","End":"02:38.030","Text":"because it\u0027s along the x-axis,"},{"Start":"02:38.030 ","End":"02:42.989","Text":"a distance of x from the origin."},{"Start":"02:43.660 ","End":"02:48.770","Text":"If we remember, our r vector,"},{"Start":"02:48.770 ","End":"02:50.060","Text":"which I\u0027ll draw in gray,"},{"Start":"02:50.060 ","End":"02:54.050","Text":"is the vector from the slice to"},{"Start":"02:54.050 ","End":"02:59.490","Text":"the point where we\u0027re trying to calculate our magnetic field."},{"Start":"03:00.830 ","End":"03:09.470","Text":"Let\u0027s start calculating what each of these vectors is with regards to our problem."},{"Start":"03:09.470 ","End":"03:15.210","Text":"Our dl vector is this over here."},{"Start":"03:15.470 ","End":"03:18.990","Text":"Our dl, because it\u0027s in the x direction,"},{"Start":"03:18.990 ","End":"03:21.780","Text":"it\u0027s equal to a size of dx,"},{"Start":"03:21.780 ","End":"03:24.600","Text":"so a small change in the x direction,"},{"Start":"03:24.600 ","End":"03:30.250","Text":"and its direction is of course in the x direction."},{"Start":"03:30.250 ","End":"03:36.320","Text":"A dl vector is just a change in x in the x direction."},{"Start":"03:36.320 ","End":"03:42.340","Text":"Now the next thing that we want to calculate is our r vector over here in gray."},{"Start":"03:42.340 ","End":"03:44.525","Text":"Because this time,"},{"Start":"03:44.525 ","End":"03:47.455","Text":"I said that my origin was over here,"},{"Start":"03:47.455 ","End":"03:54.155","Text":"and that means that my origin isn\u0027t at the point where I\u0027m measuring my magnetic field,"},{"Start":"03:54.155 ","End":"03:56.930","Text":"which means that I have to add in"},{"Start":"03:56.930 ","End":"04:02.070","Text":"a few stages in order to calculate what my vector r is equal to."},{"Start":"04:02.450 ","End":"04:06.675","Text":"I\u0027m going to do this stage by stage."},{"Start":"04:06.675 ","End":"04:11.119","Text":"This will give a general method for how to solve these types of questions,"},{"Start":"04:11.119 ","End":"04:16.165","Text":"which we\u0027ll, of course, need to use when we have more complicated examples."},{"Start":"04:16.165 ","End":"04:24.065","Text":"First of all, I\u0027m going to define the vector from the origin until our slice itself."},{"Start":"04:24.065 ","End":"04:27.835","Text":"From the origin until our slice itself is this,"},{"Start":"04:27.835 ","End":"04:31.624","Text":"and let\u0027s call this our r tag vector."},{"Start":"04:31.624 ","End":"04:34.670","Text":"Then the next vector I\u0027m going to define is from"},{"Start":"04:34.670 ","End":"04:39.770","Text":"the origin until the point where I\u0027m measuring my magnetic field,"},{"Start":"04:39.770 ","End":"04:44.225","Text":"so from the origin until the point where I\u0027m measuring my magnetic field,"},{"Start":"04:44.225 ","End":"04:49.620","Text":"and I\u0027m going to call this my r_B vector."},{"Start":"04:50.030 ","End":"04:53.940","Text":"What is my r tag vector?"},{"Start":"04:53.940 ","End":"04:58.549","Text":"My r tag vector is from the origin until my slice."},{"Start":"04:58.549 ","End":"05:04.235","Text":"We saw that my slice was a distance x away from the origin,"},{"Start":"05:04.235 ","End":"05:12.420","Text":"and this is going in the x direction."},{"Start":"05:13.330 ","End":"05:18.220","Text":"Then my r_B vector,"},{"Start":"05:18.220 ","End":"05:20.660","Text":"which is the vector from"},{"Start":"05:20.660 ","End":"05:24.960","Text":"the origin until the point where I\u0027m measuring my magnetic field."},{"Start":"05:25.760 ","End":"05:31.085","Text":"I\u0027m measuring the magnetic field at distance y away from the origin,"},{"Start":"05:31.085 ","End":"05:34.700","Text":"and it\u0027s in the positive y direction because we define"},{"Start":"05:34.700 ","End":"05:38.537","Text":"the positive y direction going downwards."},{"Start":"05:38.537 ","End":"05:41.574","Text":"So y in the y direction."},{"Start":"05:41.574 ","End":"05:45.615","Text":"Therefore, what can we say?"},{"Start":"05:45.615 ","End":"05:49.830","Text":"From here we can say that my r vector,"},{"Start":"05:49.830 ","End":"05:54.635","Text":"so from my slice until the point where I\u0027m measuring the magnetic field,"},{"Start":"05:54.635 ","End":"05:58.447","Text":"so I have to go down my r tag vector,"},{"Start":"05:58.447 ","End":"06:00.935","Text":"so I\u0027m going in the negative direction."},{"Start":"06:00.935 ","End":"06:06.050","Text":"I\u0027m going in the opposite direction to my r tag vector,"},{"Start":"06:06.050 ","End":"06:09.065","Text":"so I have negative r tag vector."},{"Start":"06:09.065 ","End":"06:18.435","Text":"Then I\u0027m going up my r_B vector in the positive direction of my r_B vector."},{"Start":"06:18.435 ","End":"06:21.090","Text":"Now I can plug in my values."},{"Start":"06:21.090 ","End":"06:31.290","Text":"I have negative x in the x direction plus y in the y direction."},{"Start":"06:31.820 ","End":"06:38.150","Text":"Now if I want to put this in vector format,"},{"Start":"06:38.150 ","End":"06:42.955","Text":"I can say that I have negative x in the x direction,"},{"Start":"06:42.955 ","End":"06:46.175","Text":"I have y in the y direction, and of course,"},{"Start":"06:46.175 ","End":"06:49.550","Text":"0 in the z direction."},{"Start":"06:49.550 ","End":"06:51.620","Text":"I didn\u0027t have to add in the z direction,"},{"Start":"06:51.620 ","End":"06:54.630","Text":"but we can do that anyway."},{"Start":"06:54.630 ","End":"06:56.285","Text":"Now in my equation,"},{"Start":"06:56.285 ","End":"07:00.635","Text":"I have my dl vector cross-product with my r vector."},{"Start":"07:00.635 ","End":"07:02.855","Text":"Let\u0027s calculate that up here."},{"Start":"07:02.855 ","End":"07:07.280","Text":"I have my dl vector cross-product with my r vector."},{"Start":"07:07.280 ","End":"07:17.300","Text":"What I have is dx in the x-direction cross multiplied"},{"Start":"07:17.300 ","End":"07:27.421","Text":"with negative x in the x direction plus y in the y direction."},{"Start":"07:27.421 ","End":"07:31.720","Text":"I can also say that this is equal to,"},{"Start":"07:31.720 ","End":"07:37.705","Text":"I have dx in the x direction 0, 0."},{"Start":"07:37.705 ","End":"07:40.360","Text":"That\u0027s this vector over here."},{"Start":"07:40.360 ","End":"07:43.990","Text":"This is equal to dx 0,"},{"Start":"07:43.990 ","End":"07:53.425","Text":"0 cross multiplied with negative xy 0,"},{"Start":"07:53.425 ","End":"07:57.440","Text":"with this vector over here."},{"Start":"07:58.230 ","End":"08:03.730","Text":"Now what we\u0027re going to do is we\u0027re going to do this cross multiplication."},{"Start":"08:03.730 ","End":"08:05.560","Text":"In order to do this,"},{"Start":"08:05.560 ","End":"08:09.490","Text":"I\u0027m going to cover the top row."},{"Start":"08:09.490 ","End":"08:12.370","Text":"So dx and negative x."},{"Start":"08:12.370 ","End":"08:14.815","Text":"Then I have 0 times 0,"},{"Start":"08:14.815 ","End":"08:18.550","Text":"which is 0 minus 0 times y, which is 0."},{"Start":"08:18.550 ","End":"08:21.385","Text":"I put a 0 in over here."},{"Start":"08:21.385 ","End":"08:25.615","Text":"Then I cover the middle rows, so my 0 and my y."},{"Start":"08:25.615 ","End":"08:28.750","Text":"Then I have 0 multiplied by negative x,"},{"Start":"08:28.750 ","End":"08:34.540","Text":"which is 0, minus dx multiplied by 0, which is 0."},{"Start":"08:34.540 ","End":"08:37.210","Text":"Then last but not least,"},{"Start":"08:37.210 ","End":"08:38.830","Text":"I cover my bottom row."},{"Start":"08:38.830 ","End":"08:40.855","Text":"So 0 and 0, I\u0027m covering."},{"Start":"08:40.855 ","End":"08:47.710","Text":"Now I have dx multiplied by y minus 0 multiplied by"},{"Start":"08:47.710 ","End":"08:55.525","Text":"negative x. I\u0027m left with dx multiplied by y."},{"Start":"08:55.525 ","End":"08:58.450","Text":"In other words, I can just write this as 0,"},{"Start":"08:58.450 ","End":"09:02.750","Text":"0 and then ydx."},{"Start":"09:04.860 ","End":"09:14.120","Text":"In other words, I have ydx in the z direction."},{"Start":"09:14.130 ","End":"09:20.290","Text":"Now what about the magnitude of my r vector?"},{"Start":"09:20.290 ","End":"09:23.290","Text":"I\u0027m just taking the magnitude of this."},{"Start":"09:23.290 ","End":"09:28.360","Text":"It\u0027s simply going to be equal to the square root of negative x^2,"},{"Start":"09:28.360 ","End":"09:32.950","Text":"which is the same as x^2 plus y^2."},{"Start":"09:32.950 ","End":"09:35.995","Text":"So classic Pythagoras,"},{"Start":"09:35.995 ","End":"09:43.900","Text":"or I can just write this as x^2 plus y^2 to the power of 1/2,"},{"Start":"09:43.900 ","End":"09:46.400","Text":"means the same thing."},{"Start":"09:46.740 ","End":"09:51.790","Text":"Now I get that my dB is equal to Mu 0 divided by"},{"Start":"09:51.790 ","End":"09:58.795","Text":"4 Pi multiplied by I multiplied by dL cross r,"},{"Start":"09:58.795 ","End":"10:02.500","Text":"ydx and the z direction, that\u0027s what I got,"},{"Start":"10:02.500 ","End":"10:08.470","Text":"divided by the magnitude of my r vector cubed."},{"Start":"10:08.470 ","End":"10:14.500","Text":"The magnitude was x^2 plus y^2 to the power of 1/2 and then because it\u0027s cubed."},{"Start":"10:14.500 ","End":"10:18.050","Text":"So I add in the cubed over here."},{"Start":"10:18.150 ","End":"10:23.995","Text":"Here\u0027s specifically I worked out in one move."},{"Start":"10:23.995 ","End":"10:31.000","Text":"I worked out both the magnitude and the direction in the z direction of my dB."},{"Start":"10:31.000 ","End":"10:34.390","Text":"But of course, I could have just worked out the magnitude and"},{"Start":"10:34.390 ","End":"10:37.510","Text":"then worked out the direction by"},{"Start":"10:37.510 ","End":"10:40.690","Text":"using the right-hand rule for this point over here"},{"Start":"10:40.690 ","End":"10:43.990","Text":"and we still would have gotten that it\u0027s going in the z direction."},{"Start":"10:43.990 ","End":"10:50.470","Text":"Notice it\u0027s in the positive z direction because we now define the y-axis going downwards,"},{"Start":"10:50.470 ","End":"10:56.470","Text":"which means that the z direction is going into the page."},{"Start":"10:56.470 ","End":"11:00.670","Text":"This would be, let\u0027s just draw it."},{"Start":"11:00.670 ","End":"11:04.210","Text":"It\u0027s going in to the page."},{"Start":"11:04.210 ","End":"11:09.205","Text":"Now of course, to find the total magnetic field at this point."},{"Start":"11:09.205 ","End":"11:14.230","Text":"All I have to do is to integrate. Let\u0027s do that."},{"Start":"11:14.230 ","End":"11:17.125","Text":"First, let\u0027s work out our integrating bounds."},{"Start":"11:17.125 ","End":"11:22.130","Text":"We define the origin as right here in the center."},{"Start":"11:22.650 ","End":"11:26.155","Text":"We\u0027re looking at this point over here,"},{"Start":"11:26.155 ","End":"11:29.290","Text":"where this is right in the middle."},{"Start":"11:29.290 ","End":"11:37.270","Text":"That means that this length up until here is L over 2 and similarly with this other half."},{"Start":"11:37.270 ","End":"11:41.350","Text":"That means that this point over here is L over"},{"Start":"11:41.350 ","End":"11:46.285","Text":"2 away from the origin in the negative x-direction."},{"Start":"11:46.285 ","End":"11:50.230","Text":"This is negative L over 2 and that this point over"},{"Start":"11:50.230 ","End":"11:54.820","Text":"here is L over 2 away from the origin in the positive x-direction."},{"Start":"11:54.820 ","End":"11:56.815","Text":"This is positive L over 2."},{"Start":"11:56.815 ","End":"12:04.160","Text":"That means we\u0027re integrating from negative L over 2 until positive L over 2."},{"Start":"12:04.470 ","End":"12:07.615","Text":"I\u0027m not going to do the integral."},{"Start":"12:07.615 ","End":"12:11.305","Text":"If you want, please pause the video and check that you get this answer."},{"Start":"12:11.305 ","End":"12:16.000","Text":"But the answer that we get for the magnetic field at this point is equal to"},{"Start":"12:16.000 ","End":"12:22.555","Text":"Mu_0 divided by 4Pi y multiplied by IL divided by,"},{"Start":"12:22.555 ","End":"12:25.810","Text":"in brackets, we have L over 2 squared plus"},{"Start":"12:25.810 ","End":"12:29.260","Text":"y^2 and then we take the square root of all of that."},{"Start":"12:29.260 ","End":"12:33.560","Text":"All of this is in the z direction."},{"Start":"12:33.870 ","End":"12:36.310","Text":"Now, a quick little note,"},{"Start":"12:36.310 ","End":"12:42.835","Text":"let\u0027s say I wanted to calculate my point of the magnetic field over here."},{"Start":"12:42.835 ","End":"12:47.440","Text":"At the same distance away from the wire, however,"},{"Start":"12:47.440 ","End":"12:54.910","Text":"this time it\u0027s located a distance of d away from the edge of the wire."},{"Start":"12:54.910 ","End":"13:03.340","Text":"What I can either do is I can rewrite and redefine these vectors of r tag and rB."},{"Start":"13:03.340 ","End":"13:09.115","Text":"However, this is going to be a slightly longer method for solving this question."},{"Start":"13:09.115 ","End":"13:16.555","Text":"All I really have to do is I can just move my origin from here to this point over here,"},{"Start":"13:16.555 ","End":"13:20.410","Text":"d. So d away from the edge."},{"Start":"13:20.410 ","End":"13:22.495","Text":"This is my origin."},{"Start":"13:22.495 ","End":"13:27.040","Text":"As we said, this is the x-direction and just like before,"},{"Start":"13:27.040 ","End":"13:29.155","Text":"this is the y-direction."},{"Start":"13:29.155 ","End":"13:31.750","Text":"Then all I have to do,"},{"Start":"13:31.750 ","End":"13:33.355","Text":"I keep everything the same,"},{"Start":"13:33.355 ","End":"13:39.880","Text":"but all I\u0027m doing is I\u0027m just changing the bounds of my integral."},{"Start":"13:39.880 ","End":"13:45.760","Text":"To this point, I\u0027m integrating from this point over here,"},{"Start":"13:45.760 ","End":"13:52.675","Text":"which is now negative d because it\u0027s d away from the origin in the negative x-direction."},{"Start":"13:52.675 ","End":"13:57.010","Text":"I\u0027m integrating from negative d in my lower bound."},{"Start":"13:57.010 ","End":"14:00.670","Text":"Then from over here, my edge."},{"Start":"14:00.670 ","End":"14:02.950","Text":"What is the value over here at the edge?"},{"Start":"14:02.950 ","End":"14:05.950","Text":"The value over here is the total length of the wire,"},{"Start":"14:05.950 ","End":"14:11.560","Text":"which is L minus where my origin is."},{"Start":"14:11.560 ","End":"14:14.230","Text":"My origin is d away from the edge of the wire."},{"Start":"14:14.230 ","End":"14:20.230","Text":"So L minus d. L minus this distance over here,"},{"Start":"14:20.230 ","End":"14:25.000","Text":"which is d. Then I just plug in L minus d over here and then"},{"Start":"14:25.000 ","End":"14:30.505","Text":"I just go over the integration and I will get my answer."},{"Start":"14:30.505 ","End":"14:33.470","Text":"That\u0027s the end of this lesson."}],"ID":21397},{"Watched":false,"Name":"Exercise 3","Duration":"27m 59s","ChapterTopicVideoID":21322,"CourseChapterTopicPlaylistID":99469,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.860","Text":"Hello. In this lesson,"},{"Start":"00:01.860 ","End":"00:05.250","Text":"we\u0027re going to be answering the following question: calculate the"},{"Start":"00:05.250 ","End":"00:09.300","Text":"magnetic fields a distance z away from the center of"},{"Start":"00:09.300 ","End":"00:13.050","Text":"a current carrying loop of radius R. The current is"},{"Start":"00:13.050 ","End":"00:17.489","Text":"traveling in an anticlockwise direction about the origin."},{"Start":"00:17.489 ","End":"00:24.795","Text":"Here we have the center of the current carrying loop."},{"Start":"00:24.795 ","End":"00:30.375","Text":"The axis is in the z-direction and it goes right through the center of the loop."},{"Start":"00:30.375 ","End":"00:34.170","Text":"Our loop we\u0027re being told is of"},{"Start":"00:34.170 ","End":"00:40.515","Text":"radius R and the current is traveling in an anticlockwise direction about the origin."},{"Start":"00:40.515 ","End":"00:44.915","Text":"That means that the current is traveling like so."},{"Start":"00:44.915 ","End":"00:47.345","Text":"We\u0027re looking at distance z above,"},{"Start":"00:47.345 ","End":"00:56.200","Text":"so let\u0027s look at this point where this point is a distance z above the origin."},{"Start":"00:56.480 ","End":"01:02.930","Text":"In order to calculate the magnetic field at this point due to this current carrying loop,"},{"Start":"01:02.930 ","End":"01:05.975","Text":"what we\u0027re going to do is we\u0027re going to split the loop up"},{"Start":"01:05.975 ","End":"01:09.593","Text":"into lots of different pieces,"},{"Start":"01:09.593 ","End":"01:14.629","Text":"where each piece is a small length."},{"Start":"01:14.629 ","End":"01:19.160","Text":"Where because we\u0027re dealing with a circle, so polar coordinates,"},{"Start":"01:19.160 ","End":"01:28.065","Text":"this length over here dl is equal to Rd Theta."},{"Start":"01:28.065 ","End":"01:30.718","Text":"Where R is a constant radius,"},{"Start":"01:30.718 ","End":"01:32.480","Text":"because we\u0027re dealing with a loop."},{"Start":"01:32.480 ","End":"01:36.750","Text":"It\u0027s capital Rd Theta."},{"Start":"01:37.670 ","End":"01:42.080","Text":"Now, in order to calculate the magnetic field at this point,"},{"Start":"01:42.080 ","End":"01:44.915","Text":"there are 2 ways that we could calculate this."},{"Start":"01:44.915 ","End":"01:49.340","Text":"The first is by using this equation exactly as it\u0027s written,"},{"Start":"01:49.340 ","End":"01:51.214","Text":"so by calculating the vectors,"},{"Start":"01:51.214 ","End":"01:54.020","Text":"working out the cross product between the 2,"},{"Start":"01:54.020 ","End":"01:57.615","Text":"and just doing this calculation as it\u0027s written."},{"Start":"01:57.615 ","End":"02:00.065","Text":"The second method is by working out"},{"Start":"02:00.065 ","End":"02:05.120","Text":"the magnitude of the magnetic field and then working out its direction."},{"Start":"02:05.120 ","End":"02:08.480","Text":"The first thing that we\u0027re going to do or"},{"Start":"02:08.480 ","End":"02:12.350","Text":"the first method that we\u0027re going to use is the vector method."},{"Start":"02:12.350 ","End":"02:15.950","Text":"Just solving it as we see it written in the equation."},{"Start":"02:15.950 ","End":"02:18.710","Text":"Then afterwards, we\u0027ll solve it via the"},{"Start":"02:18.710 ","End":"02:22.565","Text":"calculating the magnitude and then the direction."},{"Start":"02:22.565 ","End":"02:31.325","Text":"We\u0027ve already dealt with our dl vector and now we want to deal with our r vector."},{"Start":"02:31.325 ","End":"02:32.810","Text":"What is our r vector?"},{"Start":"02:32.810 ","End":"02:38.675","Text":"Our r vector is the vector pointing from the slice."},{"Start":"02:38.675 ","End":"02:40.840","Text":"That means from here,"},{"Start":"02:40.840 ","End":"02:42.823","Text":"from our piece of wire,"},{"Start":"02:42.823 ","End":"02:46.055","Text":"and up until the point where we\u0027re measuring the magnetic field."},{"Start":"02:46.055 ","End":"02:51.810","Text":"Our r vector is this."},{"Start":"02:51.810 ","End":"02:56.165","Text":"From the piece until where we\u0027re measuring the magnetic field."},{"Start":"02:56.165 ","End":"03:00.620","Text":"Now, what we want to do is we want to define our r vector."},{"Start":"03:00.620 ","End":"03:05.075","Text":"We want to know what our r vector is equal to."},{"Start":"03:05.075 ","End":"03:08.989","Text":"What I\u0027m going to do is I\u0027m going to define another 2 vectors."},{"Start":"03:08.989 ","End":"03:11.540","Text":"Because we can see that my r vector is coming"},{"Start":"03:11.540 ","End":"03:14.915","Text":"from this slice over here rather than the origin,"},{"Start":"03:14.915 ","End":"03:20.825","Text":"I\u0027m going to express my r vector with respect to vectors coming out of the origin."},{"Start":"03:20.825 ","End":"03:24.710","Text":"The first vector I\u0027m going to have is my vector from"},{"Start":"03:24.710 ","End":"03:32.225","Text":"the origin up until the point where I\u0027m measuring my magnetic field."},{"Start":"03:32.225 ","End":"03:36.235","Text":"This vector I\u0027m going to call r_B vector."},{"Start":"03:36.235 ","End":"03:40.330","Text":"It\u0027s the vector from the origin to where I\u0027m measuring my B field."},{"Start":"03:40.330 ","End":"03:47.595","Text":"The next vector is the vector from the origin until my slice of wire."},{"Start":"03:47.595 ","End":"03:55.190","Text":"This is going to be called my r tag vector from the origin until my piece dl over here."},{"Start":"03:55.190 ","End":"03:58.010","Text":"My r vector, as we can see,"},{"Start":"03:58.010 ","End":"04:03.260","Text":"in order to get from this slice over here to my point over here,"},{"Start":"04:03.260 ","End":"04:06.425","Text":"I\u0027m going to go down my r tag vector."},{"Start":"04:06.425 ","End":"04:11.795","Text":"I\u0027m going in the opposite direction to my r tag vector so I have negative r tag vector."},{"Start":"04:11.795 ","End":"04:16.115","Text":"Then I\u0027m going up my r_B vector to reach this point,"},{"Start":"04:16.115 ","End":"04:20.140","Text":"so plus r_B vector."},{"Start":"04:20.230 ","End":"04:26.030","Text":"Now let\u0027s write what our r tag vector is."},{"Start":"04:26.030 ","End":"04:29.000","Text":"Our r tag vector,"},{"Start":"04:29.000 ","End":"04:30.935","Text":"as we can see,"},{"Start":"04:30.935 ","End":"04:37.085","Text":"we\u0027re going out from the origin until our loop,"},{"Start":"04:37.085 ","End":"04:42.280","Text":"which is a distance R away from the origin because the loop is a radius R."},{"Start":"04:42.280 ","End":"04:48.283","Text":"So it\u0027s R and it\u0027s R in the radial direction,"},{"Start":"04:48.283 ","End":"04:50.495","Text":"in the r-hat direction,"},{"Start":"04:50.495 ","End":"04:52.460","Text":"or in other words,"},{"Start":"04:52.460 ","End":"04:59.290","Text":"we can also say that this is equal to (R, 0, 0)."},{"Start":"04:59.290 ","End":"05:03.140","Text":"Then let\u0027s write our vector from"},{"Start":"05:03.140 ","End":"05:08.255","Text":"the origin to where we\u0027re measuring the magnetic field, so r_B vector."},{"Start":"05:08.255 ","End":"05:12.170","Text":"Here we\u0027re going to this point over here,"},{"Start":"05:12.170 ","End":"05:18.935","Text":"which is a distance of z up in the z-direction."},{"Start":"05:18.935 ","End":"05:26.350","Text":"We can either write it like that or (0,z,0)."},{"Start":"05:26.350 ","End":"05:31.140","Text":"Therefore, we can say that our r vector is equal"},{"Start":"05:31.140 ","End":"05:36.483","Text":"to negative r tag vector,"},{"Start":"05:36.483 ","End":"05:40.116","Text":"so negative R in the r direction,"},{"Start":"05:40.116 ","End":"05:41.825","Text":"plus our r_B vector,"},{"Start":"05:41.825 ","End":"05:46.460","Text":"so plus z in the z-direction."},{"Start":"05:46.460 ","End":"05:49.100","Text":"Sorry, I made a mistake."},{"Start":"05:49.100 ","End":"05:50.780","Text":"This is z in the z-direction,"},{"Start":"05:50.780 ","End":"05:56.585","Text":"but that means that I have to put it in this position over here for the z position."},{"Start":"05:56.585 ","End":"06:01.868","Text":"Plus z in the z-direction."},{"Start":"06:01.868 ","End":"06:05.715","Text":"Therefore, I can write this as negative R,"},{"Start":"06:05.715 ","End":"06:10.210","Text":"the y nothing changes, and z."},{"Start":"06:10.370 ","End":"06:15.275","Text":"So dl magnitude we know what that is,"},{"Start":"06:15.275 ","End":"06:17.600","Text":"but let\u0027s just write our dl vector."},{"Start":"06:17.600 ","End":"06:25.485","Text":"It\u0027s just Rd Theta in the Theta direction."},{"Start":"06:25.485 ","End":"06:34.470","Text":"Now, what we want to know is what dl cross r is equal to."},{"Start":"06:34.470 ","End":"06:40.920","Text":"Let\u0027s write that. Our dl we have Rd Theta in the Theta direction,"},{"Start":"06:40.920 ","End":"06:47.385","Text":"so that\u0027s 0, then we have Rd Theta in the Theta and then 0 in the z."},{"Start":"06:47.385 ","End":"06:50.250","Text":"Then cross-product with our r vector,"},{"Start":"06:50.250 ","End":"06:52.995","Text":"which is negative R in the r position,"},{"Start":"06:52.995 ","End":"06:55.435","Text":"0 in the Theta position,"},{"Start":"06:55.435 ","End":"06:59.450","Text":"and z in the z position."},{"Start":"06:59.450 ","End":"07:01.670","Text":"To do this,"},{"Start":"07:01.670 ","End":"07:09.070","Text":"we first cover up our 0 and our negative R. Then we have Rd Theta multiplied by z,"},{"Start":"07:09.070 ","End":"07:17.205","Text":"so we have z R d Theta minus 0 times 0."},{"Start":"07:17.205 ","End":"07:19.070","Text":"Now in the next place,"},{"Start":"07:19.070 ","End":"07:21.230","Text":"we cover up our Rd Theta and our 0,"},{"Start":"07:21.230 ","End":"07:22.940","Text":"so we cover up our second row."},{"Start":"07:22.940 ","End":"07:25.910","Text":"Then we have 0 multiplied by negative R,"},{"Start":"07:25.910 ","End":"07:32.180","Text":"which is 0, minus 0 multiplied by z, which is 0."},{"Start":"07:32.180 ","End":"07:34.550","Text":"Then for our final position,"},{"Start":"07:34.550 ","End":"07:36.080","Text":"we cover up our bottom row,"},{"Start":"07:36.080 ","End":"07:38.120","Text":"so we cover up our 0 and our z."},{"Start":"07:38.120 ","End":"07:40.955","Text":"Then we have 0 times 0, which is 0,"},{"Start":"07:40.955 ","End":"07:46.490","Text":"minus Rd Theta multiplied by negative R. We have the"},{"Start":"07:46.490 ","End":"07:53.240","Text":"negative and the negative cross out and then we\u0027re just left with R^2 d Theta."},{"Start":"07:53.240 ","End":"08:00.905","Text":"In total, we\u0027re left with z Rd Theta in"},{"Start":"08:00.905 ","End":"08:10.520","Text":"the radial direction plus R^2 d Theta in the z-direction,"},{"Start":"08:10.520 ","End":"08:13.830","Text":"and we have nothing in the Theta direction."},{"Start":"08:14.240 ","End":"08:21.075","Text":"Now the next thing that we want to know is what is the magnitude of our r vector."},{"Start":"08:21.075 ","End":"08:24.405","Text":"We\u0027re taking the magnitude of this,"},{"Start":"08:24.405 ","End":"08:28.380","Text":"it\u0027s simply going to be equal to the square root of negative R squared,"},{"Start":"08:28.380 ","End":"08:34.990","Text":"which is just R squared plus 0 squared, plus z squared."},{"Start":"08:35.240 ","End":"08:43.965","Text":"This is also the same as R squared plus Z squared to the power of 1.5."},{"Start":"08:43.965 ","End":"08:48.090","Text":"Now, let\u0027s plug in all of this into our equation."},{"Start":"08:48.090 ","End":"08:55.905","Text":"We have that dB is equal to Mu naught divided by 4 Pi."},{"Start":"08:55.905 ","End":"09:00.780","Text":"Then we have I multiplied by dl cross r, which is this,"},{"Start":"09:00.780 ","End":"09:06.240","Text":"so we have I multiplied by ZR d Theta in"},{"Start":"09:06.240 ","End":"09:13.470","Text":"the r hat direction plus R squared d Theta in the z hat direction."},{"Start":"09:13.470 ","End":"09:17.924","Text":"All of this is divided by the magnitude of r cubed."},{"Start":"09:17.924 ","End":"09:23.610","Text":"That\u0027s R squared plus Z squared to the power of half cubed,"},{"Start":"09:23.610 ","End":"09:26.890","Text":"so it\u0027s to the power of 3 over 2."},{"Start":"09:27.290 ","End":"09:34.845","Text":"Now we have the magnetic field over here due to just this slice and that\u0027s equal to that."},{"Start":"09:34.845 ","End":"09:43.590","Text":"What we can see is that we have a magnetic field going in this direction."},{"Start":"09:43.590 ","End":"09:47.745","Text":"This is the direction of our dB."},{"Start":"09:47.745 ","End":"09:51.540","Text":"As we can see, it has a component in"},{"Start":"09:51.540 ","End":"09:58.830","Text":"the z-direction and a component in the radial direction."},{"Start":"09:58.830 ","End":"10:01.950","Text":"Now, something that we can also see is that there\u0027s"},{"Start":"10:01.950 ","End":"10:07.260","Text":"a 90-degree angle between our dB and our I vector. Why is that?"},{"Start":"10:07.260 ","End":"10:09.885","Text":"That we get from the equation itself."},{"Start":"10:09.885 ","End":"10:12.554","Text":"The whole point of this equation,"},{"Start":"10:12.554 ","End":"10:16.620","Text":"when we have the cross-product between 2 vectors,"},{"Start":"10:16.620 ","End":"10:22.230","Text":"the resultant vector is always perpendicular to both of those original vectors."},{"Start":"10:22.230 ","End":"10:29.890","Text":"We\u0027re going to have this 90-degree angle between our dB vector and our I vector."},{"Start":"10:29.960 ","End":"10:33.270","Text":"Of course, we have these 2 components."},{"Start":"10:33.270 ","End":"10:37.830","Text":"One thing that\u0027s very interesting to see is that from symmetry,"},{"Start":"10:37.830 ","End":"10:43.980","Text":"our radial component of our magnetic field is going to cancel out."},{"Start":"10:43.980 ","End":"10:46.335","Text":"Let\u0027s take a look at why."},{"Start":"10:46.335 ","End":"10:48.855","Text":"We can see that from this slice over here,"},{"Start":"10:48.855 ","End":"10:53.280","Text":"we have a radial component going in this direction."},{"Start":"10:53.280 ","End":"10:57.435","Text":"However, if we look at a slice over here,"},{"Start":"10:57.435 ","End":"11:02.760","Text":"let\u0027s say, which is equal and opposite to this slice."},{"Start":"11:02.760 ","End":"11:05.970","Text":"Its radial vector is going to be in this direction,"},{"Start":"11:05.970 ","End":"11:11.700","Text":"and then the dB vector over here will be in this direction."},{"Start":"11:11.700 ","End":"11:19.770","Text":"Then again, we\u0027ll have this radial component and this Z component again."},{"Start":"11:19.770 ","End":"11:25.169","Text":"We can see that the radial components are equal and opposite."},{"Start":"11:25.169 ","End":"11:27.720","Text":"That means that they will cancel each other out,"},{"Start":"11:27.720 ","End":"11:30.210","Text":"but the Z components will just add up."},{"Start":"11:30.210 ","End":"11:33.990","Text":"What we can see is that when we do this integration,"},{"Start":"11:33.990 ","End":"11:42.730","Text":"we expect the radial component to cancel out and the Z component to just add on."},{"Start":"11:42.920 ","End":"11:47.055","Text":"In order to find the total magnetic field at this point,"},{"Start":"11:47.055 ","End":"11:49.185","Text":"I\u0027m going to integrate along dB,"},{"Start":"11:49.185 ","End":"11:51.900","Text":"which means integrating along this."},{"Start":"11:51.900 ","End":"11:54.435","Text":"Let\u0027s put in our bounds."},{"Start":"11:54.435 ","End":"11:57.880","Text":"We\u0027re integrating along d theta."},{"Start":"11:58.250 ","End":"12:01.170","Text":"Because we\u0027re dealing with a circle,"},{"Start":"12:01.170 ","End":"12:07.890","Text":"that means that we\u0027re integrating from 0 radians and until 2 Pi radians."},{"Start":"12:07.890 ","End":"12:13.905","Text":"Of course, I have to do this integral in the radial direction and in the z-direction."},{"Start":"12:13.905 ","End":"12:18.280","Text":"What I can do is I can split up this integral."},{"Start":"12:18.800 ","End":"12:22.424","Text":"This is the format of our integral."},{"Start":"12:22.424 ","End":"12:24.705","Text":"Now, before we do this,"},{"Start":"12:24.705 ","End":"12:32.130","Text":"our r hat vector is changing and it\u0027s changing as a function of Theta."},{"Start":"12:32.130 ","End":"12:36.450","Text":"Our r hat will be in this direction"},{"Start":"12:36.450 ","End":"12:40.125","Text":"depending on the Theta then it can be going into the page,"},{"Start":"12:40.125 ","End":"12:41.550","Text":"then in this direction,"},{"Start":"12:41.550 ","End":"12:43.500","Text":"then coming out of the page."},{"Start":"12:43.500 ","End":"12:47.730","Text":"Our r hat vector is constantly changing depending on the Theta."},{"Start":"12:47.730 ","End":"12:50.865","Text":"What we\u0027re going to do is we\u0027re going to convert"},{"Start":"12:50.865 ","End":"12:56.650","Text":"our r hat vector into Cartesian coordinates."},{"Start":"12:56.840 ","End":"13:03.255","Text":"The r hat vector in Cartesian coordinates is equal"},{"Start":"13:03.255 ","End":"13:10.185","Text":"to cosine Theta in the x-direction and sine Theta in the y-direction."},{"Start":"13:10.185 ","End":"13:14.110","Text":"Now we\u0027re going to substitute that in."},{"Start":"13:14.720 ","End":"13:19.470","Text":"Now we can do the integration."},{"Start":"13:19.470 ","End":"13:26.055","Text":"When we integrate between 0 and 2 Pi on cosine of Theta it will equal 0."},{"Start":"13:26.055 ","End":"13:28.830","Text":"I\u0027ll get 0 in the x direction."},{"Start":"13:28.830 ","End":"13:34.170","Text":"Also integrating sine Theta between 0 and 2Pi will also give us 0."},{"Start":"13:34.170 ","End":"13:36.150","Text":"That\u0027s 0 in the y-direction,"},{"Start":"13:36.150 ","End":"13:37.530","Text":"which will give us, of course,"},{"Start":"13:37.530 ","End":"13:40.440","Text":"0 in the r hat direction."},{"Start":"13:40.440 ","End":"13:44.340","Text":"All of this expression over here becomes 0."},{"Start":"13:44.340 ","End":"13:47.310","Text":"Then we just have to integrate this where everything is,"},{"Start":"13:47.310 ","End":"13:51.165","Text":"of course, constants aside for my d Theta."},{"Start":"13:51.165 ","End":"13:54.510","Text":"The final answer, therefore,"},{"Start":"13:54.510 ","End":"13:56.130","Text":"at the end, I\u0027m just going to write it,"},{"Start":"13:56.130 ","End":"14:02.730","Text":"feel free to pause the video right now if you want to calculate it on your own."},{"Start":"14:02.730 ","End":"14:08.790","Text":"It\u0027s Mu_0I divided by 2 multiplied by R squared divided"},{"Start":"14:08.790 ","End":"14:16.635","Text":"by R squared plus Z squared to the power of 3 over 2."},{"Start":"14:16.635 ","End":"14:21.510","Text":"All of this is in the z-direction."},{"Start":"14:21.510 ","End":"14:27.120","Text":"This is the answer to this question when using the vector method."},{"Start":"14:27.120 ","End":"14:31.890","Text":"This is also the general equation for"},{"Start":"14:31.890 ","End":"14:37.980","Text":"the magnetic field along the axes of symmetry of a current carrying loop."},{"Start":"14:37.980 ","End":"14:41.085","Text":"This is something you can also write in your equation sheets."},{"Start":"14:41.085 ","End":"14:45.420","Text":"The magnetic field along the axis of symmetry for a current carrying loop."},{"Start":"14:45.420 ","End":"14:47.790","Text":"Of course, if we\u0027re trying to calculate"},{"Start":"14:47.790 ","End":"14:50.415","Text":"the magnetic field right in the center of the loop,"},{"Start":"14:50.415 ","End":"14:52.020","Text":"so at this point,"},{"Start":"14:52.020 ","End":"14:53.625","Text":"let\u0027s say over here,"},{"Start":"14:53.625 ","End":"14:57.915","Text":"that would be corresponding to Z is equal to 0,"},{"Start":"14:57.915 ","End":"14:59.760","Text":"and then you just plug in here,"},{"Start":"14:59.760 ","End":"15:02.110","Text":"Z is equal to 0."},{"Start":"15:02.810 ","End":"15:10.810","Text":"Now let\u0027s show the method using working out the magnitude and then the direction."},{"Start":"15:11.510 ","End":"15:16.230","Text":"First of all, we\u0027re going to work out the magnitude"},{"Start":"15:16.230 ","End":"15:21.885","Text":"of the B field at this point due to this slice over here."},{"Start":"15:21.885 ","End":"15:26.610","Text":"Then, we\u0027re going to calculate the direction."},{"Start":"15:26.610 ","End":"15:31.275","Text":"The first thing we want to do is we want to work out the magnitude of this cross-product."},{"Start":"15:31.275 ","End":"15:38.685","Text":"The magnitude of dL cross r. This is equal to"},{"Start":"15:38.685 ","End":"15:43.260","Text":"the magnitude of our dL vector multiplied by the magnitude"},{"Start":"15:43.260 ","End":"15:48.795","Text":"of our r vector multiplied by sine of the angle between the 2."},{"Start":"15:48.795 ","End":"15:52.185","Text":"Our dL vector is,"},{"Start":"15:52.185 ","End":"15:53.940","Text":"as we saw before,"},{"Start":"15:53.940 ","End":"15:57.510","Text":"it\u0027s equal to Rd Theta."},{"Start":"15:57.510 ","End":"16:02.880","Text":"Of course, it\u0027s traveling in the Theta direction,"},{"Start":"16:02.880 ","End":"16:06.070","Text":"just like we saw before."},{"Start":"16:08.870 ","End":"16:12.105","Text":"At this point, it\u0027s tangent to the circle,"},{"Start":"16:12.105 ","End":"16:15.225","Text":"which means that it\u0027s in the Theta direction."},{"Start":"16:15.225 ","End":"16:18.225","Text":"Our r vector,"},{"Start":"16:18.225 ","End":"16:19.770","Text":"so first of all,"},{"Start":"16:19.770 ","End":"16:24.090","Text":"let\u0027s write the magnitude of our r vector."},{"Start":"16:24.090 ","End":"16:27.690","Text":"It\u0027s simply going to be just like we saw before,"},{"Start":"16:27.690 ","End":"16:30.120","Text":"first of all, let\u0027s draw our vector."},{"Start":"16:30.120 ","End":"16:33.540","Text":"It\u0027s this over here."},{"Start":"16:33.540 ","End":"16:39.830","Text":"As we can see, we\u0027re going the distance over here,"},{"Start":"16:39.830 ","End":"16:42.350","Text":"so we\u0027re working out the hypotenuse."},{"Start":"16:42.350 ","End":"16:44.795","Text":"Just from Pythagoras,"},{"Start":"16:44.795 ","End":"16:49.095","Text":"we have R squared over here,"},{"Start":"16:49.095 ","End":"16:54.950","Text":"and then we have Z squared up in this direction."},{"Start":"16:54.950 ","End":"16:58.505","Text":"Then the square root will give us the length of the hypotenuse,"},{"Start":"16:58.505 ","End":"17:04.110","Text":"which in this case the hypotenuse is the r vector."},{"Start":"17:04.200 ","End":"17:09.370","Text":"Now we can plug in these two."},{"Start":"17:09.370 ","End":"17:18.250","Text":"Of course, from here we get that the magnitude of dl is simply Rd Theta."},{"Start":"17:18.250 ","End":"17:22.120","Text":"What is the angle between these two?"},{"Start":"17:22.120 ","End":"17:26.710","Text":"What we can do is we can draw the direction of the current."},{"Start":"17:26.710 ","End":"17:30.295","Text":"If we\u0027re looking at this point over here,"},{"Start":"17:30.295 ","End":"17:34.735","Text":"let\u0027s say, let\u0027s look at this point."},{"Start":"17:34.735 ","End":"17:38.470","Text":"We\u0027re looking at this slice of wire so we\u0027ll look at this point."},{"Start":"17:38.470 ","End":"17:45.175","Text":"We can see that the direction of current at this point is going into the page,"},{"Start":"17:45.175 ","End":"17:47.845","Text":"and then we follow the current around."},{"Start":"17:47.845 ","End":"17:50.725","Text":"If we look at the current at this point,"},{"Start":"17:50.725 ","End":"17:52.690","Text":"the opposite side of the cycle,"},{"Start":"17:52.690 ","End":"17:59.095","Text":"we can see from the direction that it\u0027s coming out of the page."},{"Start":"17:59.095 ","End":"18:04.030","Text":"Now if I draw the direction of my I vector,"},{"Start":"18:04.030 ","End":"18:06.655","Text":"if I draw it in this direction,"},{"Start":"18:06.655 ","End":"18:14.245","Text":"let\u0027s say it\u0027s obvious that the angle between the current and the I vector is 90 degrees."},{"Start":"18:14.245 ","End":"18:20.545","Text":"Of course, it\u0027s clear that if I swirl my I vector in this direction,"},{"Start":"18:20.545 ","End":"18:24.880","Text":"the current and the I vector are still perpendicular."},{"Start":"18:24.880 ","End":"18:30.160","Text":"Because the I vector is on this plane and the current is going into the page."},{"Start":"18:30.160 ","End":"18:33.160","Text":"You can even point with your pointing finger in"},{"Start":"18:33.160 ","End":"18:36.940","Text":"the direction of the current and with your thumb in the direction of the I vector,"},{"Start":"18:36.940 ","End":"18:40.300","Text":"and you\u0027ll see that they\u0027re perpendicular to one another."},{"Start":"18:40.300 ","End":"18:43.480","Text":"If our Alpha is 90 degrees,"},{"Start":"18:43.480 ","End":"18:48.310","Text":"that means that Sine of Alpha is equal to 1."},{"Start":"18:48.310 ","End":"18:52.600","Text":"In total, what we\u0027ll get is that the magnitude of"},{"Start":"18:52.600 ","End":"18:56.950","Text":"dl cross r is simply equal to the magnitude of dl,"},{"Start":"18:56.950 ","End":"19:01.405","Text":"which is Rd Theta multiplied by the magnitude of I,"},{"Start":"19:01.405 ","End":"19:09.110","Text":"which is R squared plus z squared to the power of a half multiplied by 1."},{"Start":"19:09.870 ","End":"19:16.105","Text":"Now what I can do is I can plug this all into my dB."},{"Start":"19:16.105 ","End":"19:19.120","Text":"I\u0027m just calculating the magnitude,"},{"Start":"19:19.120 ","End":"19:22.330","Text":"so I can just write it without the vector sign on top."},{"Start":"19:22.330 ","End":"19:26.920","Text":"Mu naught divided by 4 Pi multiplied by I,"},{"Start":"19:26.920 ","End":"19:30.970","Text":"multiplied by the magnitude of dl cross r,"},{"Start":"19:30.970 ","End":"19:36.880","Text":"which is equal to IRdTheta I squared plus z squared to"},{"Start":"19:36.880 ","End":"19:43.210","Text":"the power of 1.5 divided by r magnitude cubed."},{"Start":"19:43.210 ","End":"19:51.835","Text":"That\u0027s this. I have r squared plus z squared to the power of 3 over 2."},{"Start":"19:51.835 ","End":"19:55.645","Text":"Then here, I can cross this out with"},{"Start":"19:55.645 ","End":"19:59.830","Text":"one of these so that I\u0027m left over here with 2 over 2."},{"Start":"19:59.830 ","End":"20:06.160","Text":"All I\u0027m left with in the end is Mu naught divided by 4 Pi multiplied"},{"Start":"20:06.160 ","End":"20:16.190","Text":"by IRdTheta divided by r squared plus z squared."},{"Start":"20:17.040 ","End":"20:26.020","Text":"Of course, I can just write a little note in gray,"},{"Start":"20:26.020 ","End":"20:31.880","Text":"this is just equal to the magnitude of r squared."},{"Start":"20:35.160 ","End":"20:38.830","Text":"We have the magnitude of dB."},{"Start":"20:38.830 ","End":"20:40.840","Text":"Now let\u0027s look at the direction."},{"Start":"20:40.840 ","End":"20:43.465","Text":"Using the right-hand rule,"},{"Start":"20:43.465 ","End":"20:50.095","Text":"we can see that our dB will be in this direction."},{"Start":"20:50.095 ","End":"20:54.760","Text":"Of course, this angle between dB and I is going to be at 90 degrees,"},{"Start":"20:54.760 ","End":"20:57.865","Text":"just like we spoke about earlier in the lesson."},{"Start":"20:57.865 ","End":"21:01.690","Text":"Let\u0027s just speak again about the right-hand rule."},{"Start":"21:01.690 ","End":"21:07.915","Text":"What you can either do is point your thumb in the direction of the current at this point."},{"Start":"21:07.915 ","End":"21:11.785","Text":"We said that this was going into the page over here."},{"Start":"21:11.785 ","End":"21:18.595","Text":"Then you point your forefinger in the direction the r vector."},{"Start":"21:18.595 ","End":"21:24.020","Text":"Then what you\u0027ll get is that your middle finger is pointing in this direction."},{"Start":"21:24.090 ","End":"21:27.250","Text":"That is our dB direction."},{"Start":"21:27.250 ","End":"21:32.365","Text":"Another way that we can look at this is if we look at this from the side,"},{"Start":"21:32.365 ","End":"21:35.100","Text":"so this length over here is z,"},{"Start":"21:35.100 ","End":"21:37.780","Text":"this length over here is R,"},{"Start":"21:37.780 ","End":"21:45.860","Text":"and this length over here is our r vector magnitude."},{"Start":"21:47.090 ","End":"21:52.780","Text":"With our current, we said that at this point over here,"},{"Start":"21:52.800 ","End":"21:56.275","Text":"our current is going into the page."},{"Start":"21:56.275 ","End":"22:01.270","Text":"We\u0027re looking at this point from this side,"},{"Start":"22:01.270 ","End":"22:05.560","Text":"here, our current is going in."},{"Start":"22:05.560 ","End":"22:09.910","Text":"As we know the magnetic field is forming"},{"Start":"22:09.910 ","End":"22:15.670","Text":"concentric circles around the current carrying wire like so."},{"Start":"22:15.670 ","End":"22:23.430","Text":"If we use the right-hand rule so we can see that the current is traveling like so."},{"Start":"22:23.430 ","End":"22:27.990","Text":"It\u0027s going around in this direction."},{"Start":"22:27.990 ","End":"22:34.015","Text":"Now if we draw a bigger circle that meets this point over here,"},{"Start":"22:34.015 ","End":"22:41.305","Text":"the larger concentric circle is going to look something like this."},{"Start":"22:41.305 ","End":"22:48.640","Text":"Again, it\u0027s going something like so all the way around."},{"Start":"22:48.640 ","End":"22:52.405","Text":"We can see that here we have this 90-degree angle."},{"Start":"22:52.405 ","End":"22:59.440","Text":"This really is the direction of our dB at this point over here."},{"Start":"22:59.440 ","End":"23:03.820","Text":"Now what we\u0027re going to do is we\u0027re going to split"},{"Start":"23:03.820 ","End":"23:08.755","Text":"up our dB into its different components."},{"Start":"23:08.755 ","End":"23:15.580","Text":"We\u0027re going to draw our dB in the z-direction."},{"Start":"23:15.580 ","End":"23:20.905","Text":"What we\u0027re going to do is we\u0027re just going to draw its projection on the z-axis."},{"Start":"23:20.905 ","End":"23:23.690","Text":"This will be our dBz."},{"Start":"23:26.220 ","End":"23:30.895","Text":"In order to find my dBz,"},{"Start":"23:30.895 ","End":"23:33.940","Text":"so my dB in the z-direction,"},{"Start":"23:33.940 ","End":"23:38.800","Text":"I am going to take this as"},{"Start":"23:38.800 ","End":"23:44.560","Text":"a right-hand triangle and I\u0027m going to call this angle over here phi."},{"Start":"23:44.560 ","End":"23:50.365","Text":"Now what I can do is I can draw another line parallel like so."},{"Start":"23:50.365 ","End":"23:59.470","Text":"Then this angle over here is 90 minus phi. Why is that?"},{"Start":"23:59.470 ","End":"24:01.990","Text":"Because if this angle is 90,"},{"Start":"24:01.990 ","End":"24:06.355","Text":"the angle between here and here is also 90."},{"Start":"24:06.355 ","End":"24:08.395","Text":"Then to get this angle over here,"},{"Start":"24:08.395 ","End":"24:11.320","Text":"it\u0027s just 90 minus phi."},{"Start":"24:11.320 ","End":"24:16.960","Text":"Then that means that this is also phi over here because here we also have 90."},{"Start":"24:16.960 ","End":"24:19.300","Text":"This is 90 minus phi,"},{"Start":"24:19.300 ","End":"24:21.430","Text":"and then we have 90 minus 90 minus phi,"},{"Start":"24:21.430 ","End":"24:22.840","Text":"which leaves us with phi."},{"Start":"24:22.840 ","End":"24:26.050","Text":"Then, this, of course,"},{"Start":"24:26.050 ","End":"24:32.305","Text":"is also R. Then of course this dotted line over here,"},{"Start":"24:32.305 ","End":"24:35.320","Text":"is parallel to this radius over here,"},{"Start":"24:35.320 ","End":"24:40.190","Text":"which means that this angle over here is also phi."},{"Start":"24:40.530 ","End":"24:46.720","Text":"DB z is simply going to be equal to the magnitude of"},{"Start":"24:46.720 ","End":"24:52.255","Text":"dB multiplied by its projection on the z-axis."},{"Start":"24:52.255 ","End":"24:55.930","Text":"The magnitude of this and the projection over here."},{"Start":"24:55.930 ","End":"24:58.150","Text":"The projection over here,"},{"Start":"24:58.150 ","End":"25:02.170","Text":"this is the adjacent side to our angle phi."},{"Start":"25:02.170 ","End":"25:03.910","Text":"When we have the adjacent side,"},{"Start":"25:03.910 ","End":"25:07.105","Text":"we\u0027re dealing with cosine of the angle."},{"Start":"25:07.105 ","End":"25:09.880","Text":"What is cosine of the angle?"},{"Start":"25:09.880 ","End":"25:16.660","Text":"Here is our angle phi and cosine is equal to SOH CAH TOA."},{"Start":"25:16.660 ","End":"25:21.040","Text":"Cosine is adjacent over hypotenuse."},{"Start":"25:21.040 ","End":"25:25.180","Text":"The adjacent side is this r over here."},{"Start":"25:25.180 ","End":"25:27.865","Text":"We have magnitude of dB,"},{"Start":"25:27.865 ","End":"25:32.320","Text":"and then we have R divided by the hypotenuse,"},{"Start":"25:32.320 ","End":"25:37.255","Text":"which is lowercase r. Which is what we worked out over here."},{"Start":"25:37.255 ","End":"25:40.075","Text":"This is our dBz."},{"Start":"25:40.075 ","End":"25:43.705","Text":"This over here, I just plugged in all of the values."},{"Start":"25:43.705 ","End":"25:48.730","Text":"Then in order to get the total B field in the z-direction,"},{"Start":"25:48.730 ","End":"25:50.800","Text":"of course, I\u0027m just going to integrate."},{"Start":"25:50.800 ","End":"25:55.425","Text":"I\u0027m integrating along a full circle, along d Theta."},{"Start":"25:55.425 ","End":"26:00.045","Text":"I\u0027m going from 0 radians until 2 Pi radians."},{"Start":"26:00.045 ","End":"26:03.630","Text":"Everything here is a constant and I just have this d Theta,"},{"Start":"26:03.630 ","End":"26:12.580","Text":"so I\u0027m left with Mu Naught I squared multiplied by 2 Pi,"},{"Start":"26:12.580 ","End":"26:14.740","Text":"which will cancel out with the 4 Pi."},{"Start":"26:14.740 ","End":"26:24.160","Text":"Then divided by 2 R squared plus z squared to the power of 3 over 2."},{"Start":"26:24.160 ","End":"26:27.339","Text":"This is the B field in the z-direction,"},{"Start":"26:27.339 ","End":"26:29.725","Text":"which if you go back to earlier in the lesson,"},{"Start":"26:29.725 ","End":"26:32.755","Text":"this is the same answer that we got before."},{"Start":"26:32.755 ","End":"26:40.180","Text":"Now, with regards to the components of the B field that is going in this direction,"},{"Start":"26:40.180 ","End":"26:43.780","Text":"the radial direction or the x, y direction,"},{"Start":"26:43.780 ","End":"26:47.290","Text":"so dBr hat,"},{"Start":"26:47.290 ","End":"26:51.860","Text":"or rather dB in the radial direction."},{"Start":"26:52.350 ","End":"26:57.280","Text":"As we can see, this will cancel out for reasons of symmetry."},{"Start":"26:57.280 ","End":"27:00.010","Text":"If we look over here at this point,"},{"Start":"27:00.010 ","End":"27:05.050","Text":"we get that dBr is going in this rightwards direction."},{"Start":"27:05.050 ","End":"27:08.950","Text":"I\u0027ll draw this in red,"},{"Start":"27:08.950 ","End":"27:11.290","Text":"for this slice over here,"},{"Start":"27:11.290 ","End":"27:14.890","Text":"which is in the opposite side of the loop."},{"Start":"27:14.890 ","End":"27:21.115","Text":"Here, I will have dB total going in this direction,"},{"Start":"27:21.115 ","End":"27:27.625","Text":"which means that my dBr will be going in this leftwards direction."},{"Start":"27:27.625 ","End":"27:29.995","Text":"Of course, they\u0027re equal and opposite,"},{"Start":"27:29.995 ","End":"27:31.600","Text":"so they will cancel out."},{"Start":"27:31.600 ","End":"27:34.060","Text":"This of course happens all around the loop."},{"Start":"27:34.060 ","End":"27:36.205","Text":"For every piece that I choose here,"},{"Start":"27:36.205 ","End":"27:39.190","Text":"I have another opposite piece over here."},{"Start":"27:39.190 ","End":"27:42.580","Text":"My dB in the radial direction or in the x,"},{"Start":"27:42.580 ","End":"27:46.105","Text":"y direction is going to cancel out and equal to 0."},{"Start":"27:46.105 ","End":"27:50.155","Text":"This is the magnitude of my magnetic field."},{"Start":"27:50.155 ","End":"27:54.500","Text":"Of course, it\u0027s in the z-direction."},{"Start":"27:55.680 ","End":"27:59.840","Text":"That is the end of this lesson."}],"ID":21398},{"Watched":false,"Name":"Exercise 4","Duration":"14m ","ChapterTopicVideoID":21323,"CourseChapterTopicPlaylistID":99469,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.680","Text":"Hello. In this lesson we\u0027re going to be answering the following question."},{"Start":"00:04.680 ","End":"00:12.060","Text":"Calculate the magnetic field a distance of z away from the center of a disk of radius R,"},{"Start":"00:12.060 ","End":"00:14.145","Text":"charge density Sigma,"},{"Start":"00:14.145 ","End":"00:21.150","Text":"and angular velocity Omega_naught in the anticlockwise direction about the origin."},{"Start":"00:21.150 ","End":"00:24.630","Text":"So the way that we\u0027re going to do this is we\u0027re going to split"},{"Start":"00:24.630 ","End":"00:29.230","Text":"this solid disk into lots of different rings."},{"Start":"00:29.900 ","End":"00:33.000","Text":"If we draw it in blue,"},{"Start":"00:33.000 ","End":"00:34.530","Text":"will take a small ring,"},{"Start":"00:34.530 ","End":"00:37.890","Text":"calculate the magnetic field at this point, and due to this ring,"},{"Start":"00:37.890 ","End":"00:43.640","Text":"then take a larger ring and add that on over here at"},{"Start":"00:43.640 ","End":"00:47.390","Text":"the magnetic field due to the larger ring over here"},{"Start":"00:47.390 ","End":"00:51.230","Text":"and so on and so forth until we fill the entire disk."},{"Start":"00:51.230 ","End":"00:52.970","Text":"In the previous lesson,"},{"Start":"00:52.970 ","End":"00:55.620","Text":"we looked at a current carrying loop,"},{"Start":"00:55.620 ","End":"00:57.890","Text":"and we saw that the magnetic field,"},{"Start":"00:57.890 ","End":"01:01.520","Text":"a distance z away from the center of a current carrying loop,"},{"Start":"01:01.520 ","End":"01:05.120","Text":"which is just like a ring inside this disk,"},{"Start":"01:05.120 ","End":"01:08.850","Text":"the magnetic field was equal to this."},{"Start":"01:09.500 ","End":"01:14.630","Text":"What am I going to do is seeing as we already calculated this,"},{"Start":"01:14.630 ","End":"01:20.554","Text":"so we\u0027re just going to integrate this along the entire disk."},{"Start":"01:20.554 ","End":"01:26.410","Text":"Now the one thing that we can see is that over here our equation incorporates current,"},{"Start":"01:26.410 ","End":"01:29.400","Text":"but we aren\u0027t told of any current,"},{"Start":"01:29.400 ","End":"01:34.180","Text":"we\u0027re told that we have a charge density Sigma and angular velocity Omega_naught."},{"Start":"01:34.180 ","End":"01:38.030","Text":"The first thing that we have to remember about current is that"},{"Start":"01:38.030 ","End":"01:42.050","Text":"current is created from moving charged particles."},{"Start":"01:42.050 ","End":"01:44.300","Text":"If a charged particle moves,"},{"Start":"01:44.300 ","End":"01:46.405","Text":"that means that we have current."},{"Start":"01:46.405 ","End":"01:49.880","Text":"Here we have charged particles of a density"},{"Start":"01:49.880 ","End":"01:54.270","Text":"Sigma and we\u0027re being told that they\u0027re moving in circular motion."},{"Start":"01:54.800 ","End":"01:58.220","Text":"What we want to do is we want to use"},{"Start":"01:58.220 ","End":"02:03.095","Text":"this information segment Omega_naughts in order to calculate the current."},{"Start":"02:03.095 ","End":"02:05.225","Text":"There\u0027s 2 methods to doing this."},{"Start":"02:05.225 ","End":"02:08.900","Text":"We can either say that our current is equal to"},{"Start":"02:08.900 ","End":"02:13.385","Text":"the change in charge divided by the change in time,"},{"Start":"02:13.385 ","End":"02:14.735","Text":"dq by dt,"},{"Start":"02:14.735 ","End":"02:17.284","Text":"or the time derivative of the charge."},{"Start":"02:17.284 ","End":"02:22.475","Text":"Or we can use the idea of current density."},{"Start":"02:22.475 ","End":"02:25.685","Text":"So we already saw current density J,"},{"Start":"02:25.685 ","End":"02:27.560","Text":"which is equal to Rho,"},{"Start":"02:27.560 ","End":"02:29.660","Text":"the current density per unit volume,"},{"Start":"02:29.660 ","End":"02:31.910","Text":"multiplied by the velocity."},{"Start":"02:31.910 ","End":"02:39.005","Text":"Then we had k, which was the current density per unit area multiplied by the velocity,"},{"Start":"02:39.005 ","End":"02:40.775","Text":"and we had, of course,"},{"Start":"02:40.775 ","End":"02:45.620","Text":"I, which doesn\u0027t have a vector sign on top of it,"},{"Start":"02:45.620 ","End":"02:52.160","Text":"which is equal to the current density per unit length multiplied by the velocity."},{"Start":"02:52.160 ","End":"02:58.130","Text":"We already said in a previous lesson that if we\u0027re dealing with Sigma,"},{"Start":"02:58.130 ","End":"03:00.455","Text":"which is what we have in this lesson,"},{"Start":"03:00.455 ","End":"03:05.000","Text":"so this method is the easiest if we\u0027re dealing with Sigma."},{"Start":"03:05.000 ","End":"03:07.670","Text":"If we\u0027re dealing with Rho,"},{"Start":"03:07.670 ","End":"03:08.800","Text":"then this is the easiest,"},{"Start":"03:08.800 ","End":"03:11.210","Text":"and of course, if we\u0027re dealing with Lambda,"},{"Start":"03:11.210 ","End":"03:13.180","Text":"then this is the easiest."},{"Start":"03:13.180 ","End":"03:17.660","Text":"The only reason is because here with the Sigma and k,"},{"Start":"03:17.660 ","End":"03:21.170","Text":"it\u0027s a little bit difficult to figure out the exact dimensions, but of course,"},{"Start":"03:21.170 ","End":"03:22.895","Text":"you can still use this equation,"},{"Start":"03:22.895 ","End":"03:26.960","Text":"but it\u0027s a little bit easier to get confused and make a mistake."},{"Start":"03:26.960 ","End":"03:32.045","Text":"What we\u0027re going to be doing is we\u0027re going to be using these 2 methods for Sigma."},{"Start":"03:32.045 ","End":"03:34.790","Text":"Of course, this is the preferred method for Sigma,"},{"Start":"03:34.790 ","End":"03:38.760","Text":"but we\u0027re also going to go through this method."},{"Start":"03:38.960 ","End":"03:44.495","Text":"This is the first method will show and then this will be the second method."},{"Start":"03:44.495 ","End":"03:47.105","Text":"Let\u0027s begin with the first method."},{"Start":"03:47.105 ","End":"03:49.385","Text":"Let\u0027s write it over here."},{"Start":"03:49.385 ","End":"03:54.350","Text":"The first thing that we want to see is what is our dq equal to."},{"Start":"03:54.350 ","End":"03:59.810","Text":"Our dq, as we\u0027ve been seeing when we\u0027re dealing with Sigma in previous chapters,"},{"Start":"03:59.810 ","End":"04:06.320","Text":"it\u0027s equal to Sigma charged density per unit area multiplied by ds,"},{"Start":"04:06.320 ","End":"04:08.435","Text":"which is unit area."},{"Start":"04:08.435 ","End":"04:10.715","Text":"We\u0027re dealing with a full disk,"},{"Start":"04:10.715 ","End":"04:14.300","Text":"so what\u0027s the unit area of a full disk?"},{"Start":"04:14.300 ","End":"04:20.885","Text":"We\u0027re dealing with the area in polar coordinates."},{"Start":"04:20.885 ","End":"04:26.090","Text":"That is going to be equal to rdrd Theta."},{"Start":"04:26.090 ","End":"04:32.560","Text":"Please make sure that you understand why ds is equal to this in polar coordinates."},{"Start":"04:32.560 ","End":"04:34.665","Text":"Now we have our dq,"},{"Start":"04:34.665 ","End":"04:36.885","Text":"let\u0027s plug it into our I."},{"Start":"04:36.885 ","End":"04:40.385","Text":"What we have is I is equal to dq,"},{"Start":"04:40.385 ","End":"04:46.505","Text":"which is Sigma IddI Theta divided by dt."},{"Start":"04:46.505 ","End":"04:48.740","Text":"So at this stage over here,"},{"Start":"04:48.740 ","End":"04:53.990","Text":"we\u0027re meant to see that something in the numerator together with our dt and"},{"Start":"04:53.990 ","End":"05:00.245","Text":"the denominator is equal to another value that we have in our question."},{"Start":"05:00.245 ","End":"05:03.995","Text":"Over here we have d Theta by dt,"},{"Start":"05:03.995 ","End":"05:06.085","Text":"which is equal to Omega."},{"Start":"05:06.085 ","End":"05:08.609","Text":"This equal to angular velocity."},{"Start":"05:08.609 ","End":"05:11.920","Text":"Please remember that we\u0027ve also seen this in previous chapters,"},{"Start":"05:11.920 ","End":"05:14.860","Text":"so d Theta by dt is Omega."},{"Start":"05:14.860 ","End":"05:21.415","Text":"What we have is that this is equal to Sigma r Omega_naught,"},{"Start":"05:21.415 ","End":"05:23.230","Text":"because that\u0027s what we\u0027re given in the question,"},{"Start":"05:23.230 ","End":"05:29.830","Text":"multiplied by dr. Of course,"},{"Start":"05:29.830 ","End":"05:32.335","Text":"if I have a differential over here, dI,"},{"Start":"05:32.335 ","End":"05:37.320","Text":"then that must mean that this over here is dI."},{"Start":"05:37.320 ","End":"05:40.190","Text":"Now let\u0027s look at our disk from a bird\u0027s eye view,"},{"Start":"05:40.190 ","End":"05:44.439","Text":"so we\u0027ve been told that our Omega is in the anticlockwise direction,"},{"Start":"05:44.439 ","End":"05:47.945","Text":"so it\u0027s going in this direction."},{"Start":"05:47.945 ","End":"05:54.339","Text":"Of course, that means that that is the direction that our current is flowing."},{"Start":"05:54.339 ","End":"05:59.090","Text":"What we have is we have the center of our disk,"},{"Start":"05:59.090 ","End":"06:05.795","Text":"and this is our radius r. Then we have some loop around that."},{"Start":"06:05.795 ","End":"06:10.640","Text":"Then we have this very small width over here."},{"Start":"06:10.640 ","End":"06:16.250","Text":"Of course, this isn\u0027t to scale where this width is dI,"},{"Start":"06:16.250 ","End":"06:18.590","Text":"gets a small change in radius,"},{"Start":"06:18.590 ","End":"06:26.165","Text":"it\u0027s dr. What we have here is the current of this very, very small,"},{"Start":"06:26.165 ","End":"06:33.945","Text":"infinitesimal width within our disk that has a current dI flowing through and"},{"Start":"06:33.945 ","End":"06:37.760","Text":"our dI is equal to the charge density per"},{"Start":"06:37.760 ","End":"06:42.665","Text":"unit area multiplied by the radius that we\u0027re at,"},{"Start":"06:42.665 ","End":"06:46.440","Text":"the angular velocity and multiplied by this very,"},{"Start":"06:46.440 ","End":"06:49.525","Text":"very small width over here."},{"Start":"06:49.525 ","End":"06:54.980","Text":"Now we can rewrite this equation for the disk."},{"Start":"06:54.980 ","End":"06:59.330","Text":"So first of all we remember that this is for a single current carrying loop,"},{"Start":"06:59.330 ","End":"07:01.670","Text":"but we\u0027re of course taking many,"},{"Start":"07:01.670 ","End":"07:02.840","Text":"many current carrying loops,"},{"Start":"07:02.840 ","End":"07:04.340","Text":"so it\u0027s going to be dB."},{"Start":"07:04.340 ","End":"07:06.065","Text":"This is equal to,"},{"Start":"07:06.065 ","End":"07:11.210","Text":"so we have Mu_naught multiplied by I,"},{"Start":"07:11.210 ","End":"07:14.390","Text":"which as we know, is dI, which is this."},{"Start":"07:14.390 ","End":"07:21.335","Text":"Sigma r Omega_naught dr divided by 2."},{"Start":"07:21.335 ","End":"07:24.620","Text":"Then all of this is multiplied by r. However,"},{"Start":"07:24.620 ","End":"07:28.445","Text":"now our r of course because we\u0027re integrating."},{"Start":"07:28.445 ","End":"07:31.895","Text":"We saw that our r is constantly changing because we\u0027re summing up"},{"Start":"07:31.895 ","End":"07:36.515","Text":"all of the current carrying loops that we have in order to make a full disk."},{"Start":"07:36.515 ","End":"07:40.400","Text":"This I is now our variable I."},{"Start":"07:40.400 ","End":"07:45.350","Text":"It\u0027s r^2 in the z-direction"},{"Start":"07:45.350 ","End":"07:51.860","Text":"divided by the same (r^2 plus z^2)^3 over 2."},{"Start":"07:51.860 ","End":"07:55.220","Text":"Where of course, our r is a variable,"},{"Start":"07:55.220 ","End":"07:57.630","Text":"it isn\u0027t a constant."},{"Start":"07:57.940 ","End":"08:05.045","Text":"Of course, to calculate the magnetic field at point z above the center,"},{"Start":"08:05.045 ","End":"08:08.360","Text":"means that we\u0027re going to have to integrate along dB."},{"Start":"08:08.360 ","End":"08:13.205","Text":"What does that mean? We\u0027re integrating from the origin until the edge of the disc."},{"Start":"08:13.205 ","End":"08:19.070","Text":"We\u0027re integrating from 0 until radius of R. Because the disk has"},{"Start":"08:19.070 ","End":"08:26.585","Text":"a radius of R. What we\u0027re going to have to do is use integration by substitution."},{"Start":"08:26.585 ","End":"08:32.735","Text":"Let\u0027s say that t is equal to r^2 plus z^2."},{"Start":"08:32.735 ","End":"08:37.890","Text":"Then what we\u0027ll get is that dt is simply equal to 2rdr."},{"Start":"08:38.140 ","End":"08:42.185","Text":"Notice that here we have Idr."},{"Start":"08:42.185 ","End":"08:45.845","Text":"This substitution isn\u0027t very difficult."},{"Start":"08:45.845 ","End":"08:48.530","Text":"From here you can do the integral,"},{"Start":"08:48.530 ","End":"08:52.440","Text":"so I\u0027m just going to write out the answer."},{"Start":"08:52.510 ","End":"08:54.890","Text":"This is the answer,"},{"Start":"08:54.890 ","End":"08:57.740","Text":"so it\u0027s equal to the magnetic field at point z due to"},{"Start":"08:57.740 ","End":"09:04.460","Text":"this disk over here with charge density Sigma and rotating with"},{"Start":"09:04.460 ","End":"09:08.974","Text":"an angular velocity of Omega_naught"},{"Start":"09:08.974 ","End":"09:13.955","Text":"is equal to Mu_naught Sigma Omega_naught divided by 2 multiplied"},{"Start":"09:13.955 ","End":"09:17.795","Text":"by (R^2 plus z^2)^1/2"},{"Start":"09:17.795 ","End":"09:25.095","Text":"plus z^2 multiplied by (R^2 plus z^2)^ negative 1/2 minus 2z."},{"Start":"09:25.095 ","End":"09:29.075","Text":"All of this is in the z-direction."},{"Start":"09:29.075 ","End":"09:33.380","Text":"This you can even write as an equation in your equation sheets for"},{"Start":"09:33.380 ","End":"09:39.385","Text":"the magnetic field due to a disk with Sigma and Omega_naught given in the question."},{"Start":"09:39.385 ","End":"09:41.960","Text":"I suggest you write that down."},{"Start":"09:41.960 ","End":"09:45.500","Text":"This is the answer for a Method number 1."},{"Start":"09:45.500 ","End":"09:48.275","Text":"Now we\u0027re going to go over Method number 2."},{"Start":"09:48.275 ","End":"09:51.080","Text":"Of course we\u0027re going to get the exact same answer,"},{"Start":"09:51.080 ","End":"09:52.925","Text":"but it\u0027s a different method."},{"Start":"09:52.925 ","End":"09:56.190","Text":"If you want to see it, carry on watching."},{"Start":"09:56.740 ","End":"09:59.120","Text":"Method number 2,"},{"Start":"09:59.120 ","End":"10:07.025","Text":"we\u0027re using the method that k is equal to Sigma multiplied by v. First of all,"},{"Start":"10:07.025 ","End":"10:09.275","Text":"let\u0027s calculate what v is."},{"Start":"10:09.275 ","End":"10:11.225","Text":"Sigma we have in our question."},{"Start":"10:11.225 ","End":"10:16.655","Text":"We have Sigma and then we have circular or angular velocity."},{"Start":"10:16.655 ","End":"10:22.430","Text":"That is simply equal to our angular velocity Omega_naught multiplied by r. Of"},{"Start":"10:22.430 ","End":"10:29.250","Text":"course it\u0027s dependent on the radius and it\u0027s in the Theta hat direction."},{"Start":"10:29.620 ","End":"10:32.975","Text":"That wasn\u0027t very difficult to calculate."},{"Start":"10:32.975 ","End":"10:39.080","Text":"From here we know that the current is equal to the integral of kdl."},{"Start":"10:39.080 ","End":"10:42.260","Text":"This is where things get a little bit tricky and this is where you"},{"Start":"10:42.260 ","End":"10:46.110","Text":"really have to concentrate not to make a mistake."},{"Start":"10:47.050 ","End":"10:52.340","Text":"What do we have to remember is that kdl is our dI,"},{"Start":"10:52.340 ","End":"10:59.700","Text":"our change or our small amount of current coming from the specific current carrying loop."},{"Start":"11:01.000 ","End":"11:10.655","Text":"What we can see is that our k and our dl have to be perpendicular to one another."},{"Start":"11:10.655 ","End":"11:15.560","Text":"Before we saw that if this is our disc,"},{"Start":"11:15.560 ","End":"11:19.925","Text":"our k is traveling in this direction,"},{"Start":"11:19.925 ","End":"11:21.920","Text":"in the Theta direction,"},{"Start":"11:21.920 ","End":"11:27.810","Text":"which means that our dl has to be perpendicular to it."},{"Start":"11:27.850 ","End":"11:30.830","Text":"When we\u0027re dealing with the Rho option,"},{"Start":"11:30.830 ","End":"11:33.545","Text":"this is why I say it\u0027s easier Because"},{"Start":"11:33.545 ","End":"11:38.540","Text":"our dl or our ds when dealing with a Rho has to be of course,"},{"Start":"11:38.540 ","End":"11:40.205","Text":"perpendicular to the Rho."},{"Start":"11:40.205 ","End":"11:43.565","Text":"That just means we take the 2 other directions."},{"Start":"11:43.565 ","End":"11:46.595","Text":"We take the I-direction and the z-direction."},{"Start":"11:46.595 ","End":"11:48.515","Text":"However, in k,"},{"Start":"11:48.515 ","End":"11:49.955","Text":"we\u0027re dealing with dl,"},{"Start":"11:49.955 ","End":"11:56.030","Text":"which means we\u0027re using 1 direction which is perpendicular to our k. The question is,"},{"Start":"11:56.030 ","End":"11:58.970","Text":"which direction are we taking?"},{"Start":"11:58.970 ","End":"12:03.320","Text":"The z-direction, which is perpendicular to"},{"Start":"12:03.320 ","End":"12:11.090","Text":"the k. Or are we taking out I-direction which is also perpendicular to the k?"},{"Start":"12:11.090 ","End":"12:15.420","Text":"This is where it gets confusing and this is where you can make a mistake."},{"Start":"12:15.480 ","End":"12:18.340","Text":"In our case, specifically,"},{"Start":"12:18.340 ","End":"12:20.680","Text":"if we look back at our diagram,"},{"Start":"12:20.680 ","End":"12:24.995","Text":"we can see that the z-direction doesn\u0027t really make sense."},{"Start":"12:24.995 ","End":"12:29.675","Text":"Even though we\u0027re finding the magnetic field along the z-axis,"},{"Start":"12:29.675 ","End":"12:33.560","Text":"it doesn\u0027t make any sense because there\u0027s no particles along"},{"Start":"12:33.560 ","End":"12:38.300","Text":"the z-axis itself that we can be summing across."},{"Start":"12:38.300 ","End":"12:40.250","Text":"However, in the I-direction,"},{"Start":"12:40.250 ","End":"12:43.355","Text":"there particles that are causing a magnetic field"},{"Start":"12:43.355 ","End":"12:47.330","Text":"or a current rather specifically for this calculation."},{"Start":"12:47.330 ","End":"12:52.400","Text":"Therefore we\u0027re summing across or along the I-direction."},{"Start":"12:52.400 ","End":"12:54.710","Text":"We can also see it over here."},{"Start":"12:54.710 ","End":"12:58.385","Text":"We\u0027re summing across this dr over here."},{"Start":"12:58.385 ","End":"13:00.080","Text":"So in that case,"},{"Start":"13:00.080 ","End":"13:08.330","Text":"what we get is that we\u0027re integrating along kdr."},{"Start":"13:08.330 ","End":"13:11.882","Text":"What we get is that our dI, which is kdl is equal to our k"},{"Start":"13:11.882 ","End":"13:17.120","Text":"and our dl is dr, as we just said."},{"Start":"13:17.120 ","End":"13:20.550","Text":"Let\u0027s just write it over here, kdr."},{"Start":"13:21.100 ","End":"13:27.185","Text":"We got this as dI, which is the same dI that we got in our first method."},{"Start":"13:27.185 ","End":"13:33.380","Text":"Then of course we plug this in in the same exact way that we did before,"},{"Start":"13:33.380 ","End":"13:37.320","Text":"and we\u0027ll get this exact same answer."},{"Start":"13:37.600 ","End":"13:40.240","Text":"That is the end of this lesson."},{"Start":"13:40.240 ","End":"13:45.560","Text":"I hope you understand Method number 2 and that you rarely have to think about"},{"Start":"13:45.560 ","End":"13:51.140","Text":"the problem at hand and therefore think in which direction the dl is in."},{"Start":"13:51.140 ","End":"13:53.915","Text":"Remember it\u0027s always in the perpendicular direction."},{"Start":"13:53.915 ","End":"13:57.725","Text":"The question is, which option of the 2?"},{"Start":"13:57.725 ","End":"14:00.750","Text":"That\u0027s the end of this lesson."}],"ID":21399},{"Watched":false,"Name":"Exercise 5","Duration":"15m 16s","ChapterTopicVideoID":21324,"CourseChapterTopicPlaylistID":99469,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.650","Text":"Hello. In this lesson,"},{"Start":"00:01.650 ","End":"00:03.690","Text":"we\u0027re going to be answering the following question."},{"Start":"00:03.690 ","End":"00:08.085","Text":"3 infinite wires are placed parallel to the z-axis."},{"Start":"00:08.085 ","End":"00:11.325","Text":"Their positions are at 0, 0,"},{"Start":"00:11.325 ","End":"00:15.255","Text":"at 5, 2, and at 5 negative 2,"},{"Start":"00:15.255 ","End":"00:22.005","Text":"and the direction of current in each wire is I_1 is 3 amps coming out of the page,"},{"Start":"00:22.005 ","End":"00:25.040","Text":"I_2 is 5 amps going into the page,"},{"Start":"00:25.040 ","End":"00:28.953","Text":"and I_3 is 4 amps or so into the page."},{"Start":"00:28.953 ","End":"00:32.660","Text":"The question is, where along the x-axis"},{"Start":"00:32.660 ","End":"00:37.280","Text":"does the y component of the magnetic field equal 0?"},{"Start":"00:37.280 ","End":"00:40.320","Text":"We\u0027re looking somewhere along the x-axis,"},{"Start":"00:40.320 ","End":"00:42.845","Text":"so let\u0027s say over here,"},{"Start":"00:42.845 ","End":"00:45.559","Text":"where the y component,"},{"Start":"00:45.559 ","End":"00:51.440","Text":"so the y-axis is going up like so of the magnetic fields."},{"Start":"00:51.440 ","End":"00:53.825","Text":"B_y is equal to 0."},{"Start":"00:53.825 ","End":"00:58.350","Text":"We\u0027re trying to find this position along the x-axis."},{"Start":"00:59.200 ","End":"01:05.290","Text":"The first thing that we\u0027re going to do is we\u0027re going to say that this length over here"},{"Start":"01:05.290 ","End":"01:10.885","Text":"between the origin and our point along the x-axis,"},{"Start":"01:10.885 ","End":"01:14.290","Text":"let\u0027s call that a distance of x,"},{"Start":"01:14.290 ","End":"01:20.090","Text":"and we\u0027ve previously seen that the magnetic field due to an infinite wire,"},{"Start":"01:20.090 ","End":"01:22.015","Text":"so we saw this a few lessons ago,"},{"Start":"01:22.015 ","End":"01:27.190","Text":"is equal to Mu Naught multiplied by I, the current,"},{"Start":"01:27.190 ","End":"01:30.385","Text":"and divided by 2Pi r,"},{"Start":"01:30.385 ","End":"01:38.485","Text":"where r is the distance between the wire and the point that we\u0027re looking at,"},{"Start":"01:38.485 ","End":"01:40.885","Text":"that we\u0027re measuring the magnetic field at."},{"Start":"01:40.885 ","End":"01:43.730","Text":"Let\u0027s begin with B_1."},{"Start":"01:43.730 ","End":"01:51.990","Text":"This is the magnetic field at this point over here due to wire number 1."},{"Start":"01:52.070 ","End":"01:57.800","Text":"That\u0027s going to be equal to Mu Naught multiplied by the"},{"Start":"01:57.800 ","End":"02:03.215","Text":"current in wire number 1 and divided by 2Pi,"},{"Start":"02:03.215 ","End":"02:05.990","Text":"and the distance between the wire and the point that we\u0027re"},{"Start":"02:05.990 ","End":"02:10.870","Text":"measuring is this distance over here x."},{"Start":"02:11.030 ","End":"02:13.200","Text":"That\u0027s the magnitude,"},{"Start":"02:13.200 ","End":"02:15.855","Text":"and what about the direction?"},{"Start":"02:15.855 ","End":"02:17.615","Text":"From the right-hand rule,"},{"Start":"02:17.615 ","End":"02:20.570","Text":"we saw that there\u0027s 1 of the rules that we can"},{"Start":"02:20.570 ","End":"02:24.095","Text":"use as we point the thumb in the direction of the current,"},{"Start":"02:24.095 ","End":"02:27.500","Text":"and then our fingers point in the direction of the magnetic field."},{"Start":"02:27.500 ","End":"02:32.360","Text":"Of course, the magnetic field forms concentric circles around the wire."},{"Start":"02:32.360 ","End":"02:34.730","Text":"If we put our thumb towards us,"},{"Start":"02:34.730 ","End":"02:39.740","Text":"we see that the magnetic field is going in this direction."},{"Start":"02:39.740 ","End":"02:42.340","Text":"Which means that this point over here,"},{"Start":"02:42.340 ","End":"02:44.015","Text":"it\u0027s going like so,"},{"Start":"02:44.015 ","End":"02:46.310","Text":"which really means that at this point,"},{"Start":"02:46.310 ","End":"02:48.980","Text":"the magnetic field is in fact pointing in"},{"Start":"02:48.980 ","End":"02:53.240","Text":"the positive y-direction so we can add that in over here."},{"Start":"02:53.240 ","End":"02:58.470","Text":"Now let\u0027s look at B_2."},{"Start":"02:58.470 ","End":"03:00.240","Text":"B_2 as we can see,"},{"Start":"03:00.240 ","End":"03:08.970","Text":"is equal to Mu Naught I_2 divided by 2Pi r,"},{"Start":"03:08.970 ","End":"03:13.280","Text":"and we can see that we\u0027re going to have some angle over here."},{"Start":"03:13.280 ","End":"03:19.070","Text":"Remember that the direction of the magnetic field is always going"},{"Start":"03:19.070 ","End":"03:25.170","Text":"to be perpendicular to our radius."},{"Start":"03:25.170 ","End":"03:28.475","Text":"Here, our radius was along the x-axis and"},{"Start":"03:28.475 ","End":"03:32.030","Text":"our magnetic field was only in the y-direction,"},{"Start":"03:32.030 ","End":"03:37.640","Text":"and of course the y-direction is perpendicular to the x direction."},{"Start":"03:37.640 ","End":"03:42.060","Text":"Now, if we again use the right-hand rule,"},{"Start":"03:42.320 ","End":"03:47.450","Text":"so our thumb is pointing inside the page and here we can see that"},{"Start":"03:47.450 ","End":"03:52.085","Text":"our magnetic field is going like so in this direction."},{"Start":"03:52.085 ","End":"04:00.485","Text":"If we complete the circle like so and imagine that this is a perfect circle."},{"Start":"04:00.485 ","End":"04:07.950","Text":"We can see that the magnetic field is going to be something in this direction."},{"Start":"04:07.950 ","End":"04:11.430","Text":"This is our B_2 and of course,"},{"Start":"04:11.430 ","End":"04:15.645","Text":"this over here is"},{"Start":"04:15.645 ","End":"04:22.400","Text":"our r. The angle between our magnetic field and our r is 90 degrees."},{"Start":"04:22.400 ","End":"04:24.080","Text":"This is what we just said,"},{"Start":"04:24.080 ","End":"04:26.720","Text":"it\u0027s always going to be 90 degrees."},{"Start":"04:26.720 ","End":"04:30.590","Text":"Let\u0027s label this angle over here, Alpha,"},{"Start":"04:30.590 ","End":"04:32.914","Text":"so if this is Alpha and this is 90 degrees,"},{"Start":"04:32.914 ","End":"04:37.190","Text":"this angle over here is 90 minus Alpha."},{"Start":"04:37.190 ","End":"04:44.300","Text":"If I take this over here and I make this a 90-degree angle as well,"},{"Start":"04:44.300 ","End":"04:46.265","Text":"this is a right angle triangle,"},{"Start":"04:46.265 ","End":"04:49.990","Text":"so this angle over here is also Alpha."},{"Start":"04:49.990 ","End":"04:54.685","Text":"I know that this length over here is 2 because I\u0027ve given that,"},{"Start":"04:54.685 ","End":"04:56.990","Text":"and over here I have x."},{"Start":"04:56.990 ","End":"05:03.440","Text":"But the total distance from the origin until this point along the x-axis is 5 which"},{"Start":"05:03.440 ","End":"05:10.555","Text":"means that the distance over here is going to be 5 minus x,"},{"Start":"05:10.555 ","End":"05:15.195","Text":"and then I can use my trig identities."},{"Start":"05:15.195 ","End":"05:17.445","Text":"First of all, my B_2,"},{"Start":"05:17.445 ","End":"05:19.995","Text":"what direction is this?"},{"Start":"05:19.995 ","End":"05:22.300","Text":"I can see if this is Alpha,"},{"Start":"05:22.300 ","End":"05:25.340","Text":"my x-direction is going in the negative x-direction,"},{"Start":"05:25.340 ","End":"05:28.880","Text":"so it\u0027s going to be negative cosine of Alpha in"},{"Start":"05:28.880 ","End":"05:34.650","Text":"the x-direction and then the y-direction is plus sine of Alpha."},{"Start":"05:34.760 ","End":"05:39.410","Text":"Of course, all I care about is my y-direction because I\u0027m"},{"Start":"05:39.410 ","End":"05:44.520","Text":"trying to find where my y component of the B field is equal to 0."},{"Start":"05:44.740 ","End":"05:48.940","Text":"Now let\u0027s see what the y component is."},{"Start":"05:48.940 ","End":"05:51.020","Text":"I have sine of Alpha,"},{"Start":"05:51.020 ","End":"05:52.190","Text":"so as I know,"},{"Start":"05:52.190 ","End":"05:56.790","Text":"that\u0027s equal to the opposite over the hypotenuse."},{"Start":"05:57.920 ","End":"06:01.205","Text":"If this is my angle Alpha,"},{"Start":"06:01.205 ","End":"06:07.789","Text":"so the side opposite is this side over here so the opposite is 5 minus x."},{"Start":"06:07.789 ","End":"06:11.540","Text":"It\u0027s opposite to my Alpha and divided by the hypotenuse,"},{"Start":"06:11.540 ","End":"06:16.820","Text":"which is of course my distance r. Then from the same triangle,"},{"Start":"06:16.820 ","End":"06:19.535","Text":"I can use Pythagoras to calculate what r is,"},{"Start":"06:19.535 ","End":"06:21.065","Text":"so r is, of course,"},{"Start":"06:21.065 ","End":"06:25.940","Text":"equal to this side squared so I have 5 minus x^2"},{"Start":"06:25.940 ","End":"06:31.565","Text":"plus this side squared, so plus 2^2."},{"Start":"06:31.565 ","End":"06:35.600","Text":"Now, I can pop this in to this equation over here,"},{"Start":"06:35.600 ","End":"06:38.689","Text":"and of course I just want my y components."},{"Start":"06:38.689 ","End":"06:47.080","Text":"I\u0027ll say that B_2y is equal to Mu Naught multiplied by I_2 divided by 2Pir."},{"Start":"06:47.080 ","End":"06:49.185","Text":"I\u0027ll soon substitute this in,"},{"Start":"06:49.185 ","End":"06:53.270","Text":"and this is multiplied by just the y component,"},{"Start":"06:53.270 ","End":"06:59.415","Text":"which is sine Alpha in the y-direction so that is equal to"},{"Start":"06:59.415 ","End":"07:07.940","Text":"5 minus x divided by r. Then here I have in the denominator r^2,"},{"Start":"07:07.940 ","End":"07:13.360","Text":"so this is simply equal to Mu Naught I_2,"},{"Start":"07:13.360 ","End":"07:20.520","Text":"5 minus x divided by 2Pi r^2,"},{"Start":"07:20.520 ","End":"07:28.600","Text":"where I^2 is 5 minus x^2 plus 2^2, which is 4."},{"Start":"07:28.790 ","End":"07:34.090","Text":"Now, let\u0027s move on to B_3."},{"Start":"07:34.090 ","End":"07:41.280","Text":"I\u0027m just going to rub out all the details for our B_2."},{"Start":"07:41.280 ","End":"07:45.655","Text":"Now I\u0027m looking at this y over here."},{"Start":"07:45.655 ","End":"07:48.220","Text":"Again, I point my thumb in the direction of the current,"},{"Start":"07:48.220 ","End":"07:50.590","Text":"which means inside of the page."},{"Start":"07:50.590 ","End":"07:56.395","Text":"That means that my magnetic field is going in this direction again."},{"Start":"07:56.395 ","End":"07:58.060","Text":"Now what I\u0027m going to do,"},{"Start":"07:58.060 ","End":"08:00.940","Text":"is I\u0027m going to complete the circle for"},{"Start":"08:00.940 ","End":"08:04.990","Text":"the magnetic field and of course imagine that this is a perfect circle."},{"Start":"08:04.990 ","End":"08:08.965","Text":"But what we can see is that at this point over here,"},{"Start":"08:08.965 ","End":"08:16.580","Text":"the direction of the magnetic field is in this direction, like so."},{"Start":"08:18.390 ","End":"08:22.990","Text":"First of all, let\u0027s again draw our r,"},{"Start":"08:22.990 ","End":"08:25.960","Text":"so this is our r. Again,"},{"Start":"08:25.960 ","End":"08:28.075","Text":"from the definition of the equation,"},{"Start":"08:28.075 ","End":"08:31.480","Text":"we know that the angle between the magnetic field and"},{"Start":"08:31.480 ","End":"08:35.905","Text":"the radius is 90 degrees, and of course,"},{"Start":"08:35.905 ","End":"08:38.665","Text":"because of the symmetry in the question,"},{"Start":"08:38.665 ","End":"08:43.900","Text":"so our I_2 and I_3 are located at the exact same positions."},{"Start":"08:43.900 ","End":"08:47.485","Text":"Just 1 is above the x-axis and the other 1 is below."},{"Start":"08:47.485 ","End":"08:50.665","Text":"We know that this angle is also going to be Alpha,"},{"Start":"08:50.665 ","End":"08:59.010","Text":"just like this angle was because the vectors are just mirror images of 1 another."},{"Start":"08:59.010 ","End":"09:04.045","Text":"What we can see is that we\u0027re going to get the same equation for B_3,"},{"Start":"09:04.045 ","End":"09:07.090","Text":"but the only difference is first of all,"},{"Start":"09:07.090 ","End":"09:10.360","Text":"we\u0027re going to be using I_3 and the other difference is"},{"Start":"09:10.360 ","End":"09:13.765","Text":"that the x-component is going to be a positive,"},{"Start":"09:13.765 ","End":"09:17.800","Text":"because we can see that the x-component is pointing in the positive x-direction."},{"Start":"09:17.800 ","End":"09:20.440","Text":"But anyway, it doesn\u0027t matter, we just want the y component,"},{"Start":"09:20.440 ","End":"09:23.110","Text":"so we can just write B_3y,"},{"Start":"09:23.110 ","End":"09:27.760","Text":"which is exactly like B_2y side from the current."},{"Start":"09:27.760 ","End":"09:31.240","Text":"We have Mu Naught multiplied by I_3 this time,"},{"Start":"09:31.240 ","End":"09:33.640","Text":"divided by 2Pi r,"},{"Start":"09:33.640 ","End":"09:36.910","Text":"multiplied by, so the sine Alpha is going to be"},{"Start":"09:36.910 ","End":"09:41.890","Text":"the exact same thing because this distance is still going to be 5 minus x."},{"Start":"09:41.890 ","End":"09:45.745","Text":"The radius is the same just because of this symmetry that we have."},{"Start":"09:45.745 ","End":"09:50.665","Text":"If we didn\u0027t have the symmetry than we would have to calculate it again."},{"Start":"09:50.665 ","End":"09:54.880","Text":"What we\u0027re left with is Mu Naught I_3,"},{"Start":"09:54.880 ","End":"09:59.450","Text":"5 minus x divided by"},{"Start":"09:59.450 ","End":"10:06.800","Text":"2Pi 5 minus x^2 plus 4."},{"Start":"10:08.160 ","End":"10:14.275","Text":"First of all, we can see that we chose this random point over here in the middle,"},{"Start":"10:14.275 ","End":"10:18.415","Text":"but we could have chosen a point more in the negative x-direction"},{"Start":"10:18.415 ","End":"10:23.605","Text":"or in the further along the x-axis in the positive direction."},{"Start":"10:23.605 ","End":"10:25.600","Text":"We just chose this as a random point,"},{"Start":"10:25.600 ","End":"10:27.190","Text":"but as you\u0027re going to see now,"},{"Start":"10:27.190 ","End":"10:28.615","Text":"it doesn\u0027t even matter,"},{"Start":"10:28.615 ","End":"10:31.315","Text":"we\u0027re still going to be able to solve this."},{"Start":"10:31.315 ","End":"10:37.885","Text":"The sum of all of the magnetic fields in the y direction is equal to."},{"Start":"10:37.885 ","End":"10:43.045","Text":"First of all, we can see that everywhere we have Mu Naught divided by 2Pi."},{"Start":"10:43.045 ","End":"10:48.700","Text":"Let\u0027s just take that out over here and this is multiplied by."},{"Start":"10:48.700 ","End":"10:52.780","Text":"First of all, we have I_1 divided by x."},{"Start":"10:52.780 ","End":"10:58.930","Text":"Then over here we have I_2,"},{"Start":"10:58.930 ","End":"11:02.065","Text":"and then we have multiplied by all of this and I_3."},{"Start":"11:02.065 ","End":"11:09.745","Text":"What we\u0027re going to do is we\u0027re going to have I_2 plus I_3."},{"Start":"11:09.745 ","End":"11:13.435","Text":"Because I_2 and I_3 have the exact same coefficients."},{"Start":"11:13.435 ","End":"11:23.260","Text":"I_2 and I_3 are multiplied by 5 minus x divided by 5 minus x^2 plus 4."},{"Start":"11:23.260 ","End":"11:25.765","Text":"If you open the brackets, you get the same answer."},{"Start":"11:25.765 ","End":"11:28.180","Text":"Now let\u0027s plug in our values,"},{"Start":"11:28.180 ","End":"11:31.780","Text":"so we have Mu Naught divided by 2Pi,"},{"Start":"11:31.780 ","End":"11:35.140","Text":"and then our I_1 is equal to 3 amps."},{"Start":"11:35.140 ","End":"11:42.910","Text":"We have 3 divided by x plus I_2 is 5 amps plus I_3 is 4."},{"Start":"11:42.910 ","End":"11:45.400","Text":"We have 5 plus 4, which, of course,"},{"Start":"11:45.400 ","End":"11:54.830","Text":"we know is 9 multiplied by 5 minus x divided by 5 -minus x^2 plus 4."},{"Start":"11:55.350 ","End":"11:58.630","Text":"In the question, I\u0027m trying to find the point where"},{"Start":"11:58.630 ","End":"12:03.025","Text":"the magnetic field over here in the y direction is equal to zero."},{"Start":"12:03.025 ","End":"12:05.095","Text":"I\u0027m setting that is equal to zero."},{"Start":"12:05.095 ","End":"12:09.850","Text":"Of course, my Mu Naught divided by 2 pi doesn\u0027t affect anything."},{"Start":"12:09.850 ","End":"12:14.470","Text":"What I need is that everything inside the square brackets will be equal to zero."},{"Start":"12:14.470 ","End":"12:17.800","Text":"I can just rewrite this,"},{"Start":"12:17.800 ","End":"12:20.710","Text":"where I make a common denominator."},{"Start":"12:20.710 ","End":"12:31.285","Text":"What I\u0027m going to be left with is 3 multiplied by 5 minus x^2 plus 12 plus"},{"Start":"12:31.285 ","End":"12:38.065","Text":"9x multiplied by 5 minus x divided by"},{"Start":"12:38.065 ","End":"12:47.470","Text":"x multiplied by 5 minus x^2 plus 4 is equal to 0."},{"Start":"12:47.470 ","End":"12:55.645","Text":"I just made a common denominator and didn\u0027t simplify the numerator."},{"Start":"12:55.645 ","End":"13:01.330","Text":"What I can do is I can see that my denominator can just stay this,"},{"Start":"13:01.330 ","End":"13:04.930","Text":"all I need is that my numerator will be equal to 0,"},{"Start":"13:04.930 ","End":"13:06.475","Text":"or will cancel out."},{"Start":"13:06.475 ","End":"13:10.990","Text":"If I multiply both sides by the denominator, that also works."},{"Start":"13:10.990 ","End":"13:19.460","Text":"But either way, I can see that I have a quadratic equation in my numerator."},{"Start":"13:20.340 ","End":"13:23.530","Text":"Let\u0027s open up all of our brackets."},{"Start":"13:23.530 ","End":"13:28.600","Text":"What we get is 3 times 5^2."},{"Start":"13:28.600 ","End":"13:30.520","Text":"We\u0027re just going to open all of this up."},{"Start":"13:30.520 ","End":"13:39.325","Text":"What we\u0027re going to have is 75 minus 30x plus 3x^2."},{"Start":"13:39.325 ","End":"13:45.070","Text":"I just opened up this bracket and multiplied it by 3 plus 12."},{"Start":"13:45.070 ","End":"13:51.680","Text":"Then I\u0027m going to multiply this bracket by 9x plus 45x minus 9x^2."},{"Start":"13:54.360 ","End":"13:57.835","Text":"This is of course equal to 0."},{"Start":"13:57.835 ","End":"14:01.750","Text":"Now if I cancel out all my like terms,"},{"Start":"14:01.750 ","End":"14:04.900","Text":"so what I\u0027m left with is negative 6x^2 plus"},{"Start":"14:04.900 ","End":"14:12.770","Text":"15x plus 87 is equal to 0."},{"Start":"14:13.260 ","End":"14:20.785","Text":"Now what we can do is we can plug all of this in to our quadratic equation solver,"},{"Start":"14:20.785 ","End":"14:23.620","Text":"that formula for solving quadratic equations."},{"Start":"14:23.620 ","End":"14:31.884","Text":"The first answer for x_1 is equal to negative 2.76 to 2 decimal places,"},{"Start":"14:31.884 ","End":"14:36.925","Text":"and x_2 is equal to 5.26,"},{"Start":"14:36.925 ","End":"14:39.830","Text":"again to 2 decimal places."},{"Start":"14:39.900 ","End":"14:44.860","Text":"What we can see is that if we\u0027re looking at X_1,"},{"Start":"14:44.860 ","End":"14:49.090","Text":"so that\u0027s somewhere over here and the negative realm."},{"Start":"14:49.090 ","End":"14:51.025","Text":"X_1 is over here,"},{"Start":"14:51.025 ","End":"14:53.470","Text":"and if we look at this point over here,"},{"Start":"14:53.470 ","End":"14:56.665","Text":"the y component of our magnetic field will cancel out,"},{"Start":"14:56.665 ","End":"15:00.265","Text":"or alternatively, if we look over here,"},{"Start":"15:00.265 ","End":"15:04.375","Text":"so this is X_2 at 5.26."},{"Start":"15:04.375 ","End":"15:07.000","Text":"Just after these two wires, so again,"},{"Start":"15:07.000 ","End":"15:09.790","Text":"the y component of the magnetic field will cancel out,"},{"Start":"15:09.790 ","End":"15:13.480","Text":"which is exactly what we were being asked to find."},{"Start":"15:13.480 ","End":"15:17.240","Text":"That is the end of our lesson."}],"ID":21400},{"Watched":false,"Name":"Force Between Two Current Carrying Wires","Duration":"10m 43s","ChapterTopicVideoID":21325,"CourseChapterTopicPlaylistID":99469,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.210","Text":"Hello. In this lesson we\u0027re going to be deriving"},{"Start":"00:03.210 ","End":"00:06.975","Text":"the force between 2 current carrying wires."},{"Start":"00:06.975 ","End":"00:13.065","Text":"Here we have 2 infinitely long wires that are carrying currents."},{"Start":"00:13.065 ","End":"00:17.115","Text":"In here we have I_1 and then the bus and why we have I_2,"},{"Start":"00:17.115 ","End":"00:19.890","Text":"and the distance between the 2 wires is"},{"Start":"00:19.890 ","End":"00:25.410","Text":"d. We\u0027re going to derive the equation for the force between these 2 wires."},{"Start":"00:25.410 ","End":"00:29.625","Text":"The equation that will get is that dF by dl."},{"Start":"00:29.625 ","End":"00:36.420","Text":"The force per unit length is equal to Mu naught multiplied by I_1,"},{"Start":"00:36.420 ","End":"00:40.560","Text":"I_2 divided by 2Pi and d,"},{"Start":"00:40.560 ","End":"00:44.550","Text":"where d is of course the distance between the 2 wires."},{"Start":"00:44.930 ","End":"00:53.405","Text":"We can either work out the force that this wire is exerting on this wire."},{"Start":"00:53.405 ","End":"01:00.230","Text":"What we will take is we will find the force on this wire per unit length."},{"Start":"01:00.230 ","End":"01:02.420","Text":"Here this is dl,"},{"Start":"01:02.420 ","End":"01:11.150","Text":"and the force over here due to this wire over here and vice versa."},{"Start":"01:11.150 ","End":"01:17.780","Text":"What we\u0027ll see is if both of the currents are in the same direction,"},{"Start":"01:17.780 ","End":"01:21.530","Text":"then we have an attraction between the 2 wires,"},{"Start":"01:21.530 ","End":"01:26.180","Text":"and if the directions of the current is opposed to one another,"},{"Start":"01:26.180 ","End":"01:30.880","Text":"then we have repulsion."},{"Start":"01:31.370 ","End":"01:35.205","Text":"Let\u0027s begin deriving this equation."},{"Start":"01:35.205 ","End":"01:37.790","Text":"First of all from Ampere\u0027s Law,"},{"Start":"01:37.790 ","End":"01:44.510","Text":"we know that the magnetic field caused by this wire with"},{"Start":"01:44.510 ","End":"01:48.050","Text":"current I_1 traveling through it is equal"},{"Start":"01:48.050 ","End":"01:53.195","Text":"to Mu naught multiplied by the current flowing through it,"},{"Start":"01:53.195 ","End":"01:57.270","Text":"I_1 divided by 2Pir."},{"Start":"01:59.570 ","End":"02:04.640","Text":"This is the magnetic field due to this upper wire."},{"Start":"02:04.640 ","End":"02:06.350","Text":"Let\u0027s call this wire,"},{"Start":"02:06.350 ","End":"02:09.170","Text":"wire number 1 and this wire, wire number 2."},{"Start":"02:09.170 ","End":"02:14.635","Text":"Now we want to know the direction of the magnetic field."},{"Start":"02:14.635 ","End":"02:17.225","Text":"We\u0027re going to use the right-hand rule,"},{"Start":"02:17.225 ","End":"02:21.430","Text":"where we point our thumb in the direction of the current."},{"Start":"02:21.430 ","End":"02:26.555","Text":"Then our fingers curl in the direction of the magnetic field."},{"Start":"02:26.555 ","End":"02:31.430","Text":"Here we\u0027ll have our thumb pointing in this direction,"},{"Start":"02:31.430 ","End":"02:40.500","Text":"and then we have our fingers curling in like so."},{"Start":"02:42.070 ","End":"02:44.080","Text":"What we can see,"},{"Start":"02:44.080 ","End":"02:46.720","Text":"so this is the direction of course of our current,"},{"Start":"02:46.720 ","End":"02:49.345","Text":"and this is the direction of the magnetic field."},{"Start":"02:49.345 ","End":"02:50.830","Text":"What does that mean?"},{"Start":"02:50.830 ","End":"02:54.715","Text":"That means that we know the magnetic field around"},{"Start":"02:54.715 ","End":"02:59.995","Text":"a current carrying wire is in concentric circles."},{"Start":"02:59.995 ","End":"03:07.605","Text":"That means that the magnetic field is traveling in this direction around the wire."},{"Start":"03:07.605 ","End":"03:11.305","Text":"Because our fingers are curling towards us at the top"},{"Start":"03:11.305 ","End":"03:15.985","Text":"and curling back in towards the page at the bottom."},{"Start":"03:15.985 ","End":"03:19.475","Text":"Now if we look at I_2,"},{"Start":"03:19.475 ","End":"03:22.890","Text":"not I_2 rather at wire number 2,"},{"Start":"03:22.890 ","End":"03:24.720","Text":"at this wire over here."},{"Start":"03:24.720 ","End":"03:28.700","Text":"We can see that the wire is located below wire number 1."},{"Start":"03:28.700 ","End":"03:31.835","Text":"Which means that the magnetic field,"},{"Start":"03:31.835 ","End":"03:35.690","Text":"which is acting on wire number 2,"},{"Start":"03:35.690 ","End":"03:40.595","Text":"is acting in this direction into the page."},{"Start":"03:40.595 ","End":"03:42.290","Text":"This of course is B_1."},{"Start":"03:42.290 ","End":"03:47.415","Text":"Don\u0027t get confused,"},{"Start":"03:47.415 ","End":"03:53.180","Text":"B_1 is the magnetic field caused by wire number 1."},{"Start":"03:53.180 ","End":"03:57.040","Text":"But of course it\u0027s acting on wire number 2."},{"Start":"03:57.040 ","End":"04:04.550","Text":"Now what I want to do is I want to calculate exactly what my magnetic field is over here."},{"Start":"04:04.550 ","End":"04:09.125","Text":"Of course, instead of r I\u0027m substituting in the distance,"},{"Start":"04:09.125 ","End":"04:12.080","Text":"r is the distance from the wire."},{"Start":"04:12.080 ","End":"04:16.760","Text":"The distance from the wire where I\u0027m measuring the magnetic field is this distance d,"},{"Start":"04:16.760 ","End":"04:27.340","Text":"so r is equal to d. This is just equal to Mu naught I_1 divided by 2Pid."},{"Start":"04:27.500 ","End":"04:35.650","Text":"Now what I want to do is I want to calculate the force on wire number 2,"},{"Start":"04:35.650 ","End":"04:40.850","Text":"so dF, and then I\u0027m working out the magnitude first of all."},{"Start":"04:40.850 ","End":"04:46.610","Text":"We know that the force on a wire from Lawrence\u0027s law,"},{"Start":"04:46.610 ","End":"04:50.390","Text":"it\u0027s equal to the current on that wire."},{"Start":"04:50.390 ","End":"04:56.600","Text":"Here it\u0027s I_2 multiplied by the unit length."},{"Start":"04:56.600 ","End":"05:00.230","Text":"Here we\u0027re measuring the force on wire number 2,"},{"Start":"05:00.230 ","End":"05:03.680","Text":"which is of course due to wire number 1."},{"Start":"05:03.680 ","End":"05:07.765","Text":"Our unit of length is this over here on wire number 2."},{"Start":"05:07.765 ","End":"05:09.670","Text":"This is dl_2."},{"Start":"05:09.670 ","End":"05:19.745","Text":"This I_2 multiplied by dl_2 cross multiplied with the magnetic field at that point,"},{"Start":"05:19.745 ","End":"05:22.910","Text":"which is of course B_1."},{"Start":"05:22.910 ","End":"05:25.240","Text":"Which we know is a vector."},{"Start":"05:25.240 ","End":"05:31.290","Text":"This is simply going to be equal 2."},{"Start":"05:31.290 ","End":"05:34.130","Text":"We have I_2,"},{"Start":"05:34.130 ","End":"05:35.585","Text":"which is a constant,"},{"Start":"05:35.585 ","End":"05:44.425","Text":"and then we have dl_2 multiplied by the magnitude of B_1,"},{"Start":"05:44.425 ","End":"05:51.070","Text":"which we said was equal to Mu naught I_1 divided by 2Pid."},{"Start":"05:51.070 ","End":"05:58.955","Text":"Then we have to multiply by sine of the angle between our dl and our B_1."},{"Start":"05:58.955 ","End":"06:03.755","Text":"Our dl, we can define this direction as the x direction."},{"Start":"06:03.755 ","End":"06:06.650","Text":"Our B_1 is going into the page."},{"Start":"06:06.650 ","End":"06:08.900","Text":"We can say that if something is going into"},{"Start":"06:08.900 ","End":"06:11.630","Text":"the pages in the z direction or the y direction,"},{"Start":"06:11.630 ","End":"06:13.085","Text":"it doesn\u0027t really make a difference."},{"Start":"06:13.085 ","End":"06:16.500","Text":"It\u0027s perpendicular to our dl."},{"Start":"06:16.500 ","End":"06:22.390","Text":"That\u0027s what we can see. We\u0027re multiplying by sine of 90 degrees,"},{"Start":"06:22.390 ","End":"06:26.465","Text":"and sine of 90 degrees is of course 1."},{"Start":"06:26.465 ","End":"06:29.125","Text":"We just multiply all of this by 1."},{"Start":"06:29.125 ","End":"06:32.825","Text":"Now all I have to do is divide by dl."},{"Start":"06:32.825 ","End":"06:37.850","Text":"What I have is dF magnitude divided by dl,"},{"Start":"06:37.850 ","End":"06:45.585","Text":"which is simply going to be equal to Mu naught I_1 multiplied by I_2."},{"Start":"06:45.585 ","End":"06:50.860","Text":"All of this is divided by 2Pid."},{"Start":"06:51.740 ","End":"06:58.460","Text":"Of course here, because we know from Newton\u0027s law that the force acting on"},{"Start":"06:58.460 ","End":"07:05.040","Text":"wire number 2 is going to be equal and opposite to the force acting on wire number 1."},{"Start":"07:05.040 ","End":"07:08.990","Text":"That means that the magnitude of the force acting on"},{"Start":"07:08.990 ","End":"07:12.740","Text":"wire number 2 is equal to the force acting on wire number 1,"},{"Start":"07:12.740 ","End":"07:16.470","Text":"just the direction will be opposite."},{"Start":"07:16.700 ","End":"07:21.125","Text":"That\u0027s why, because here I have the magnitude of the force."},{"Start":"07:21.125 ","End":"07:24.590","Text":"I could just leave it as dl instead of dl_2."},{"Start":"07:24.590 ","End":"07:28.580","Text":"Because of course, if I worked out the force on wire number 1,"},{"Start":"07:28.580 ","End":"07:31.925","Text":"the magnitude of the force would be this exact same thing,"},{"Start":"07:31.925 ","End":"07:34.470","Text":"but just dF divided by dl_1,"},{"Start":"07:34.470 ","End":"07:42.300","Text":"and then we could equate the 2 and we\u0027ll see that dF by dl_2 is equal to dF by dl_1."},{"Start":"07:42.300 ","End":"07:44.070","Text":"We would get the same answer."},{"Start":"07:44.070 ","End":"07:48.980","Text":"That\u0027s why I could take away this 2 over here and just write it like so."},{"Start":"07:48.980 ","End":"07:52.115","Text":"First of all, we\u0027ve derived this equation."},{"Start":"07:52.115 ","End":"07:57.070","Text":"But now what we want to do is we want to see the direction."},{"Start":"07:57.070 ","End":"08:00.585","Text":"From the right hand rule,"},{"Start":"08:00.585 ","End":"08:07.460","Text":"what we we\u0027ll have is we have the direction of the force is dl cross B,"},{"Start":"08:07.460 ","End":"08:11.285","Text":"so we can use two versions of the right-hand rule."},{"Start":"08:11.285 ","End":"08:14.464","Text":"But another thing that we can do is we can curl"},{"Start":"08:14.464 ","End":"08:19.235","Text":"our fingers from the x-direction into the z direction."},{"Start":"08:19.235 ","End":"08:24.954","Text":"Then we will get that our force is pointing upwards."},{"Start":"08:24.954 ","End":"08:28.095","Text":"Of course this is dF by dl,"},{"Start":"08:28.095 ","End":"08:31.120","Text":"so it\u0027s pointing upwards."},{"Start":"08:32.180 ","End":"08:38.295","Text":"Or we have dl cross B."},{"Start":"08:38.295 ","End":"08:45.825","Text":"We have our thumb pointing in the dl direction."},{"Start":"08:45.825 ","End":"08:48.170","Text":"Here, dl_2,"},{"Start":"08:48.170 ","End":"08:53.015","Text":"we have our forefinger pointing in the direction of B."},{"Start":"08:53.015 ","End":"08:55.845","Text":"That\u0027s going into the page."},{"Start":"08:55.845 ","End":"09:00.350","Text":"This is in the direction of our B_ 1."},{"Start":"09:00.750 ","End":"09:04.825","Text":"Then that means that we have our middle finger,"},{"Start":"09:04.825 ","End":"09:07.245","Text":"which is of course,"},{"Start":"09:07.245 ","End":"09:12.550","Text":"this will look something like this."},{"Start":"09:12.550 ","End":"09:15.755","Text":"Excuse my drawing skills."},{"Start":"09:15.755 ","End":"09:19.015","Text":"This is the other fingers that you have in your hand."},{"Start":"09:19.015 ","End":"09:26.540","Text":"This is going to be the direction of our force pointing upwards,"},{"Start":"09:26.540 ","End":"09:34.020","Text":"like so we can say that it\u0027s in the y direction."},{"Start":"09:34.060 ","End":"09:38.365","Text":"Of course we can see that the button wire,"},{"Start":"09:38.365 ","End":"09:42.625","Text":"is being attracted in this positive y-direction."},{"Start":"09:42.625 ","End":"09:44.980","Text":"From Newton\u0027s third law,"},{"Start":"09:44.980 ","End":"09:50.830","Text":"we know that this wire number 1 is going to experience an equal and opposite force."},{"Start":"09:50.830 ","End":"09:55.765","Text":"Wire number 1 is going to be pulled in this direction,"},{"Start":"09:55.765 ","End":"09:59.485","Text":"so an equal magnitude but in opposite direction."},{"Start":"09:59.485 ","End":"10:04.824","Text":"That\u0027s why we can see when both of the currents are in the same direction,"},{"Start":"10:04.824 ","End":"10:07.990","Text":"we have attraction, so they\u0027re attracted to one another."},{"Start":"10:07.990 ","End":"10:11.070","Text":"Of course, if we flipped one of these currents around,"},{"Start":"10:11.070 ","End":"10:15.405","Text":"then we would get opposite directions for our force."},{"Start":"10:15.405 ","End":"10:18.665","Text":"Then of course we would get repulsion."},{"Start":"10:18.665 ","End":"10:22.775","Text":"Feel free to pause this video and just practice that."},{"Start":"10:22.775 ","End":"10:28.480","Text":"Try and figure out what the force on I_1 would be or on wire number 1,"},{"Start":"10:28.480 ","End":"10:32.185","Text":"due to wire number 2 in this configuration."},{"Start":"10:32.185 ","End":"10:35.150","Text":"Then try flipping the direction of one of"},{"Start":"10:35.150 ","End":"10:40.525","Text":"the currents and then see if you really do get this repulsion."},{"Start":"10:40.525 ","End":"10:43.660","Text":"That\u0027s the end of this lesson."}],"ID":21401},{"Watched":false,"Name":"Exercise 6","Duration":"9m 19s","ChapterTopicVideoID":21326,"CourseChapterTopicPlaylistID":99469,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.980","Text":"Hello. In this lesson,"},{"Start":"00:01.980 ","End":"00:04.874","Text":"we\u0027re going to be answering the following question:"},{"Start":"00:04.874 ","End":"00:14.085","Text":"Two very long wires are hung from a ceiling via 2 strings of equal and unknown length."},{"Start":"00:14.085 ","End":"00:20.169","Text":"A current of 100 amps flows through each wire but in the opposite direction."},{"Start":"00:20.169 ","End":"00:22.835","Text":"Here the current is flowing into the page,"},{"Start":"00:22.835 ","End":"00:25.925","Text":"and here the current is flowing out of the page."},{"Start":"00:25.925 ","End":"00:30.260","Text":"There is a 45-degree angle between the strings and their mass"},{"Start":"00:30.260 ","End":"00:34.805","Text":"per unit length is 2 grams per meter."},{"Start":"00:34.805 ","End":"00:41.160","Text":"What we want to do is we want to find the distance between the wires."},{"Start":"00:42.410 ","End":"00:45.020","Text":"If we\u0027re trying to find the distance,"},{"Start":"00:45.020 ","End":"00:48.754","Text":"then we\u0027re expecting that the distance is a constant."},{"Start":"00:48.754 ","End":"00:51.755","Text":"How will this distance be constant?"},{"Start":"00:51.755 ","End":"00:56.670","Text":"That is if these 2 wires are at rest."},{"Start":"00:56.750 ","End":"01:01.760","Text":"They\u0027re both position like so and they\u0027re not moving."},{"Start":"01:01.760 ","End":"01:04.325","Text":"We know that if something is at rest,"},{"Start":"01:04.325 ","End":"01:09.725","Text":"that means that the sum of all of the forces on it is equal to 0."},{"Start":"01:09.725 ","End":"01:15.505","Text":"Let\u0027s write out my free flow diagram for my forces."},{"Start":"01:15.505 ","End":"01:22.055","Text":"Here, I\u0027m going to have the tension in the string going down in this direction."},{"Start":"01:22.055 ","End":"01:24.035","Text":"I have my mg,"},{"Start":"01:24.035 ","End":"01:31.040","Text":"and as we saw in the lesson dealing with the force between 2 current-carrying wires."},{"Start":"01:31.040 ","End":"01:35.255","Text":"We saw that if the currents are in opposing directions,"},{"Start":"01:35.255 ","End":"01:39.065","Text":"then the force between them is a repulsive force,"},{"Start":"01:39.065 ","End":"01:45.900","Text":"which means that the force is going to be in this direction."},{"Start":"01:46.310 ","End":"01:50.940","Text":"Now I can write out my force equations."},{"Start":"01:50.940 ","End":"01:58.835","Text":"I can see that my tension over here has components in both the x and the y direction."},{"Start":"01:58.835 ","End":"02:03.020","Text":"What I\u0027m going to do is I\u0027m going to draw this right-angle triangle over"},{"Start":"02:03.020 ","End":"02:07.292","Text":"here that goes straight through this 45-degree angle,"},{"Start":"02:07.292 ","End":"02:11.180","Text":"and I\u0027m going to call this angle over here Theta."},{"Start":"02:11.180 ","End":"02:16.820","Text":"Now, I\u0027ll write out the sum of all the forces in the x-direction."},{"Start":"02:16.820 ","End":"02:23.780","Text":"What I have is F in the positive x-direction minus T,"},{"Start":"02:23.780 ","End":"02:27.840","Text":"and then cosine of Theta,"},{"Start":"02:28.420 ","End":"02:32.060","Text":"as we can see, is in the negative x-direction,"},{"Start":"02:32.060 ","End":"02:35.460","Text":"so that\u0027s why I have a minus over here."},{"Start":"02:35.460 ","End":"02:42.320","Text":"The sum of all of my forces in the y-direction is going to"},{"Start":"02:42.320 ","End":"02:48.620","Text":"be equal to T multiplied by sine of Theta in the positive y direction,"},{"Start":"02:48.620 ","End":"02:53.520","Text":"and mg in the negative y direction, so negative mg."},{"Start":"02:53.800 ","End":"02:55.910","Text":"In the previous lesson,"},{"Start":"02:55.910 ","End":"03:00.680","Text":"we saw that the equation for the force on a wire"},{"Start":"03:00.680 ","End":"03:06.485","Text":"per unit length was equal to Mu naught I_ 1,"},{"Start":"03:06.485 ","End":"03:11.325","Text":"I_2 divided by 2Pid."},{"Start":"03:11.325 ","End":"03:12.995","Text":"Of course over here,"},{"Start":"03:12.995 ","End":"03:17.570","Text":"because everything is constant as we can see,"},{"Start":"03:17.570 ","End":"03:21.810","Text":"so we can take out the d\u0027s over here."},{"Start":"03:21.810 ","End":"03:27.800","Text":"We can just leave this as F divided by l, force per length."},{"Start":"03:27.800 ","End":"03:30.715","Text":"Because this length is some constant length."},{"Start":"03:30.715 ","End":"03:33.155","Text":"In order to find out the force,"},{"Start":"03:33.155 ","End":"03:37.445","Text":"it\u0027s just equal to Mu naught I_1,"},{"Start":"03:37.445 ","End":"03:44.220","Text":"I_2 divided by 2Pid and multiplied by l,"},{"Start":"03:44.220 ","End":"03:47.555","Text":"which is the length of the wire."},{"Start":"03:47.555 ","End":"03:50.045","Text":"Now of course we can\u0027t write this."},{"Start":"03:50.045 ","End":"03:52.156","Text":"We\u0027re dealing with a very long wire,"},{"Start":"03:52.156 ","End":"03:54.455","Text":"so it\u0027s like dealing with an infinite wire."},{"Start":"03:54.455 ","End":"03:58.469","Text":"We\u0027re just going to leave it as l for the meantime."},{"Start":"03:58.750 ","End":"04:01.880","Text":"Of course, as we said before,"},{"Start":"04:01.880 ","End":"04:05.885","Text":"the sum of the forces is equal to 0 because everything\u0027s at rest."},{"Start":"04:05.885 ","End":"04:09.780","Text":"This and this are both equal to 0."},{"Start":"04:09.780 ","End":"04:14.195","Text":"Now we\u0027ll substitute in our force into this equation and we can already move"},{"Start":"04:14.195 ","End":"04:19.370","Text":"the negative T cosine Theta and the negative mg to the other side."},{"Start":"04:19.370 ","End":"04:27.740","Text":"What we\u0027re going to get is our first equation is going to be Mu naught I_1,"},{"Start":"04:27.740 ","End":"04:33.290","Text":"I_2l divided by 2Pid."},{"Start":"04:33.290 ","End":"04:37.730","Text":"That\u0027s equal to T multiplied by cosine of Theta."},{"Start":"04:37.730 ","End":"04:46.760","Text":"Our second equation is going to be equal to mg,"},{"Start":"04:46.760 ","End":"04:51.320","Text":"which is equal to of course T sine of Theta."},{"Start":"04:51.320 ","End":"04:54.845","Text":"First of all, I can know what my Theta is."},{"Start":"04:54.845 ","End":"04:56.795","Text":"This I know is 90 degrees."},{"Start":"04:56.795 ","End":"05:00.140","Text":"This because we\u0027ve split this right"},{"Start":"05:00.140 ","End":"05:03.800","Text":"over here due to symmetry in the middle of the 45-degree angle,"},{"Start":"05:03.800 ","End":"05:08.335","Text":"so this angle is 22.5 degrees."},{"Start":"05:08.335 ","End":"05:16.275","Text":"Which means that Theta is equal to 67.5 degrees."},{"Start":"05:16.275 ","End":"05:18.335","Text":"That will substitute in soon."},{"Start":"05:18.335 ","End":"05:22.280","Text":"But what I want to do is I want to cancel out my T. What I\u0027m going"},{"Start":"05:22.280 ","End":"05:26.660","Text":"to do is I\u0027m going to do equation 2 divided by equation 1."},{"Start":"05:26.660 ","End":"05:28.490","Text":"What this will give us,"},{"Start":"05:28.490 ","End":"05:29.780","Text":"so the T\u0027s will cancel out."},{"Start":"05:29.780 ","End":"05:32.950","Text":"Then we have sine Theta divided by cosine Theta,"},{"Start":"05:32.950 ","End":"05:34.995","Text":"which is equal to tan Theta."},{"Start":"05:34.995 ","End":"05:42.615","Text":"This is equal to mg divided by Mu naught I_1,"},{"Start":"05:42.615 ","End":"05:49.090","Text":"I_2 l and multiply it by 2Pid."},{"Start":"05:50.210 ","End":"05:52.970","Text":"Now we want to know what the mass is."},{"Start":"05:52.970 ","End":"05:55.100","Text":"The mass of course comes from the string."},{"Start":"05:55.100 ","End":"06:04.195","Text":"We\u0027re told that the mass of the string is per unit length is 2 grams per meter."},{"Start":"06:04.195 ","End":"06:06.440","Text":"Let\u0027s write over here,"},{"Start":"06:06.440 ","End":"06:12.185","Text":"m is equal to this Mu multiplied by the length."},{"Start":"06:12.185 ","End":"06:14.870","Text":"Now we can substitute that in here."},{"Start":"06:14.870 ","End":"06:21.548","Text":"We have Mu length multiplied by g divided by Mu naught,"},{"Start":"06:21.548 ","End":"06:24.880","Text":"don\u0027t get confused between the Mu in the Mu naught,"},{"Start":"06:25.040 ","End":"06:30.915","Text":"I_1, I_2 l multiplied by 2Pid."},{"Start":"06:30.915 ","End":"06:35.530","Text":"Now we can see that our l\u0027s cancel out, so that\u0027s great."},{"Start":"06:35.530 ","End":"06:38.690","Text":"Now what we can do is we can isolate out our d,"},{"Start":"06:38.690 ","End":"06:40.310","Text":"because that\u0027s what we\u0027re trying to find,"},{"Start":"06:40.310 ","End":"06:42.245","Text":"the distance between the strings."},{"Start":"06:42.245 ","End":"06:47.095","Text":"We have that d is equal to Mu naught I_1,"},{"Start":"06:47.095 ","End":"06:54.375","Text":"I_2 multiplied by tan of Theta."},{"Start":"06:54.375 ","End":"06:58.740","Text":"It\u0027s divided by 2Pi Mu,"},{"Start":"06:58.740 ","End":"07:02.640","Text":"which is 2 grams per meter,"},{"Start":"07:02.640 ","End":"07:07.220","Text":"multiplied by g. First of all,"},{"Start":"07:07.220 ","End":"07:09.560","Text":"let\u0027s substitute in all of the values,"},{"Start":"07:09.560 ","End":"07:12.990","Text":"and then we\u0027ll use the calculator and we\u0027ll calculate d."},{"Start":"07:13.280 ","End":"07:19.855","Text":"Mu naught is something that you should have written in your equation sheets."},{"Start":"07:19.855 ","End":"07:24.866","Text":"Mu naught is equal to 4Pi times 10 to"},{"Start":"07:24.866 ","End":"07:32.025","Text":"the power of negative 7 newtons per amp squared."},{"Start":"07:32.025 ","End":"07:37.750","Text":"It gave that a square because you\u0027re meant to write this in your equation books."},{"Start":"07:37.750 ","End":"07:41.174","Text":"Then, what else do we have?"},{"Start":"07:41.174 ","End":"07:43.570","Text":"Now we can start plugging everything in."},{"Start":"07:43.570 ","End":"07:45.805","Text":"What we have is Mu naught."},{"Start":"07:45.805 ","End":"07:50.260","Text":"We have 4Pi times 10 to the negative 7."},{"Start":"07:50.260 ","End":"07:53.395","Text":"Then we have I_1 multiplied by I_2."},{"Start":"07:53.395 ","End":"07:56.165","Text":"Each one has a current of 100 amps."},{"Start":"07:56.165 ","End":"08:00.830","Text":"We can just multiply this by 100 squared and then multiplied"},{"Start":"08:00.830 ","End":"08:05.705","Text":"by tan of Theta where of course Theta is 67.5 degrees."},{"Start":"08:05.705 ","End":"08:12.915","Text":"Tan of 67.5 degrees is approximately equal to 2.41."},{"Start":"08:12.915 ","End":"08:19.265","Text":"Then all of this is divided by 2Pi multiplied by our Mu."},{"Start":"08:19.265 ","End":"08:28.400","Text":"Now notice our Mu from the question is equal to 2 grams per meter."},{"Start":"08:28.400 ","End":"08:33.350","Text":"But we want this in standard units,"},{"Start":"08:33.350 ","End":"08:35.510","Text":"so we want to convert this to kilograms."},{"Start":"08:35.510 ","End":"08:38.810","Text":"What we\u0027re going to do is we\u0027re just going to divide it by 1,000."},{"Start":"08:38.810 ","End":"08:46.930","Text":"What we have is 0.002 and then the units are kilograms per meters, so that\u0027s great."},{"Start":"08:46.930 ","End":"08:50.130","Text":"Now we\u0027re multiplying this by Mu in kilograms,"},{"Start":"08:50.130 ","End":"08:57.425","Text":"so 0.002 kilograms per meters,"},{"Start":"08:57.425 ","End":"08:59.165","Text":"and then multiplied by g,"},{"Start":"08:59.165 ","End":"09:01.510","Text":"which is approximately 10."},{"Start":"09:01.510 ","End":"09:04.580","Text":"Once we plug this into our calculator,"},{"Start":"09:04.580 ","End":"09:10.880","Text":"we\u0027ll get that this is equal to approximately 0.241 and of course,"},{"Start":"09:10.880 ","End":"09:13.210","Text":"the units are meters."},{"Start":"09:13.210 ","End":"09:20.039","Text":"That\u0027s it. That is the distance between the 2 wires and that is the end of the question."}],"ID":21402},{"Watched":false,"Name":"Exercise 7","Duration":"12m 49s","ChapterTopicVideoID":21327,"CourseChapterTopicPlaylistID":99469,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.785","Text":"Hello. In this lesson,"},{"Start":"00:01.785 ","End":"00:04.290","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.290 ","End":"00:06.360","Text":"A current I flows through"},{"Start":"00:06.360 ","End":"00:12.030","Text":"a regular n-sided polygon that is enclosed by circle of radius I."},{"Start":"00:12.030 ","End":"00:13.455","Text":"Question number 1 is,"},{"Start":"00:13.455 ","End":"00:17.820","Text":"what is the magnetic field at the center of the polygon?"},{"Start":"00:17.820 ","End":"00:20.685","Text":"First of all, a regular n-sided polygon."},{"Start":"00:20.685 ","End":"00:26.850","Text":"That means that every side of the polygon has an equal length"},{"Start":"00:26.850 ","End":"00:34.275","Text":"and is n-side so we\u0027re just taking some number that we\u0027re not specifying."},{"Start":"00:34.275 ","End":"00:39.545","Text":"I\u0027m just going to draw some random n-sided polygon."},{"Start":"00:39.545 ","End":"00:42.020","Text":"This has n sides."},{"Start":"00:42.020 ","End":"00:44.750","Text":"I\u0027m not even going to count how many sides that is."},{"Start":"00:44.750 ","End":"00:48.230","Text":"This is enclosed by a circle of"},{"Start":"00:48.230 ","End":"00:53.160","Text":"radius R. I\u0027m just going to redraw this so that it looks a bit better."},{"Start":"00:54.140 ","End":"00:58.024","Text":"Here is the center of our circle,"},{"Start":"00:58.024 ","End":"01:02.600","Text":"and it is a radius R. Here we have n sides,"},{"Start":"01:02.600 ","End":"01:05.285","Text":"it doesn\u0027t really matter how many."},{"Start":"01:05.285 ","End":"01:09.140","Text":"We\u0027re being told that a current I is"},{"Start":"01:09.140 ","End":"01:14.120","Text":"flowing through so let\u0027s say that the current is flowing"},{"Start":"01:14.120 ","End":"01:23.815","Text":"through in this direction so through each one of these sides."},{"Start":"01:23.815 ","End":"01:26.419","Text":"In one of the previous lessons,"},{"Start":"01:26.419 ","End":"01:31.310","Text":"we learned the equation for the magnetic field"},{"Start":"01:31.310 ","End":"01:36.560","Text":"of some current carrying wire,"},{"Start":"01:36.560 ","End":"01:40.670","Text":"of length L at some point over here,"},{"Start":"01:40.670 ","End":"01:43.550","Text":"a distance y below."},{"Start":"01:43.550 ","End":"01:50.120","Text":"We saw from Biot–Savart\u0027s law that the magnetic field for one such current"},{"Start":"01:50.120 ","End":"01:57.230","Text":"carrying wire is equal to Mu_naught multiplied by I multiplied by L,"},{"Start":"01:57.230 ","End":"01:59.525","Text":"the length of the wire,"},{"Start":"01:59.525 ","End":"02:02.005","Text":"divided by 4 Pi,"},{"Start":"02:02.005 ","End":"02:04.310","Text":"and the distance between the point where we\u0027re"},{"Start":"02:04.310 ","End":"02:07.160","Text":"measuring the magnetic fields so here it was y."},{"Start":"02:07.160 ","End":"02:13.109","Text":"Then all of this was multiplied by L divided by 2^2"},{"Start":"02:13.109 ","End":"02:22.090","Text":"plus y^2 and all of this to the power of 1/2."},{"Start":"02:23.000 ","End":"02:30.600","Text":"This is the magnetic field at a certain point due to"},{"Start":"02:30.600 ","End":"02:33.410","Text":"1 wire and then of course we\u0027re just going to multiply it by"},{"Start":"02:33.410 ","End":"02:37.745","Text":"n sides to get the final magnetic field."},{"Start":"02:37.745 ","End":"02:40.375","Text":"Let\u0027s just deal with the direction."},{"Start":"02:40.375 ","End":"02:42.230","Text":"Again, if we use the right-hand rule,"},{"Start":"02:42.230 ","End":"02:45.530","Text":"we point our thumb in the direction of"},{"Start":"02:45.530 ","End":"02:51.005","Text":"the current and our fingers curl around in the direction of the magnetic field."},{"Start":"02:51.005 ","End":"02:55.190","Text":"When we do that, we\u0027ll see that the magnetic field at the center where we\u0027re"},{"Start":"02:55.190 ","End":"03:01.235","Text":"measuring the magnetic field is coming out of the page and"},{"Start":"03:01.235 ","End":"03:08.520","Text":"towards us because our thumb is pointing in this direction"},{"Start":"03:08.520 ","End":"03:17.280","Text":"and then our fingers are going to curl in like so."},{"Start":"03:17.870 ","End":"03:22.315","Text":"Something like this so what that means that in the center,"},{"Start":"03:22.315 ","End":"03:25.870","Text":"the magnetic field is pointing outwards."},{"Start":"03:25.870 ","End":"03:27.490","Text":"We have this equation."},{"Start":"03:27.490 ","End":"03:31.300","Text":"We know the direction of the magnetic field."},{"Start":"03:31.300 ","End":"03:38.540","Text":"What we want to know is what length this L is and what length is our y."},{"Start":"03:39.140 ","End":"03:47.370","Text":"The L is of course the length of each wire so let\u0027s draw that in blue."},{"Start":"03:47.370 ","End":"03:49.215","Text":"L is of course,"},{"Start":"03:49.215 ","End":"03:54.250","Text":"this length over here and y is"},{"Start":"03:54.250 ","End":"03:56.665","Text":"the distance between the wire to"},{"Start":"03:56.665 ","End":"03:59.680","Text":"where we\u0027re measuring the magnetic field, which is over here."},{"Start":"03:59.680 ","End":"04:02.660","Text":"It\u0027s from the center"},{"Start":"04:05.090 ","End":"04:10.535","Text":"of the wire until this point over here."},{"Start":"04:10.535 ","End":"04:14.900","Text":"Of course we know that from the center to this corner over here,"},{"Start":"04:14.900 ","End":"04:19.310","Text":"this length is R. What we can see is that here we have"},{"Start":"04:19.310 ","End":"04:25.960","Text":"a right angle triangle and here we can say that this angle over here is Alpha."},{"Start":"04:25.960 ","End":"04:31.670","Text":"Of course this length over here in blue is y."},{"Start":"04:31.670 ","End":"04:38.660","Text":"In that case, we can say that y=R"},{"Start":"04:38.660 ","End":"04:46.520","Text":"multiplied by cosine of this angle Alpha and that half of this length,"},{"Start":"04:46.520 ","End":"04:48.350","Text":"just this side over here,"},{"Start":"04:48.350 ","End":"04:51.610","Text":"which is equal to L divided by 2,"},{"Start":"04:51.610 ","End":"04:57.645","Text":"is equal to R multiplied by sine of Alpha."},{"Start":"04:57.645 ","End":"05:01.245","Text":"Let\u0027s see now what Alpha is."},{"Start":"05:01.245 ","End":"05:07.715","Text":"We of course know that this over here is also the radius equal to I."},{"Start":"05:07.715 ","End":"05:09.965","Text":"Then we can say,"},{"Start":"05:09.965 ","End":"05:12.150","Text":"let\u0027s see, let\u0027s use pink."},{"Start":"05:12.150 ","End":"05:18.080","Text":"We can say that this angle over here is 2 Alpha,"},{"Start":"05:18.080 ","End":"05:20.875","Text":"because it\u0027s Alpha multiplied by 2."},{"Start":"05:20.875 ","End":"05:26.960","Text":"Of course, we\u0027re going to have n of these."},{"Start":"05:26.960 ","End":"05:32.260","Text":"Every single one of these many triangles that we make over here,"},{"Start":"05:32.260 ","End":"05:36.985","Text":"so that\u0027s 2 Alpha plus 2 Alpha plus 2 Alpha n times."},{"Start":"05:36.985 ","End":"05:47.665","Text":"What we have is 2 Alpha multiplied by n is equal to the degrees in a circle,"},{"Start":"05:47.665 ","End":"05:51.655","Text":"which is 360, or rather,"},{"Start":"05:51.655 ","End":"05:54.740","Text":"let\u0027s write that as 2 Pi."},{"Start":"05:54.740 ","End":"06:01.717","Text":"The degrees of a point can divide both sides by 2, sorry,"},{"Start":"06:01.717 ","End":"06:06.775","Text":"and then what we get is that Alpha is equal to Pi divided by n,"},{"Start":"06:06.775 ","End":"06:12.140","Text":"where of course n is what we\u0027re given in the question."},{"Start":"06:12.140 ","End":"06:17.840","Text":"If so, we can substitute in this Alpha into here so what we get is"},{"Start":"06:17.840 ","End":"06:22.715","Text":"that y=R cosine of Pi divided by"},{"Start":"06:22.715 ","End":"06:29.562","Text":"n. We get that L divided by 2 is equal to R"},{"Start":"06:29.562 ","End":"06:34.080","Text":"sine of Pi divided by"},{"Start":"06:34.080 ","End":"06:40.445","Text":"n. Now let\u0027s substitute this into our equation for B."},{"Start":"06:40.445 ","End":"06:47.270","Text":"The magnetic fields we saw is equal to Mu_naught I multiplied by L. We\u0027re just going to"},{"Start":"06:47.270 ","End":"06:54.960","Text":"multiply this by 2 so multiplied by 2 sin(Pi over n),"},{"Start":"06:54.960 ","End":"06:57.595","Text":"because this is L divided by 2,"},{"Start":"06:57.595 ","End":"07:05.585","Text":"and then all of this is divided by 4Pi multiplied by y so y is"},{"Start":"07:05.585 ","End":"07:15.255","Text":"R cosine of Pi divided by n. Then all of this is multiplied by L divided by 2^2."},{"Start":"07:15.255 ","End":"07:17.715","Text":"This is L divided by 2."},{"Start":"07:17.715 ","End":"07:25.635","Text":"We just square it, so R^2 sine^2 Pi over n plus y^2"},{"Start":"07:25.635 ","End":"07:35.905","Text":"plus R^2 cosine^2 of Pi over n and all of this is to the power of 1/2."},{"Start":"07:35.905 ","End":"07:38.480","Text":"Sorry, and of course over here,"},{"Start":"07:38.480 ","End":"07:46.740","Text":"we have to multiply by R. I forgot to multiply by R. This R and this R can cancel out,"},{"Start":"07:46.740 ","End":"07:51.225","Text":"this 2 can cancel out with this 4 to make a 2 over here."},{"Start":"07:51.225 ","End":"07:54.030","Text":"Now we can carry on."},{"Start":"07:54.030 ","End":"07:59.950","Text":"This, what we have over here in this brackets is simply R^2 multiplied by"},{"Start":"07:59.950 ","End":"08:07.565","Text":"sine^2 of Pi over n plus cosine^2 of Pi over n,"},{"Start":"08:07.565 ","End":"08:08.900","Text":"which as we know,"},{"Start":"08:08.900 ","End":"08:14.795","Text":"is equal to 1. sine^2 plus cosine^2 of an angle is equal to 1."},{"Start":"08:14.795 ","End":"08:20.930","Text":"What we have is (R^2)^1/2 so that\u0027s just taking the square root of R^2,"},{"Start":"08:20.930 ","End":"08:30.335","Text":"which is just R. What we\u0027re left with is Mu_naught I multiplied by sine of Pi divided by"},{"Start":"08:30.335 ","End":"08:36.851","Text":"n divided by 2Pi cos(Pi over"},{"Start":"08:36.851 ","End":"08:45.210","Text":"n) multiplied by R. Again,"},{"Start":"08:45.210 ","End":"08:47.290","Text":"this is just for 1 wire and reminding you,"},{"Start":"08:47.290 ","End":"08:51.010","Text":"and this can be rewritten as,"},{"Start":"08:51.010 ","End":"08:59.845","Text":"let\u0027s carry it on over here as Mu_naught I divided by 2PiR."},{"Start":"08:59.845 ","End":"09:04.015","Text":"Then we have sin(Pi over n) divided by cos(Pi over n),"},{"Start":"09:04.015 ","End":"09:09.095","Text":"which is just tan(Pi over n)."},{"Start":"09:09.095 ","End":"09:11.295","Text":"This is for 1 wire,"},{"Start":"09:11.295 ","End":"09:12.975","Text":"but we have n wires,"},{"Start":"09:12.975 ","End":"09:15.240","Text":"corresponding to n sides."},{"Start":"09:15.240 ","End":"09:21.145","Text":"All we have to do is we have to multiply it by n. Let\u0027s say now for"},{"Start":"09:21.145 ","End":"09:30.300","Text":"n wires so B is simply going to be this multiplied by n over here."},{"Start":"09:30.580 ","End":"09:34.730","Text":"This is the answer to Question number 1."},{"Start":"09:34.730 ","End":"09:38.825","Text":"Now, let\u0027s go on to Question number 2,"},{"Start":"09:38.825 ","End":"09:41.870","Text":"which is what is the magnetic field at the center of"},{"Start":"09:41.870 ","End":"09:46.280","Text":"the polygon but this time when n is approaching infinity,"},{"Start":"09:46.280 ","End":"09:50.780","Text":"so we have an infinite amount of sides."},{"Start":"09:51.200 ","End":"09:56.330","Text":"What\u0027s going to happen as the number of sides approaches infinity?"},{"Start":"09:56.330 ","End":"10:03.890","Text":"That means that our n-sided polygon is going to resemble more and more a circle."},{"Start":"10:03.890 ","End":"10:10.595","Text":"That means that we\u0027ll be expecting to measure magnetic field in the center of the circle,"},{"Start":"10:10.595 ","End":"10:14.375","Text":"similar to that of a current carrying loop,"},{"Start":"10:14.375 ","End":"10:17.250","Text":"which we\u0027ve seen in previous lessons."},{"Start":"10:17.560 ","End":"10:21.140","Text":"We\u0027re expecting to get the equation for"},{"Start":"10:21.140 ","End":"10:24.230","Text":"the magnetic field due to a current carrying loop."},{"Start":"10:24.230 ","End":"10:27.620","Text":"But let\u0027s see how we can get that from this equation"},{"Start":"10:27.620 ","End":"10:31.730","Text":"when we set our n to be approaching infinity."},{"Start":"10:31.730 ","End":"10:33.855","Text":"What we have over here,"},{"Start":"10:33.855 ","End":"10:41.960","Text":"is here Pi divided by n. Let\u0027s deal with the tangents first."},{"Start":"10:41.960 ","End":"10:45.840","Text":"When n is approaching infinity,"},{"Start":"10:45.840 ","End":"10:47.540","Text":"Pi divided by infinity,"},{"Start":"10:47.540 ","End":"10:49.975","Text":"it will be approaching 0."},{"Start":"10:49.975 ","End":"10:53.385","Text":"As we know, tan of 0=0."},{"Start":"10:53.385 ","End":"10:56.435","Text":"If we take the zeroth order of this,"},{"Start":"10:56.435 ","End":"10:58.910","Text":"the magnetic field that we\u0027ll get in"},{"Start":"10:58.910 ","End":"11:03.665","Text":"the center of a circle or of a ring will be equal to 0,"},{"Start":"11:03.665 ","End":"11:05.150","Text":"which we know isn\u0027t the case."},{"Start":"11:05.150 ","End":"11:07.475","Text":"We know that there will be a magnetic field"},{"Start":"11:07.475 ","End":"11:11.320","Text":"that is different to 0 in the center over here."},{"Start":"11:11.320 ","End":"11:17.190","Text":"What we\u0027re going to do is we\u0027re going to take our first order."},{"Start":"11:17.570 ","End":"11:20.240","Text":"The equation for the first-order,"},{"Start":"11:20.240 ","End":"11:24.050","Text":"and we\u0027ll try it out and we\u0027ll see if it works,tan(x),"},{"Start":"11:24.050 ","End":"11:31.935","Text":"when x is approaching 0 is approximately equal to for the first-order, x."},{"Start":"11:31.935 ","End":"11:39.165","Text":"Here our x is of course Pi over n. Let\u0027s try that."},{"Start":"11:39.165 ","End":"11:46.490","Text":"What we\u0027ll have is n multiplied by Mu_naught multiplied by I divided by"},{"Start":"11:46.490 ","End":"11:51.980","Text":"2PiR multiplied by tan(Pi over n)"},{"Start":"11:51.980 ","End":"11:59.700","Text":"as Pi over n is approaching infinity so multiply it by Pi over n,"},{"Start":"12:03.080 ","End":"12:05.235","Text":"that we got from this."},{"Start":"12:05.235 ","End":"12:07.770","Text":"Pi over n is x."},{"Start":"12:07.770 ","End":"12:10.820","Text":"Then we can see that the n\u0027s cancel out,"},{"Start":"12:10.820 ","End":"12:14.105","Text":"which is great because the n\u0027s are approaching infinity."},{"Start":"12:14.105 ","End":"12:16.175","Text":"The Pi cancels out."},{"Start":"12:16.175 ","End":"12:20.630","Text":"What we\u0027re in the end left with is that the magnetic field at"},{"Start":"12:20.630 ","End":"12:26.390","Text":"the center of this circle or this current carrying loop,"},{"Start":"12:26.390 ","End":"12:33.510","Text":"is equal to Mu_naught I divided by 2R."},{"Start":"12:34.610 ","End":"12:37.640","Text":"If you remember or you have this written down,"},{"Start":"12:37.640 ","End":"12:40.850","Text":"this is exactly the equation for"},{"Start":"12:40.850 ","End":"12:44.980","Text":"the magnetic field at the center of a current carrying loop."},{"Start":"12:44.980 ","End":"12:47.705","Text":"That\u0027s great. We got what we expected,"},{"Start":"12:47.705 ","End":"12:50.670","Text":"and that is the end of this lesson."}],"ID":21403},{"Watched":false,"Name":"Exercise 8","Duration":"12m 27s","ChapterTopicVideoID":21328,"CourseChapterTopicPlaylistID":99469,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.845","Text":"Hello. In this lesson,"},{"Start":"00:01.845 ","End":"00:04.815","Text":"we\u0027re going to be answering the following question,"},{"Start":"00:04.815 ","End":"00:08.430","Text":"2 infinite wires are parallel to one another and"},{"Start":"00:08.430 ","End":"00:11.980","Text":"have a charged distribution of Lambda and negative Lambda."},{"Start":"00:11.980 ","End":"00:17.740","Text":"The wires are pulled in the rightwards direction at a constant velocity of v_0,"},{"Start":"00:17.740 ","End":"00:22.260","Text":"calculate the magnitude of the velocity such"},{"Start":"00:22.260 ","End":"00:27.850","Text":"that the magnetic force will cancel out with the electric force."},{"Start":"00:29.060 ","End":"00:37.190","Text":"Let\u0027s first start speaking about the electric field and the electric force,"},{"Start":"00:37.190 ","End":"00:40.860","Text":"let\u0027s look at this wire over here."},{"Start":"00:40.860 ","End":"00:44.960","Text":"As we know, we\u0027re going to use Gauss\u0027s law,"},{"Start":"00:44.960 ","End":"00:49.970","Text":"we\u0027re here forming some Gaussian surface,"},{"Start":"00:49.970 ","End":"00:56.270","Text":"which is a cylinder going around this wire and as we know,"},{"Start":"00:56.270 ","End":"00:59.990","Text":"the electric field is going to be coming out."},{"Start":"00:59.990 ","End":"01:01.730","Text":"If you can\u0027t remember this,"},{"Start":"01:01.730 ","End":"01:04.490","Text":"please go back to the chapters dealing with"},{"Start":"01:04.490 ","End":"01:08.880","Text":"Gauss\u0027s law and calculating the electric field and of course,"},{"Start":"01:08.880 ","End":"01:10.860","Text":"the length of the cylinder is"},{"Start":"01:10.860 ","End":"01:18.160","Text":"L. We have that the electric field multiplied by the surface area of"},{"Start":"01:18.160 ","End":"01:19.795","Text":"the cylinder so that is"},{"Start":"01:19.795 ","End":"01:26.935","Text":"2PirL=1 divided by Epsilon_naught"},{"Start":"01:26.935 ","End":"01:31.210","Text":"multiplied by the charge enclosed inside."},{"Start":"01:31.210 ","End":"01:35.395","Text":"The charge inside is equal to the charge distribution"},{"Start":"01:35.395 ","End":"01:41.320","Text":"multiplied by the length so Lambda L. Therefore,"},{"Start":"01:41.320 ","End":"01:43.330","Text":"we can see that our L\u0027s cancel out,"},{"Start":"01:43.330 ","End":"01:49.570","Text":"and what we get is that the electric field is equal to Lambda divided"},{"Start":"01:49.570 ","End":"01:55.880","Text":"by 2Pi Epsilon_naught r. Now,"},{"Start":"01:55.880 ","End":"01:58.355","Text":"we want to find the force,"},{"Start":"01:58.355 ","End":"02:03.575","Text":"the electric force is going to be acting over here on this wire,"},{"Start":"02:03.575 ","End":"02:05.990","Text":"that\u0027s where we\u0027re measuring."},{"Start":"02:05.990 ","End":"02:09.743","Text":"This wire is of length L,"},{"Start":"02:09.743 ","End":"02:12.995","Text":"these Ls don\u0027t have to be equal,"},{"Start":"02:12.995 ","End":"02:16.910","Text":"but this is some length L and so"},{"Start":"02:16.910 ","End":"02:21.605","Text":"the force exerted by the electric field created by this wire,"},{"Start":"02:21.605 ","End":"02:25.640","Text":"on this wire, or specifically this section of wire,"},{"Start":"02:25.640 ","End":"02:32.405","Text":"is equal to q multiplied by the electric field at that point so q,"},{"Start":"02:32.405 ","End":"02:36.875","Text":"of course, refers to the charge of this section of wire."},{"Start":"02:36.875 ","End":"02:43.830","Text":"That is of course equal to negative Lambda multiplied by"},{"Start":"02:43.830 ","End":"02:47.330","Text":"this length L so the charge distribution"},{"Start":"02:47.330 ","End":"02:51.830","Text":"multiplied by the length so that is the charge of this section of wire,"},{"Start":"02:51.830 ","End":"02:56.330","Text":"and then multiplied by the E field at this point,"},{"Start":"02:56.330 ","End":"02:58.310","Text":"which is caused by this wire."},{"Start":"02:58.310 ","End":"03:08.310","Text":"That\u0027s Lambda divided by 2Pi Epsilon_naught r. Usually,"},{"Start":"03:08.310 ","End":"03:11.650","Text":"in these cases, we want to get the force per unit length"},{"Start":"03:11.650 ","End":"03:16.050","Text":"so we\u0027re just going to divide both sides by L,"},{"Start":"03:16.050 ","End":"03:19.880","Text":"and then what we\u0027re left with is negative."},{"Start":"03:19.880 ","End":"03:24.520","Text":"Then we have Lambda^2divided"},{"Start":"03:24.520 ","End":"03:30.180","Text":"by 2Pi Epsilon_naught r and of course,"},{"Start":"03:30.180 ","End":"03:36.410","Text":"this is a negative so we can see that this is an attractive force."},{"Start":"03:36.410 ","End":"03:42.160","Text":"The wires are pulling each other closer together."},{"Start":"03:42.160 ","End":"03:50.440","Text":"Now let\u0027s look at the magnetic field and force."},{"Start":"03:51.680 ","End":"03:55.400","Text":"The moment where we have current flowing,"},{"Start":"03:55.400 ","End":"03:59.900","Text":"that means that we\u0027re going to have some magnetic field and of course,"},{"Start":"03:59.900 ","End":"04:07.175","Text":"we know that current is formed by the movement of charged particles,"},{"Start":"04:07.175 ","End":"04:08.870","Text":"which is exactly what we have here."},{"Start":"04:08.870 ","End":"04:11.990","Text":"We have some charge distribution that is"},{"Start":"04:11.990 ","End":"04:15.470","Text":"moving with a velocity so we know that we\u0027re going to have current,"},{"Start":"04:15.470 ","End":"04:18.395","Text":"and therefore we know that we\u0027re going to have a magnetic field."},{"Start":"04:18.395 ","End":"04:21.425","Text":"There\u0027s 2 ways we can calculate current."},{"Start":"04:21.425 ","End":"04:27.035","Text":"We can either use the fact that it is equal to dq by dt,"},{"Start":"04:27.035 ","End":"04:30.650","Text":"the change in charges in the change of time,"},{"Start":"04:30.650 ","End":"04:36.740","Text":"or the amount of charges that pass an area in a certain time interval or we"},{"Start":"04:36.740 ","End":"04:43.360","Text":"can use the equation that I is equal to Lambda multiplied by v,"},{"Start":"04:43.360 ","End":"04:47.290","Text":"where of course v is velocity."},{"Start":"04:49.910 ","End":"04:52.660","Text":"Remember here we have Lambda,"},{"Start":"04:52.660 ","End":"04:55.250","Text":"so we could use this example,"},{"Start":"04:55.250 ","End":"05:02.450","Text":"but remember if you have Sigma or Rho you can also work out the current density."},{"Start":"05:02.450 ","End":"05:05.120","Text":"If we remember to previous lessons,"},{"Start":"05:05.120 ","End":"05:07.820","Text":"k is equal to Sigma v, again,"},{"Start":"05:07.820 ","End":"05:14.370","Text":"v is the velocity and J is equal to Rho v. Again,"},{"Start":"05:14.370 ","End":"05:16.185","Text":"v is the velocity."},{"Start":"05:16.185 ","End":"05:25.145","Text":"If we\u0027re going down the dq root so we know that dq is equal to"},{"Start":"05:25.145 ","End":"05:30.590","Text":"either Lambda multiplied by the unit length so dl or"},{"Start":"05:30.590 ","End":"05:36.590","Text":"Sigma multiplied by unit area so ds or Rho multiplied by unit volume,"},{"Start":"05:36.590 ","End":"05:41.285","Text":"or dV, and of course,"},{"Start":"05:41.285 ","End":"05:43.399","Text":"this V over here is volume,"},{"Start":"05:43.399 ","End":"05:45.289","Text":"so don\u0027t get confused."},{"Start":"05:45.289 ","End":"05:51.080","Text":"What we can do is we can either use this and then just divide it by dt,"},{"Start":"05:51.080 ","End":"05:54.480","Text":"or we can use this."},{"Start":"05:55.040 ","End":"06:00.830","Text":"Let\u0027s imagine that this is the x-direction."},{"Start":"06:00.830 ","End":"06:06.520","Text":"We can say that I is equal to dq by dt."},{"Start":"06:06.520 ","End":"06:11.030","Text":"I\u0027m just going to show you that both of these methods are the same,"},{"Start":"06:11.030 ","End":"06:20.870","Text":"so dq here we\u0027re dealing with Lambda so dq will be Lambda dl divided by dt and"},{"Start":"06:20.870 ","End":"06:24.530","Text":"specifically our changes in length as we defined"},{"Start":"06:24.530 ","End":"06:28.190","Text":"it is in the x-direction because the wires are in"},{"Start":"06:28.190 ","End":"06:32.360","Text":"the x-direction so this length is on the x-axis so that will be"},{"Start":"06:32.360 ","End":"06:37.755","Text":"Lambda dx divided by dt and of course,"},{"Start":"06:37.755 ","End":"06:42.403","Text":"dx by dt is exactly the velocity,"},{"Start":"06:42.403 ","End":"06:45.335","Text":"what we\u0027ll be left with is Lambda v,"},{"Start":"06:45.335 ","End":"06:48.420","Text":"which is this equation over here."},{"Start":"06:49.670 ","End":"06:53.494","Text":"Now, we have our current,"},{"Start":"06:53.494 ","End":"06:59.600","Text":"our current from this wire is exerting a magnetic field on this wire."},{"Start":"06:59.600 ","End":"07:03.380","Text":"We\u0027ve already seen that also this wire is of course going to"},{"Start":"07:03.380 ","End":"07:08.245","Text":"be exerting a magnetic field and the magnetic force on this wire."},{"Start":"07:08.245 ","End":"07:11.630","Text":"Let\u0027s first of all look at our charge distributions."},{"Start":"07:11.630 ","End":"07:18.170","Text":"Here we see we have a negative charge distribution moving in the rightwards direction,"},{"Start":"07:18.170 ","End":"07:20.990","Text":"which is exactly the same as saying that we have"},{"Start":"07:20.990 ","End":"07:26.405","Text":"a positive charge distribution moving in the leftwards direction."},{"Start":"07:26.405 ","End":"07:30.545","Text":"In that case, let\u0027s use pink."},{"Start":"07:30.545 ","End":"07:36.380","Text":"We can say that in this wire we have a current I traveling in"},{"Start":"07:36.380 ","End":"07:42.170","Text":"this direction and here we have positive charges moving in the rightwards direction."},{"Start":"07:42.170 ","End":"07:44.195","Text":"On the bottom wire,"},{"Start":"07:44.195 ","End":"07:48.720","Text":"we just have a current moving in the rightwards direction."},{"Start":"07:49.400 ","End":"07:52.640","Text":"While we speak about Lawrence\u0027s law,"},{"Start":"07:52.640 ","End":"08:00.320","Text":"we saw that the force per unit length between 2 infinite current carrying wires."},{"Start":"08:00.320 ","End":"08:05.540","Text":"We saw that, that so F divided by L force per unit length is"},{"Start":"08:05.540 ","End":"08:10.595","Text":"equal to Mu_naught multiplied by the current in 1 wire,"},{"Start":"08:10.595 ","End":"08:14.120","Text":"multiplied by the current in the second wire,"},{"Start":"08:14.120 ","End":"08:20.110","Text":"divided by 2Pir and this comes from Lawrence\u0027s law."},{"Start":"08:20.420 ","End":"08:23.360","Text":"In our case over here,"},{"Start":"08:23.360 ","End":"08:27.207","Text":"what we have is Mu_naught,"},{"Start":"08:27.207 ","End":"08:29.870","Text":"the current in both wires is equal."},{"Start":"08:29.870 ","End":"08:35.750","Text":"Just one is in the different direction or in the opposite direction so I\u0027m just going to"},{"Start":"08:35.750 ","End":"08:41.970","Text":"write Lambda^2multiplied by v squared."},{"Start":"08:41.970 ","End":"08:43.835","Text":"This is the current we saw,"},{"Start":"08:43.835 ","End":"08:45.995","Text":"and they\u0027re both just in opposite direction."},{"Start":"08:45.995 ","End":"08:49.460","Text":"I\u0027ll show that when we speak about the direction of"},{"Start":"08:49.460 ","End":"08:54.050","Text":"the force and all of this is divided by, of course,"},{"Start":"08:54.050 ","End":"08:57.980","Text":"2Pir, and the force per"},{"Start":"08:57.980 ","End":"09:02.960","Text":"unit length is of course repulsive because in one of the previous lessons,"},{"Start":"09:02.960 ","End":"09:09.260","Text":"we saw that if the direction of current is opposite to one another then"},{"Start":"09:09.260 ","End":"09:13.130","Text":"the force will be repulsive and that we can"},{"Start":"09:13.130 ","End":"09:18.299","Text":"see also because how charged distributions oppositely charged."},{"Start":"09:19.370 ","End":"09:22.900","Text":"Because here I wrote that this is attractive,"},{"Start":"09:22.900 ","End":"09:25.690","Text":"I can just take away this minus sign,"},{"Start":"09:25.690 ","End":"09:29.250","Text":"because the minus just shows that it\u0027s attractive but I said that it\u0027s"},{"Start":"09:29.250 ","End":"09:34.340","Text":"attractive and here I gave the direction that it\u0027s a repulsive force."},{"Start":"09:34.340 ","End":"09:38.750","Text":"Now, what we have is the force per unit length when we\u0027re dealing"},{"Start":"09:38.750 ","End":"09:42.800","Text":"with electric force and when we\u0027re dealing with magnetic force."},{"Start":"09:42.800 ","End":"09:47.180","Text":"The question was, to calculate the magnitude of"},{"Start":"09:47.180 ","End":"09:52.430","Text":"the velocity such that the magnetic force will cancel out with the electric force."},{"Start":"09:52.430 ","End":"09:58.310","Text":"What I want to do is I want to show that Lambda^2divided by"},{"Start":"09:58.310 ","End":"10:03.600","Text":"2Pi Epsilon_naught r minus"},{"Start":"10:03.600 ","End":"10:09.960","Text":"Mu_naught Lambda^2v^2divided by 2Pir=0."},{"Start":"10:09.960 ","End":"10:12.279","Text":"In other words, I can just equate them,"},{"Start":"10:12.279 ","End":"10:15.545","Text":"I\u0027m just equating my force per unit length."},{"Start":"10:15.545 ","End":"10:25.490","Text":"I have Mu_naught Lambda^2v^2divided by 2Pir is equal to"},{"Start":"10:25.490 ","End":"10:31.280","Text":"Lambda^2divided by 2Pi Epsilon_naught r."},{"Start":"10:31.580 ","End":"10:36.560","Text":"We can see that our charge distributions cancel out."},{"Start":"10:36.560 ","End":"10:41.180","Text":"We can see that 2Pir cancels out."},{"Start":"10:41.180 ","End":"10:45.380","Text":"Now, what we want to do is we just want to isolate out our velocity."},{"Start":"10:45.380 ","End":"10:52.825","Text":"We get that our velocity is equal to 1 divided by Epsilon_naught,"},{"Start":"10:52.825 ","End":"10:57.465","Text":"Mu_naught, and the square root of this."},{"Start":"10:57.465 ","End":"11:00.755","Text":"What we could see is that the electric force is"},{"Start":"11:00.755 ","End":"11:05.375","Text":"an attractive force and the magnetic force is a repulsive force."},{"Start":"11:05.375 ","End":"11:11.315","Text":"We wanted these 2 forces to cancel out such that our wires will remain"},{"Start":"11:11.315 ","End":"11:15.860","Text":"this certain distance away from one another and won\u0027t"},{"Start":"11:15.860 ","End":"11:20.995","Text":"either move further away from one another or come closer together."},{"Start":"11:20.995 ","End":"11:25.320","Text":"This is what we were looking for and so"},{"Start":"11:25.320 ","End":"11:29.510","Text":"this first of all is the answer but something that we can see"},{"Start":"11:29.510 ","End":"11:33.170","Text":"if we were to plug in the values for Epsilon_naught and"},{"Start":"11:33.170 ","End":"11:38.060","Text":"Mu_naught and then work out the square root of all of this,"},{"Start":"11:38.060 ","End":"11:42.560","Text":"we would get that this is equal to c,"},{"Start":"11:42.560 ","End":"11:45.440","Text":"where c is the speed of light, which is,"},{"Start":"11:45.440 ","End":"11:53.050","Text":"of course, 3 times 10^8 meters per second."},{"Start":"11:53.050 ","End":"11:54.803","Text":"This is the answer,"},{"Start":"11:54.803 ","End":"11:59.600","Text":"if we are moving these 2 wires at the speed of light,"},{"Start":"11:59.600 ","End":"12:03.905","Text":"then the electric force will cancel out with"},{"Start":"12:03.905 ","End":"12:09.530","Text":"the magnetic force and this is correct when we\u0027re working this out theoretically,"},{"Start":"12:09.530 ","End":"12:11.345","Text":"of course, in practice,"},{"Start":"12:11.345 ","End":"12:16.760","Text":"we have not yet managed to travel or move anything at the speed of light."},{"Start":"12:16.760 ","End":"12:21.845","Text":"In practice, this obviously is not possible but theoretically,"},{"Start":"12:21.845 ","End":"12:25.100","Text":"this is the magnitude of the velocity."},{"Start":"12:25.100 ","End":"12:27.960","Text":"That\u0027s the end of this lesson."}],"ID":21404},{"Watched":false,"Name":"Exercise 9","Duration":"28m 44s","ChapterTopicVideoID":21329,"CourseChapterTopicPlaylistID":99469,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.620","Text":"Hello. In this lesson,"},{"Start":"00:01.620 ","End":"00:03.930","Text":"we\u0027re going to be answering the following question."},{"Start":"00:03.930 ","End":"00:09.360","Text":"Wire ACDFG includes a circular section of"},{"Start":"00:09.360 ","End":"00:14.895","Text":"radius R and 2 infinitely long straight regions."},{"Start":"00:14.895 ","End":"00:18.765","Text":"The continuation of line AC cuts through"},{"Start":"00:18.765 ","End":"00:24.510","Text":"the center of the circle\u0027s radius and the current I flows through the wire."},{"Start":"00:24.510 ","End":"00:25.935","Text":"Question number 1 is,"},{"Start":"00:25.935 ","End":"00:30.285","Text":"what is the magnetic field at the center of the circular region?"},{"Start":"00:30.285 ","End":"00:33.615","Text":"Let\u0027s begin by solving question Number 1."},{"Start":"00:33.615 ","End":"00:36.945","Text":"We\u0027re going to do that by using superposition."},{"Start":"00:36.945 ","End":"00:41.170","Text":"We\u0027re going to find what Section AC,"},{"Start":"00:41.170 ","End":"00:44.735","Text":"the magnetic field that is produced at this point from AC,"},{"Start":"00:44.735 ","End":"00:49.490","Text":"then we\u0027re going to add onto that the magnetic fields produced by the circle."},{"Start":"00:49.490 ","End":"00:51.589","Text":"Notice this isn\u0027t a full circle,"},{"Start":"00:51.589 ","End":"00:55.705","Text":"so we have some arc over here which is missing."},{"Start":"00:55.705 ","End":"00:58.850","Text":"The magnetic field produced by"},{"Start":"00:58.850 ","End":"01:03.960","Text":"this region at this point and then we\u0027re going to add them all up."},{"Start":"01:04.190 ","End":"01:08.160","Text":"Let\u0027s begin with region AC."},{"Start":"01:08.160 ","End":"01:11.000","Text":"What we have is over here,"},{"Start":"01:11.000 ","End":"01:13.025","Text":"we have A,"},{"Start":"01:13.025 ","End":"01:15.319","Text":"here we have C,"},{"Start":"01:15.319 ","End":"01:20.310","Text":"and then here, this distance is R. Here,"},{"Start":"01:20.310 ","End":"01:24.780","Text":"we\u0027re trying to work out the magnetic field, okay?"},{"Start":"01:24.780 ","End":"01:28.055","Text":"This distance over here,"},{"Start":"01:28.055 ","End":"01:33.250","Text":"of course, is R divided by 2."},{"Start":"01:33.470 ","End":"01:39.619","Text":"What we have over here is a wire causing a magnetic field over here."},{"Start":"01:39.619 ","End":"01:43.490","Text":"Now we\u0027ve already seen that an equation for the magnetic field"},{"Start":"01:43.490 ","End":"01:48.055","Text":"due to a wire is given like so."},{"Start":"01:48.055 ","End":"01:51.905","Text":"Let\u0027s imagine that we have"},{"Start":"01:51.905 ","End":"01:57.655","Text":"our wire over here and we\u0027re trying to measure the magnetic field over here."},{"Start":"01:57.655 ","End":"02:02.660","Text":"What we did is we defined this angle over here as"},{"Start":"02:02.660 ","End":"02:07.460","Text":"Alpha 1 and this angle over here as Alpha 2,"},{"Start":"02:07.460 ","End":"02:13.700","Text":"with this distance between the point and the wire is a distance a,"},{"Start":"02:13.700 ","End":"02:16.160","Text":"and this is a 90-degree angle."},{"Start":"02:16.160 ","End":"02:21.380","Text":"Then the equation for the magnetic field was given as Mu naught I,"},{"Start":"02:21.380 ","End":"02:23.360","Text":"again, of course,"},{"Start":"02:23.360 ","End":"02:28.405","Text":"as a current I traveling through Mu naught I divided by"},{"Start":"02:28.405 ","End":"02:33.485","Text":"4 Pi a multiplied by cosine of"},{"Start":"02:33.485 ","End":"02:39.600","Text":"Alpha 1 plus cosine of Alpha 2."},{"Start":"02:40.940 ","End":"02:48.180","Text":"Now let\u0027s see what our Alpha 1 is and what our Alpha 2 is in our question over here."},{"Start":"02:48.180 ","End":"02:52.790","Text":"First thing we have to remember is that AC is"},{"Start":"02:52.790 ","End":"03:00.470","Text":"just a small portion of the infinitely long straight region of the wire."},{"Start":"03:00.470 ","End":"03:06.020","Text":"That\u0027s why I have the 3 dots over here representing that this is infinitely long."},{"Start":"03:06.020 ","End":"03:13.250","Text":"If I were to draw a line connecting the edge of this wire,"},{"Start":"03:13.250 ","End":"03:14.540","Text":"which is at infinity,"},{"Start":"03:14.540 ","End":"03:16.220","Text":"to this point over here."},{"Start":"03:16.220 ","End":"03:18.560","Text":"We would see that our angle over here,"},{"Start":"03:18.560 ","End":"03:20.285","Text":"which is Alpha 1,"},{"Start":"03:20.285 ","End":"03:23.455","Text":"would be approaching 0."},{"Start":"03:23.455 ","End":"03:28.905","Text":"Alpha 1 is approaching 0 because our y is infinitely long."},{"Start":"03:28.905 ","End":"03:33.290","Text":"This angle just becomes smaller and smaller and smaller, it\u0027s approaching 0."},{"Start":"03:33.290 ","End":"03:43.105","Text":"Therefore, cosine of Alpha 1 is approaching 1."},{"Start":"03:43.105 ","End":"03:48.530","Text":"The next thing that we know is what a is,"},{"Start":"03:48.530 ","End":"03:52.130","Text":"a we saw as the distance between the wire and"},{"Start":"03:52.130 ","End":"03:58.280","Text":"the point when we\u0027re just looking at vertical distance."},{"Start":"03:58.280 ","End":"04:06.935","Text":"For us, a is equal to this vertical distance over here,"},{"Start":"04:06.935 ","End":"04:10.580","Text":"which is just R divided by 2."},{"Start":"04:10.580 ","End":"04:14.570","Text":"It doesn\u0027t matter that the point is further away along the x axis,"},{"Start":"04:14.570 ","End":"04:17.330","Text":"but along this vertical axis,"},{"Start":"04:17.330 ","End":"04:20.220","Text":"the distance is r divided by 2."},{"Start":"04:20.220 ","End":"04:25.250","Text":"Now what we want to do is we want to calculate what Alpha 2 is equal to,"},{"Start":"04:25.250 ","End":"04:30.270","Text":"or more specifically, what cosine of Alpha 2 is equal to."},{"Start":"04:31.280 ","End":"04:36.060","Text":"This angle over here is our Alpha 2."},{"Start":"04:36.060 ","End":"04:39.920","Text":"Then we have this length over here."},{"Start":"04:39.920 ","End":"04:45.350","Text":"Let\u0027s call this x and this angle over here is equal"},{"Start":"04:45.350 ","End":"04:52.440","Text":"to Pi or a 180 degrees minus Alpha 2."},{"Start":"04:52.440 ","End":"04:55.160","Text":"From our trig identities,"},{"Start":"04:55.160 ","End":"04:59.600","Text":"we know that cosine of some angle is"},{"Start":"04:59.600 ","End":"05:05.180","Text":"equal to the adjacent side divided by the hypotenuse."},{"Start":"05:05.180 ","End":"05:08.690","Text":"In our example, the adjacent side is of"},{"Start":"05:08.690 ","End":"05:14.480","Text":"length x and the hypotenuse is this side over here,"},{"Start":"05:14.480 ","End":"05:21.830","Text":"which is R. Now what we want to do is we want to know what length x is."},{"Start":"05:21.830 ","End":"05:28.260","Text":"From Pythagoras, we know that the hypotenuse R^2 is equal"},{"Start":"05:28.260 ","End":"05:37.635","Text":"to the x^2 plus R divided by 2^2."},{"Start":"05:37.635 ","End":"05:40.325","Text":"Therefore, in order to get x,"},{"Start":"05:40.325 ","End":"05:49.130","Text":"we get that x is equal to the square root of R^2 minus R divided by 2^2."},{"Start":"05:49.130 ","End":"05:53.660","Text":"Which we can then further simplify to be equal"},{"Start":"05:53.660 ","End":"05:59.705","Text":"to R multiplied by the square root of 1 minus 1/4."},{"Start":"05:59.705 ","End":"06:02.435","Text":"Which once you\u0027ve thought this out,"},{"Start":"06:02.435 ","End":"06:09.940","Text":"is just equal to R multiplied by the square root of 3 divided by 2."},{"Start":"06:09.940 ","End":"06:15.050","Text":"What we have is that cosine of this angle,"},{"Start":"06:15.050 ","End":"06:17.825","Text":"so let\u0027s plug in what our angle is."},{"Start":"06:17.825 ","End":"06:25.140","Text":"Cosine of Pi minus"},{"Start":"06:25.140 ","End":"06:34.460","Text":"Alpha 2 is equal to R multiplied by root 3 divided by 2 divided by R,"},{"Start":"06:34.460 ","End":"06:40.560","Text":"which is simply equal to root 3 divided by 2."},{"Start":"06:40.580 ","End":"06:47.690","Text":"Now we\u0027re going to use the identity that says that cosine of Pi minus"},{"Start":"06:47.690 ","End":"06:54.570","Text":"some angle is equal to negative cosine of that angle."},{"Start":"06:54.570 ","End":"07:03.710","Text":"Therefore, what we\u0027ll have is that cosine of Pi minus Alpha 2 is simply equal to"},{"Start":"07:03.710 ","End":"07:10.190","Text":"negative cosine of Alpha 2 which all we have to"},{"Start":"07:10.190 ","End":"07:19.885","Text":"do is we already found what cosine of Pi minus Alpha 2 is equal to."},{"Start":"07:19.885 ","End":"07:21.520","Text":"This is just negative,"},{"Start":"07:21.520 ","End":"07:26.530","Text":"so it\u0027s equal to negative root 3 over 2."},{"Start":"07:26.530 ","End":"07:29.380","Text":"Sorry, of course this is without the minus."},{"Start":"07:29.380 ","End":"07:33.850","Text":"But what we do is we know that therefore cosine of"},{"Start":"07:33.850 ","End":"07:39.440","Text":"Alpha 2 is equal to negative root 3 over 2."},{"Start":"07:39.920 ","End":"07:45.385","Text":"Now we can plug everything into our equation so what we get is that B is"},{"Start":"07:45.385 ","End":"07:51.600","Text":"equal to Mu naught I divided by 4 Pi a,"},{"Start":"07:51.600 ","End":"07:57.295","Text":"which is R divided by 2 multiplied by cosine of Alpha 1,"},{"Start":"07:57.295 ","End":"08:02.805","Text":"which is 1 plus cosine of Alpha 2,"},{"Start":"08:02.805 ","End":"08:08.230","Text":"which is negative root 3 over 2."},{"Start":"08:08.360 ","End":"08:13.230","Text":"This 4 and this 2 can cancel over here."},{"Start":"08:13.230 ","End":"08:14.675","Text":"Then what we\u0027ll do is we\u0027ll make"},{"Start":"08:14.675 ","End":"08:17.690","Text":"a common denominator so what we\u0027ll get is that this is equal to"},{"Start":"08:17.690 ","End":"08:23.100","Text":"Mu naught I multiplied by 2 minus"},{"Start":"08:23.100 ","End":"08:26.460","Text":"root 3 divided by"},{"Start":"08:26.460 ","End":"08:36.865","Text":"4 Pi R. Because then the 2 over here from the denominator will cancel out."},{"Start":"08:36.865 ","End":"08:40.795","Text":"We\u0027ll just multiply together with these two."},{"Start":"08:40.795 ","End":"08:48.980","Text":"This is the magnitude so far of the magnetic field due to this section of the wire."},{"Start":"08:49.470 ","End":"08:53.230","Text":"This infinite, the long section."},{"Start":"08:53.230 ","End":"08:54.790","Text":"Now we want the direction."},{"Start":"08:54.790 ","End":"08:58.600","Text":"We\u0027re just going to use our right hand rule where we point"},{"Start":"08:58.600 ","End":"09:05.950","Text":"our thumb of the right hand in the direction of the current,"},{"Start":"09:05.950 ","End":"09:09.700","Text":"and then our fingers curl in the direction of the magnetic field."},{"Start":"09:09.700 ","End":"09:13.000","Text":"What we\u0027ll see if we\u0027re looking at this point over here,"},{"Start":"09:13.000 ","End":"09:19.707","Text":"so we can see that the magnetic field over here is going into the page."},{"Start":"09:19.707 ","End":"09:23.714","Text":"Or rather the magnetic field at this point due to this section of the wire,"},{"Start":"09:23.714 ","End":"09:24.985","Text":"the straight section,"},{"Start":"09:24.985 ","End":"09:27.850","Text":"is going into the page."},{"Start":"09:27.850 ","End":"09:31.855","Text":"We can write that over here. It\u0027s going into the page."},{"Start":"09:31.855 ","End":"09:33.650","Text":"If we give axis,"},{"Start":"09:33.650 ","End":"09:36.790","Text":"we already see x and y like so."},{"Start":"09:36.790 ","End":"09:43.215","Text":"That means that the z axis is coming out of the page,"},{"Start":"09:43.215 ","End":"09:47.640","Text":"but we saw that our magnetic field over here is going into the page."},{"Start":"09:47.640 ","End":"09:54.370","Text":"Therefore, the magnetic field is in the negative z direction."},{"Start":"09:54.370 ","End":"09:57.579","Text":"This is the magnetic field for this section,"},{"Start":"09:57.579 ","End":"10:01.195","Text":"and now let\u0027s do the next section."},{"Start":"10:01.195 ","End":"10:08.860","Text":"Let\u0027s calculate the magnetic field in the center due to just this circular region."},{"Start":"10:08.860 ","End":"10:12.805","Text":"We know that the magnetic field of a ring,"},{"Start":"10:12.805 ","End":"10:18.280","Text":"we know that the equation for a full ring is equal"},{"Start":"10:18.280 ","End":"10:25.015","Text":"to Mu naught multiplied by I and divided it by 2."},{"Start":"10:25.015 ","End":"10:29.440","Text":"This we\u0027ve seen in 1 of the previous lessons."},{"Start":"10:29.440 ","End":"10:36.340","Text":"Notice this is speaking about a full ring where the angle of the ring,"},{"Start":"10:36.340 ","End":"10:43.020","Text":"the angle at a point is 360 degrees or 2Pi."},{"Start":"10:43.020 ","End":"10:46.919","Text":"As we\u0027ll notice over here, we have this."},{"Start":"10:46.919 ","End":"10:51.205","Text":"Our angle is this angle over here Theta,"},{"Start":"10:51.205 ","End":"10:55.880","Text":"and we\u0027re missing this section over here."},{"Start":"10:56.760 ","End":"11:00.410","Text":"This is for a full ring."},{"Start":"11:00.570 ","End":"11:04.315","Text":"Now, let\u0027s work out for this."},{"Start":"11:04.315 ","End":"11:08.350","Text":"If we multiply B,"},{"Start":"11:08.350 ","End":"11:13.055","Text":"or let\u0027s work out magnetic field for pottering."},{"Start":"11:13.055 ","End":"11:18.660","Text":"The magnetic field due to a part of a ring is going to be equal to the magnetic field of"},{"Start":"11:18.660 ","End":"11:23.890","Text":"a full ring Mu_0I divided by 2R,"},{"Start":"11:23.890 ","End":"11:27.670","Text":"and it\u0027s going to be multiplied by the proportion that it fills."},{"Start":"11:27.670 ","End":"11:34.210","Text":"Here, it\u0027s filling up Theta degrees out of 2Pi,"},{"Start":"11:34.210 ","End":"11:39.080","Text":"or Theta radians out of 2Pi radians."},{"Start":"11:39.540 ","End":"11:42.820","Text":"This is the proportion that it is."},{"Start":"11:42.820 ","End":"11:48.126","Text":"It\u0027s Theta degrees divided by 360 degrees of a full circle,"},{"Start":"11:48.126 ","End":"11:52.810","Text":"multiplied by the magnetic field of a full circle."},{"Start":"11:52.810 ","End":"11:57.115","Text":"Now let\u0027s see what our Theta is equal to."},{"Start":"11:57.115 ","End":"12:01.610","Text":"Up over here, we got that cosine."},{"Start":"12:03.240 ","End":"12:06.655","Text":"Here we\u0027re taking this from over here."},{"Start":"12:06.655 ","End":"12:11.860","Text":"We got that cosine of Pi minus Alpha 2 was"},{"Start":"12:11.860 ","End":"12:17.373","Text":"equal to root 3 over 2,"},{"Start":"12:17.373 ","End":"12:24.170","Text":"and of course this angle our Alpha 2 was this angle over here."},{"Start":"12:24.690 ","End":"12:28.504","Text":"If we remember the diagram,"},{"Start":"12:28.504 ","End":"12:30.820","Text":"this was our Alpha 2,"},{"Start":"12:30.820 ","End":"12:33.550","Text":"which is exactly this."},{"Start":"12:33.550 ","End":"12:36.130","Text":"This is our Alpha 2."},{"Start":"12:36.130 ","End":"12:41.840","Text":"We know that this total angle is 90 degrees."},{"Start":"12:42.390 ","End":"12:45.550","Text":"What we\u0027re going to do is we\u0027re going I cos"},{"Start":"12:45.550 ","End":"12:49.495","Text":"both sides and then we\u0027re going to isolate out our Alpha 2."},{"Start":"12:49.495 ","End":"12:53.540","Text":"Let\u0027s just do this in the calculator."},{"Start":"12:53.910 ","End":"13:02.395","Text":"Alpha 2 is 5Pi divided by 6,"},{"Start":"13:02.395 ","End":"13:03.460","Text":"and over here,"},{"Start":"13:03.460 ","End":"13:05.515","Text":"we see we have a right angle."},{"Start":"13:05.515 ","End":"13:08.499","Text":"Right angle is 90 degrees,"},{"Start":"13:08.499 ","End":"13:14.210","Text":"or in other words, Pi over 2."},{"Start":"13:14.820 ","End":"13:18.250","Text":"Let\u0027s call this angle over here Phi."},{"Start":"13:18.250 ","End":"13:25.480","Text":"What we\u0027ll get is that Phi is equal to 90 degrees minus our Alpha 2,"},{"Start":"13:25.480 ","End":"13:32.440","Text":"which was 5Pi divided by 6 which is equal to."},{"Start":"13:32.670 ","End":"13:35.500","Text":"This has a negative angle,"},{"Start":"13:35.500 ","End":"13:41.230","Text":"but that\u0027s just because we\u0027re going into the negative quadrant."},{"Start":"13:41.230 ","End":"13:44.350","Text":"This just shows the direction."},{"Start":"13:44.350 ","End":"13:46.480","Text":"But in actual fact,"},{"Start":"13:46.480 ","End":"13:51.625","Text":"this just corresponds to Pi over 3 radians,"},{"Start":"13:51.625 ","End":"13:56.890","Text":"which of course, is equal to 60 degrees."},{"Start":"13:56.890 ","End":"14:03.820","Text":"That means that our Phi is equal to 60 degrees,"},{"Start":"14:03.820 ","End":"14:05.890","Text":"but what we want is Theta."},{"Start":"14:05.890 ","End":"14:08.440","Text":"Theta, because we want here,"},{"Start":"14:08.440 ","End":"14:14.575","Text":"so we have Theta which is equal to just 360 degrees."},{"Start":"14:14.575 ","End":"14:20.515","Text":"All of the angles in the circle minus this Phi over here,"},{"Start":"14:20.515 ","End":"14:23.095","Text":"so minus 60 degrees,"},{"Start":"14:23.095 ","End":"14:27.165","Text":"which gives us that Theta is 300 degrees."},{"Start":"14:27.165 ","End":"14:32.365","Text":"300 degrees is equal to,"},{"Start":"14:32.365 ","End":"14:34.135","Text":"of course in radians."},{"Start":"14:34.135 ","End":"14:40.090","Text":"We have 2Pi minus Pi over"},{"Start":"14:40.090 ","End":"14:46.450","Text":"3 which is equal to 5Pi over 3."},{"Start":"14:46.450 ","End":"14:49.060","Text":"Therefore, we can plug this in over here."},{"Start":"14:49.060 ","End":"14:56.140","Text":"What we get is Mu naught I multiplied by 5Pi divided by"},{"Start":"14:56.140 ","End":"15:05.990","Text":"3 multiplied by 2R multiplied by 2Pi."},{"Start":"15:07.710 ","End":"15:12.100","Text":"The Pi\u0027s cancel out,"},{"Start":"15:12.100 ","End":"15:15.010","Text":"and then what we have is,"},{"Start":"15:15.010 ","End":"15:23.515","Text":"so here we have 3 times 2 is 6 times 2 is 12."},{"Start":"15:23.515 ","End":"15:28.735","Text":"What we have is Mu naught I,"},{"Start":"15:28.735 ","End":"15:35.120","Text":"multiplied by 5 divided by 12R."},{"Start":"15:36.450 ","End":"15:39.145","Text":"That\u0027s the magnitude."},{"Start":"15:39.145 ","End":"15:41.095","Text":"Now we want to know the direction."},{"Start":"15:41.095 ","End":"15:44.560","Text":"Again, we use the right hand rule where we point our thumb"},{"Start":"15:44.560 ","End":"15:48.580","Text":"in the direction of the current which is of course in this direction."},{"Start":"15:48.580 ","End":"15:52.015","Text":"Then we see that our fingers curl in the direction"},{"Start":"15:52.015 ","End":"15:57.235","Text":"that the magnetic field is coming in at this point over here."},{"Start":"15:57.235 ","End":"16:01.870","Text":"Then when we look in this occasion the magnetic field is coming"},{"Start":"16:01.870 ","End":"16:07.105","Text":"out of the page towards us which is exactly the z direction."},{"Start":"16:07.105 ","End":"16:09.340","Text":"Specifically, here,"},{"Start":"16:09.340 ","End":"16:13.825","Text":"this is traveling in the z direction."},{"Start":"16:13.825 ","End":"16:18.730","Text":"Now we\u0027re going to work out the addition"},{"Start":"16:18.730 ","End":"16:24.250","Text":"to the magnetic field over here due to this section of the wire."},{"Start":"16:24.250 ","End":"16:27.580","Text":"Now we\u0027re looking at FG,"},{"Start":"16:27.580 ","End":"16:34.525","Text":"and what we have is an infinitely long section of wire that looks something like this."},{"Start":"16:34.525 ","End":"16:40.315","Text":"It\u0027s not infinitely long over here,"},{"Start":"16:40.315 ","End":"16:46.180","Text":"infinitely long, and then at this point over here, we\u0027re measuring B."},{"Start":"16:46.180 ","End":"16:49.450","Text":"If we remember B also survives law."},{"Start":"16:49.450 ","End":"16:51.160","Text":"It goes like this,"},{"Start":"16:51.160 ","End":"16:58.765","Text":"dB is equal to Mu naught I multiplied by dl cross"},{"Start":"16:58.765 ","End":"17:08.050","Text":"r vector divided by 4Pi r cubed."},{"Start":"17:08.050 ","End":"17:13.240","Text":"Of course, our current is flowing upwards in this direction,"},{"Start":"17:13.240 ","End":"17:19.780","Text":"which means this vector over here,"},{"Start":"17:19.780 ","End":"17:23.860","Text":"dl is also in the upwards direction,"},{"Start":"17:23.860 ","End":"17:26.755","Text":"and our vector r, what\u0027s our vector r?"},{"Start":"17:26.755 ","End":"17:30.835","Text":"It\u0027s this. This is our r vector."},{"Start":"17:30.835 ","End":"17:35.515","Text":"First of all, we can see that our dl and our r vectors are opposite to 1 another."},{"Start":"17:35.515 ","End":"17:40.750","Text":"But we can also see that they are parallel to 1 another."},{"Start":"17:40.750 ","End":"17:44.875","Text":"As we know, when we do the cross product of something,"},{"Start":"17:44.875 ","End":"17:48.970","Text":"it\u0027s the equivalent of taking dl vector,"},{"Start":"17:48.970 ","End":"17:52.135","Text":"the magnitude, multiplied by r vector,"},{"Start":"17:52.135 ","End":"17:58.190","Text":"its magnitude multiplied by sine of the angle between the 2."},{"Start":"18:00.600 ","End":"18:03.100","Text":"This is r vector."},{"Start":"18:03.100 ","End":"18:04.870","Text":"Now what\u0027s the angle over here?"},{"Start":"18:04.870 ","End":"18:07.495","Text":"The angle over here is Pi."},{"Start":"18:07.495 ","End":"18:13.420","Text":"We\u0027re taking our dl magnitude multiplied by r magnitude,"},{"Start":"18:13.420 ","End":"18:18.270","Text":"multiplied by sine(Pi),"},{"Start":"18:18.270 ","End":"18:20.130","Text":"and sine(Pi), of course,"},{"Start":"18:20.130 ","End":"18:22.080","Text":"is equal to 0."},{"Start":"18:22.080 ","End":"18:24.725","Text":"That means that our db,"},{"Start":"18:24.725 ","End":"18:27.730","Text":"the magnetic field of this section,"},{"Start":"18:27.730 ","End":"18:31.210","Text":"and then of this section we added all up,"},{"Start":"18:31.210 ","End":"18:36.850","Text":"is always going to be multiplied by 0 due to this cross-product."},{"Start":"18:36.850 ","End":"18:43.089","Text":"Which means that the magnetic field over here due to FG section,"},{"Start":"18:43.089 ","End":"18:44.425","Text":"this section over here,"},{"Start":"18:44.425 ","End":"18:47.155","Text":"is going to always be equal to 0."},{"Start":"18:47.155 ","End":"18:48.955","Text":"That\u0027s something to know."},{"Start":"18:48.955 ","End":"18:50.980","Text":"If you have some wire,"},{"Start":"18:50.980 ","End":"18:55.165","Text":"and somewhere parallel along the length of the wire."},{"Start":"18:55.165 ","End":"19:00.880","Text":"If you want to measure the magnetic field at any of these points,"},{"Start":"19:00.880 ","End":"19:07.400","Text":"if you do that, you\u0027ll always see that the magnetic field here is equal to 0."},{"Start":"19:07.860 ","End":"19:10.450","Text":"Exactly because of what we just did,"},{"Start":"19:10.450 ","End":"19:15.340","Text":"because cross-product is going to be equal to 0."},{"Start":"19:15.340 ","End":"19:19.690","Text":"You\u0027ll only be able to measure the magnetic field at"},{"Start":"19:19.690 ","End":"19:25.610","Text":"points that aren\u0027t in line with the wire."},{"Start":"19:26.790 ","End":"19:31.735","Text":"Now, we\u0027re going to work out the superposition of both."},{"Start":"19:31.735 ","End":"19:38.080","Text":"Let\u0027s calculate the total magnetic field over here at the center."},{"Start":"19:38.080 ","End":"19:44.965","Text":"We have 0 plus 5 Mu naught I"},{"Start":"19:44.965 ","End":"19:50.120","Text":"divided by 12R in the z-direction."},{"Start":"19:50.120 ","End":"19:52.980","Text":"All of this is, of course, and the z-direction,"},{"Start":"19:52.980 ","End":"19:56.205","Text":"negative because here it\u0027s in the negative z-direction."},{"Start":"19:56.205 ","End":"19:58.515","Text":"Mu naught I,"},{"Start":"19:58.515 ","End":"20:06.650","Text":"multiplied by 2 minus root 3 divided by 4Pi R,"},{"Start":"20:06.650 ","End":"20:11.845","Text":"and if we plug this into our calculator to work out,"},{"Start":"20:11.845 ","End":"20:21.910","Text":"this is equal to 0.396 Mu naught I divided by R,"},{"Start":"20:21.910 ","End":"20:26.170","Text":"and of course, it\u0027s in the z-direction,"},{"Start":"20:26.170 ","End":"20:28.940","Text":"I can just write this like so."},{"Start":"20:29.970 ","End":"20:34.675","Text":"This is the answer to question number 1."},{"Start":"20:34.675 ","End":"20:38.240","Text":"Now let\u0027s move on to question number 2."},{"Start":"20:38.280 ","End":"20:41.125","Text":"Question number 2 is,"},{"Start":"20:41.125 ","End":"20:45.910","Text":"a charged particle travels through the center of the circular region."},{"Start":"20:45.910 ","End":"20:50.350","Text":"The particle\u0027s trajectory changes due to the magnetic field,"},{"Start":"20:50.350 ","End":"20:53.215","Text":"and the trajectory appears in the diagram."},{"Start":"20:53.215 ","End":"20:57.115","Text":"Here is our center where we just calculated the magnetic field,"},{"Start":"20:57.115 ","End":"21:01.450","Text":"and then the classical begins bending like so,"},{"Start":"21:01.450 ","End":"21:05.095","Text":"and we\u0027re being asked what is the particle\u0027s charge?"},{"Start":"21:05.095 ","End":"21:08.020","Text":"First of all, we know from this,"},{"Start":"21:08.020 ","End":"21:11.950","Text":"that the particle is going to be traveling like so towards"},{"Start":"21:11.950 ","End":"21:16.840","Text":"the center of the circle over here."},{"Start":"21:16.840 ","End":"21:20.049","Text":"Which means that at this point in the center,"},{"Start":"21:20.049 ","End":"21:25.285","Text":"the velocity is like so in this direction in a straight line."},{"Start":"21:25.285 ","End":"21:29.230","Text":"But due to the magnetic field over here,"},{"Start":"21:29.230 ","End":"21:33.490","Text":"our particle\u0027s trajectory bends and changes."},{"Start":"21:33.490 ","End":"21:39.445","Text":"What\u0027s actually happening is that the magnetic field is applying a force,"},{"Start":"21:39.445 ","End":"21:43.359","Text":"a magnetic force FB in this downwards direction,"},{"Start":"21:43.359 ","End":"21:47.620","Text":"which is what is causing our charged particle to"},{"Start":"21:47.620 ","End":"21:53.420","Text":"move in this downwards direction to change the direction of its trajectory."},{"Start":"21:53.520 ","End":"21:56.095","Text":"I made some space."},{"Start":"21:56.095 ","End":"21:59.665","Text":"What we\u0027re going to do is we\u0027re going to use Lawrence\u0027s law."},{"Start":"21:59.665 ","End":"22:06.835","Text":"We know that the magnetic force is equal to q multiplied by v,"},{"Start":"22:06.835 ","End":"22:12.250","Text":"the velocity cross-product with the magnetic field."},{"Start":"22:12.250 ","End":"22:15.895","Text":"If we use the right-hand rule,"},{"Start":"22:15.895 ","End":"22:20.980","Text":"so we know that the thumb points in the direction of the velocity."},{"Start":"22:20.980 ","End":"22:23.395","Text":"As you can see, it\u0027s going in the x-direction."},{"Start":"22:23.395 ","End":"22:29.080","Text":"This is our thumb going like so in the direction of the velocity."},{"Start":"22:29.080 ","End":"22:34.765","Text":"Then we have our forefinger pointing in the direction of the magnetic field."},{"Start":"22:34.765 ","End":"22:37.780","Text":"We saw that the total magnetic field or"},{"Start":"22:37.780 ","End":"22:43.660","Text":"the overall magnetic field due to all of the parts is in the positive z-direction,"},{"Start":"22:43.660 ","End":"22:46.900","Text":"which as we said, is coming towards us."},{"Start":"22:46.900 ","End":"22:50.140","Text":"Here is our pointing finger,"},{"Start":"22:50.140 ","End":"22:52.060","Text":"and it\u0027s coming towards us,"},{"Start":"22:52.060 ","End":"22:56.605","Text":"and this represents the magnetic field coming out towards us,"},{"Start":"22:56.605 ","End":"22:58.299","Text":"and then in that case,"},{"Start":"22:58.299 ","End":"22:59.965","Text":"if you draw this,"},{"Start":"22:59.965 ","End":"23:01.645","Text":"it\u0027s a bit difficult to draw."},{"Start":"23:01.645 ","End":"23:09.860","Text":"The middle finger, which represents the force is therefore pointing upwards."},{"Start":"23:10.560 ","End":"23:13.390","Text":"This is the middle finger."},{"Start":"23:13.390 ","End":"23:21.020","Text":"However, we know that the force is pointing downwards because we can see the trajectory."},{"Start":"23:21.510 ","End":"23:23.590","Text":"From the trajectory,"},{"Start":"23:23.590 ","End":"23:26.965","Text":"we see that this downwards force is being applied,"},{"Start":"23:26.965 ","End":"23:30.700","Text":"which means that the force is in the wrong direction."},{"Start":"23:30.700 ","End":"23:32.500","Text":"How can we change this?"},{"Start":"23:32.500 ","End":"23:36.865","Text":"That means that our charge q has to be negative."},{"Start":"23:36.865 ","End":"23:43.250","Text":"From this, we get that q must be less than 0."},{"Start":"23:43.710 ","End":"23:47.800","Text":"The charge is negative because that\u0027s the only way that"},{"Start":"23:47.800 ","End":"23:51.355","Text":"given the current situation and what we\u0027re given in the question,"},{"Start":"23:51.355 ","End":"23:54.760","Text":"we can get that our force will be pointing downwards in"},{"Start":"23:54.760 ","End":"24:00.490","Text":"conjunction with the change in direction of the trajectory."},{"Start":"24:00.490 ","End":"24:04.520","Text":"That\u0027s it. That\u0027s the answer to question number 2."},{"Start":"24:05.370 ","End":"24:12.715","Text":"In another experiment, a non-uniform magnetic field is present in the region of y,"},{"Start":"24:12.715 ","End":"24:15.865","Text":"which is between R and 2R."},{"Start":"24:15.865 ","End":"24:23.725","Text":"We have a non-uniform magnetic field in this region over here."},{"Start":"24:23.725 ","End":"24:30.895","Text":"Part of the FG section of the wire is in the region with the non-uniform magnetic field."},{"Start":"24:30.895 ","End":"24:33.895","Text":"The magnetic field in this region is given to us."},{"Start":"24:33.895 ","End":"24:41.290","Text":"It\u0027s B=(0,0,ay^2), so 0 in the x and y-direction,"},{"Start":"24:41.290 ","End":"24:45.970","Text":"and ay^2 in the z-direction,"},{"Start":"24:45.970 ","End":"24:48.115","Text":"where a is given to us,"},{"Start":"24:48.115 ","End":"24:54.310","Text":"and we\u0027re being asked what magnetic force does this field apply to the wire?"},{"Start":"24:54.310 ","End":"24:59.605","Text":"We\u0027re looking for the force on this wire over here."},{"Start":"24:59.605 ","End":"25:02.995","Text":"The force on the wire is given to us,"},{"Start":"25:02.995 ","End":"25:07.720","Text":"dF is equal to the current"},{"Start":"25:07.720 ","End":"25:14.330","Text":"multiplied by dl cross B."},{"Start":"25:15.240 ","End":"25:18.145","Text":"This we\u0027ve seen in a previous lesson."},{"Start":"25:18.145 ","End":"25:22.720","Text":"This magnetic field over here is causing some force,"},{"Start":"25:22.720 ","End":"25:25.030","Text":"is applying a force to the wire,"},{"Start":"25:25.030 ","End":"25:27.865","Text":"and this is the equation to work this out."},{"Start":"25:27.865 ","End":"25:31.570","Text":"Our dl, so we\u0027re choosing this,"},{"Start":"25:31.570 ","End":"25:34.345","Text":"so this is our dl,"},{"Start":"25:34.345 ","End":"25:38.530","Text":"and its direction is of course in the direction of the current."},{"Start":"25:38.530 ","End":"25:40.555","Text":"In this section of the wire,"},{"Start":"25:40.555 ","End":"25:45.655","Text":"the current is pointing upwards over here in the y-direction."},{"Start":"25:45.655 ","End":"25:56.090","Text":"dl is equal to some change in y. dy in the y-direction."},{"Start":"25:56.310 ","End":"25:58.990","Text":"dl is equal to dy."},{"Start":"25:58.990 ","End":"26:03.100","Text":"Then we have our magnetic field,"},{"Start":"26:03.100 ","End":"26:08.050","Text":"which is of course equal to ay^2 in the z-direction."},{"Start":"26:08.050 ","End":"26:10.180","Text":"I\u0027ve got that in the question."},{"Start":"26:10.180 ","End":"26:16.450","Text":"Therefore, our dF is simply going to be"},{"Start":"26:16.450 ","End":"26:22.180","Text":"equal to I multiplied by dy in the y-direction,"},{"Start":"26:22.180 ","End":"26:28.495","Text":"cross-product with ay^2 in the z-direction."},{"Start":"26:28.495 ","End":"26:34.140","Text":"What we have is I, ay^2 dy."},{"Start":"26:34.140 ","End":"26:36.525","Text":"I put all the constants to 1 side,"},{"Start":"26:36.525 ","End":"26:39.300","Text":"and then we have y cross z,"},{"Start":"26:39.300 ","End":"26:44.170","Text":"which we know is in the x-direction."},{"Start":"26:44.170 ","End":"26:46.509","Text":"This I just know, but otherwise,"},{"Start":"26:46.509 ","End":"26:54.445","Text":"you can just work out the cross-product with 010 crossed with 001."},{"Start":"26:54.445 ","End":"26:56.755","Text":"This is something in the y-direction,"},{"Start":"26:56.755 ","End":"26:58.585","Text":"this is y hat and this is z hat,"},{"Start":"26:58.585 ","End":"27:00.670","Text":"and then you can just work it out,"},{"Start":"27:00.670 ","End":"27:04.700","Text":"and then you\u0027ll get x in the end."},{"Start":"27:04.920 ","End":"27:07.495","Text":"Now we have the dF."},{"Start":"27:07.495 ","End":"27:14.035","Text":"The total force is simply going to be the integral on dF,"},{"Start":"27:14.035 ","End":"27:18.265","Text":"which is the integral on I,"},{"Start":"27:18.265 ","End":"27:23.485","Text":"ay^2 dy in the x-direction,"},{"Start":"27:23.485 ","End":"27:25.840","Text":"and it\u0027s between the bounds of,"},{"Start":"27:25.840 ","End":"27:27.070","Text":"where told in the question,"},{"Start":"27:27.070 ","End":"27:31.100","Text":"between R and 2R."},{"Start":"27:31.100 ","End":"27:34.000","Text":"We\u0027re even told that in the question."},{"Start":"27:34.000 ","End":"27:38.215","Text":"What we\u0027re going to get is I,"},{"Start":"27:38.215 ","End":"27:40.090","Text":"a, our constants,"},{"Start":"27:40.090 ","End":"27:44.830","Text":"and then we have y^3 divided by 3,"},{"Start":"27:44.830 ","End":"27:47.860","Text":"and all of this is in the x-direction,"},{"Start":"27:47.860 ","End":"27:52.900","Text":"and then we have to plug in our bounds of R and 2R."},{"Start":"27:52.900 ","End":"27:55.870","Text":"Once we plug that in,"},{"Start":"27:55.870 ","End":"28:00.685","Text":"what we have is Ia divided by 3,"},{"Start":"28:00.685 ","End":"28:03.400","Text":"and this is multiplied by,"},{"Start":"28:03.400 ","End":"28:05.935","Text":"we have 2R^3,"},{"Start":"28:05.935 ","End":"28:09.640","Text":"that is simply 8R^3,"},{"Start":"28:09.640 ","End":"28:11.200","Text":"because it\u0027s 2R,"},{"Start":"28:11.200 ","End":"28:16.285","Text":"and all of that is cubed minus R^3,"},{"Start":"28:16.285 ","End":"28:19.600","Text":"which is equal to 7R^3,"},{"Start":"28:19.600 ","End":"28:28.400","Text":"what we have is 7Ia R^3 divided by 3."},{"Start":"28:28.890 ","End":"28:38.350","Text":"This is the magnetic force that is being applied to the wire in this region,"},{"Start":"28:38.350 ","End":"28:42.235","Text":"and of course, it\u0027s in the x-direction."},{"Start":"28:42.235 ","End":"28:45.110","Text":"That\u0027s the end of this lesson."}],"ID":21405},{"Watched":false,"Name":"Exercise 10","Duration":"25m 18s","ChapterTopicVideoID":21330,"CourseChapterTopicPlaylistID":99469,"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":"00:04.560","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.560 ","End":"00:12.135","Text":"A spherical shell of radius R has charge Q evenly distributed on the surface."},{"Start":"00:12.135 ","End":"00:18.705","Text":"The shell rotates about its axis of symmetry at a constant angular velocity of Omega."},{"Start":"00:18.705 ","End":"00:23.535","Text":"Calculate the magnetic field at the center of the shell,"},{"Start":"00:23.535 ","End":"00:28.440","Text":"and assume that the rotation does not affect the charge distribution."},{"Start":"00:28.440 ","End":"00:34.080","Text":"Here is the center of the shell and we\u0027re trying to calculate,"},{"Start":"00:34.080 ","End":"00:37.710","Text":"over here what the magnetic field is."},{"Start":"00:37.710 ","End":"00:42.245","Text":"First of all, let\u0027s see how the magnetic field over here is formed."},{"Start":"00:42.245 ","End":"00:46.065","Text":"We know that the magnetic field is formed by some"},{"Start":"00:46.065 ","End":"00:51.015","Text":"current and current is formed by charges moving."},{"Start":"00:51.015 ","End":"00:55.400","Text":"We can see that this shell has charge evenly distributed across it"},{"Start":"00:55.400 ","End":"01:00.050","Text":"and because it is rotating about itself,"},{"Start":"01:00.050 ","End":"01:01.460","Text":"we can see that,"},{"Start":"01:01.460 ","End":"01:05.030","Text":"that rotation is causing the charges to move,"},{"Start":"01:05.030 ","End":"01:11.095","Text":"thereby creating current and thereby creating a magnetic field."},{"Start":"01:11.095 ","End":"01:14.485","Text":"What we can see is that,"},{"Start":"01:14.485 ","End":"01:18.460","Text":"if we have a particle over here,"},{"Start":"01:18.460 ","End":"01:20.021","Text":"and it has some charge,"},{"Start":"01:20.021 ","End":"01:24.005","Text":"it\u0027s going to rotate in the direction of"},{"Start":"01:24.005 ","End":"01:29.530","Text":"Omega and it\u0027s going to rotate where eventually it will be over here."},{"Start":"01:29.530 ","End":"01:33.465","Text":"We can see that its radius of rotation,"},{"Start":"01:33.465 ","End":"01:38.615","Text":"is this, or rather let me draw it a bit lower down."},{"Start":"01:38.615 ","End":"01:43.940","Text":"It\u0027s traveling in this direction and it will eventually reach this point."},{"Start":"01:43.940 ","End":"01:46.280","Text":"We can see that its distance,"},{"Start":"01:46.280 ","End":"01:54.995","Text":"or its radius from the axis of rotation is this lowercase r over here."},{"Start":"01:54.995 ","End":"02:00.865","Text":"We can see that it is different to the radius of the spherical shell."},{"Start":"02:00.865 ","End":"02:06.400","Text":"Lowercase r represents the distance or the radius from the axis of"},{"Start":"02:06.400 ","End":"02:13.345","Text":"rotation until the specific charge that we\u0027re looking at."},{"Start":"02:13.345 ","End":"02:18.760","Text":"Okay. If we label this angle over here as Phi."},{"Start":"02:18.760 ","End":"02:23.965","Text":"We could say that lowercase r is equal to"},{"Start":"02:23.965 ","End":"02:30.385","Text":"the radius of the spherical shell multiplied by sine of Phi,"},{"Start":"02:30.385 ","End":"02:33.070","Text":"because R is the opposite side,"},{"Start":"02:33.070 ","End":"02:34.240","Text":"that\u0027s opposite to the angle,"},{"Start":"02:34.240 ","End":"02:36.500","Text":"so that is sine of Phi."},{"Start":"02:36.840 ","End":"02:45.080","Text":"Now the velocity, which of course is a vector of each piece."},{"Start":"02:45.080 ","End":"02:53.490","Text":"We can see that the pieces are moving in this circular motion around the spherical shell."},{"Start":"02:54.680 ","End":"03:03.635","Text":"It is equal to omega r. The angular velocity multiplied by the radius of rotation,"},{"Start":"03:03.635 ","End":"03:07.460","Text":"where the radius of rotation of each piece is obviously going to be"},{"Start":"03:07.460 ","End":"03:15.150","Text":"different and of course it\u0027s in the Theta direction, like so."},{"Start":"03:16.730 ","End":"03:19.365","Text":"This is of course,"},{"Start":"03:19.365 ","End":"03:25.065","Text":"equal to Omega multiplied by r,"},{"Start":"03:25.065 ","End":"03:33.330","Text":"which is r sine of Phi and this is of course in the Theta direction."},{"Start":"03:34.220 ","End":"03:39.200","Text":"As we thought, this is what is producing the current."},{"Start":"03:39.200 ","End":"03:42.875","Text":"Let\u0027s look, we\u0027re dealing with a spherical shell,"},{"Start":"03:42.875 ","End":"03:49.580","Text":"which means that we have current density k because what we have,"},{"Start":"03:49.580 ","End":"03:52.385","Text":"if we\u0027re dealing with a spherical shell with charge Q,"},{"Start":"03:52.385 ","End":"03:57.040","Text":"then it has charge density per unit area Sigma."},{"Start":"03:57.040 ","End":"04:03.680","Text":"K is equal to Sigma multiplied by V,"},{"Start":"04:03.680 ","End":"04:07.720","Text":"the velocity that the charges are moving at."},{"Start":"04:07.720 ","End":"04:12.964","Text":"What is Sigma? Sigma is equal to the charge,"},{"Start":"04:12.964 ","End":"04:17.015","Text":"which is Q, divided by the total area."},{"Start":"04:17.015 ","End":"04:20.540","Text":"The total area is that of a spherical shell,"},{"Start":"04:20.540 ","End":"04:28.235","Text":"4 Pi R squared where we\u0027re using capital R because the radius of the sphere"},{"Start":"04:28.235 ","End":"04:35.885","Text":"is capital R. Let\u0027s plug in our values for k. K is equal to Sigma,"},{"Start":"04:35.885 ","End":"04:43.425","Text":"which is Q, divided by 4 Pi R squared multiplied by V,"},{"Start":"04:43.425 ","End":"04:48.860","Text":"which is Omega R sine of Phi in"},{"Start":"04:48.860 ","End":"04:55.415","Text":"the Theta direction and we can cancel out one of these Rs."},{"Start":"04:55.415 ","End":"04:57.860","Text":"The direction of k,"},{"Start":"04:57.860 ","End":"04:59.725","Text":"if we\u0027re looking over here,"},{"Start":"04:59.725 ","End":"05:03.415","Text":"k is into the page."},{"Start":"05:03.415 ","End":"05:07.100","Text":"But if we\u0027re looking over here on this side,"},{"Start":"05:07.100 ","End":"05:13.650","Text":"k is coming out of the page because it\u0027s going around in the Theta direction."},{"Start":"05:14.420 ","End":"05:18.470","Text":"Now what we want to do is we want to calculate the B field."},{"Start":"05:18.470 ","End":"05:20.029","Text":"We have 2 options."},{"Start":"05:20.029 ","End":"05:28.085","Text":"The first option is to sum up the current contributions from each infinitesimal charge,"},{"Start":"05:28.085 ","End":"05:33.152","Text":"we see the current contribution from this charge,and"},{"Start":"05:33.152 ","End":"05:38.845","Text":"then we sum it up for everything along the surface of the spherical shell."},{"Start":"05:38.845 ","End":"05:42.980","Text":"Then we use Biot-savart\u0027s equation for"},{"Start":"05:42.980 ","End":"05:49.675","Text":"calculating the magnetic field given a certain current."},{"Start":"05:49.675 ","End":"05:55.790","Text":"This option will take relatively a long time or option number 2,"},{"Start":"05:55.790 ","End":"06:02.225","Text":"we can see that each circle over here represents a current carrying loop."},{"Start":"06:02.225 ","End":"06:12.280","Text":"I can look at this as some current carrying loop at some height Z. I can say that"},{"Start":"06:12.280 ","End":"06:17.870","Text":"this height above the center over here is"},{"Start":"06:17.870 ","End":"06:23.780","Text":"some height to Z and that this is the Z axis."},{"Start":"06:23.780 ","End":"06:31.430","Text":"Then I can use the equation for the B field due to a CCL,"},{"Start":"06:31.430 ","End":"06:33.830","Text":"due to a current carrying loop."},{"Start":"06:33.830 ","End":"06:42.180","Text":"Then all I will do is I will sum up all of that for all of the loops."},{"Start":"06:42.190 ","End":"06:49.450","Text":"What we\u0027re going to do is we\u0027re going to calculate this via option number 2."},{"Start":"06:49.450 ","End":"06:56.375","Text":"We saw that the equation for the magnetic field,"},{"Start":"06:56.375 ","End":"06:59.645","Text":"and here it\u0027s going to be along the Z axes,"},{"Start":"06:59.645 ","End":"07:04.940","Text":"caused by a current-carrying loop,"},{"Start":"07:04.940 ","End":"07:11.690","Text":"or a current-carrying ring is equal to Mioknotes multiplied by I,"},{"Start":"07:11.690 ","End":"07:14.795","Text":"multiplied by R squared."},{"Start":"07:14.795 ","End":"07:17.935","Text":"We\u0027re going to speak about this I in a second,"},{"Start":"07:17.935 ","End":"07:28.950","Text":"divided by 2 multiplied by Z squared plus I squared to the power of 3 over 2."},{"Start":"07:29.030 ","End":"07:33.935","Text":"This is an equation that we saw in one of the previous lessons."},{"Start":"07:33.935 ","End":"07:35.345","Text":"Now, first of all,"},{"Start":"07:35.345 ","End":"07:37.460","Text":"what is this capital R?"},{"Start":"07:37.460 ","End":"07:45.890","Text":"Do not get confused between the R over here and the radius of the sphere in the question."},{"Start":"07:45.890 ","End":"07:53.545","Text":"Over here, we are speaking about the radius of the current carrying loop."},{"Start":"07:53.545 ","End":"07:57.500","Text":"The R in this equation is the radius of"},{"Start":"07:57.500 ","End":"08:04.265","Text":"the current carrying loop and the Z that we are speaking about over here"},{"Start":"08:04.265 ","End":"08:08.580","Text":"is speaking about the heights"},{"Start":"08:09.670 ","End":"08:17.880","Text":"where we\u0027re measuring the magnetic field along the axis of symmetry."},{"Start":"08:18.390 ","End":"08:22.390","Text":"That is the position along the axis of symmetry where we"},{"Start":"08:22.390 ","End":"08:26.425","Text":"are measuring the magnetic field."},{"Start":"08:26.425 ","End":"08:30.050","Text":"Over here, this is our z."},{"Start":"08:30.540 ","End":"08:34.000","Text":"That\u0027s over here. Now let\u0027s see,"},{"Start":"08:34.000 ","End":"08:37.900","Text":"our radius over here is going to be equal to,"},{"Start":"08:37.900 ","End":"08:40.630","Text":"so it\u0027s the radius of the current carrying loop,"},{"Start":"08:40.630 ","End":"08:47.575","Text":"which as we said, is this r and our r,"},{"Start":"08:47.575 ","End":"08:50.050","Text":"as we calculated over here,"},{"Start":"08:50.050 ","End":"08:54.715","Text":"is equal to R sine of Phi,"},{"Start":"08:54.715 ","End":"09:00.640","Text":"where of course, this is the radius of the sphere."},{"Start":"09:00.640 ","End":"09:10.240","Text":"Our z, in this case is this distance over here."},{"Start":"09:10.240 ","End":"09:16.550","Text":"Now, notice that z is the position along the axis of symmetry where we\u0027re measuring B."},{"Start":"09:17.580 ","End":"09:22.900","Text":"This point is the center of the current carrying loop,"},{"Start":"09:22.900 ","End":"09:28.495","Text":"and we\u0027re measuring the distance away from this center where we\u0027re measuring B."},{"Start":"09:28.495 ","End":"09:32.560","Text":"Here specifically we\u0027re measuring B a distance of"},{"Start":"09:32.560 ","End":"09:37.675","Text":"z below the center of the current carrying loop."},{"Start":"09:37.675 ","End":"09:48.700","Text":"In that case, z over here is equal to negative R cosine of Phi."},{"Start":"09:48.700 ","End":"09:54.280","Text":"Because here we\u0027re dealing with the adjacent side to our angle Phi,"},{"Start":"09:54.280 ","End":"09:58.000","Text":"so that\u0027s why cosine and the negative because we\u0027re"},{"Start":"09:58.000 ","End":"10:04.130","Text":"located below the origin of the current carrying loop."},{"Start":"10:04.650 ","End":"10:07.600","Text":"I added in here the position along"},{"Start":"10:07.600 ","End":"10:10.600","Text":"the axis of symmetry where we are measuring the magnetic field"},{"Start":"10:10.600 ","End":"10:16.495","Text":"relative to the center of the specific current carrying loop that we\u0027re dealing with."},{"Start":"10:16.495 ","End":"10:21.050","Text":"That\u0027s why here, this is a negative."},{"Start":"10:21.120 ","End":"10:27.955","Text":"Now we want to calculate our current I. I as we know,"},{"Start":"10:27.955 ","End":"10:31.405","Text":"is the integral along dI,"},{"Start":"10:31.405 ","End":"10:35.035","Text":"where of course, what is dI."},{"Start":"10:35.035 ","End":"10:42.670","Text":"This is going to be the integral along k,"},{"Start":"10:42.670 ","End":"10:46.090","Text":"which we have over here,"},{"Start":"10:46.090 ","End":"10:50.570","Text":"and when we have k, it\u0027s k dl."},{"Start":"10:51.510 ","End":"10:57.680","Text":"Now what we want to know is what exactly is our dl?"},{"Start":"10:58.080 ","End":"11:00.145","Text":"I\u0027ll draw it in pink."},{"Start":"11:00.145 ","End":"11:01.630","Text":"But what we can see is that we have"},{"Start":"11:01.630 ","End":"11:06.205","Text":"our current carrying loop where our current is traveling,"},{"Start":"11:06.205 ","End":"11:08.185","Text":"of course, like this,"},{"Start":"11:08.185 ","End":"11:12.085","Text":"through the loop around in a circle."},{"Start":"11:12.085 ","End":"11:14.680","Text":"Which means that if we look over here,"},{"Start":"11:14.680 ","End":"11:22.165","Text":"the current is passing through this arc that we have over here."},{"Start":"11:22.165 ","End":"11:24.685","Text":"What is this arc?"},{"Start":"11:24.685 ","End":"11:27.369","Text":"We\u0027re dealing with spherical coordinates."},{"Start":"11:27.369 ","End":"11:35.690","Text":"First of all we can see that it is dependent first of all on the radius of the sphere."},{"Start":"11:37.590 ","End":"11:44.935","Text":"But that\u0027s of course a constant and it is dependent on this width over here of our loop."},{"Start":"11:44.935 ","End":"11:46.855","Text":"What is the width of the loop?"},{"Start":"11:46.855 ","End":"11:51.265","Text":"It\u0027s this change in this angle Phi over here."},{"Start":"11:51.265 ","End":"11:56.305","Text":"We have Phi starting over here at the beginning of the loop,"},{"Start":"11:56.305 ","End":"11:59.860","Text":"and then if we move this line a bit over here,"},{"Start":"11:59.860 ","End":"12:01.735","Text":"we move this line up,"},{"Start":"12:01.735 ","End":"12:05.050","Text":"which means that we decrease this angle Phi,"},{"Start":"12:05.050 ","End":"12:08.545","Text":"then we get the edge of this arc."},{"Start":"12:08.545 ","End":"12:12.860","Text":"What we have is that dl"},{"Start":"12:13.290 ","End":"12:20.170","Text":"is equal to RdPhi."},{"Start":"12:20.170 ","End":"12:22.870","Text":"Now if we substitute in our k,"},{"Start":"12:22.870 ","End":"12:30.535","Text":"so we saw our k is equal to Q divided by 4Pi,"},{"Start":"12:30.535 ","End":"12:35.575","Text":"and then I\u0027ll write the Q over here 4PiR,"},{"Start":"12:35.575 ","End":"12:37.345","Text":"only the squares canceled off,"},{"Start":"12:37.345 ","End":"12:42.835","Text":"multiplied by Omega sine of Phi."},{"Start":"12:42.835 ","End":"12:45.790","Text":"We\u0027ll leave the direction out."},{"Start":"12:45.790 ","End":"12:50.660","Text":"Then this is multiplied by RdPhi."},{"Start":"12:51.330 ","End":"12:56.410","Text":"This was the tricky part to get over this over here"},{"Start":"12:56.410 ","End":"13:01.600","Text":"because it\u0027s difficult when working with k to work out what the dl is."},{"Start":"13:01.600 ","End":"13:04.470","Text":"Another way that you could have solved it,"},{"Start":"13:04.470 ","End":"13:08.400","Text":"which is a little bit simpler when using k is using the idea"},{"Start":"13:08.400 ","End":"13:12.375","Text":"that the current is equal to dq by dt,"},{"Start":"13:12.375 ","End":"13:17.410","Text":"and then you substitute in what dq is."},{"Start":"13:17.490 ","End":"13:20.800","Text":"Over here all of the charges on the surface,"},{"Start":"13:20.800 ","End":"13:27.445","Text":"so dq is just Sigma ds and divided by dt,"},{"Start":"13:27.445 ","End":"13:33.415","Text":"which of course, our Sigma is equal to"},{"Start":"13:33.415 ","End":"13:40.220","Text":"the total charge Q divided by the surface area of a sphere so 4PiR^2."},{"Start":"13:42.990 ","End":"13:46.870","Text":"We\u0027re working in spherical coordinates where of course,"},{"Start":"13:46.870 ","End":"13:49.165","Text":"our radius is constant."},{"Start":"13:49.165 ","End":"13:54.490","Text":"If we had a non-constant radius then we\u0027d be dealing with volume, of course."},{"Start":"13:54.490 ","End":"13:57.010","Text":"But here we\u0027re dealing with surface area so we have"},{"Start":"13:57.010 ","End":"14:00.520","Text":"a constant radius so we multiply this by"},{"Start":"14:00.520 ","End":"14:06.025","Text":"R^2 multiplied by sine"},{"Start":"14:06.025 ","End":"14:11.395","Text":"of Phi d Theta d Phi."},{"Start":"14:11.395 ","End":"14:16.615","Text":"This is in spherical coordinates and of course divided by the dt."},{"Start":"14:16.615 ","End":"14:20.725","Text":"First of all, the R squares cancel out,"},{"Start":"14:20.725 ","End":"14:26.110","Text":"and then usually something in the numerator will"},{"Start":"14:26.110 ","End":"14:32.300","Text":"sort itself out with the dt to give us this equation over here."},{"Start":"14:33.390 ","End":"14:37.525","Text":"What we can see is that Omega,"},{"Start":"14:37.525 ","End":"14:45.325","Text":"the angular velocity, is equal to d Theta by dt."},{"Start":"14:45.325 ","End":"14:51.670","Text":"I can say that this is equal to q multiplied by d Theta dt"},{"Start":"14:51.670 ","End":"14:58.420","Text":"so Omega multiplied by sine of Phi d Phi,"},{"Start":"14:58.420 ","End":"15:02.335","Text":"and all of this is divided by 4Pi,"},{"Start":"15:02.335 ","End":"15:03.760","Text":"which if we look over here,"},{"Start":"15:03.760 ","End":"15:09.220","Text":"we can of course cancel out these I\u0027s and we get the exact same equation."},{"Start":"15:09.220 ","End":"15:13.180","Text":"This is an easier method to use when we\u0027re dealing with"},{"Start":"15:13.180 ","End":"15:22.120","Text":"k. Let\u0027s scroll down and let\u0027s do the calculation."},{"Start":"15:22.120 ","End":"15:25.555","Text":"Now we know what our current is."},{"Start":"15:25.555 ","End":"15:28.480","Text":"This is of course dI,"},{"Start":"15:28.480 ","End":"15:38.890","Text":"and of course this is also dI so we have that B over here is equal"},{"Start":"15:38.890 ","End":"15:43.870","Text":"to Mu Naught multiplied by I where of course here we\u0027re dealing with"},{"Start":"15:43.870 ","End":"15:49.645","Text":"dI because this is an infinitesimal current loop."},{"Start":"15:49.645 ","End":"15:53.650","Text":"We\u0027re summing on dI, which means that this is also dB."},{"Start":"15:53.650 ","End":"15:56.905","Text":"Because this is a loop that we\u0027re about to sum up,"},{"Start":"15:56.905 ","End":"15:58.885","Text":"we know that it\u0027s dI,"},{"Start":"15:58.885 ","End":"16:05.560","Text":"and if we have some partial derivative on one side,"},{"Start":"16:05.560 ","End":"16:09.985","Text":"then we have to have a partial derivative on the other side."},{"Start":"16:09.985 ","End":"16:17.020","Text":"DB_z is equal to Mu Naught multiplied by dI as we just said."},{"Start":"16:17.020 ","End":"16:23.530","Text":"That\u0027s this over here so we have Q Omega sine of"},{"Start":"16:23.530 ","End":"16:34.000","Text":"Phi d Phi divided by 4Pi,"},{"Start":"16:34.000 ","End":"16:37.015","Text":"and then multiplied by R^2,"},{"Start":"16:37.015 ","End":"16:38.990","Text":"where this is R in the equation."},{"Start":"16:38.990 ","End":"16:41.940","Text":"Remember this is the radius of the current carrying loop,"},{"Start":"16:41.940 ","End":"16:46.015","Text":"so this is actually multiplied by"},{"Start":"16:46.015 ","End":"16:53.290","Text":"R sine of Phi^2."},{"Start":"16:53.290 ","End":"17:00.175","Text":"Then this is divided by 2. Then we have z^2."},{"Start":"17:00.175 ","End":"17:03.190","Text":"Z is this over here,"},{"Start":"17:03.190 ","End":"17:11.095","Text":"which is negative rcos(Phi) squared plus r squared."},{"Start":"17:11.095 ","End":"17:17.980","Text":"Again, it\u0027s this r from the equation plus rsine(Phi) because"},{"Start":"17:17.980 ","End":"17:25.900","Text":"this is that r squared and all of this to the power of 3/2."},{"Start":"17:25.900 ","End":"17:28.630","Text":"Now, a quick note,"},{"Start":"17:28.630 ","End":"17:37.135","Text":"you have to take the dI that we calculated over here and plug that into this equation,"},{"Start":"17:37.135 ","End":"17:42.625","Text":"because what we want to do is we want to calculate the magnetic field at this point,"},{"Start":"17:42.625 ","End":"17:48.580","Text":"which is a summation of all of these tiny loops."},{"Start":"17:48.580 ","End":"17:51.400","Text":"Whereas if we take the dI,"},{"Start":"17:51.400 ","End":"17:53.650","Text":"integrate for the total current,"},{"Start":"17:53.650 ","End":"17:56.860","Text":"and then substitute the I into this equation,"},{"Start":"17:56.860 ","End":"17:59.875","Text":"we will get the wrong answer."},{"Start":"17:59.875 ","End":"18:05.050","Text":"Remember, you take the dI from the loop and then you plug that"},{"Start":"18:05.050 ","End":"18:09.985","Text":"into this equation and then you integrate on this equation as a whole."},{"Start":"18:09.985 ","End":"18:11.740","Text":"Because you\u0027re looking at this point,"},{"Start":"18:11.740 ","End":"18:15.280","Text":"the magnetic field of this point due to each loop,"},{"Start":"18:15.280 ","End":"18:18.295","Text":"and then you sum up for all of the loops."},{"Start":"18:18.295 ","End":"18:22.160","Text":"That\u0027s the whole point of this equation."},{"Start":"18:23.610 ","End":"18:27.700","Text":"Now what we\u0027re going to do is we\u0027re going to simplify this."},{"Start":"18:27.700 ","End":"18:31.420","Text":"We have Mu_0, Q Omega,"},{"Start":"18:31.420 ","End":"18:37.765","Text":"and then we have over here sine(Phi),"},{"Start":"18:37.765 ","End":"18:43.540","Text":"and then multiplied by sine^2(Phi)."},{"Start":"18:43.540 ","End":"18:45.715","Text":"We have sine^3(Phi),"},{"Start":"18:45.715 ","End":"18:50.995","Text":"and then we have r squared, and then dPhi."},{"Start":"18:50.995 ","End":"18:56.320","Text":"All of this is divided by 4 times 2, which is 8Pi."},{"Start":"18:56.320 ","End":"18:58.750","Text":"Then we have inside the brackets,"},{"Start":"18:58.750 ","End":"19:08.665","Text":"so we have r^2cos^2(Phi) plus r squared sine squared Phi,"},{"Start":"19:08.665 ","End":"19:11.395","Text":"and all of this to the power of 3/2."},{"Start":"19:11.395 ","End":"19:16.255","Text":"Of course, cos^2(Phi) plus sine^2(Phi) is equal to 1."},{"Start":"19:16.255 ","End":"19:22.555","Text":"Then what we have inside the brackets is just r^2,"},{"Start":"19:22.555 ","End":"19:25.960","Text":"and then we have r^2, of course,"},{"Start":"19:25.960 ","End":"19:28.270","Text":"to the power of 3 over 2,"},{"Start":"19:28.270 ","End":"19:32.740","Text":"which is just the square root of r^2 to the power of 3,"},{"Start":"19:32.740 ","End":"19:36.865","Text":"that is simply equal to r^3."},{"Start":"19:36.865 ","End":"19:46.915","Text":"Now what we can do is we can write this out as the total magnetic field at this point."},{"Start":"19:46.915 ","End":"19:49.315","Text":"We\u0027re just going to integrate along all of this."},{"Start":"19:49.315 ","End":"19:53.450","Text":"First of all, we\u0027ll take out all of our constants."},{"Start":"19:54.450 ","End":"19:56.710","Text":"Actually, before we do that,"},{"Start":"19:56.710 ","End":"20:01.075","Text":"we have this r^2 in the numerator divided by r^3."},{"Start":"20:01.075 ","End":"20:03.550","Text":"The r^3 is in the denominator,"},{"Start":"20:03.550 ","End":"20:07.825","Text":"so we can cancel this out and cancel this cubed over here."},{"Start":"20:07.825 ","End":"20:15.430","Text":"What we have is Mu_0 Q Omega are all constants divided by"},{"Start":"20:15.430 ","End":"20:24.040","Text":"8Pi r. Then we\u0027re integrating along this over here,"},{"Start":"20:24.040 ","End":"20:27.680","Text":"which is just sine^3(Phi)dPhi."},{"Start":"20:31.230 ","End":"20:36.850","Text":"The borders, notice we\u0027re integrating along Phi."},{"Start":"20:36.850 ","End":"20:41.470","Text":"Phi goes from here where it\u0027s 0 degrees up until here,"},{"Start":"20:41.470 ","End":"20:44.410","Text":"where it\u0027s Pi or Pi radians."},{"Start":"20:44.410 ","End":"20:49.880","Text":"We\u0027re going from 0 until Pi or 0-180 degrees."},{"Start":"20:50.070 ","End":"20:53.170","Text":"Those are our bounds."},{"Start":"20:53.170 ","End":"20:55.060","Text":"Let\u0027s carry this on over here."},{"Start":"20:55.060 ","End":"20:59.965","Text":"We have Mu_0 Q Omega divided by"},{"Start":"20:59.965 ","End":"21:07.600","Text":"8Pi r. Now what we want to do is we want to somehow simplify this."},{"Start":"21:07.600 ","End":"21:11.365","Text":"Again, we\u0027re going from 0 until Pi,"},{"Start":"21:11.365 ","End":"21:13.645","Text":"and what we can do,"},{"Start":"21:13.645 ","End":"21:16.520","Text":"let\u0027s just scroll down a little bit more."},{"Start":"21:18.840 ","End":"21:26.800","Text":"Sine^3(Phi) is equal to sine^2(Phi) multiplied by sine(Phi)."},{"Start":"21:26.800 ","End":"21:35.910","Text":"Now, sine^2(Phi) is equal to 1 minus cos^2(Phi)."},{"Start":"21:35.910 ","End":"21:39.160","Text":"1 minus cos^2(Phi),"},{"Start":"21:39.160 ","End":"21:46.240","Text":"notice this is the same identity that we used over here,"},{"Start":"21:46.240 ","End":"21:51.175","Text":"where cos^2(Phi) plus sine^2(Phi) is equal to 1."},{"Start":"21:51.175 ","End":"21:56.455","Text":"1 minus cos^2(Phi) is sine^2(Phi)."},{"Start":"21:56.455 ","End":"22:01.815","Text":"Then we\u0027re just going to multiply this by sine(Phi)."},{"Start":"22:01.815 ","End":"22:05.625","Text":"Now what we can do is we can define x."},{"Start":"22:05.625 ","End":"22:07.665","Text":"Let\u0027s just center it."},{"Start":"22:07.665 ","End":"22:13.750","Text":"X, we can say is equal to cos(Phi)."},{"Start":"22:13.750 ","End":"22:23.540","Text":"Therefore dx is equal to negative sine(Phi)dPhi."},{"Start":"22:23.540 ","End":"22:26.180","Text":"What we\u0027re going to have here,"},{"Start":"22:26.340 ","End":"22:37.930","Text":"we\u0027ll have an integral of 1 minus cos^2(Phi) multiplied by sine(Phi)dPhi,"},{"Start":"22:37.930 ","End":"22:41.860","Text":"which is simply this."},{"Start":"22:41.860 ","End":"22:44.530","Text":"We can rewrite from here."},{"Start":"22:44.530 ","End":"22:53.020","Text":"We\u0027re going to have 1 minus cos^2(Phi), 1 minus x^2."},{"Start":"22:53.020 ","End":"22:56.440","Text":"Then we have sine(Phi)dPhi,"},{"Start":"22:56.440 ","End":"22:59.380","Text":"which is just equal to negative dx."},{"Start":"22:59.380 ","End":"23:03.475","Text":"We can have a negative over here, dx."},{"Start":"23:03.475 ","End":"23:09.410","Text":"Then this is very easy to solve."},{"Start":"23:09.990 ","End":"23:12.400","Text":"Then when I plug this in,"},{"Start":"23:12.400 ","End":"23:15.640","Text":"I just get this over here."},{"Start":"23:15.640 ","End":"23:21.595","Text":"I\u0027m leaving out the bounds because I\u0027ve put in my xs and dx,"},{"Start":"23:21.595 ","End":"23:24.295","Text":"I\u0027ll add in the bounds soon."},{"Start":"23:24.295 ","End":"23:31.210","Text":"This is simply going to be equal to Mu_0 Q Omega divided by 8Pi r,"},{"Start":"23:31.210 ","End":"23:35.665","Text":"and then we\u0027re integrating 1dx."},{"Start":"23:35.665 ","End":"23:41.785","Text":"This is going to be multiplied by just x minus x^2dx,"},{"Start":"23:41.785 ","End":"23:45.595","Text":"which is x^3, divided by 3."},{"Start":"23:45.595 ","End":"23:51.400","Text":"Now we\u0027re going to plug in what our x is equal to,"},{"Start":"23:51.400 ","End":"23:56.815","Text":"so we have Mu_0 Q Omega divided by 8Pi r,"},{"Start":"23:56.815 ","End":"23:59.770","Text":"and then we\u0027ll also plug in our bounds."},{"Start":"23:59.770 ","End":"24:09.550","Text":"We have x, which we already saw was cos(Phi) minus x^3."},{"Start":"24:09.550 ","End":"24:14.900","Text":"Cos^3(Phi) divided by 3 between the bounds of 0 and Pi."},{"Start":"24:16.260 ","End":"24:26.695","Text":"Over here, what we\u0027ll have is cos(Phi) is negative 1 minus Cos^3(Phi),"},{"Start":"24:26.695 ","End":"24:28.705","Text":"which is negative 1^3,"},{"Start":"24:28.705 ","End":"24:31.060","Text":"which is just negative 1 divided by 3,"},{"Start":"24:31.060 ","End":"24:36.190","Text":"so negative, negative 1/3."},{"Start":"24:36.190 ","End":"24:41.590","Text":"Then we have minus cos(0),"},{"Start":"24:41.590 ","End":"24:46.285","Text":"which is 1 minus cos^3(0),"},{"Start":"24:46.285 ","End":"24:55.810","Text":"which is 1^3 divided by 3, so 1/3."},{"Start":"24:55.810 ","End":"25:01.239","Text":"Once you add all of this up and put it in to the whole equation,"},{"Start":"25:01.239 ","End":"25:03.580","Text":"this is the answer."},{"Start":"25:03.580 ","End":"25:08.740","Text":"The magnetic field in the center of a spherical shell is equal to"},{"Start":"25:08.740 ","End":"25:18.380","Text":"negative Mu_0 Q Omega divided by 6Pi r. That\u0027s the end of this lesson."}],"ID":21406}],"Thumbnail":null,"ID":99469}]

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