[{"Name":"Introduction to Lorentz Law","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Intro to Lorentz Law","Duration":"18m 10s","ChapterTopicVideoID":21524,"CourseChapterTopicPlaylistID":99484,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/21524.jpeg","UploadDate":"2020-04-21T15:16:02.6400000","DurationForVideoObject":"PT18M10S","Description":null,"MetaTitle":"Intro to Lorentz Law: Video + Workbook | Proprep","MetaDescription":"Lorentz Law and Force on Current Carrying Conductor - Introduction to Lorentz 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/lorentz-law-and-force-on-current-carrying-conductor/introduction-to-lorentz-law/vid22379","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.815","Text":"Hello. In this lesson,"},{"Start":"00:01.815 ","End":"00:04.665","Text":"we\u0027re going to be speaking about Lorentz\u0027s law,"},{"Start":"00:04.665 ","End":"00:07.455","Text":"or in other words, Lorentz\u0027s force."},{"Start":"00:07.455 ","End":"00:09.630","Text":"This is speaking about the force on"},{"Start":"00:09.630 ","End":"00:14.205","Text":"a charged particle traveling through a magnetic field."},{"Start":"00:14.205 ","End":"00:19.380","Text":"The force is equal to the charge of the particle multiplied by"},{"Start":"00:19.380 ","End":"00:26.950","Text":"the velocity at which it is traveling and that cross-product by the magnetic field."},{"Start":"00:27.050 ","End":"00:31.695","Text":"Let\u0027s say we have this particle over here,"},{"Start":"00:31.695 ","End":"00:33.420","Text":"and let\u0027s call it q,"},{"Start":"00:33.420 ","End":"00:38.360","Text":"and it is traveling in this rightwards direction with a velocity of v_0."},{"Start":"00:38.360 ","End":"00:41.599","Text":"Then into the page,"},{"Start":"00:41.599 ","End":"00:45.740","Text":"these x\u0027s is a symbol for something going into the page,"},{"Start":"00:45.740 ","End":"00:48.170","Text":"we have a magnetic field,"},{"Start":"00:48.170 ","End":"00:50.910","Text":"B naught, or B_0."},{"Start":"00:50.910 ","End":"00:55.745","Text":"The magnitude on the force on this particle due to"},{"Start":"00:55.745 ","End":"01:03.030","Text":"this magnetic field is going to be equal to q multiplied by the velocity v_0."},{"Start":"01:03.040 ","End":"01:08.165","Text":"Then multiplied by the magnetic field B_0,"},{"Start":"01:08.165 ","End":"01:15.120","Text":"assuming of course that our v_0 and B_0 are at right angles to one another."},{"Start":"01:15.290 ","End":"01:17.595","Text":"This is the magnitude,"},{"Start":"01:17.595 ","End":"01:21.380","Text":"and if we wanted to find the direction of the force,"},{"Start":"01:21.380 ","End":"01:23.900","Text":"then we have to use the right-hand rule."},{"Start":"01:23.900 ","End":"01:27.190","Text":"Let\u0027s speak a little bit about the right-hand rule."},{"Start":"01:27.190 ","End":"01:29.645","Text":"When using the right-hand rule,"},{"Start":"01:29.645 ","End":"01:31.355","Text":"we have two options."},{"Start":"01:31.355 ","End":"01:34.700","Text":"The first option is obviously to take"},{"Start":"01:34.700 ","End":"01:39.152","Text":"your right hand and you hold each one of these fingers."},{"Start":"01:39.152 ","End":"01:40.670","Text":"So your thumb, your pointing finger,"},{"Start":"01:40.670 ","End":"01:43.760","Text":"and your middle finger perpendicular to one another,"},{"Start":"01:43.760 ","End":"01:46.860","Text":"pointing in the directions that you\u0027re given."},{"Start":"01:46.860 ","End":"01:48.200","Text":"For instance, in this question,"},{"Start":"01:48.200 ","End":"01:53.535","Text":"we were given the direction of our velocity and our magnetic field."},{"Start":"01:53.535 ","End":"01:59.960","Text":"If you hold your right hand where our pointing finger is pointing inside to the screen,"},{"Start":"01:59.960 ","End":"02:03.819","Text":"your thumb is pointing in the rightwards direction."},{"Start":"02:03.819 ","End":"02:08.095","Text":"So what you will get is that our force,"},{"Start":"02:08.095 ","End":"02:10.065","Text":"let\u0027s draw it in pink,"},{"Start":"02:10.065 ","End":"02:16.820","Text":"F_B is going to be pointing in this upwards direction."},{"Start":"02:17.000 ","End":"02:19.400","Text":"That\u0027s the first. Obviously,"},{"Start":"02:19.400 ","End":"02:25.625","Text":"if you\u0027re being asked about the direction of the magnetic field given F and v,"},{"Start":"02:25.625 ","End":"02:29.855","Text":"or in any other combination,"},{"Start":"02:29.855 ","End":"02:33.800","Text":"you can just point your fingers in the directions that you\u0027re given."},{"Start":"02:33.800 ","End":"02:41.555","Text":"Then you\u0027ll find the direction of the missing variable or alternatively,"},{"Start":"02:41.555 ","End":"02:46.490","Text":"you can use this version of the right-hand rule where this over"},{"Start":"02:46.490 ","End":"02:52.155","Text":"here is the direction of the force."},{"Start":"02:52.155 ","End":"02:53.730","Text":"I\u0027ll draw it like here."},{"Start":"02:53.730 ","End":"03:03.210","Text":"This is the direction of the force and this over here is the direction of the velocity."},{"Start":"03:03.210 ","End":"03:07.430","Text":"In this case, we\u0027d have our fingers"},{"Start":"03:07.430 ","End":"03:11.660","Text":"pointing in this rightwards direction because that\u0027s the direction of the velocity,"},{"Start":"03:11.660 ","End":"03:19.890","Text":"and then the fingers curl in the direction of the B field."},{"Start":"03:19.890 ","End":"03:24.395","Text":"In our case, the velocity would be going like this in the rightwards directions,"},{"Start":"03:24.395 ","End":"03:26.660","Text":"they would curl into the screen,"},{"Start":"03:26.660 ","End":"03:30.230","Text":"which is the direction of our B field in this question."},{"Start":"03:30.230 ","End":"03:34.585","Text":"Then that would give us that the force is pointing upwards like so."},{"Start":"03:34.585 ","End":"03:38.810","Text":"The important thing over here is not to get confused"},{"Start":"03:38.810 ","End":"03:43.325","Text":"between what each finger or each direction represents."},{"Start":"03:43.325 ","End":"03:47.630","Text":"A good way to remember this version of the right-hand rule,"},{"Start":"03:47.630 ","End":"03:50.120","Text":"and this is correct, of course,"},{"Start":"03:50.120 ","End":"03:53.870","Text":"for every equation where you have a cross-product is to"},{"Start":"03:53.870 ","End":"03:58.310","Text":"remember that you go according to the order of the equation."},{"Start":"03:58.310 ","End":"04:02.735","Text":"If you put your right hand and you look at it like so,"},{"Start":"04:02.735 ","End":"04:07.275","Text":"you\u0027ll see that your thumb is the first finger pointing out."},{"Start":"04:07.275 ","End":"04:10.880","Text":"That corresponds to the first element in"},{"Start":"04:10.880 ","End":"04:16.085","Text":"the equation v. Then your pointing finger or your fourth finger,"},{"Start":"04:16.085 ","End":"04:19.985","Text":"is the second finger that is pointing outwards,"},{"Start":"04:19.985 ","End":"04:27.030","Text":"and that corresponds to the second element in your cross-product equation B."},{"Start":"04:27.560 ","End":"04:35.793","Text":"Then the third finger pointing out corresponds to the answer of the cross-product."},{"Start":"04:35.793 ","End":"04:38.660","Text":"In this case, it\u0027s just the direction."},{"Start":"04:38.660 ","End":"04:42.305","Text":"Your middle finger is the answer,"},{"Start":"04:42.305 ","End":"04:48.550","Text":"and the other fingers correspond to the order in which the equation is written."},{"Start":"04:48.550 ","End":"04:51.540","Text":"A note, later on in this course,"},{"Start":"04:51.540 ","End":"04:57.500","Text":"we\u0027ll get cross-products equations where the answer is the magnetic field."},{"Start":"04:57.500 ","End":"05:00.440","Text":"Then, remember that the B will go for"},{"Start":"05:00.440 ","End":"05:04.730","Text":"the middle finger because the middle finger is the answer."},{"Start":"05:04.730 ","End":"05:09.560","Text":"It\u0027s better in that sense to remember the right-hand rule"},{"Start":"05:09.560 ","End":"05:14.660","Text":"as the thumb is the first finger and represents the first element,"},{"Start":"05:14.660 ","End":"05:18.575","Text":"the pointing finger is the second finger and represents the second element,"},{"Start":"05:18.575 ","End":"05:21.090","Text":"and the third finger is the answer."},{"Start":"05:21.090 ","End":"05:27.260","Text":"Then you can match the fingers to any cross-products equation that you"},{"Start":"05:27.260 ","End":"05:35.430","Text":"get where you have 3 of these variables or any variables, but 3 of them."},{"Start":"05:36.500 ","End":"05:40.000","Text":"Here we have a similar situation."},{"Start":"05:40.000 ","End":"05:45.490","Text":"Your fingers, all 4 fingers point in the direction of the first element,"},{"Start":"05:45.490 ","End":"05:50.455","Text":"and then they curl in the direction of the second element."},{"Start":"05:50.455 ","End":"05:52.540","Text":"If you do it right now,"},{"Start":"05:52.540 ","End":"05:54.903","Text":"you put out your right hand."},{"Start":"05:54.903 ","End":"05:58.750","Text":"So your fingers are going to point in the rightwards direction and then you\u0027ll"},{"Start":"05:58.750 ","End":"06:03.220","Text":"bend them inwards because the magnetic field is pointing into the screen,"},{"Start":"06:03.220 ","End":"06:05.830","Text":"and then you\u0027ll see that your thumb is pointing upwards,"},{"Start":"06:05.830 ","End":"06:08.500","Text":"which is the direction of the force."},{"Start":"06:08.500 ","End":"06:13.105","Text":"Alternatively, if our B field was coming out of the page,"},{"Start":"06:13.105 ","End":"06:17.380","Text":"so I\u0027m reminding you the symbol for coming out of the page is this."},{"Start":"06:17.380 ","End":"06:20.075","Text":"If it was coming out of the page,"},{"Start":"06:20.075 ","End":"06:21.950","Text":"then what you would do is you would point"},{"Start":"06:21.950 ","End":"06:26.240","Text":"your 4 fingers in this rightwards direction for the velocity."},{"Start":"06:26.240 ","End":"06:32.810","Text":"But then you\u0027d have to flip your hand around so that the thumb would be pointing"},{"Start":"06:32.810 ","End":"06:40.110","Text":"downwards in order to bend your fingers outside of the screen."},{"Start":"06:40.110 ","End":"06:48.120","Text":"What you\u0027ll have is that you\u0027ll have your pinky on top and then your other 4 fingers."},{"Start":"06:49.430 ","End":"06:53.900","Text":"Notice the fingernails, you can\u0027t see them right now,"},{"Start":"06:53.900 ","End":"07:01.286","Text":"and then your thumb will be pointing downwards."},{"Start":"07:01.286 ","End":"07:04.480","Text":"This is what your hand would look like if"},{"Start":"07:04.480 ","End":"07:09.430","Text":"the B field was pointing out of the page but of course, this isn\u0027t the case."},{"Start":"07:09.430 ","End":"07:11.634","Text":"Now the B field is pointing inwards,"},{"Start":"07:11.634 ","End":"07:15.175","Text":"so this is what your hand should look like."},{"Start":"07:15.175 ","End":"07:19.554","Text":"Of course, this is true for every cross-products equation."},{"Start":"07:19.554 ","End":"07:28.045","Text":"Just remember the direction that your fingers are pointing represents the first element."},{"Start":"07:28.045 ","End":"07:31.120","Text":"The direction that your fingers curl represent"},{"Start":"07:31.120 ","End":"07:37.105","Text":"the second element and then a thumbs-up because you got your answer."},{"Start":"07:37.105 ","End":"07:41.060","Text":"The thumb represents the answer."},{"Start":"07:41.280 ","End":"07:46.795","Text":"Now let\u0027s go back to our force and let\u0027s take a look at what\u0027s happening."},{"Start":"07:46.795 ","End":"07:49.210","Text":"What we can see is that we have"},{"Start":"07:49.210 ","End":"07:56.500","Text":"this 90 degree angle between the force and the direction of the velocity."},{"Start":"07:56.500 ","End":"08:02.920","Text":"When the force and velocity are at 90 degrees or perpendicular to one another,"},{"Start":"08:02.920 ","End":"08:08.830","Text":"then the force is just going to cause a change in the direction of the velocity."},{"Start":"08:08.830 ","End":"08:12.280","Text":"It\u0027s not going to change the size of the velocity,"},{"Start":"08:12.280 ","End":"08:15.920","Text":"it\u0027s just going to change the direction."},{"Start":"08:16.170 ","End":"08:19.300","Text":"If the force is perpendicular,"},{"Start":"08:19.300 ","End":"08:25.030","Text":"this means perpendicular because we can see that we have 2 right angles to the velocity."},{"Start":"08:25.030 ","End":"08:30.850","Text":"That means that the force changes the direction of the velocity,"},{"Start":"08:30.850 ","End":"08:33.265","Text":"but not the magnitude of the velocity."},{"Start":"08:33.265 ","End":"08:35.695","Text":"Just the direction of the velocity changes."},{"Start":"08:35.695 ","End":"08:37.840","Text":"If we look 1 second later,"},{"Start":"08:37.840 ","End":"08:40.660","Text":"our particle q will be here,"},{"Start":"08:40.660 ","End":"08:44.620","Text":"the direction of velocity will be like this and of course,"},{"Start":"08:44.620 ","End":"08:47.906","Text":"the force always has to be perpendicular."},{"Start":"08:47.906 ","End":"08:52.690","Text":"So the direction of the force is also going to change."},{"Start":"08:52.690 ","End":"08:57.700","Text":"They are still perpendicular to one another but in"},{"Start":"08:57.700 ","End":"08:59.650","Text":"order to stay perpendicular to one another"},{"Start":"08:59.650 ","End":"09:02.350","Text":"when the direction of the velocity has changed,"},{"Start":"09:02.350 ","End":"09:06.620","Text":"the direction of the force also has to change a bit."},{"Start":"09:06.900 ","End":"09:09.745","Text":"If we continue with this,"},{"Start":"09:09.745 ","End":"09:11.845","Text":"a few moments later,"},{"Start":"09:11.845 ","End":"09:14.620","Text":"our particle will be over here."},{"Start":"09:14.620 ","End":"09:19.330","Text":"The velocity will be in this direction, and therefore,"},{"Start":"09:19.330 ","End":"09:24.050","Text":"we know that the force has to be in this direction."},{"Start":"09:24.050 ","End":"09:26.950","Text":"Of course, these are still perpendicular."},{"Start":"09:26.950 ","End":"09:32.990","Text":"What we can see is that when the force is perpendicular to the velocity,"},{"Start":"09:32.990 ","End":"09:42.060","Text":"that will result in our particle doing or exhibiting circular motion."},{"Start":"09:42.060 ","End":"09:45.480","Text":"If it\u0027s exhibiting circular motion,"},{"Start":"09:45.480 ","End":"09:55.360","Text":"then the force on a particle moving in a circular motion is equal to mV^2 divided by R,"},{"Start":"09:55.360 ","End":"10:00.560","Text":"the radius of the circle that it is mapping out."},{"Start":"10:00.810 ","End":"10:04.555","Text":"If F is perpendicular to V,"},{"Start":"10:04.555 ","End":"10:06.580","Text":"we get circular motion."},{"Start":"10:06.580 ","End":"10:12.025","Text":"That means that this is the equivalent of circular motion given this equation."},{"Start":"10:12.025 ","End":"10:15.400","Text":"Which means that we can also say it\u0027s equal to this, and therefore,"},{"Start":"10:15.400 ","End":"10:19.630","Text":"we can isolate out the radius of the circle that is being mapped out."},{"Start":"10:19.630 ","End":"10:23.980","Text":"That is equal to mV_0, in this case,"},{"Start":"10:23.980 ","End":"10:30.220","Text":"divided by qB_0 or in general,"},{"Start":"10:30.220 ","End":"10:33.520","Text":"the radius of the circle being mapped"},{"Start":"10:33.520 ","End":"10:36.820","Text":"out is the mass of the particle multiplied by its velocity,"},{"Start":"10:36.820 ","End":"10:38.260","Text":"divided by its charge,"},{"Start":"10:38.260 ","End":"10:41.360","Text":"multiplied by the magnetic field."},{"Start":"10:41.430 ","End":"10:47.155","Text":"Of course, you can write this as an equation in your equation worksheets for the radius"},{"Start":"10:47.155 ","End":"10:52.525","Text":"of this mapped out circle given this type of system."},{"Start":"10:52.525 ","End":"10:57.670","Text":"However, you definitely have to write out this equation and remember"},{"Start":"10:57.670 ","End":"11:03.250","Text":"how you can just derive this equation given this type of system."},{"Start":"11:03.250 ","End":"11:09.410","Text":"Just remember your equation for the force on a particle traveling in circular motion."},{"Start":"11:09.630 ","End":"11:15.324","Text":"Another thing that is very important to note is that because"},{"Start":"11:15.324 ","End":"11:22.285","Text":"our B field is perpendicular to our velocity."},{"Start":"11:22.285 ","End":"11:26.019","Text":"Therefore, it\u0027s always perpendicular"},{"Start":"11:26.019 ","End":"11:30.174","Text":"to our velocity as we can see throughout the motion of the particle."},{"Start":"11:30.174 ","End":"11:39.910","Text":"That means that the work done by the force of the magnetic field is always equal to 0,"},{"Start":"11:39.910 ","End":"11:42.920","Text":"and this is always."},{"Start":"11:43.350 ","End":"11:48.895","Text":"The work done by the magnetic force is equal to 0"},{"Start":"11:48.895 ","End":"11:54.175","Text":"always because it\u0027s always going to be perpendicular to the velocity,"},{"Start":"11:54.175 ","End":"11:59.785","Text":"and therefore, it does no work on the particle."},{"Start":"11:59.785 ","End":"12:02.800","Text":"Definitely remember this or write this"},{"Start":"12:02.800 ","End":"12:05.800","Text":"out in your equation sheets if you think you\u0027ll forget."},{"Start":"12:05.800 ","End":"12:12.895","Text":"Now let\u0027s look at another example where our velocity isn\u0027t perpendicular to our B field."},{"Start":"12:12.895 ","End":"12:18.265","Text":"Let\u0027s say that this is the direction of our B field."},{"Start":"12:18.265 ","End":"12:23.695","Text":"It\u0027s traveling in this rightwards direction and this is our B field."},{"Start":"12:23.695 ","End":"12:28.585","Text":"Then our particle is over here, it has a charge of q,"},{"Start":"12:28.585 ","End":"12:33.109","Text":"and its direction of the velocity is in this direction,"},{"Start":"12:33.109 ","End":"12:34.660","Text":"this is V naught."},{"Start":"12:34.660 ","End":"12:43.370","Text":"We know that the angle of V naught relative to our B field is Alpha."},{"Start":"12:44.880 ","End":"12:53.005","Text":"What I\u0027m going to do is I\u0027m going to split up the components of this velocity."},{"Start":"12:53.005 ","End":"13:00.010","Text":"I\u0027ll have my velocity in the perpendicular direction to the magnetic field."},{"Start":"13:00.010 ","End":"13:02.710","Text":"Then in this direction,"},{"Start":"13:02.710 ","End":"13:10.480","Text":"I have my component of the velocity which is parallel to my B field,"},{"Start":"13:10.480 ","End":"13:14.630","Text":"so it\u0027s in the same direction as my B field."},{"Start":"13:15.120 ","End":"13:25.030","Text":"We can first of all see that the force due to the parallel component of our velocity."},{"Start":"13:25.030 ","End":"13:30.370","Text":"The component which is parallel to a B field isn\u0027t going to change the force."},{"Start":"13:30.370 ","End":"13:33.250","Text":"Because when we do the cross-product between"},{"Start":"13:33.250 ","End":"13:37.840","Text":"the parallel component and the B field because they\u0027re in the same direction,"},{"Start":"13:37.840 ","End":"13:42.535","Text":"the cross product is going to be equal to 0."},{"Start":"13:42.535 ","End":"13:46.870","Text":"Then if we look at the perpendicular component,"},{"Start":"13:46.870 ","End":"13:52.270","Text":"the component of velocity which is perpendicular to the magnetic field,"},{"Start":"13:52.270 ","End":"13:56.170","Text":"we\u0027re going to get the same force as what we had before,"},{"Start":"13:56.170 ","End":"14:03.200","Text":"which is equal to qv perpendicular multiplied by B."},{"Start":"14:04.050 ","End":"14:10.100","Text":"This is the magnetic force which is acting on our particle and we"},{"Start":"14:10.100 ","End":"14:15.395","Text":"can see that the parallel component doesn\u0027t affect the force whatsoever,"},{"Start":"14:15.395 ","End":"14:20.630","Text":"only the perpendicular component or in other words,"},{"Start":"14:20.630 ","End":"14:29.465","Text":"we can write that this force is equal to q multiplied by V_0, the velocity."},{"Start":"14:29.465 ","End":"14:31.310","Text":"Then if we have cross B,"},{"Start":"14:31.310 ","End":"14:37.540","Text":"we know that this is equal to just B naught multiplied by sine of the angle."},{"Start":"14:37.540 ","End":"14:42.695","Text":"This is the same equation as what we have over here and specifically,"},{"Start":"14:42.695 ","End":"14:46.500","Text":"we\u0027ll get just this equation in the end."},{"Start":"14:47.070 ","End":"14:56.360","Text":"We know that because V_0 multiplied by sine of the angle is just V perpendicular."},{"Start":"14:56.360 ","End":"15:00.960","Text":"V perpendicular is V sine of Alpha."},{"Start":"15:01.470 ","End":"15:04.145","Text":"Now using the right-hand rule,"},{"Start":"15:04.145 ","End":"15:07.055","Text":"we point our thumb in this direction,"},{"Start":"15:07.055 ","End":"15:09.170","Text":"or rather in this direction,"},{"Start":"15:09.170 ","End":"15:12.830","Text":"because this is the component that is affecting the force."},{"Start":"15:12.830 ","End":"15:18.320","Text":"We point our pointing finger in this direction,"},{"Start":"15:18.320 ","End":"15:21.740","Text":"the rightwards direction and what we get is that"},{"Start":"15:21.740 ","End":"15:27.440","Text":"the direction of the force is coming out of the page."},{"Start":"15:27.440 ","End":"15:30.380","Text":"This is the direction of F_B coming out of"},{"Start":"15:30.380 ","End":"15:34.110","Text":"the page or out of the screen and into your eye."},{"Start":"15:34.260 ","End":"15:39.155","Text":"What will be the motion of this particle?"},{"Start":"15:39.155 ","End":"15:42.740","Text":"First of all, we can see that from our V"},{"Start":"15:42.740 ","End":"15:47.165","Text":"perpendicular and our B field and the direction of the force."},{"Start":"15:47.165 ","End":"15:48.920","Text":"We can see that again,"},{"Start":"15:48.920 ","End":"15:51.350","Text":"we\u0027re going to have this circular motion,"},{"Start":"15:51.350 ","End":"15:53.690","Text":"but in a different plane."},{"Start":"15:53.690 ","End":"16:00.220","Text":"It\u0027s a circular motion in this plane like so,"},{"Start":"16:00.220 ","End":"16:03.100","Text":"which is the V_F plane."},{"Start":"16:03.100 ","End":"16:07.985","Text":"So it\u0027s just going to be traveling like so but then,"},{"Start":"16:07.985 ","End":"16:12.185","Text":"because we have this parallel component of the velocity"},{"Start":"16:12.185 ","End":"16:17.615","Text":"that is of course not affecting the force but it\u0027s still exists."},{"Start":"16:17.615 ","End":"16:23.855","Text":"What we\u0027re going to have is we\u0027re going to have some spiral going along."},{"Start":"16:23.855 ","End":"16:29.795","Text":"We\u0027re still going to be doing the circular motion along the V_F plane."},{"Start":"16:29.795 ","End":"16:35.630","Text":"However, it\u0027s also going to be coming forwards and"},{"Start":"16:35.630 ","End":"16:38.120","Text":"traveling in the direction of the B field in"},{"Start":"16:38.120 ","End":"16:42.520","Text":"the rightwards direction due to this parallel component."},{"Start":"16:42.520 ","End":"16:47.705","Text":"It is as if we have some screw over here and around the screw,"},{"Start":"16:47.705 ","End":"16:50.900","Text":"you have that helical ridge."},{"Start":"16:50.900 ","End":"16:56.640","Text":"It\u0027s as if the particle is just traveling along that helical ridge."},{"Start":"16:56.640 ","End":"17:03.635","Text":"We\u0027re just moving in the direction of the B field and just going around like so."},{"Start":"17:03.635 ","End":"17:08.420","Text":"That will be the movement of the particle in this case."},{"Start":"17:08.420 ","End":"17:13.790","Text":"Now just one last note for this equation for Lorentz law, or Lorentz force."},{"Start":"17:13.790 ","End":"17:19.085","Text":"Here, we\u0027ve been speaking about the case where we just have a magnetic field."},{"Start":"17:19.085 ","End":"17:25.810","Text":"However, in some questions you\u0027ll also have an E field in this whole system."},{"Start":"17:25.810 ","End":"17:31.420","Text":"The full equation for Lorentz\u0027s force is to do this cross-product,"},{"Start":"17:31.420 ","End":"17:39.660","Text":"and then at the end, add the charge of the particle multiplied by the E field."},{"Start":"17:40.200 ","End":"17:43.055","Text":"Just remember that and of course,"},{"Start":"17:43.055 ","End":"17:49.340","Text":"none of these vector components are in the same direction or they might be,"},{"Start":"17:49.340 ","End":"17:51.473","Text":"but they aren\u0027t necessarily,"},{"Start":"17:51.473 ","End":"17:53.950","Text":"so that\u0027s just something to remember."},{"Start":"17:53.950 ","End":"17:55.505","Text":"You do this cross-product,"},{"Start":"17:55.505 ","End":"17:58.340","Text":"just what we\u0027ve been doing this entire lesson and then you"},{"Start":"17:58.340 ","End":"18:01.535","Text":"just add on Q multiplied by the E field."},{"Start":"18:01.535 ","End":"18:03.650","Text":"If there is an E field and the question,"},{"Start":"18:03.650 ","End":"18:07.550","Text":"and then you get the total force on that particle."},{"Start":"18:07.550 ","End":"18:10.440","Text":"That\u0027s the end of this lesson."}],"ID":22379},{"Watched":false,"Name":"Exercise 1","Duration":"5m 43s","ChapterTopicVideoID":21525,"CourseChapterTopicPlaylistID":99484,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.755","Text":"Hello, in this lesson,"},{"Start":"00:01.755 ","End":"00:04.155","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.155 ","End":"00:09.660","Text":"A charged particle with a charge q moves at a velocity v in a region where"},{"Start":"00:09.660 ","End":"00:13.560","Text":"the magnetic field of B is equal to negative"},{"Start":"00:13.560 ","End":"00:17.985","Text":"2 in the x-direction plus 3 in the y-direction Tesla."},{"Start":"00:17.985 ","End":"00:23.116","Text":"Calculate the magnetic force applied to the bicycle given that;"},{"Start":"00:23.116 ","End":"00:24.910","Text":"so in question number 1,"},{"Start":"00:24.910 ","End":"00:30.465","Text":"we\u0027re being told that the velocity is 2 in the x direction plus 3 in the y direction,"},{"Start":"00:30.465 ","End":"00:33.980","Text":"and that the charge of the particle is 2 coulombs."},{"Start":"00:33.980 ","End":"00:41.880","Text":"First of all we remember that the equation for force is equal to qv cross B,"},{"Start":"00:41.880 ","End":"00:46.125","Text":"so q is a constant and it isn\u0027t a vector,"},{"Start":"00:46.125 ","End":"00:52.950","Text":"and v cross B is what we have to calculate first and then we\u0027ll multiply it by q."},{"Start":"00:53.060 ","End":"00:56.325","Text":"We have v cross B."},{"Start":"00:56.325 ","End":"00:57.870","Text":"Let\u0027s write this out,"},{"Start":"00:57.870 ","End":"01:02.310","Text":"v cross B is equal to."},{"Start":"01:02.310 ","End":"01:05.990","Text":"We\u0027ll do this as a vector in Cartesian"},{"Start":"01:05.990 ","End":"01:10.865","Text":"coordinates with coordinate z as well because we can see in the next question we have z,"},{"Start":"01:10.865 ","End":"01:15.240","Text":"and also we\u0027ll make this cross-product much easier to do."},{"Start":"01:15.240 ","End":"01:17.115","Text":"We have our v,"},{"Start":"01:17.115 ","End":"01:19.230","Text":"which is 2 in the x direction,"},{"Start":"01:19.230 ","End":"01:20.510","Text":"3 in the y direction,"},{"Start":"01:20.510 ","End":"01:22.460","Text":"and 0 in the z direction."},{"Start":"01:22.460 ","End":"01:25.805","Text":"If we\u0027re not given a z component, then it\u0027s just 0."},{"Start":"01:25.805 ","End":"01:29.150","Text":"Cross-product with our B,"},{"Start":"01:29.150 ","End":"01:32.720","Text":"a magnetic field which is negative 2 in the x-direction,"},{"Start":"01:32.720 ","End":"01:34.190","Text":"3 in the y-direction."},{"Start":"01:34.190 ","End":"01:37.045","Text":"We\u0027re not given a z components, so 0."},{"Start":"01:37.045 ","End":"01:41.025","Text":"Then what we do is we do this cross-product."},{"Start":"01:41.025 ","End":"01:46.490","Text":"We put our thumb over the top rows, the top numbers,"},{"Start":"01:46.490 ","End":"01:50.795","Text":"and then we have 3 times 0 minus 0 times 3,"},{"Start":"01:50.795 ","End":"01:54.260","Text":"which is 0 in the x-direction."},{"Start":"01:54.260 ","End":"01:57.755","Text":"Then we crossover the middle row and we switch it up."},{"Start":"01:57.755 ","End":"02:05.150","Text":"What we have is 0 multiplied by negative 2 minus 2 times 0,"},{"Start":"02:05.150 ","End":"02:08.600","Text":"so we have 0 in the y-direction,"},{"Start":"02:08.600 ","End":"02:12.210","Text":"and then we cross it out the bottom numbers and then we do 2"},{"Start":"02:12.210 ","End":"02:16.200","Text":"times 3 minus 3 times minus 2."},{"Start":"02:16.200 ","End":"02:19.920","Text":"What we have is 2 times 3,"},{"Start":"02:19.920 ","End":"02:24.960","Text":"and then we subtract minus,"},{"Start":"02:24.960 ","End":"02:30.690","Text":"and then we have 3 times negative 2 in the z direction."},{"Start":"02:30.690 ","End":"02:37.245","Text":"What we get is that our v cross B is equal to 12 in the z direction."},{"Start":"02:37.245 ","End":"02:42.695","Text":"Therefore, our force is equal to q multiplied by this,"},{"Start":"02:42.695 ","End":"02:44.975","Text":"where q is 2 coulombs."},{"Start":"02:44.975 ","End":"02:51.180","Text":"We have 2 coulombs multiplied by 12,"},{"Start":"02:51.180 ","End":"02:54.746","Text":"and then this is velocity,"},{"Start":"02:54.746 ","End":"02:58.910","Text":"so we have meters per second cross-product with the magnetic fields,"},{"Start":"02:58.910 ","End":"03:00.920","Text":"so multiplied by Tesla,"},{"Start":"03:00.920 ","End":"03:03.650","Text":"and this is of course in the z direction."},{"Start":"03:03.650 ","End":"03:06.305","Text":"We\u0027re dealing with base coordinates."},{"Start":"03:06.305 ","End":"03:10.115","Text":"What we\u0027ll get is that the coordinates for forces of course in newtons."},{"Start":"03:10.115 ","End":"03:11.945","Text":"We have 2 times 12,"},{"Start":"03:11.945 ","End":"03:13.460","Text":"which is 24,"},{"Start":"03:13.460 ","End":"03:20.110","Text":"and then this is newtons in the z direction."},{"Start":"03:20.390 ","End":"03:25.695","Text":"Now we can answer question number 2."},{"Start":"03:25.695 ","End":"03:30.540","Text":"Here again, we\u0027ll do our v cross B."},{"Start":"03:30.540 ","End":"03:34.950","Text":"Here our v is negative 1,"},{"Start":"03:34.950 ","End":"03:40.410","Text":"0 in the y direction and 2 in the z direction."},{"Start":"03:40.410 ","End":"03:42.560","Text":"Then our B field is the same,"},{"Start":"03:42.560 ","End":"03:44.090","Text":"we have negative 2,"},{"Start":"03:44.090 ","End":"03:45.815","Text":"3, and 0."},{"Start":"03:45.815 ","End":"03:48.230","Text":"First we cross off the top numbers."},{"Start":"03:48.230 ","End":"03:53.235","Text":"What we have is 0 multiplied by 0 minus 2 times 3,"},{"Start":"03:53.235 ","End":"03:58.025","Text":"so that is negative 6 in the x direction."},{"Start":"03:58.025 ","End":"04:02.180","Text":"Then we cross out the middle row and then we have 2 times"},{"Start":"04:02.180 ","End":"04:06.860","Text":"negative 2 minus negative 1 times 0."},{"Start":"04:06.860 ","End":"04:13.665","Text":"We have negative 4 in the y direction."},{"Start":"04:13.665 ","End":"04:17.150","Text":"Then we cross out the bottom numbers and we have negative 1"},{"Start":"04:17.150 ","End":"04:20.630","Text":"times 3 minus 0 times minus 2."},{"Start":"04:20.630 ","End":"04:26.610","Text":"Then we have a negative 3 in the z direction."},{"Start":"04:26.610 ","End":"04:31.054","Text":"Then our force is equal to our charge,"},{"Start":"04:31.054 ","End":"04:39.766","Text":"which is negative 1 micro coulombs multiplied by this over here,"},{"Start":"04:39.766 ","End":"04:43.979","Text":"so we have negative 6 in the x direction,"},{"Start":"04:43.979 ","End":"04:46.755","Text":"minus 4 in the y-direction,"},{"Start":"04:46.755 ","End":"04:50.190","Text":"minus 3 in the z directions,"},{"Start":"04:50.190 ","End":"04:51.455","Text":"and this is of course,"},{"Start":"04:51.455 ","End":"04:55.820","Text":"meters per second multiplied by Tesla."},{"Start":"04:57.320 ","End":"05:00.540","Text":"We have a minus 1 multiplying all of this,"},{"Start":"05:00.540 ","End":"05:04.730","Text":"so we can just turn this all into pluses over here,"},{"Start":"05:04.730 ","End":"05:08.960","Text":"and then we have the coulomb meters per second Tesla,"},{"Start":"05:08.960 ","End":"05:11.011","Text":"which we already saw is equal to newtons,"},{"Start":"05:11.011 ","End":"05:15.215","Text":"but then we have micro over here which is 10^minus 6."},{"Start":"05:15.215 ","End":"05:18.110","Text":"We could either write that we have all of this times"},{"Start":"05:18.110 ","End":"05:21.520","Text":"10^minus 6 or we could just leave this in micro coulombs,"},{"Start":"05:21.520 ","End":"05:27.500","Text":"so we could get that our force is equal to 6 in the x direction"},{"Start":"05:27.500 ","End":"05:33.915","Text":"plus 4 in the y direction plus 3 in the z direction,"},{"Start":"05:33.915 ","End":"05:39.095","Text":"and all of this in units of micro newtons."},{"Start":"05:39.095 ","End":"05:43.620","Text":"That\u0027s it. That is the end of this question."}],"ID":22380},{"Watched":false,"Name":"Exercise 2","Duration":"8m 45s","ChapterTopicVideoID":21315,"CourseChapterTopicPlaylistID":99484,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.920","Text":"Hello. In this lesson,"},{"Start":"00:01.920 ","End":"00:04.620","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.620 ","End":"00:09.195","Text":"The following system describes the Dempster mass spectrometer,"},{"Start":"00:09.195 ","End":"00:13.185","Text":"whose purpose is to separate particles of different masses."},{"Start":"00:13.185 ","End":"00:17.985","Text":"Positive charge particles are released from a state of rest near"},{"Start":"00:17.985 ","End":"00:21.330","Text":"the positive capacitor plates and accelerated by"},{"Start":"00:21.330 ","End":"00:26.580","Text":"a voltage source V applied across the capacitor."},{"Start":"00:26.580 ","End":"00:32.295","Text":"The particles pass through the negative plate and"},{"Start":"00:32.295 ","End":"00:37.425","Text":"enter a uniform magnetic field going into the page,"},{"Start":"00:37.425 ","End":"00:42.440","Text":"Find the radius of rotation as a function of particle mass."},{"Start":"00:42.440 ","End":"00:44.960","Text":"We are given the values for the B field,"},{"Start":"00:44.960 ","End":"00:46.655","Text":"the charge of the particle,"},{"Start":"00:46.655 ","End":"00:49.385","Text":"and the voltage of the voltage source."},{"Start":"00:49.385 ","End":"00:54.890","Text":"Here is our voltage source across the capacitor plates."},{"Start":"00:54.890 ","End":"00:57.140","Text":"We can see that this is the long side,"},{"Start":"00:57.140 ","End":"00:59.540","Text":"so this is the positive side,"},{"Start":"00:59.540 ","End":"01:02.165","Text":"and this is the negative side of the voltage source,"},{"Start":"01:02.165 ","End":"01:05.850","Text":"which means that this is going to be the positive capacitor plate,"},{"Start":"01:05.850 ","End":"01:08.830","Text":"and this is the negative capacitor plate."},{"Start":"01:08.830 ","End":"01:12.860","Text":"What we see, is we have a particle which"},{"Start":"01:12.860 ","End":"01:19.030","Text":"travels and is accelerated between the positive plates and the negative plate,"},{"Start":"01:19.030 ","End":"01:21.725","Text":"and then it passes through the negative plate,"},{"Start":"01:21.725 ","End":"01:26.930","Text":"and then it travels over here in circular motion."},{"Start":"01:26.930 ","End":"01:30.440","Text":"Just like we learned in previous lessons"},{"Start":"01:30.440 ","End":"01:35.600","Text":"when the velocity is perpendicular to the magnetic field,"},{"Start":"01:35.600 ","End":"01:41.090","Text":"which we can see the velocity is on this plane over here,"},{"Start":"01:41.090 ","End":"01:42.740","Text":"the plane of the screen,"},{"Start":"01:42.740 ","End":"01:45.420","Text":"and the magnetic field is going into the screen,"},{"Start":"01:45.420 ","End":"01:49.010","Text":"so the magnetic field is always perpendicular to the velocity."},{"Start":"01:49.010 ","End":"01:51.890","Text":"We saw in the previous lessons that,"},{"Start":"01:51.890 ","End":"01:55.565","Text":"that means that the particle will travel in circular motion."},{"Start":"01:55.565 ","End":"01:57.200","Text":"What we\u0027re trying to do,"},{"Start":"01:57.200 ","End":"02:01.205","Text":"is we\u0027re trying to find the radius of this circle"},{"Start":"02:01.205 ","End":"02:06.970","Text":"mapped out by the particle dependent on the particle\u0027s mass."},{"Start":"02:07.760 ","End":"02:10.070","Text":"What I want to do is,"},{"Start":"02:10.070 ","End":"02:19.055","Text":"I want to find the velocity at which my particle is exiting the negatively charged plate."},{"Start":"02:19.055 ","End":"02:24.575","Text":"First of all, don\u0027t get confused between the voltage and the velocity."},{"Start":"02:24.575 ","End":"02:27.485","Text":"Every time I write out voltage,"},{"Start":"02:27.485 ","End":"02:31.865","Text":"I\u0027ll write it like this so that it\u0027s clear that this is a V for voltage,"},{"Start":"02:31.865 ","End":"02:35.195","Text":"and my velocity will be like so,"},{"Start":"02:35.195 ","End":"02:39.800","Text":"v with the vector sign on top."},{"Start":"02:39.860 ","End":"02:46.550","Text":"If I run and calculate my velocity of the particle exiting the negative plate,"},{"Start":"02:46.550 ","End":"02:47.860","Text":"what I\u0027m going to use,"},{"Start":"02:47.860 ","End":"02:51.604","Text":"is I\u0027m going to use the idea of conservation of energy."},{"Start":"02:51.604 ","End":"02:55.190","Text":"I know that I have an electric force"},{"Start":"02:55.190 ","End":"02:59.480","Text":"accelerating the particle from the positive to the negative plate,"},{"Start":"02:59.480 ","End":"03:03.020","Text":"and the electric force is a conservative force,"},{"Start":"03:03.020 ","End":"03:07.560","Text":"which means that I can use the idea of energy conservation."},{"Start":"03:08.540 ","End":"03:12.305","Text":"The energy in general of this particle,"},{"Start":"03:12.305 ","End":"03:18.605","Text":"is going to be equal to the kinetic energy plus the potential energy."},{"Start":"03:18.605 ","End":"03:20.480","Text":"The kinetic energy is,"},{"Start":"03:20.480 ","End":"03:24.685","Text":"as we know, 1/2 mv^2,"},{"Start":"03:24.685 ","End":"03:32.470","Text":"and the potential energy is the charge multiplied by the potential at that point."},{"Start":"03:32.470 ","End":"03:35.780","Text":"Let\u0027s take a look at what the potential is."},{"Start":"03:35.780 ","End":"03:39.845","Text":"We know that the voltage of the voltage source is V. Now,"},{"Start":"03:39.845 ","End":"03:43.640","Text":"voltage is equal to the potential difference."},{"Start":"03:43.640 ","End":"03:49.070","Text":"The potential difference between these 2 sides of the voltage source, or in other words,"},{"Start":"03:49.070 ","End":"03:51.230","Text":"the potential difference between"},{"Start":"03:51.230 ","End":"03:56.835","Text":"this side of the capacitor and this side of the capacitor."},{"Start":"03:56.835 ","End":"04:01.490","Text":"We know that the difference between this side and this side is v,"},{"Start":"04:01.490 ","End":"04:06.500","Text":"so we can say that the potential on this side is equal to 0 and that"},{"Start":"04:06.500 ","End":"04:11.870","Text":"the potential on this side is equal to this V. Then,"},{"Start":"04:11.870 ","End":"04:15.230","Text":"when we take the potential difference between this side and this side,"},{"Start":"04:15.230 ","End":"04:16.985","Text":"we have V minus 0,"},{"Start":"04:16.985 ","End":"04:19.950","Text":"which is V, the voltage."},{"Start":"04:22.670 ","End":"04:26.735","Text":"Our initial kinetic energy is equal to 0,"},{"Start":"04:26.735 ","End":"04:35.895","Text":"because we\u0027re told that the charged particle is at rest at the beginning."},{"Start":"04:35.895 ","End":"04:37.755","Text":"That\u0027s what it says over here,"},{"Start":"04:37.755 ","End":"04:43.955","Text":"the charged particles are released from a state of rest at the positive capacitor,"},{"Start":"04:43.955 ","End":"04:48.095","Text":"so 1/2 m multiplied by 0 is 0,"},{"Start":"04:48.095 ","End":"04:51.260","Text":"and the kinetic energy is the charge of"},{"Start":"04:51.260 ","End":"04:55.535","Text":"the particle multiplied by the potential over here,"},{"Start":"04:55.535 ","End":"04:59.920","Text":"where we already said that the potential at this point is"},{"Start":"04:59.920 ","End":"05:05.820","Text":"V. The final energy is as we know,"},{"Start":"05:05.820 ","End":"05:07.500","Text":"so will have kinetic energy,"},{"Start":"05:07.500 ","End":"05:09.620","Text":"that\u0027s what we\u0027re trying to find the velocity over here,"},{"Start":"05:09.620 ","End":"05:14.360","Text":"so 1/2mv^2 and the potential energy"},{"Start":"05:14.360 ","End":"05:18.575","Text":"we have q multiplied by the potential at the negative plate,"},{"Start":"05:18.575 ","End":"05:22.310","Text":"which we just saw is equal to 0 so that\u0027s like so."},{"Start":"05:22.310 ","End":"05:27.300","Text":"Now, we can equate the two because we\u0027re using conservation of energy."},{"Start":"05:27.300 ","End":"05:31.254","Text":"So the initial energy is equal to the final energy."},{"Start":"05:31.254 ","End":"05:38.605","Text":"So we have qV is equal to 1/2 mv^2,"},{"Start":"05:38.605 ","End":"05:42.010","Text":"where of course this v is the velocity."},{"Start":"05:43.160 ","End":"05:46.340","Text":"Now in the previous lesson,"},{"Start":"05:46.340 ","End":"05:49.430","Text":"we saw that the Lawrence force,"},{"Start":"05:49.430 ","End":"05:53.293","Text":"so the force on a particle moving through a magnetic field,"},{"Start":"05:53.293 ","End":"05:59.530","Text":"was equal to q multiplied by v cross B."},{"Start":"05:59.780 ","End":"06:07.767","Text":"Because we saw that the velocity is perpendicular to the magnetic field,"},{"Start":"06:07.767 ","End":"06:10.540","Text":"therefore, what we got,"},{"Start":"06:10.540 ","End":"06:12.040","Text":"and let\u0027s just write a note,"},{"Start":"06:12.040 ","End":"06:16.915","Text":"the velocity is perpendicular to the B field."},{"Start":"06:16.915 ","End":"06:20.120","Text":"Therefore, we got that this is equal to qvB."},{"Start":"06:21.300 ","End":"06:27.235","Text":"Then we saw that when the velocity is perpendicular to the magnetic field,"},{"Start":"06:27.235 ","End":"06:28.870","Text":"we get circular motion,"},{"Start":"06:28.870 ","End":"06:32.440","Text":"and the force on a particle in circular motion is equal"},{"Start":"06:32.440 ","End":"06:38.100","Text":"to mv^2 divided R by the radius of the circle,"},{"Start":"06:38.100 ","End":"06:41.215","Text":"and then what we did, is we isolate out our R,"},{"Start":"06:41.215 ","End":"06:49.250","Text":"and we got that R was equal to mv divided by qB."},{"Start":"06:51.650 ","End":"06:53.900","Text":"Now, all we\u0027re going to do,"},{"Start":"06:53.900 ","End":"06:55.130","Text":"is we\u0027re going to plug this in,"},{"Start":"06:55.130 ","End":"06:58.085","Text":"so we have m divided by qB,"},{"Start":"06:58.085 ","End":"06:59.780","Text":"which are constants,"},{"Start":"06:59.780 ","End":"07:04.590","Text":"and then we\u0027re multiplying this by V. What\u0027s our v?"},{"Start":"07:04.590 ","End":"07:05.940","Text":"Let\u0027s just isolate v,"},{"Start":"07:05.940 ","End":"07:10.490","Text":"our velocity is equal to the square root of"},{"Start":"07:10.490 ","End":"07:17.265","Text":"2q voltage divided by mass."},{"Start":"07:17.265 ","End":"07:19.680","Text":"We\u0027re just going to multiply that by this,"},{"Start":"07:19.680 ","End":"07:26.625","Text":"so we have 2q voltage divided by mass."},{"Start":"07:26.625 ","End":"07:28.270","Text":"Now what we want,"},{"Start":"07:28.270 ","End":"07:32.630","Text":"is we wanted the radius of rotation,"},{"Start":"07:32.630 ","End":"07:38.930","Text":"so R as a function of the mass m. Let\u0027s just simplify everything and put"},{"Start":"07:38.930 ","End":"07:45.485","Text":"this in terms of a constant multiplied by our variable m. What we can do,"},{"Start":"07:45.485 ","End":"07:51.150","Text":"is we can rewrite this as we can put this into the square root sign,"},{"Start":"07:51.150 ","End":"07:55.020","Text":"so here we\u0027ll have m^2 divided by m,"},{"Start":"07:55.020 ","End":"07:58.550","Text":"so we\u0027re multiplying this by square root"},{"Start":"07:58.550 ","End":"08:02.240","Text":"of m. Then we\u0027ll put everything into the square roots."},{"Start":"08:02.240 ","End":"08:06.860","Text":"Here we have q^2 in the denominator and B^2,"},{"Start":"08:06.860 ","End":"08:11.070","Text":"which this q will cancel out with this q^2,"},{"Start":"08:11.070 ","End":"08:19.689","Text":"so what we\u0027ll have is 2 multiplied by the voltage divided by B^2."},{"Start":"08:19.689 ","End":"08:23.130","Text":"Of course, the q here in the denominator."},{"Start":"08:23.130 ","End":"08:25.430","Text":"If we know the voltage source,"},{"Start":"08:25.430 ","End":"08:28.310","Text":"the charge of the particle and the magnetic fields,"},{"Start":"08:28.310 ","End":"08:30.470","Text":"this is just a constant."},{"Start":"08:30.470 ","End":"08:34.960","Text":"And then our radius is as a function of the mass of the particles."},{"Start":"08:34.960 ","End":"08:37.505","Text":"If we know the mass of the particle,"},{"Start":"08:37.505 ","End":"08:42.610","Text":"we take the square root and then we can just get the radius of the circle."},{"Start":"08:42.610 ","End":"08:46.340","Text":"That\u0027s it. That\u0027s the end of this question."}],"ID":21407},{"Watched":false,"Name":"Exercise 3","Duration":"10m 22s","ChapterTopicVideoID":21526,"CourseChapterTopicPlaylistID":99484,"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":"00:04.425","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.425 ","End":"00:09.540","Text":"A charge is in linear motion within a parallel plate capacitor."},{"Start":"00:09.540 ","End":"00:15.825","Text":"It has a constant velocity of v_0 and travels parallel to the capacitor plates."},{"Start":"00:15.825 ","End":"00:19.595","Text":"Inside and only inside the capacitor,"},{"Start":"00:19.595 ","End":"00:25.275","Text":"a uniform known electric field E is present."},{"Start":"00:25.275 ","End":"00:27.945","Text":"On exiting the capacitor,"},{"Start":"00:27.945 ","End":"00:34.155","Text":"the charge travels in an upwards circular direction like so."},{"Start":"00:34.155 ","End":"00:37.635","Text":"Both inside and outside the capacitor,"},{"Start":"00:37.635 ","End":"00:44.750","Text":"a uniform magnetic field B is present and its magnitude and direction are unknown."},{"Start":"00:44.750 ","End":"00:47.355","Text":"We don\u0027t know what B is."},{"Start":"00:47.355 ","End":"00:51.335","Text":"Disregard the gravitational force acting on the charge."},{"Start":"00:51.335 ","End":"00:53.990","Text":"Question number 1 is,"},{"Start":"00:53.990 ","End":"00:57.545","Text":"what is the charge polarity?"},{"Start":"00:57.545 ","End":"01:01.769","Text":"Is this a positive charge or a negative charge?"},{"Start":"01:02.020 ","End":"01:07.220","Text":"The first thing that we have to notice is that our particle reaches this point."},{"Start":"01:07.220 ","End":"01:11.240","Text":"Here, it begins doing this circular motion in"},{"Start":"01:11.240 ","End":"01:15.965","Text":"this direction and here it\u0027s only subject to a magnetic field."},{"Start":"01:15.965 ","End":"01:18.440","Text":"If we look at that,"},{"Start":"01:18.440 ","End":"01:27.635","Text":"we can see that the magnetic force has to be acting in this direction, like so."},{"Start":"01:27.635 ","End":"01:34.484","Text":"This is our F_B pointing upwards."},{"Start":"01:34.484 ","End":"01:37.370","Text":"Then because we\u0027re being told that both inside and"},{"Start":"01:37.370 ","End":"01:40.670","Text":"outside the capacitor we have a uniform magnetic field,"},{"Start":"01:40.670 ","End":"01:48.410","Text":"that must mean that over here our F_B is also pointing in this direction."},{"Start":"01:48.410 ","End":"01:53.210","Text":"The next thing we know is that inside the capacitor plates,"},{"Start":"01:53.210 ","End":"01:59.850","Text":"the charge is traveling in a straight line with a constant velocity."},{"Start":"01:59.850 ","End":"02:02.615","Text":"If we have a constant velocity,"},{"Start":"02:02.615 ","End":"02:07.760","Text":"so v_0 is equal to a constant,"},{"Start":"02:07.760 ","End":"02:11.525","Text":"then that means that the acceleration"},{"Start":"02:11.525 ","End":"02:16.535","Text":"is equal to 0 and if the acceleration is equal to 0,"},{"Start":"02:16.535 ","End":"02:21.214","Text":"that means that the sum of all the forces on the particle"},{"Start":"02:21.214 ","End":"02:28.080","Text":"inside the parallel plate capacitor has to be equal to 0."},{"Start":"02:28.220 ","End":"02:36.950","Text":"The force being applied to this particle is equal to what we saw in Lorentz\u0027s law."},{"Start":"02:36.950 ","End":"02:41.720","Text":"That is equal to q and then we can open the brackets and we have"},{"Start":"02:41.720 ","End":"02:48.215","Text":"the velocity cross-product with the magnetic field plus our E field."},{"Start":"02:48.215 ","End":"02:51.500","Text":"Remember this is the equation for Lorentz\u0027s law when we have"},{"Start":"02:51.500 ","End":"02:55.850","Text":"both a magnetic and electric field acting and we know"},{"Start":"02:55.850 ","End":"03:00.875","Text":"that this has to be equal to 0 because of what we wrote above."},{"Start":"03:00.875 ","End":"03:03.800","Text":"Therefore, what is inside these brackets because we"},{"Start":"03:03.800 ","End":"03:07.180","Text":"know the charge isn\u0027t going to be equal to 0."},{"Start":"03:07.180 ","End":"03:12.140","Text":"Inside the brackets, this has to be equal to 0."},{"Start":"03:12.140 ","End":"03:21.810","Text":"Therefore, we get that the E field is equal to negative V cross B."},{"Start":"03:22.460 ","End":"03:28.250","Text":"We can see that my E field is in the negative direction."},{"Start":"03:28.250 ","End":"03:34.265","Text":"What I have to see from this relationship over here is that the force from"},{"Start":"03:34.265 ","End":"03:41.285","Text":"my E field has to be in the opposite direction to the force from my B field."},{"Start":"03:41.285 ","End":"03:45.030","Text":"That means that this is F_E."},{"Start":"03:46.460 ","End":"03:51.125","Text":"Our force due to the electric field is pointing downwards."},{"Start":"03:51.125 ","End":"03:53.420","Text":"But we can see that our E field from"},{"Start":"03:53.420 ","End":"03:57.895","Text":"the diagram is pointing upwards in the opposite direction."},{"Start":"03:57.895 ","End":"04:06.280","Text":"We know that our force due to our electric field is equal to q multiplied by e,"},{"Start":"04:06.280 ","End":"04:08.640","Text":"as we can see over here."},{"Start":"04:08.640 ","End":"04:13.010","Text":"In order for the direction of our force due to"},{"Start":"04:13.010 ","End":"04:18.605","Text":"the electric field to be in the opposite direction to our E field,"},{"Start":"04:18.605 ","End":"04:23.430","Text":"our charge has to be a negative charge."},{"Start":"04:23.870 ","End":"04:26.840","Text":"If our charge is negative,"},{"Start":"04:26.840 ","End":"04:31.740","Text":"then our force will be pointing in this opposite direction."},{"Start":"04:32.480 ","End":"04:38.985","Text":"Our force due to the electric field is in the opposite direction to the electric field."},{"Start":"04:38.985 ","End":"04:48.700","Text":"Therefore, my charge has to be a negatively charged charge."},{"Start":"04:48.710 ","End":"04:52.970","Text":"This is the answer to question number 1."},{"Start":"04:52.970 ","End":"04:57.175","Text":"This is the polarity of the charge is negative."},{"Start":"04:57.175 ","End":"05:04.745","Text":"Now question number 2 is to find the direction and magnitude of the magnetic field."},{"Start":"05:04.745 ","End":"05:10.195","Text":"Let\u0027s first deal with the direction of the magnetic field."},{"Start":"05:10.195 ","End":"05:14.200","Text":"First of all, we can see that our F_B,"},{"Start":"05:14.200 ","End":"05:17.395","Text":"our force due to the magnetic field,"},{"Start":"05:17.395 ","End":"05:20.710","Text":"is pointing in the upwards direction."},{"Start":"05:20.710 ","End":"05:22.930","Text":"Then of course we do circular motion,"},{"Start":"05:22.930 ","End":"05:24.955","Text":"so it slightly changes direction."},{"Start":"05:24.955 ","End":"05:26.410","Text":"But in theory over here,"},{"Start":"05:26.410 ","End":"05:29.420","Text":"it\u0027s in the upwards direction."},{"Start":"05:29.420 ","End":"05:34.715","Text":"Our F_B is equal to, as we know,"},{"Start":"05:34.715 ","End":"05:41.645","Text":"the charge multiplied by the velocity cross-product with the magnetic field."},{"Start":"05:41.645 ","End":"05:45.830","Text":"Now we know that this charge is a negative charge."},{"Start":"05:45.830 ","End":"05:47.330","Text":"In our case,"},{"Start":"05:47.330 ","End":"05:53.250","Text":"we have q minus multiplied by V cross B."},{"Start":"05:53.420 ","End":"05:58.140","Text":"If we draw our magnetic field,"},{"Start":"05:58.140 ","End":"06:02.015","Text":"let\u0027s draw it in blue as going into the page."},{"Start":"06:02.015 ","End":"06:08.190","Text":"If we use the right-hand rule of V cross B,"},{"Start":"06:08.190 ","End":"06:14.045","Text":"we get really that our force due to the magnetic field is pointing upwards."},{"Start":"06:14.045 ","End":"06:18.185","Text":"However, once we multiply by the negative charge,"},{"Start":"06:18.185 ","End":"06:20.995","Text":"so we\u0027ll have a negative coefficient over here,"},{"Start":"06:20.995 ","End":"06:26.000","Text":"our F_B will point in the opposite direction."},{"Start":"06:26.000 ","End":"06:27.905","Text":"It will point downwards."},{"Start":"06:27.905 ","End":"06:31.220","Text":"That we\u0027ve already seen is incorrect."},{"Start":"06:31.220 ","End":"06:33.320","Text":"It has to point upwards."},{"Start":"06:33.320 ","End":"06:39.980","Text":"It all stands with this being a negatively charged particle."},{"Start":"06:39.980 ","End":"06:43.415","Text":"However, if we imagine that"},{"Start":"06:43.415 ","End":"06:49.835","Text":"our magnetic field is coming out of the page when we do V cross B,"},{"Start":"06:49.835 ","End":"06:52.970","Text":"so we get that our F_B is pointing downwards."},{"Start":"06:52.970 ","End":"06:57.620","Text":"But then once we multiply it by the negative charge over here,"},{"Start":"06:57.620 ","End":"07:02.089","Text":"we really get that our F_B is pointing upwards."},{"Start":"07:02.089 ","End":"07:07.760","Text":"Therefore, we can say that"},{"Start":"07:07.760 ","End":"07:15.090","Text":"the direction of our B field is pointing out of the page."},{"Start":"07:15.140 ","End":"07:17.855","Text":"Let\u0027s just go over that again."},{"Start":"07:17.855 ","End":"07:24.460","Text":"This cross with this will give us some direction."},{"Start":"07:24.460 ","End":"07:28.430","Text":"But then we have to remember that we have to multiply it by a"},{"Start":"07:28.430 ","End":"07:32.330","Text":"negative so we get the opposite direction because of this negative charge."},{"Start":"07:32.330 ","End":"07:35.360","Text":"That means that we have to take the opposite B field"},{"Start":"07:35.360 ","End":"07:39.760","Text":"in order to keep our F_B pointing upwards."},{"Start":"07:39.770 ","End":"07:45.010","Text":"Now let\u0027s find the magnitude of the magnetic field."},{"Start":"07:45.010 ","End":"07:49.015","Text":"In the question, I know my value for the E field,"},{"Start":"07:49.015 ","End":"07:52.615","Text":"and I know the velocity of my particle."},{"Start":"07:52.615 ","End":"07:55.510","Text":"Here, I have an equation linking the 2."},{"Start":"07:55.510 ","End":"08:03.745","Text":"What I can see from this equation is that the magnitude of my E field is equal to,"},{"Start":"08:03.745 ","End":"08:06.460","Text":"without taking in the direction just the magnitude,"},{"Start":"08:06.460 ","End":"08:09.610","Text":"is equal to the magnitude of the velocity,"},{"Start":"08:09.610 ","End":"08:14.635","Text":"which is V_0, multiplied by the magnitude of the B field."},{"Start":"08:14.635 ","End":"08:20.095","Text":"Therefore, I can just isolate out my B and I get that the B field is equal to"},{"Start":"08:20.095 ","End":"08:26.970","Text":"the magnitude of my E field divided by the magnitude of my velocity."},{"Start":"08:27.140 ","End":"08:32.210","Text":"This is the answer to question number 2 and that\u0027s the end of the lesson."},{"Start":"08:32.210 ","End":"08:34.280","Text":"Just before I switch off,"},{"Start":"08:34.280 ","End":"08:38.000","Text":"I just want to explain this question a little bit more."},{"Start":"08:38.000 ","End":"08:43.670","Text":"This system is what we use in order to separate out velocities."},{"Start":"08:43.670 ","End":"08:48.845","Text":"What we can see is that only for a velocity"},{"Start":"08:48.845 ","End":"08:55.040","Text":"which is equal to the electric field divided by the magnetic field,"},{"Start":"08:55.040 ","End":"08:57.815","Text":"so for these magnitudes,"},{"Start":"08:57.815 ","End":"09:01.525","Text":"and this is of course velocity."},{"Start":"09:01.525 ","End":"09:06.050","Text":"Only for a velocity which is equal to E divided by B,"},{"Start":"09:06.050 ","End":"09:09.440","Text":"where we get this straight line over here,"},{"Start":"09:09.440 ","End":"09:14.545","Text":"this motion in a straight line between the capacitor plates."},{"Start":"09:14.545 ","End":"09:18.725","Text":"If my E field is greater than"},{"Start":"09:18.725 ","End":"09:22.130","Text":"my force due to the E field will be"},{"Start":"09:22.130 ","End":"09:26.300","Text":"greater and then the particle will travel towards the bottom plates."},{"Start":"09:26.300 ","End":"09:33.185","Text":"If my B field is greater than my force due to the magnetic field,"},{"Start":"09:33.185 ","End":"09:35.105","Text":"F_B will be greater,"},{"Start":"09:35.105 ","End":"09:39.610","Text":"in which case my particles will move upwards like so."},{"Start":"09:39.610 ","End":"09:43.220","Text":"The only time that my particle just travels straight"},{"Start":"09:43.220 ","End":"09:46.640","Text":"through the capacitor plates and then have"},{"Start":"09:46.640 ","End":"09:50.870","Text":"the circular motion is if the relationship between"},{"Start":"09:50.870 ","End":"09:56.505","Text":"my E and B field is equal to a specific velocity."},{"Start":"09:56.505 ","End":"10:00.920","Text":"What that means is that I can play around with the values of"},{"Start":"10:00.920 ","End":"10:05.405","Text":"my E and B field and then I can know what velocity"},{"Start":"10:05.405 ","End":"10:09.530","Text":"my particle has because only then will it reach"},{"Start":"10:09.530 ","End":"10:14.365","Text":"this point exactly and then exhibit this circular motion."},{"Start":"10:14.365 ","End":"10:20.285","Text":"That\u0027s how we can program the velocity at which a particle will travel."},{"Start":"10:20.285 ","End":"10:23.220","Text":"That\u0027s the end of this lesson."}],"ID":22381},{"Watched":false,"Name":"Exercise 4","Duration":"5m 41s","ChapterTopicVideoID":21527,"CourseChapterTopicPlaylistID":99484,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.270","Text":"Hello. In this lesson we\u0027re going to be learning about"},{"Start":"00:03.270 ","End":"00:06.885","Text":"the force on a current carrying wire."},{"Start":"00:06.885 ","End":"00:12.090","Text":"Imagine we have a long wire and we take a small piece of it."},{"Start":"00:12.090 ","End":"00:17.265","Text":"I\u0027m going to draw it larger just so that we have a close-up view."},{"Start":"00:17.265 ","End":"00:19.430","Text":"But here we have a small, small,"},{"Start":"00:19.430 ","End":"00:24.270","Text":"length of wire and its length is dl."},{"Start":"00:25.310 ","End":"00:32.445","Text":"Through this wire, we have a current and let\u0027s say it\u0027s going in this direction."},{"Start":"00:32.445 ","End":"00:35.785","Text":"Then we also have a magnetic field,"},{"Start":"00:35.785 ","End":"00:40.385","Text":"and let\u0027s just assume that the magnetic field is into the page."},{"Start":"00:40.385 ","End":"00:43.260","Text":"It doesn\u0027t really matter."},{"Start":"00:44.660 ","End":"00:47.035","Text":"What exactly is current?"},{"Start":"00:47.035 ","End":"00:48.965","Text":"We\u0027ve spoken about this before."},{"Start":"00:48.965 ","End":"00:56.615","Text":"It means that there\u0027s just charges traveling like so in the direction of the current."},{"Start":"00:56.615 ","End":"01:00.870","Text":"That\u0027s what\u0027s current means, there\u0027s charges flowing."},{"Start":"01:00.870 ","End":"01:08.045","Text":"As we know, if there\u0027s a charge and it is in a region of a magnetic field,"},{"Start":"01:08.045 ","End":"01:14.660","Text":"that means that some kind of magnetic force is going to be acting on the charges."},{"Start":"01:14.660 ","End":"01:19.860","Text":"That means that there\u0027s a magnetic force acting on the wire."},{"Start":"01:20.030 ","End":"01:23.315","Text":"As we know from Lorentz\u0027s law,"},{"Start":"01:23.315 ","End":"01:27.455","Text":"the force on each charge F_B,"},{"Start":"01:27.455 ","End":"01:35.225","Text":"is equal to the charge multiplied by the velocity cross product with the magnetic field."},{"Start":"01:35.225 ","End":"01:37.820","Text":"This is the force on 1 charge."},{"Start":"01:37.820 ","End":"01:40.205","Text":"However, inside this wire,"},{"Start":"01:40.205 ","End":"01:42.335","Text":"there\u0027s lots of charges."},{"Start":"01:42.335 ","End":"01:46.940","Text":"How do we calculate the force on this wire that has lots of charges?"},{"Start":"01:46.940 ","End":"01:50.830","Text":"What do we do is we do some manipulation"},{"Start":"01:50.830 ","End":"01:56.415","Text":"where we sum up all of the charges in the certain length,"},{"Start":"01:56.415 ","End":"01:59.390","Text":"and what we get after we do"},{"Start":"01:59.390 ","End":"02:04.490","Text":"this mathematical manipulation to take into account all of the charges and not just 1,"},{"Start":"02:04.490 ","End":"02:07.680","Text":"we get this equation over here."},{"Start":"02:07.680 ","End":"02:12.980","Text":"These 2 equations are identical."},{"Start":"02:12.980 ","End":"02:17.995","Text":"Just Lorentz\u0027s law refers to the force on 1 charge,"},{"Start":"02:17.995 ","End":"02:24.080","Text":"and this equation just refers to all of the charges in this section of wire,"},{"Start":"02:24.080 ","End":"02:26.465","Text":"but they basically mean the same thing."},{"Start":"02:26.465 ","End":"02:30.590","Text":"Let\u0027s imagine that we have a long wire like"},{"Start":"02:30.590 ","End":"02:34.990","Text":"so where we have this current I flowing through."},{"Start":"02:34.990 ","End":"02:38.375","Text":"What we\u0027ll do is we\u0027ll cut the wire into"},{"Start":"02:38.375 ","End":"02:42.625","Text":"a thin slice where there are slices of length dl."},{"Start":"02:42.625 ","End":"02:50.165","Text":"Then in order to find the total force along this entire wire,"},{"Start":"02:50.165 ","End":"02:56.500","Text":"we\u0027re just going to sum up the force on these dl\u0027s along the wire."},{"Start":"02:56.500 ","End":"02:59.690","Text":"What does that mean? Because we have a dF over here,"},{"Start":"02:59.690 ","End":"03:01.775","Text":"that means we\u0027re going to integrate."},{"Start":"03:01.775 ","End":"03:04.655","Text":"The total force is just going to be"},{"Start":"03:04.655 ","End":"03:10.055","Text":"this dF and integrating along it for the full length of the wire,"},{"Start":"03:10.055 ","End":"03:11.585","Text":"so we can just substitute that in,"},{"Start":"03:11.585 ","End":"03:17.670","Text":"that\u0027s equal to I dl cross B."},{"Start":"03:18.260 ","End":"03:25.495","Text":"Let\u0027s imagine that our magnetic field in this example is in this direction."},{"Start":"03:25.495 ","End":"03:32.575","Text":"We have this angle Alpha"},{"Start":"03:32.575 ","End":"03:38.125","Text":"between our magnetic field and the direction of the current."},{"Start":"03:38.125 ","End":"03:42.100","Text":"Let\u0027s imagine that everything I,"},{"Start":"03:42.100 ","End":"03:43.960","Text":"the length, of course,"},{"Start":"03:43.960 ","End":"03:47.395","Text":"and the magnetic fields are all constants."},{"Start":"03:47.395 ","End":"03:50.800","Text":"There\u0027s this angle Alpha over here between the 2."},{"Start":"03:50.800 ","End":"03:55.495","Text":"But let\u0027s imagine that the current and the magnetic field are constants."},{"Start":"03:55.495 ","End":"04:05.835","Text":"Therefore, what we can say is that we\u0027ll get the integral of I dl multiplied by B,"},{"Start":"04:05.835 ","End":"04:07.340","Text":"and then when we have the cross products,"},{"Start":"04:07.340 ","End":"04:14.190","Text":"so we multiply this by sine of the angle between dl and B."},{"Start":"04:14.870 ","End":"04:19.160","Text":"Now again, if everything over here is a constant,"},{"Start":"04:19.160 ","End":"04:24.185","Text":"what we will have is that we\u0027re just integrating along the length of the wire."},{"Start":"04:24.185 ","End":"04:34.690","Text":"Let\u0027s say the length of the wire is l. What we have is ILB multiplied by sine of Alpha."},{"Start":"04:35.950 ","End":"04:38.135","Text":"A lot of the time,"},{"Start":"04:38.135 ","End":"04:40.820","Text":"if we\u0027re dealing with everything as a constant,"},{"Start":"04:40.820 ","End":"04:45.410","Text":"we\u0027ll get that our magnetic force on this wire is going to be,"},{"Start":"04:45.410 ","End":"04:47.800","Text":"it\u0027s often written as BIL."},{"Start":"04:47.800 ","End":"04:50.795","Text":"B, the magnetic field multiplied by the current,"},{"Start":"04:50.795 ","End":"04:53.015","Text":"multiplied by the length of the wire,"},{"Start":"04:53.015 ","End":"04:59.600","Text":"multiplied by sine of the angle between the magnetic field and"},{"Start":"04:59.600 ","End":"05:07.630","Text":"the direction of the current or the direction of the wire."},{"Start":"05:09.050 ","End":"05:12.010","Text":"This is the equation that you need to remember,"},{"Start":"05:12.010 ","End":"05:15.650","Text":"and remember that\u0027s just for the force on this section of wire."},{"Start":"05:15.650 ","End":"05:18.545","Text":"If you want to know the force on the entire wire, you of course,"},{"Start":"05:18.545 ","End":"05:22.460","Text":"have to integrate because we have a differential over here."},{"Start":"05:22.460 ","End":"05:27.090","Text":"If everything is a constant,"},{"Start":"05:27.090 ","End":"05:34.645","Text":"then you can just substitute into this equation, F=BIL sine Alpha."},{"Start":"05:34.645 ","End":"05:42.000","Text":"That\u0027s the end of the lesson and this is the force on a current carrying wire."}],"ID":22382},{"Watched":false,"Name":"Deriving Equation for Force on Current Carrying Wire","Duration":"5m 36s","ChapterTopicVideoID":24813,"CourseChapterTopicPlaylistID":99484,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.830","Text":"Hello. In this lesson,"},{"Start":"00:01.830 ","End":"00:07.230","Text":"we\u0027re going to be deriving the equation for force on a current carrying wire."},{"Start":"00:07.230 ","End":"00:08.865","Text":"In the previous lesson,"},{"Start":"00:08.865 ","End":"00:15.990","Text":"we saw that we can integrate along df to get the total force acting on"},{"Start":"00:15.990 ","End":"00:24.465","Text":"a current carrying wire where df was equal to the current multiplied by the length,"},{"Start":"00:24.465 ","End":"00:30.770","Text":"the small length of the wire cross-product with the magnetic field."},{"Start":"00:30.770 ","End":"00:34.800","Text":"Now we\u0027re going to be deriving this equation."},{"Start":"00:36.260 ","End":"00:46.399","Text":"The force on one charge due to the magnetic field as we know is equal to q."},{"Start":"00:46.399 ","End":"00:53.480","Text":"The charge multiplied by its velocity cross product with the magnetic field."},{"Start":"00:53.480 ","End":"00:56.450","Text":"This is the force on one charge but we"},{"Start":"00:56.450 ","End":"00:59.840","Text":"can see that in the wire we have more than one charge,"},{"Start":"00:59.840 ","End":"01:06.840","Text":"so let\u0027s multiply it by n where this is the number of charges."},{"Start":"01:06.980 ","End":"01:16.450","Text":"This is the force on this slice of wire due to the total number of charges in the wire."},{"Start":"01:16.450 ","End":"01:18.530","Text":"What is this N?"},{"Start":"01:18.530 ","End":"01:22.410","Text":"How do we calculate the total number of charges in the wire?"},{"Start":"01:22.410 ","End":"01:29.195","Text":"This is equal to lowercase n which is the density of charges per"},{"Start":"01:29.195 ","End":"01:35.715","Text":"unit volume multiplied by the unit volume,"},{"Start":"01:35.715 ","End":"01:42.670","Text":"so dv where this v is of course volume."},{"Start":"01:43.100 ","End":"01:48.620","Text":"So dv, I\u0027ve written the v like this so that we don\u0027t get confused with the velocity,"},{"Start":"01:48.620 ","End":"01:52.140","Text":"so this dv is a unit volume."},{"Start":"01:52.310 ","End":"01:55.575","Text":"How do we calculate this unit volume?"},{"Start":"01:55.575 ","End":"02:02.075","Text":"As we know, the length of this piece of wire is dl and then"},{"Start":"02:02.075 ","End":"02:08.910","Text":"what we have is over here we have the cross-sectional area of the wire,"},{"Start":"02:08.910 ","End":"02:14.280","Text":"so it\u0027s a circular cross-sectional area in this case over here but it can be anything."},{"Start":"02:14.280 ","End":"02:18.065","Text":"The area is equal to s,"},{"Start":"02:18.065 ","End":"02:23.315","Text":"so this is the cross sectional area of the wire."},{"Start":"02:23.315 ","End":"02:29.275","Text":"Therefore we can say that N"},{"Start":"02:29.275 ","End":"02:35.060","Text":"is equal to charge density per unit volume multiplied by the unit volume,"},{"Start":"02:35.060 ","End":"02:40.550","Text":"so the unit volume will be the cross-sectional area which"},{"Start":"02:40.550 ","End":"02:47.160","Text":"is S multiplied by this unit length dl."},{"Start":"02:48.260 ","End":"02:52.420","Text":"Now let\u0027s plug all of this back into this equation."},{"Start":"02:52.420 ","End":"02:57.520","Text":"I have that my force on the wire is equal to"},{"Start":"02:57.520 ","End":"03:03.445","Text":"the number of charges in the wire which we\u0027ve already seen is equal to n,"},{"Start":"03:03.445 ","End":"03:08.590","Text":"the charge density per unit volume multiplied by the cross-sectional area of the wire,"},{"Start":"03:08.590 ","End":"03:11.125","Text":"multiplied by the length of the wire,"},{"Start":"03:11.125 ","End":"03:16.015","Text":"so this is n multiplied by q,"},{"Start":"03:16.015 ","End":"03:23.420","Text":"multiplied by the velocity cross-product by my magnetic field."},{"Start":"03:24.420 ","End":"03:28.645","Text":"Now I\u0027m just going to write this in a slightly different order."},{"Start":"03:28.645 ","End":"03:36.350","Text":"I\u0027m going to write this as nq multiplied by the velocity"},{"Start":"03:36.570 ","End":"03:46.865","Text":"multiplied by Sdl cross-product with the B field."},{"Start":"03:46.865 ","End":"03:49.625","Text":"The equation has remained the exact same."},{"Start":"03:49.625 ","End":"03:51.830","Text":"I\u0027ve just written this in a different order,"},{"Start":"03:51.830 ","End":"03:55.520","Text":"and if you notice nq multiplied by"},{"Start":"03:55.520 ","End":"04:01.295","Text":"the velocity is equal to our current density which we learned about which is J,"},{"Start":"04:01.295 ","End":"04:03.775","Text":"remember our current density."},{"Start":"04:03.775 ","End":"04:10.780","Text":"Then of course when we have J multiplied by S,"},{"Start":"04:12.770 ","End":"04:15.525","Text":"let\u0027s write this over here."},{"Start":"04:15.525 ","End":"04:21.930","Text":"This is of course equal to the current."},{"Start":"04:21.930 ","End":"04:27.070","Text":"The current density multiplied by the surface area is equal to the current,"},{"Start":"04:27.070 ","End":"04:37.160","Text":"so therefore nqvS is equal to Idl cross B."},{"Start":"04:39.410 ","End":"04:43.420","Text":"Then the vector on the dl comes from"},{"Start":"04:43.420 ","End":"04:49.630","Text":"the direction of our velocity because of course the velocity is"},{"Start":"04:49.630 ","End":"04:53.485","Text":"traveling along the length of the wire and the direction of the"},{"Start":"04:53.485 ","End":"04:56.480","Text":"current and of course current isn\u0027t"},{"Start":"04:56.480 ","End":"05:00.110","Text":"a vector quantity so we just put the vector on the dl."},{"Start":"05:00.110 ","End":"05:02.900","Text":"The direction of the dl is in"},{"Start":"05:02.900 ","End":"05:06.020","Text":"the direction of the current and the direction of the current is in"},{"Start":"05:06.020 ","End":"05:09.230","Text":"the direction of the charges and the direction of the charges"},{"Start":"05:09.230 ","End":"05:13.345","Text":"is of course given by the direction of their velocity."},{"Start":"05:13.345 ","End":"05:18.470","Text":"Now the last thing is because we have a dl over here,"},{"Start":"05:18.470 ","End":"05:23.270","Text":"so we know this is a differential so we have to put a d over here,"},{"Start":"05:23.270 ","End":"05:29.420","Text":"so this is df and that\u0027s how we get to this final equation over here"},{"Start":"05:29.420 ","End":"05:36.960","Text":"and that is the force on a small section of the wire and that is the end of this lesson."}],"ID":25726},{"Watched":false,"Name":"Force Acting on Bent Wire","Duration":"24m 5s","ChapterTopicVideoID":21529,"CourseChapterTopicPlaylistID":99484,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.890","Text":"Hello. In this lesson,"},{"Start":"00:01.890 ","End":"00:04.470","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.470 ","End":"00:08.624","Text":"A conducting wire is bent as shown in the diagram."},{"Start":"00:08.624 ","End":"00:12.525","Text":"The rounded section forming a quadrant of radius R."},{"Start":"00:12.525 ","End":"00:15.540","Text":"A uniform magnetic field is present coming out of"},{"Start":"00:15.540 ","End":"00:19.455","Text":"the page and the current I flows through the wire."},{"Start":"00:19.455 ","End":"00:24.560","Text":"Length L value for the magnetic field B,"},{"Start":"00:24.560 ","End":"00:28.335","Text":"and the current I and the radius R are given."},{"Start":"00:28.335 ","End":"00:33.465","Text":"We\u0027re being asked to find the resultant force acting on the wire."},{"Start":"00:33.465 ","End":"00:40.390","Text":"The total force acting on the wire is what we\u0027re trying to calculate."},{"Start":"00:40.640 ","End":"00:44.465","Text":"We\u0027re not told this length over here,"},{"Start":"00:44.465 ","End":"00:47.315","Text":"or this length over here,"},{"Start":"00:47.315 ","End":"00:50.840","Text":"this diagonal but we are told that the length of"},{"Start":"00:50.840 ","End":"00:55.850","Text":"the straight wire here and here is equal to L, each 1."},{"Start":"00:55.850 ","End":"00:59.000","Text":"What we\u0027re going to do is we\u0027re going to calculate"},{"Start":"00:59.000 ","End":"01:02.840","Text":"the force on the wire over here in Region 1,"},{"Start":"01:02.840 ","End":"01:08.795","Text":"the force on the wire in Region 2 on this arc and Region 3 in this diagonal,"},{"Start":"01:08.795 ","End":"01:11.844","Text":"and in Region 4 the straight line."},{"Start":"01:11.844 ","End":"01:16.370","Text":"Then we\u0027re going to add them up to find the total force."},{"Start":"01:16.370 ","End":"01:22.990","Text":"We remember that our equation for the force on a current-carrying wire,"},{"Start":"01:22.990 ","End":"01:28.010","Text":"dF is equal to the current multiplied"},{"Start":"01:28.010 ","End":"01:34.015","Text":"by the length of the wire cross product with the magnetic field."},{"Start":"01:34.015 ","End":"01:35.950","Text":"For Region 1,"},{"Start":"01:35.950 ","End":"01:37.895","Text":"let\u0027s call this F_1."},{"Start":"01:37.895 ","End":"01:41.840","Text":"What we have is a magnetic field,"},{"Start":"01:41.840 ","End":"01:44.330","Text":"which we know its value and its constant,"},{"Start":"01:44.330 ","End":"01:49.400","Text":"so we have B multiplied by the current,"},{"Start":"01:49.400 ","End":"01:53.225","Text":"multiplied by the length of this section of wire,"},{"Start":"01:53.225 ","End":"02:01.715","Text":"which is L. Then we can see that the angle between the length of the wire which is"},{"Start":"02:01.715 ","End":"02:03.410","Text":"traveling let\u0027s say that this is"},{"Start":"02:03.410 ","End":"02:11.600","Text":"the x-direction and we can see that our magnetic field is coming out of the page."},{"Start":"02:11.600 ","End":"02:14.645","Text":"We can see that that is,"},{"Start":"02:14.645 ","End":"02:18.460","Text":"we can say in the z-direction,"},{"Start":"02:18.460 ","End":"02:22.005","Text":"rather along the z axes,"},{"Start":"02:22.005 ","End":"02:26.060","Text":"and then we can see if we use the right-hand rule,"},{"Start":"02:26.060 ","End":"02:30.500","Text":"that our force due to the magnetic field on"},{"Start":"02:30.500 ","End":"02:37.504","Text":"the wire is going to be pointing in this downwards direction."},{"Start":"02:37.504 ","End":"02:42.980","Text":"If we say that this is the y-direction,"},{"Start":"02:42.980 ","End":"02:47.615","Text":"so we can see that this will be pointing in the negative y-direction,"},{"Start":"02:47.615 ","End":"02:51.845","Text":"so you just point your thumb in the direction of the current,"},{"Start":"02:51.845 ","End":"02:54.920","Text":"point your fourth finger in the direction of the B field,"},{"Start":"02:54.920 ","End":"03:01.530","Text":"and you\u0027ll see that your middle finger representing the force, is pointing downwards."},{"Start":"03:02.170 ","End":"03:08.750","Text":"Now usually, we would have to multiply this by sine of the angle between"},{"Start":"03:08.750 ","End":"03:12.170","Text":"the direction of the current and the direction of"},{"Start":"03:12.170 ","End":"03:17.000","Text":"the magnetic field but we see that the current is perpendicular to the magnetic field,"},{"Start":"03:17.000 ","End":"03:21.470","Text":"which means that we\u0027re multiplying this by sine of 90 degrees,"},{"Start":"03:21.470 ","End":"03:23.645","Text":"which is just equal to 1."},{"Start":"03:23.645 ","End":"03:28.160","Text":"This is what we\u0027re left with for the force on the wire in Section 1."},{"Start":"03:28.160 ","End":"03:33.755","Text":"If this is the force on Region 1,"},{"Start":"03:33.755 ","End":"03:37.115","Text":"Region 1 is identical to Region 4."},{"Start":"03:37.115 ","End":"03:44.490","Text":"It\u0027s of the same length and they\u0027re lined up in the exact same fashion."},{"Start":"03:44.490 ","End":"03:51.120","Text":"That means that this is also equal to the force over here."},{"Start":"03:51.120 ","End":"03:53.580","Text":"Now let\u0027s do Region 3,"},{"Start":"03:53.580 ","End":"03:55.580","Text":"then we\u0027ll go on to Region 2,"},{"Start":"03:55.580 ","End":"03:58.525","Text":"which is slightly more complicated."},{"Start":"03:58.525 ","End":"04:02.590","Text":"We know that this is a radius R,"},{"Start":"04:02.590 ","End":"04:06.410","Text":"which means that if we draw a line from here to here,"},{"Start":"04:06.410 ","End":"04:12.919","Text":"this is also radius R. We know that this angle over here is 30 degrees,"},{"Start":"04:12.919 ","End":"04:15.664","Text":"and this is 90 degrees."},{"Start":"04:15.664 ","End":"04:19.985","Text":"We can see that from the diagram and from what we\u0027re being told in the question."},{"Start":"04:19.985 ","End":"04:24.035","Text":"What we have is that the opposite side is of length R,"},{"Start":"04:24.035 ","End":"04:26.215","Text":"which is given to us in the question,"},{"Start":"04:26.215 ","End":"04:30.170","Text":"and what we\u0027re trying to find is less length over here,"},{"Start":"04:30.170 ","End":"04:33.035","Text":"which is the length of the hypotenuse."},{"Start":"04:33.035 ","End":"04:39.755","Text":"What we have is opposite over hypotenuse and from our SOHCAHTOA,"},{"Start":"04:39.755 ","End":"04:44.750","Text":"we know that this is equal to sine of the angle."},{"Start":"04:44.750 ","End":"04:48.200","Text":"Therefore, we can say that sine of our angle,"},{"Start":"04:48.200 ","End":"04:50.825","Text":"which is 30 degrees,"},{"Start":"04:50.825 ","End":"04:52.654","Text":"is equal to the opposite,"},{"Start":"04:52.654 ","End":"04:54.410","Text":"which is our R,"},{"Start":"04:54.410 ","End":"04:58.315","Text":"divided by this length over here."},{"Start":"04:58.315 ","End":"05:03.330","Text":"Let\u0027s call it H, the hypotenuse."},{"Start":"05:03.330 ","End":"05:09.555","Text":"Sine of 30 is equal to 1.5."},{"Start":"05:09.555 ","End":"05:19.070","Text":"What we have therefore is that 1.5 is equal to R divided by H and therefore,"},{"Start":"05:19.070 ","End":"05:27.120","Text":"this length H over"},{"Start":"05:27.120 ","End":"05:34.305","Text":"here is equal to R multiplied by 2,"},{"Start":"05:34.305 ","End":"05:37.300","Text":"or in other words, 2R."},{"Start":"05:37.580 ","End":"05:45.190","Text":"The length of Region 3 is equal to 2R."},{"Start":"05:45.530 ","End":"05:50.185","Text":"Let\u0027s start filling out F_3."},{"Start":"05:50.185 ","End":"05:54.125","Text":"F_3 is, we have the same magnetic field,"},{"Start":"05:54.125 ","End":"06:01.350","Text":"the same current, and our length we\u0027ve seen is equal to 2R."},{"Start":"06:02.060 ","End":"06:07.970","Text":"Now what we want to do is we want to see the direction."},{"Start":"06:07.970 ","End":"06:11.285","Text":"First of all, from the right-hand rule,"},{"Start":"06:11.285 ","End":"06:14.570","Text":"we can see that the force,"},{"Start":"06:14.570 ","End":"06:16.245","Text":"so this is F_B_1,"},{"Start":"06:16.245 ","End":"06:18.820","Text":"so here we have"},{"Start":"06:19.120 ","End":"06:24.350","Text":"F_B_3 is going to be in"},{"Start":"06:24.350 ","End":"06:30.785","Text":"this direction where there\u0027s a 90-degree angle between this diagonal,"},{"Start":"06:30.785 ","End":"06:35.910","Text":"and between the wire and the direction of our force."},{"Start":"06:35.910 ","End":"06:40.970","Text":"It\u0027s for the exact same reason that we got this as the direction of our force over here."},{"Start":"06:40.970 ","End":"06:43.670","Text":"I just use the right-hand rule and you\u0027ll see that it\u0027s in"},{"Start":"06:43.670 ","End":"06:46.865","Text":"this direction at 90 degrees to the wire."},{"Start":"06:46.865 ","End":"06:51.920","Text":"Now what we want to do is we want to see what angle this"},{"Start":"06:51.920 ","End":"06:57.630","Text":"is relative to the x and relative to the y axes."},{"Start":"06:57.890 ","End":"07:01.005","Text":"Let\u0027s just draw out this section."},{"Start":"07:01.005 ","End":"07:05.060","Text":"What we have is we have over here,"},{"Start":"07:05.060 ","End":"07:08.660","Text":"we have the diagonal of the wire where this angle over here is"},{"Start":"07:08.660 ","End":"07:14.615","Text":"30 degrees and then in this direction over here,"},{"Start":"07:14.615 ","End":"07:17.885","Text":"we have our magnetic force,"},{"Start":"07:17.885 ","End":"07:22.235","Text":"like so, where here we have a 90-degree angle."},{"Start":"07:22.235 ","End":"07:25.925","Text":"Now, we know that over here,"},{"Start":"07:25.925 ","End":"07:29.615","Text":"if we draw a line parallel to this upper wire,"},{"Start":"07:29.615 ","End":"07:33.470","Text":"that from Z angles or alternate angles,"},{"Start":"07:33.470 ","End":"07:38.679","Text":"that means that this angle over here is also 30 degrees."},{"Start":"07:38.679 ","End":"07:44.720","Text":"What we have, is we have that the total angle all around here is,"},{"Start":"07:44.720 ","End":"07:48.380","Text":"of course, that of a straight line, so that\u0027s 180."},{"Start":"07:48.380 ","End":"07:54.650","Text":"Therefore, this angle over here is going to be equal to 180 minus"},{"Start":"07:54.650 ","End":"08:01.025","Text":"90 minus 30 so 180 minus 90 degrees minus 30 degrees."},{"Start":"08:01.025 ","End":"08:03.050","Text":"That will give us this angle over here,"},{"Start":"08:03.050 ","End":"08:08.504","Text":"which is equal to 60 degrees."},{"Start":"08:08.504 ","End":"08:12.975","Text":"Now we can see that our force,"},{"Start":"08:12.975 ","End":"08:16.710","Text":"F_B is going to have some component."},{"Start":"08:16.710 ","End":"08:21.345","Text":"Let\u0027s draw it in gray in the x direction, like so."},{"Start":"08:21.345 ","End":"08:24.525","Text":"This is F_B in the x-direction,"},{"Start":"08:24.525 ","End":"08:29.955","Text":"and like so in the y direction, so F_By."},{"Start":"08:29.955 ","End":"08:32.280","Text":"Now let\u0027s just write this out."},{"Start":"08:32.280 ","End":"08:36.975","Text":"This is the magnitude and the direction is going to be,"},{"Start":"08:36.975 ","End":"08:40.890","Text":"so we have for the x,"},{"Start":"08:40.890 ","End":"08:44.895","Text":"this is the adjacent side to our angle."},{"Start":"08:44.895 ","End":"08:54.525","Text":"That\u0027s going to be multiplied by cosine of 60 degrees in the x-direction."},{"Start":"08:54.525 ","End":"08:59.160","Text":"Then we have this side over here,"},{"Start":"08:59.160 ","End":"09:02.430","Text":"which represents the y-direction."},{"Start":"09:02.430 ","End":"09:04.680","Text":"First of all, we can see that it\u0027s pointing down,"},{"Start":"09:04.680 ","End":"09:07.770","Text":"which is the opposite direction to y-axis,"},{"Start":"09:07.770 ","End":"09:10.050","Text":"so we\u0027re going to have minus,"},{"Start":"09:10.050 ","End":"09:13.345","Text":"and this is the opposite side to the angle."},{"Start":"09:13.345 ","End":"09:16.850","Text":"This is minus sine of the angle,"},{"Start":"09:16.850 ","End":"09:22.020","Text":"which is 60 degrees in the y direction."},{"Start":"09:22.150 ","End":"09:26.279","Text":"Cosine of 60 degrees,"},{"Start":"09:26.279 ","End":"09:28.140","Text":"we can just rewrite this."},{"Start":"09:28.140 ","End":"09:29.790","Text":"We have, first of all,"},{"Start":"09:29.790 ","End":"09:39.615","Text":"2 multiplied by BIR and then cosine of 60 is"},{"Start":"09:39.615 ","End":"09:45.225","Text":"half in the x-direction and sine of 60 is root 3/2"},{"Start":"09:45.225 ","End":"09:53.418","Text":"so negative root 3/2 in the y direction."},{"Start":"09:53.418 ","End":"09:59.340","Text":"Now let\u0027s deal with region Number 2 where we have this arc."},{"Start":"09:59.340 ","End":"10:03.479","Text":"What we\u0027re going to do is we\u0027re going to split the wire at this section"},{"Start":"10:03.479 ","End":"10:09.880","Text":"into little lengths where these lengths are of length dl."},{"Start":"10:10.910 ","End":"10:15.540","Text":"This is tracing a quarter of a circle."},{"Start":"10:15.540 ","End":"10:20.115","Text":"The length when dealing with polar coordinates,"},{"Start":"10:20.115 ","End":"10:25.950","Text":"so dl in polar coordinates is simply equal to the radius,"},{"Start":"10:25.950 ","End":"10:30.555","Text":"which here is a constant multiplied by d Theta,"},{"Start":"10:30.555 ","End":"10:33.120","Text":"the angle that we\u0027re at."},{"Start":"10:33.120 ","End":"10:37.035","Text":"Now we\u0027re going to plug this into this equation over here."},{"Start":"10:37.035 ","End":"10:44.055","Text":"What we\u0027ll get there for is that dF is equal to,"},{"Start":"10:44.055 ","End":"10:45.675","Text":"so we have I,"},{"Start":"10:45.675 ","End":"10:48.615","Text":"the current multiplied by dl,"},{"Start":"10:48.615 ","End":"10:54.390","Text":"which over here is Id Theta and then cross-product with the B field,"},{"Start":"10:54.390 ","End":"10:55.545","Text":"which is a constant."},{"Start":"10:55.545 ","End":"11:05.295","Text":"We can write here B and then multiply it by the angle between the dl and the B field."},{"Start":"11:05.295 ","End":"11:09.240","Text":"Of course, we can see that our dl is in"},{"Start":"11:09.240 ","End":"11:13.110","Text":"the xy plane and the B field is along the z axis."},{"Start":"11:13.110 ","End":"11:15.735","Text":"They are 90 degrees to one another."},{"Start":"11:15.735 ","End":"11:19.140","Text":"When we multiply this by sine of the angle between the two,"},{"Start":"11:19.140 ","End":"11:20.850","Text":"we\u0027re multiplying by sine of 90,"},{"Start":"11:20.850 ","End":"11:23.350","Text":"which is just equal to 1."},{"Start":"11:23.660 ","End":"11:26.415","Text":"That\u0027s what we\u0027re left with,"},{"Start":"11:26.415 ","End":"11:30.490","Text":"IRd Theta multiplied by B."},{"Start":"11:30.860 ","End":"11:34.635","Text":"This is the magnitude of the force."},{"Start":"11:34.635 ","End":"11:37.875","Text":"Now what we want to do is we want to find the direction."},{"Start":"11:37.875 ","End":"11:41.130","Text":"Again, using the right-hand rule,"},{"Start":"11:41.130 ","End":"11:47.640","Text":"we can see that the direction of the force is going to be in this direction,"},{"Start":"11:47.640 ","End":"11:51.420","Text":"so this is our dF."},{"Start":"11:51.420 ","End":"11:53.370","Text":"It\u0027s in this direction,"},{"Start":"11:53.370 ","End":"11:56.235","Text":"which is the radial direction."},{"Start":"11:56.235 ","End":"12:01.680","Text":"What do we want to do is we want to figure out what the radial direction"},{"Start":"12:01.680 ","End":"12:08.800","Text":"is with respect to x and y because that\u0027s what we\u0027ve been using up until now."},{"Start":"12:08.900 ","End":"12:14.190","Text":"If we carry on this line over here,"},{"Start":"12:14.190 ","End":"12:20.790","Text":"and we take a vector from the origin of the circle until our piece over here,"},{"Start":"12:20.790 ","End":"12:28.890","Text":"so this is of course radius r and let\u0027s call this angle over here, angle Theta."},{"Start":"12:28.890 ","End":"12:34.800","Text":"Now if I draw a parallel line to the dotted line that I drew,"},{"Start":"12:34.800 ","End":"12:39.480","Text":"here we can see z angles or alternate angles."},{"Start":"12:39.480 ","End":"12:43.740","Text":"That means that this angle over here is also angle Theta."},{"Start":"12:43.740 ","End":"12:50.800","Text":"Then of course, that means that this angle over here is also Theta."},{"Start":"12:52.910 ","End":"12:58.890","Text":"Because here we can see that angles at a point add up to 360 and this Theta is"},{"Start":"12:58.890 ","End":"13:02.130","Text":"exactly opposite to this angle Theta over"},{"Start":"13:02.130 ","End":"13:06.355","Text":"here with this parallel line and the line crossing it."},{"Start":"13:06.355 ","End":"13:12.950","Text":"In that case, I can say that my dF in the x-direction,"},{"Start":"13:12.950 ","End":"13:19.440","Text":"so my x component of the force is going to be like so."},{"Start":"13:20.230 ","End":"13:24.185","Text":"It\u0027s going to be in this direction."},{"Start":"13:24.185 ","End":"13:28.395","Text":"This is my dF in the x-direction."},{"Start":"13:28.395 ","End":"13:32.460","Text":"We can see that it\u0027s pointing in this leftwards direction,"},{"Start":"13:32.460 ","End":"13:34.455","Text":"which is the negative direction,"},{"Start":"13:34.455 ","End":"13:38.400","Text":"because we said that the x-axis is pointing in the right direction."},{"Start":"13:38.400 ","End":"13:45.870","Text":"That means we put a minus over here and because we\u0027re taking this side over here,"},{"Start":"13:45.870 ","End":"13:50.010","Text":"which is the same as if we just move this vector up over here,"},{"Start":"13:50.010 ","End":"13:54.480","Text":"we can see that it\u0027s the adjacent side to our angle Theta."},{"Start":"13:54.480 ","End":"14:03.760","Text":"That means that our dF_x is equal to negative dF multiplied by cosine of Theta."},{"Start":"14:04.580 ","End":"14:09.015","Text":"Negative because it\u0027s pointing in the negative x-direction,"},{"Start":"14:09.015 ","End":"14:12.120","Text":"dF is the total magnitude."},{"Start":"14:12.120 ","End":"14:15.165","Text":"Then we\u0027re taking its projection along the x-axis,"},{"Start":"14:15.165 ","End":"14:17.010","Text":"which is over here,"},{"Start":"14:17.010 ","End":"14:21.405","Text":"the adjacent side so we\u0027re using cosine of Theta."},{"Start":"14:21.405 ","End":"14:24.570","Text":"Then similarly in the y-direction,"},{"Start":"14:24.570 ","End":"14:29.380","Text":"so the y-direction is going to be pointing like this, dF_y."},{"Start":"14:29.960 ","End":"14:33.900","Text":"Then we can say that the y components, so dF_y,"},{"Start":"14:33.900 ","End":"14:37.995","Text":"so this is also pointing in the negative direction,"},{"Start":"14:37.995 ","End":"14:40.935","Text":"negative y-direction so we have a negative."},{"Start":"14:40.935 ","End":"14:43.410","Text":"Then this vector over here,"},{"Start":"14:43.410 ","End":"14:45.700","Text":"we can move it over here."},{"Start":"14:45.770 ","End":"14:48.840","Text":"Then we can see it\u0027s the opposite sides,"},{"Start":"14:48.840 ","End":"14:53.580","Text":"so we\u0027ll use sine but of course,"},{"Start":"14:53.580 ","End":"14:55.950","Text":"first we have to multiply it by the total magnitude."},{"Start":"14:55.950 ","End":"14:58.950","Text":"So dF and its projection on the y-axis,"},{"Start":"14:58.950 ","End":"15:00.450","Text":"which is going to be of course,"},{"Start":"15:00.450 ","End":"15:06.100","Text":"on the opposite side so multiplied by sine of Theta."},{"Start":"15:06.500 ","End":"15:09.780","Text":"Now let\u0027s substitute in our dF."},{"Start":"15:09.780 ","End":"15:15.030","Text":"We have that dF in the x-direction is equal to negative."},{"Start":"15:15.030 ","End":"15:19.650","Text":"Then we have IRB and then we"},{"Start":"15:19.650 ","End":"15:25.335","Text":"have cosine of Theta and then multiplied by d Theta."},{"Start":"15:25.335 ","End":"15:27.690","Text":"I\u0027ve just substituted in what dF is."},{"Start":"15:27.690 ","End":"15:32.685","Text":"Then my dF in the y direction is equal to"},{"Start":"15:32.685 ","End":"15:37.320","Text":"negative IRB"},{"Start":"15:37.320 ","End":"15:44.263","Text":"sine Theta d Theta."},{"Start":"15:44.263 ","End":"15:46.180","Text":"I just move them like this."},{"Start":"15:46.180 ","End":"15:48.520","Text":"Now, this is of course,"},{"Start":"15:48.520 ","End":"15:58.630","Text":"case of the x component region 2 and the y component for region 2, this arc."},{"Start":"15:58.630 ","End":"16:02.455","Text":"If I want to find F_2x,"},{"Start":"16:02.455 ","End":"16:07.330","Text":"the total force in the x-direction in region 2,"},{"Start":"16:07.330 ","End":"16:12.284","Text":"so all I have to do is integrate like so where"},{"Start":"16:12.284 ","End":"16:15.490","Text":"my integrating bounds are going to be"},{"Start":"16:15.490 ","End":"16:20.095","Text":"from this point over here until this point over here."},{"Start":"16:20.095 ","End":"16:22.190","Text":"We\u0027re integrating along Theta."},{"Start":"16:22.190 ","End":"16:26.323","Text":"My angle theta over here is equal to 0 and"},{"Start":"16:26.323 ","End":"16:31.480","Text":"my angle Theta over here in radians is Pi divided by 2."},{"Start":"16:31.480 ","End":"16:36.505","Text":"It\u0027s 90 degrees or in radians Pi divided by 2 and what we\u0027ll"},{"Start":"16:36.505 ","End":"16:42.620","Text":"get is that this is equal to negative BIR."},{"Start":"16:44.100 ","End":"16:48.340","Text":"Then the same for the total force in region"},{"Start":"16:48.340 ","End":"16:52.240","Text":"2 in the y direction so we\u0027re just going to integrate."},{"Start":"16:52.240 ","End":"16:56.740","Text":"The bounds are again from this point here at an angle Theta and"},{"Start":"16:56.740 ","End":"17:01.405","Text":"this point over here at an angle of 90 or in radians Pi divided by 2,"},{"Start":"17:01.405 ","End":"17:08.635","Text":"which when we substitute this in we again have this minus BIR."},{"Start":"17:08.635 ","End":"17:13.735","Text":"What is the total force?"},{"Start":"17:13.735 ","End":"17:15.520","Text":"Let\u0027s scroll down."},{"Start":"17:15.520 ","End":"17:22.030","Text":"F_total, we can say let\u0027s first add the x-direction."},{"Start":"17:22.030 ","End":"17:25.765","Text":"F_1 and F_4 are in the y-direction."},{"Start":"17:25.765 ","End":"17:33.835","Text":"Then we have F_3 in the x-direction and F_2 also in the x-direction."},{"Start":"17:33.835 ","End":"17:44.140","Text":"We have 2BIR minus,"},{"Start":"17:44.140 ","End":"17:51.074","Text":"over here, BIR,"},{"Start":"17:51.074 ","End":"17:57.880","Text":"and this is of course divided by 2 over here."},{"Start":"17:57.880 ","End":"18:00.775","Text":"This is in the x-direction."},{"Start":"18:00.775 ","End":"18:03.815","Text":"Then in the y-direction,"},{"Start":"18:03.815 ","End":"18:06.610","Text":"we have F_1 plus F_4."},{"Start":"18:06.610 ","End":"18:16.135","Text":"We have 2BIL so BIL for F_1 and BIL for F_4."},{"Start":"18:16.135 ","End":"18:19.570","Text":"Then we have 4F_3."},{"Start":"18:19.570 ","End":"18:22.555","Text":"This is of course, in the negative y-direction."},{"Start":"18:22.555 ","End":"18:29.830","Text":"Then our F_3 is also with a negative and this is going to be root 3 divided by 2,"},{"Start":"18:29.830 ","End":"18:32.005","Text":"which cancels out with this 2."},{"Start":"18:32.005 ","End":"18:37.190","Text":"Let\u0027s just write it root 3 over 2 multiplied by 2BIR,"},{"Start":"18:38.190 ","End":"18:40.735","Text":"what we have over here."},{"Start":"18:40.735 ","End":"18:50.530","Text":"Then again, we have this negative BIR and all of this is in the y-direction."},{"Start":"18:50.530 ","End":"18:52.990","Text":"This 2 divided by 2 cancels out."},{"Start":"18:52.990 ","End":"18:58.975","Text":"Then we have BIR minus BIR so we have 0 in the x-direction."},{"Start":"18:58.975 ","End":"19:06.830","Text":"Then here, we have negative 2BIL minus."},{"Start":"19:07.530 ","End":"19:10.315","Text":"Rather let\u0027s do this slow."},{"Start":"19:10.315 ","End":"19:14.470","Text":"This 2 cancels out with this 2 over"},{"Start":"19:14.470 ","End":"19:21.385","Text":"here and then we can move out the minus outside so we have negative."},{"Start":"19:21.385 ","End":"19:24.970","Text":"Then we see that we have BI in all of these."},{"Start":"19:24.970 ","End":"19:28.281","Text":"We have negative BI."},{"Start":"19:28.281 ","End":"19:34.720","Text":"Then here we have 2L then plus,"},{"Start":"19:34.720 ","End":"19:36.220","Text":"because we move the minus out,"},{"Start":"19:36.220 ","End":"19:38.140","Text":"negative root 3R,"},{"Start":"19:38.140 ","End":"19:43.600","Text":"so plus root 3R over here."},{"Start":"19:43.600 ","End":"19:48.550","Text":"Then we have R plus R,"},{"Start":"19:48.550 ","End":"19:51.680","Text":"and all of this is in the y-direction."},{"Start":"19:51.750 ","End":"19:54.925","Text":"This is the final answer."},{"Start":"19:54.925 ","End":"20:00.970","Text":"What we can see is that all of the force is in the negative y-direction."},{"Start":"20:00.970 ","End":"20:04.615","Text":"We don\u0027t have any component in the x-direction."},{"Start":"20:04.615 ","End":"20:08.350","Text":"We can see that the force that we got from"},{"Start":"20:08.350 ","End":"20:10.630","Text":"this weird shaped wire is"},{"Start":"20:10.630 ","End":"20:15.610","Text":"the exact same force that we would get if we just had 1 straight wire."},{"Start":"20:15.610 ","End":"20:22.247","Text":"As we can see this is R. That means that this length over here is also R. This is"},{"Start":"20:22.247 ","End":"20:26.725","Text":"R. Then from trigonometry we can see"},{"Start":"20:26.725 ","End":"20:32.035","Text":"that this length over here would be equal to root 3R."},{"Start":"20:32.035 ","End":"20:35.680","Text":"Here you can just use trig identities and you can calculate that."},{"Start":"20:35.680 ","End":"20:40.360","Text":"What we can see is that the total length of the wire is 2L"},{"Start":"20:40.360 ","End":"20:47.510","Text":"plus R multiplied by 1 plus root 3."},{"Start":"20:48.210 ","End":"20:53.920","Text":"We got BI, which is what we would get here,"},{"Start":"20:53.920 ","End":"20:56.560","Text":"multiplied by the length of the wire,"},{"Start":"20:56.560 ","End":"21:03.250","Text":"which has 2L multiplied by R of 1 plus root 3 which is exactly this."},{"Start":"21:03.250 ","End":"21:07.195","Text":"Of course, it\u0027s in the negative y direction from the right hand rule,"},{"Start":"21:07.195 ","End":"21:13.700","Text":"which is what we saw earlier that the force is pointing in this direction."},{"Start":"21:13.920 ","End":"21:19.975","Text":"That means that instead of doing this whole long calculation where we"},{"Start":"21:19.975 ","End":"21:25.629","Text":"calculated for each region and we have to separate into x components and y components,"},{"Start":"21:25.629 ","End":"21:28.855","Text":"all you have to do is find the total displacement"},{"Start":"21:28.855 ","End":"21:32.800","Text":"from the beginning of the wire until the end of the wire."},{"Start":"21:32.800 ","End":"21:40.420","Text":"Find the total length if we draw a straight line between the beginning and the end."},{"Start":"21:40.420 ","End":"21:46.090","Text":"Then you just plug that into the DL and multiply that by I and B."},{"Start":"21:46.090 ","End":"21:49.570","Text":"Then just use the right-hand rule to see which direction it sends."},{"Start":"21:49.570 ","End":"21:52.390","Text":"Here, specifically the force is pointing down in"},{"Start":"21:52.390 ","End":"21:55.600","Text":"the negative y-direction so you just plug that in."},{"Start":"21:55.600 ","End":"22:00.880","Text":"In other words, if we had a wire that does something like"},{"Start":"22:00.880 ","End":"22:07.325","Text":"so and it has a current flowing through and a B field presence somewhere,"},{"Start":"22:07.325 ","End":"22:08.845","Text":"you take the start,"},{"Start":"22:08.845 ","End":"22:10.285","Text":"this is point A,"},{"Start":"22:10.285 ","End":"22:11.590","Text":"we take the end,"},{"Start":"22:11.590 ","End":"22:18.400","Text":"this is point B, we draw a straight line between A and B."},{"Start":"22:18.400 ","End":"22:22.540","Text":"Let\u0027s not get confused."},{"Start":"22:22.540 ","End":"22:27.910","Text":"Let\u0027s call this Z, from A to Z because they\u0027re B field, so don\u0027t get confused."},{"Start":"22:27.910 ","End":"22:32.248","Text":"We draw a straight line from the start to the finish from A to Z."},{"Start":"22:32.248 ","End":"22:36.985","Text":"We know the magnetic field acting in the region,"},{"Start":"22:36.985 ","End":"22:39.625","Text":"so we can say that the total force is equal to"},{"Start":"22:39.625 ","End":"22:43.735","Text":"the magnetic field acting in the region multiplied by the current,"},{"Start":"22:43.735 ","End":"22:47.455","Text":"multiplied by this length from A to B,"},{"Start":"22:47.455 ","End":"22:51.205","Text":"so multiplied by this Delta x in this case."},{"Start":"22:51.205 ","End":"22:54.220","Text":"Then multiplied by sine of the angle"},{"Start":"22:54.220 ","End":"22:59.810","Text":"between the current or the direction of the wire and the B field."},{"Start":"23:00.630 ","End":"23:08.380","Text":"Here, the wire is along the x-axes and let\u0027s say that this is going into the page,"},{"Start":"23:08.380 ","End":"23:12.760","Text":"so this would just be in the positive y-direction then."},{"Start":"23:12.760 ","End":"23:16.090","Text":"You find the total displacement of the wire."},{"Start":"23:16.090 ","End":"23:19.660","Text":"A straight line from A to B that is your length."},{"Start":"23:19.660 ","End":"23:21.310","Text":"Here it\u0027s Delta x."},{"Start":"23:21.310 ","End":"23:24.850","Text":"Then using the right-hand rule you see the direction of the force,"},{"Start":"23:24.850 ","End":"23:29.920","Text":"which if the B field is pointing into the page and the wire is in"},{"Start":"23:29.920 ","End":"23:36.880","Text":"this direction along the x-axis so the force will be in the positive y-direction."},{"Start":"23:36.880 ","End":"23:42.505","Text":"This is of course, only correct if your B field is uniform throughout."},{"Start":"23:42.505 ","End":"23:50.035","Text":"If this whole area or different parts of the wire are experiencing different B fields,"},{"Start":"23:50.035 ","End":"23:52.433","Text":"so B isn\u0027t a constant throughout"},{"Start":"23:52.433 ","End":"23:54.925","Text":"then you\u0027ll have to do the calculation that we did before."},{"Start":"23:54.925 ","End":"23:57.430","Text":"But if B is constant as we were told in"},{"Start":"23:57.430 ","End":"24:02.485","Text":"this question then we can just do this simple method."},{"Start":"24:02.485 ","End":"24:05.360","Text":"That\u0027s the end of this lesson."}],"ID":22384},{"Watched":false,"Name":"Exercise 5","Duration":"18m 54s","ChapterTopicVideoID":21316,"CourseChapterTopicPlaylistID":99484,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.965","Text":"Hello. In this lesson,"},{"Start":"00:01.965 ","End":"00:04.875","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.875 ","End":"00:10.170","Text":"A particle with mass m and charge of q is bigger than 0,"},{"Start":"00:10.170 ","End":"00:13.980","Text":"so a positive charge enters the center of"},{"Start":"00:13.980 ","End":"00:19.695","Text":"a parallel plate capacitor with a velocity of v_0 in the x-direction."},{"Start":"00:19.695 ","End":"00:26.294","Text":"The capacitor plates are parallel to the xy-plane and their separation distance"},{"Start":"00:26.294 ","End":"00:33.765","Text":"is d. Let\u0027s imagine that this is the direction of the y-axis."},{"Start":"00:33.765 ","End":"00:39.880","Text":"In other words, it\u0027s going into the page like so."},{"Start":"00:40.430 ","End":"00:50.105","Text":"We can see that each plate is parallel like so to the xy-plane."},{"Start":"00:50.105 ","End":"00:56.825","Text":"Also over here the same thing and our particle is moving right in the middle."},{"Start":"00:56.825 ","End":"01:00.995","Text":"The capacitor is connected to a voltage source V,"},{"Start":"01:00.995 ","End":"01:04.270","Text":"where the upper plate is at the higher potential."},{"Start":"01:04.270 ","End":"01:07.070","Text":"Here we have the positive side of"},{"Start":"01:07.070 ","End":"01:11.520","Text":"the voltage source which is pumping positive charges over here."},{"Start":"01:11.520 ","End":"01:15.005","Text":"The top plate is positively charged."},{"Start":"01:15.005 ","End":"01:21.595","Text":"That means that the bottom plate is negatively charged, like so."},{"Start":"01:21.595 ","End":"01:25.520","Text":"If our plates have this charge at separation,"},{"Start":"01:25.520 ","End":"01:29.275","Text":"that means that there\u0027s an E field between them."},{"Start":"01:29.275 ","End":"01:37.160","Text":"As we remember, the equation for the E field between a parallel plate capacitor is equal"},{"Start":"01:37.160 ","End":"01:40.460","Text":"to the voltage across the capacitor divided"},{"Start":"01:40.460 ","End":"01:45.200","Text":"by the distance between the plates or the plate separation."},{"Start":"01:45.200 ","End":"01:50.510","Text":"Question number 1 is to find the distance from the edge of the capacitor."},{"Start":"01:50.510 ","End":"01:54.755","Text":"Let\u0027s call this point x=0,"},{"Start":"01:54.755 ","End":"01:57.650","Text":"where the charge will strike the plate."},{"Start":"01:57.650 ","End":"02:03.425","Text":"We\u0027re not being told if the charge will strike the plate at the top or at the bottom."},{"Start":"02:03.425 ","End":"02:09.130","Text":"But let\u0027s just assume that it will strike the plates somewhere over here at the bottom."},{"Start":"02:09.130 ","End":"02:16.100","Text":"This E field is going to exert a force on this particle,"},{"Start":"02:16.100 ","End":"02:21.140","Text":"which is what is going to make the particle strike 1 of the plates."},{"Start":"02:21.140 ","End":"02:24.485","Text":"This force is of course a vector and it has a direction,"},{"Start":"02:24.485 ","End":"02:30.000","Text":"and it\u0027s equal to the charge multiplied by the E field."},{"Start":"02:30.000 ","End":"02:32.135","Text":"What is the direction of the E field?"},{"Start":"02:32.135 ","End":"02:38.295","Text":"The E field is always in the direction of the plus to the minus."},{"Start":"02:38.295 ","End":"02:42.270","Text":"This is the direction of the E field inside the capacitor."},{"Start":"02:42.270 ","End":"02:45.950","Text":"Of course, it is uniform throughout."},{"Start":"02:45.950 ","End":"02:50.659","Text":"If you want to be reminded how we got this equation and why it\u0027s uniform."},{"Start":"02:50.659 ","End":"02:53.420","Text":"Please go back to the chapter where we speak about"},{"Start":"02:53.420 ","End":"02:58.510","Text":"capacitors and specifically parallel plate capacitors."},{"Start":"02:59.210 ","End":"03:03.980","Text":"Now we can plug in our values over here."},{"Start":"03:03.980 ","End":"03:07.670","Text":"We have that this is equal to q multiplied by the E field,"},{"Start":"03:07.670 ","End":"03:14.360","Text":"which is v divided by d. We can see because it\u0027s going from positive to negative,"},{"Start":"03:14.360 ","End":"03:19.455","Text":"it\u0027s in the negative z-direction."},{"Start":"03:19.455 ","End":"03:22.114","Text":"Now if we have a force,"},{"Start":"03:22.114 ","End":"03:27.020","Text":"we know that we have to have acceleration because force is equal to ma."},{"Start":"03:27.020 ","End":"03:31.595","Text":"We can say that the sum of all of the forces, as we know,"},{"Start":"03:31.595 ","End":"03:36.955","Text":"is equal to mass times acceleration as we saw in mechanics."},{"Start":"03:36.955 ","End":"03:43.145","Text":"We can see that we only have a force in the negative z-direction."},{"Start":"03:43.145 ","End":"03:47.735","Text":"Therefore, we can say that our force in the z-direction"},{"Start":"03:47.735 ","End":"03:53.005","Text":"is equal to mass multiplied by the acceleration in the z-direction."},{"Start":"03:53.005 ","End":"04:00.030","Text":"Let\u0027s write that out. We have qv in the negative divided"},{"Start":"04:00.030 ","End":"04:07.610","Text":"by d is equal to our mass multiplied by our acceleration in the z-direction."},{"Start":"04:07.610 ","End":"04:11.065","Text":"A force in the z-axis is this."},{"Start":"04:11.065 ","End":"04:13.760","Text":"We already are saying this is in the z-direction,"},{"Start":"04:13.760 ","End":"04:15.425","Text":"so we don\u0027t have to include this."},{"Start":"04:15.425 ","End":"04:20.570","Text":"Therefore we can isolate out our acceleration in the z-direction."},{"Start":"04:20.570 ","End":"04:29.395","Text":"Then we get that it\u0027s equal to negative qv divided by md."},{"Start":"04:29.395 ","End":"04:33.530","Text":"Now I\u0027m just going to remind you that this is our V for voltage."},{"Start":"04:33.530 ","End":"04:37.790","Text":"I\u0027m just going to put these lines on top to accentuate that this"},{"Start":"04:37.790 ","End":"04:43.315","Text":"is a V for voltage and not to be confused with velocity."},{"Start":"04:43.315 ","End":"04:48.565","Text":"What we can see is that all of these values over here are constant."},{"Start":"04:48.565 ","End":"04:54.694","Text":"What we can see is that we have constant acceleration and we have an equation for"},{"Start":"04:54.694 ","End":"05:02.240","Text":"the velocity or the position along the z-axis when dealing with constant acceleration."},{"Start":"05:02.240 ","End":"05:06.560","Text":"The position or the displacement along the z axis"},{"Start":"05:06.560 ","End":"05:10.870","Text":"as a function of time is equal to z_naught,"},{"Start":"05:10.870 ","End":"05:13.775","Text":"the initial position along the z-axis,"},{"Start":"05:13.775 ","End":"05:18.185","Text":"plus our velocity in the z-direction_naught."},{"Start":"05:18.185 ","End":"05:22.625","Text":"Our initial velocity in the z-direction multiplied by t"},{"Start":"05:22.625 ","End":"05:30.450","Text":"plus 1/2 the acceleration in the z-direction multiplied by t^2."},{"Start":"05:30.830 ","End":"05:34.865","Text":"I said that this side over here is x is equal to naught."},{"Start":"05:34.865 ","End":"05:38.870","Text":"But let\u0027s imagine that this is in fact the origin over here,"},{"Start":"05:38.870 ","End":"05:45.500","Text":"so 0 in the z-direction is right over here between the 2 plates."},{"Start":"05:45.500 ","End":"05:50.165","Text":"In that case, the initial position is equal to 0."},{"Start":"05:50.165 ","End":"05:55.280","Text":"Our initial velocity in the z-direction is equal to 0."},{"Start":"05:55.280 ","End":"05:58.990","Text":"We only have a velocity in the x-direction?"},{"Start":"05:58.990 ","End":"06:01.410","Text":"In the z-direction is equal to 0."},{"Start":"06:01.410 ","End":"06:04.190","Text":"What we\u0027re left with is the position in"},{"Start":"06:04.190 ","End":"06:07.440","Text":"the z-direction with respect to time is equal to this."},{"Start":"06:07.440 ","End":"06:11.960","Text":"We have 1/2 multiplied by the acceleration, multiplied by t^2."},{"Start":"06:11.960 ","End":"06:14.240","Text":"The acceleration is a negative,"},{"Start":"06:14.240 ","End":"06:19.190","Text":"so we have negative 1/2q multiplied"},{"Start":"06:19.190 ","End":"06:25.350","Text":"by the voltage divided by md multiplied by t^2."},{"Start":"06:26.270 ","End":"06:29.840","Text":"We\u0027ve already seen that our force is in"},{"Start":"06:29.840 ","End":"06:34.565","Text":"the negative z-direction and our acceleration is also in the negative z-direction."},{"Start":"06:34.565 ","End":"06:37.370","Text":"We can see that our charge q will in"},{"Start":"06:37.370 ","End":"06:40.925","Text":"fact be hitting the bottom plate rather than the top."},{"Start":"06:40.925 ","End":"06:45.350","Text":"What we want to see is at what time that will happen."},{"Start":"06:45.350 ","End":"06:50.065","Text":"Then we can figure out the exposition that it happens at."},{"Start":"06:50.065 ","End":"06:54.110","Text":"What we want is we want to find the time that z is"},{"Start":"06:54.110 ","End":"06:58.115","Text":"equal to negative d divided by 2. Why is that?"},{"Start":"06:58.115 ","End":"07:05.630","Text":"The bottom plate is located over here at z is equal to negative d divided by 2."},{"Start":"07:05.630 ","End":"07:08.360","Text":"If z=0 over here,"},{"Start":"07:08.360 ","End":"07:10.985","Text":"and the total distance is d,"},{"Start":"07:10.985 ","End":"07:13.760","Text":"then that means that if this is 0,"},{"Start":"07:13.760 ","End":"07:18.030","Text":"this is negative d divided by 2."},{"Start":"07:18.410 ","End":"07:24.950","Text":"We want that z will be equal to negative d divided by 2."},{"Start":"07:24.950 ","End":"07:27.590","Text":"Then we\u0027ll just equate this to this equation over"},{"Start":"07:27.590 ","End":"07:30.410","Text":"here and isolate out our t. This is equal to"},{"Start":"07:30.410 ","End":"07:38.930","Text":"negative 1/2q multiplied by the voltage divided by md t^2."},{"Start":"07:38.930 ","End":"07:45.790","Text":"First of all, we can see that the halves cancel out and the negatives cancel out."},{"Start":"07:45.790 ","End":"07:48.770","Text":"Now we\u0027re just going to isolate out t. What we get is"},{"Start":"07:48.770 ","End":"07:52.220","Text":"that t is equal to the square root of"},{"Start":"07:52.220 ","End":"08:00.580","Text":"m multiplied by d^2 divided by q multiplied by the voltage."},{"Start":"08:01.430 ","End":"08:05.750","Text":"We know that it takes this time for the particle"},{"Start":"08:05.750 ","End":"08:09.120","Text":"to travel from here and hit the bottom plate."},{"Start":"08:09.120 ","End":"08:15.890","Text":"What we want to know is what distance in the x-direction this is until it hits the plate."},{"Start":"08:15.890 ","End":"08:24.820","Text":"What we know is that displacement is equal to velocity multiplied by time."},{"Start":"08:24.820 ","End":"08:28.970","Text":"In the x-direction, we\u0027re looking for x displacement;"},{"Start":"08:28.970 ","End":"08:30.949","Text":"its velocity in the x-direction,"},{"Start":"08:30.949 ","End":"08:35.030","Text":"which is v_naught multiplied by time which we have here."},{"Start":"08:35.030 ","End":"08:40.025","Text":"What we have is v_naught multiplied by the square root"},{"Start":"08:40.025 ","End":"08:46.735","Text":"of md^2 divided by q multiplied by the voltage,"},{"Start":"08:46.735 ","End":"08:49.515","Text":"and the square root of all of this."},{"Start":"08:49.515 ","End":"08:53.990","Text":"This is the distance away from the edge of"},{"Start":"08:53.990 ","End":"08:59.100","Text":"the plate that our particle will hit the capacitor plate."},{"Start":"08:59.450 ","End":"09:02.745","Text":"That\u0027s the answer to Question number 1."},{"Start":"09:02.745 ","End":"09:05.170","Text":"Now let\u0027s answer Question number 2."},{"Start":"09:05.170 ","End":"09:07.490","Text":"Now we\u0027re assuming that the capacitor is"},{"Start":"09:07.490 ","End":"09:12.305","Text":"discharged and it\u0027s not connected to the voltage source."},{"Start":"09:12.305 ","End":"09:15.510","Text":"Let\u0027s just rub all of this out."},{"Start":"09:15.700 ","End":"09:18.410","Text":"If our capacitor is discharged,"},{"Start":"09:18.410 ","End":"09:21.395","Text":"there\u0027s also no E field over here."},{"Start":"09:21.395 ","End":"09:25.610","Text":"Now we\u0027re being told that a uniform magnetic field B is equal"},{"Start":"09:25.610 ","End":"09:29.990","Text":"to B_0 in the y-direction is present in the space."},{"Start":"09:29.990 ","End":"09:33.935","Text":"The y-direction, as we saw is going into the page."},{"Start":"09:33.935 ","End":"09:36.690","Text":"Let\u0027s draw it in blue."},{"Start":"09:36.690 ","End":"09:44.515","Text":"We have a uniform B field going like so into the page which is the y-direction."},{"Start":"09:44.515 ","End":"09:50.200","Text":"Find the distance from the edge of the capacitor where the charge will strike the plate."},{"Start":"09:50.200 ","End":"09:53.510","Text":"We\u0027re answering the same question just this time we have"},{"Start":"09:53.510 ","End":"09:57.940","Text":"a magnetic field rather than an electric field."},{"Start":"09:57.940 ","End":"10:00.710","Text":"If we have a magnetic field,"},{"Start":"10:00.710 ","End":"10:06.585","Text":"that means that we have a magnetic force acting on the particle."},{"Start":"10:06.585 ","End":"10:08.735","Text":"The equation for this force is,"},{"Start":"10:08.735 ","End":"10:10.985","Text":"which we saw earlier in this chapter."},{"Start":"10:10.985 ","End":"10:16.820","Text":"It\u0027s equal to q multiplied by v the velocity,"},{"Start":"10:16.820 ","End":"10:23.810","Text":"cross-product with the B field."},{"Start":"10:23.810 ","End":"10:26.180","Text":"In our case we have q,"},{"Start":"10:26.180 ","End":"10:30.215","Text":"our velocity is v_0 in the x-direction."},{"Start":"10:30.215 ","End":"10:32.630","Text":"We have v_0 in the x-direction,"},{"Start":"10:32.630 ","End":"10:39.375","Text":"cross-product with our B field which is equal to B_naught in the y-direction."},{"Start":"10:39.375 ","End":"10:43.160","Text":"Now what we can do is we can put all of our constants on 1 side."},{"Start":"10:43.160 ","End":"10:46.805","Text":"We have qv_naught, B_naught,"},{"Start":"10:46.805 ","End":"10:50.315","Text":"and then we have x cross"},{"Start":"10:50.315 ","End":"10:58.395","Text":"y and x-hat cross y-hat gives us z-hat."},{"Start":"10:58.395 ","End":"11:07.385","Text":"Therefore, what we get is that our force is equal to qv_naught,"},{"Start":"11:07.385 ","End":"11:13.115","Text":"B_naught in the z-direction."},{"Start":"11:13.115 ","End":"11:18.640","Text":"Here we see that our force is in the positive z-direction,"},{"Start":"11:18.640 ","End":"11:21.404","Text":"which means that here,"},{"Start":"11:21.404 ","End":"11:25.469","Text":"rather than our particle being deflected downwards,"},{"Start":"11:25.469 ","End":"11:27.255","Text":"which is what we saw with the E field."},{"Start":"11:27.255 ","End":"11:32.500","Text":"It\u0027s actually going to be deflected upwards here now"},{"Start":"11:32.500 ","End":"11:37.825","Text":"that we have a magnetic field and so that\u0027s the first difference in this question,"},{"Start":"11:37.825 ","End":"11:40.900","Text":"the second difference is that with the E field,"},{"Start":"11:40.900 ","End":"11:45.655","Text":"we saw that we had a constant acceleration."},{"Start":"11:45.655 ","End":"11:49.060","Text":"Now, the difference that we have here with the B field,"},{"Start":"11:49.060 ","End":"11:51.010","Text":"which you have to remember all the time."},{"Start":"11:51.010 ","End":"11:53.695","Text":"We spoke about it in the past few lessons."},{"Start":"11:53.695 ","End":"11:57.400","Text":"It\u0027s that our force is constant."},{"Start":"11:57.400 ","End":"11:59.395","Text":"As we can see, however,"},{"Start":"11:59.395 ","End":"12:03.910","Text":"what we have is that a particle moving in a magnetic field is"},{"Start":"12:03.910 ","End":"12:08.380","Text":"going to exhibit circular motion and we spoke about that because"},{"Start":"12:08.380 ","End":"12:12.745","Text":"the B field is always going to be perpendicular to the direction"},{"Start":"12:12.745 ","End":"12:18.010","Text":"of the velocity so our particle is going to move in circular motion."},{"Start":"12:18.010 ","End":"12:23.290","Text":"We saw that the radius of this circle being mapped out is equal to M"},{"Start":"12:23.290 ","End":"12:29.710","Text":"multiplied by the velocity divided by q multiplied by the B field."},{"Start":"12:29.710 ","End":"12:34.315","Text":"In our case, the velocity is v_naught and the B field is B_naught."},{"Start":"12:34.315 ","End":"12:38.560","Text":"Of course, the velocity over here is the component of"},{"Start":"12:38.560 ","End":"12:42.400","Text":"the velocity which is perpendicular to the magnetic field."},{"Start":"12:42.400 ","End":"12:45.820","Text":"Here specifically, our velocity is only in"},{"Start":"12:45.820 ","End":"12:50.290","Text":"the x-direction and our magnetic field is only in the y-direction,"},{"Start":"12:50.290 ","End":"12:55.120","Text":"which means that our velocity is already perpendicular to the b field."},{"Start":"12:55.120 ","End":"12:58.735","Text":"But if it had a component also in the y-direction,"},{"Start":"12:58.735 ","End":"13:04.400","Text":"so we would only take this x component which is perpendicular."},{"Start":"13:04.980 ","End":"13:09.850","Text":"Seeing as this is constantly changing direction,"},{"Start":"13:09.850 ","End":"13:12.880","Text":"which means that the force is constantly changing direction."},{"Start":"13:12.880 ","End":"13:17.330","Text":"The force is going to be acting like so."},{"Start":"13:18.000 ","End":"13:27.535","Text":"We\u0027re going to see that our acceleration isn\u0027t constant because it is changing direction."},{"Start":"13:27.535 ","End":"13:33.140","Text":"That means that we\u0027re going to use this equation in order to solve this question."},{"Start":"13:33.750 ","End":"13:40.990","Text":"Here we have the radius of our circular motion due to this B field."},{"Start":"13:40.990 ","End":"13:44.170","Text":"Now what we want to do is we want to"},{"Start":"13:44.170 ","End":"13:48.970","Text":"calculate this displacement in the x-direction over here."},{"Start":"13:48.970 ","End":"13:56.875","Text":"First of all, let\u0027s imagine that this radius is greater than d divided by 2."},{"Start":"13:56.875 ","End":"14:00.175","Text":"From here until here,"},{"Start":"14:00.175 ","End":"14:03.325","Text":"we have a distance of d divided by 2."},{"Start":"14:03.325 ","End":"14:07.000","Text":"Let\u0027s say that our radius is up until here."},{"Start":"14:07.000 ","End":"14:11.035","Text":"This is our radius I,"},{"Start":"14:11.035 ","End":"14:13.900","Text":"which we calculated over here."},{"Start":"14:13.900 ","End":"14:21.830","Text":"Let\u0027s say that this over here is the center of the circle."},{"Start":"14:22.590 ","End":"14:31.885","Text":"We have a radius r and this is traveling like so in a circle around."},{"Start":"14:31.885 ","End":"14:35.965","Text":"I know this doesn\u0027t look exactly like a circle but let\u0027s just"},{"Start":"14:35.965 ","End":"14:40.960","Text":"imagine that this is a perfect circle where here is the center of the circle."},{"Start":"14:40.960 ","End":"14:44.770","Text":"What we know is that this distance"},{"Start":"14:44.770 ","End":"14:48.729","Text":"over here is what we\u0027re trying to find out what this distances,"},{"Start":"14:48.729 ","End":"14:51.265","Text":"and currently we\u0027re calling it x."},{"Start":"14:51.265 ","End":"14:57.670","Text":"If we draw a line from our point over here to the center of the circle,"},{"Start":"14:57.670 ","End":"14:59.425","Text":"we know that this is of course,"},{"Start":"14:59.425 ","End":"15:02.690","Text":"the radius of the circle."},{"Start":"15:04.080 ","End":"15:07.510","Text":"We also know that this height,"},{"Start":"15:07.510 ","End":"15:09.640","Text":"this distance over here,"},{"Start":"15:09.640 ","End":"15:15.370","Text":"is equal to the radius minus d divided by 2."},{"Start":"15:15.370 ","End":"15:18.310","Text":"This length over here is r minus d"},{"Start":"15:18.310 ","End":"15:22.550","Text":"divided by 2 and then we\u0027re left with this space over here."},{"Start":"15:23.220 ","End":"15:29.155","Text":"Now from Pythagoras, we can say that this length over here,"},{"Start":"15:29.155 ","End":"15:33.790","Text":"x^2 is equal to the hypotenuse squared,"},{"Start":"15:33.790 ","End":"15:38.620","Text":"which is I^2 minus this side over here squared"},{"Start":"15:38.620 ","End":"15:45.235","Text":"minus r minus d divided by 2^2."},{"Start":"15:45.235 ","End":"15:50.905","Text":"All we have to do now is plug in our values for r and of course, all of this,"},{"Start":"15:50.905 ","End":"15:54.670","Text":"we\u0027re given in the question and d divided by 2,"},{"Start":"15:54.670 ","End":"16:00.440","Text":"we\u0027re also given and then we just isolate out x and we have our answer."},{"Start":"16:01.650 ","End":"16:08.900","Text":"I just took the square root of everything and that is our value for x."},{"Start":"16:08.900 ","End":"16:12.045","Text":"Now let\u0027s take a look at Question number 3."},{"Start":"16:12.045 ","End":"16:17.190","Text":"In which direction will the charge be deflected if the capacitor is"},{"Start":"16:17.190 ","End":"16:23.710","Text":"connected to the voltage source and the magnetic field is present."},{"Start":"16:24.680 ","End":"16:29.730","Text":"Now we have this B field going in the y-direction,"},{"Start":"16:29.730 ","End":"16:36.375","Text":"and we also have our E field going in the negative z-direction."},{"Start":"16:36.375 ","End":"16:45.090","Text":"We already saw from Question number 1 that the force due to the E field is pointing in"},{"Start":"16:45.090 ","End":"16:51.490","Text":"this direction and we already saw in Question Number 2 that the force"},{"Start":"16:51.490 ","End":"16:59.330","Text":"due to the B field is pointing in this direction."},{"Start":"17:00.570 ","End":"17:06.010","Text":"Now what we can do is we can see that both forces are along these"},{"Start":"17:06.010 ","End":"17:10.915","Text":"z axis so we can write out an equation for the sum"},{"Start":"17:10.915 ","End":"17:16.270","Text":"of all of the forces along the z axis and"},{"Start":"17:16.270 ","End":"17:22.285","Text":"this is going to be equal to the force in the positive z-direction,"},{"Start":"17:22.285 ","End":"17:23.890","Text":"which we saw is this."},{"Start":"17:23.890 ","End":"17:26.635","Text":"We have qv_naught,"},{"Start":"17:26.635 ","End":"17:32.680","Text":"B_naught minus the force over here,"},{"Start":"17:32.680 ","End":"17:34.945","Text":"which is in the negative z-direction."},{"Start":"17:34.945 ","End":"17:37.615","Text":"That\u0027s why it\u0027s with a minus."},{"Start":"17:37.615 ","End":"17:46.435","Text":"We see that this is equal to so negative q multiplied by the voltage"},{"Start":"17:46.435 ","End":"17:54.895","Text":"divided by d. The sum of the forces on the z-axis is the magnetic field."},{"Start":"17:54.895 ","End":"18:00.160","Text":"The force due to the magnetic field plus the force due to the electric field."},{"Start":"18:00.160 ","End":"18:02.380","Text":"Where the force due to the electric field is"},{"Start":"18:02.380 ","End":"18:06.055","Text":"a negative force because it\u0027s in the negative direction."},{"Start":"18:06.055 ","End":"18:08.845","Text":"Now we can take q out."},{"Start":"18:08.845 ","End":"18:12.685","Text":"What we have is q multiplied V_0,"},{"Start":"18:12.685 ","End":"18:20.275","Text":"B_0 minus the voltage divided by d. From this equation,"},{"Start":"18:20.275 ","End":"18:28.060","Text":"we can see that if the voltage divided by the distance between the 2 plates is bigger,"},{"Start":"18:28.060 ","End":"18:31.465","Text":"the magnitude of this is bigger than V_0, B_0,"},{"Start":"18:31.465 ","End":"18:34.690","Text":"the charge will go down because that means that will"},{"Start":"18:34.690 ","End":"18:38.830","Text":"go into the negative realm and no charge will go like so."},{"Start":"18:38.830 ","End":"18:43.930","Text":"But if the voltage divided by the distance is smaller than V_0,"},{"Start":"18:43.930 ","End":"18:47.110","Text":"B_0, then the charge will go,"},{"Start":"18:47.110 ","End":"18:50.350","Text":"will be deflected upwards and that\u0027s it."},{"Start":"18:50.350 ","End":"18:55.340","Text":"That\u0027s the answer to Question Number 3 and that is the end of this lesson."}],"ID":21408},{"Watched":false,"Name":"Exercise 6","Duration":"5m 4s","ChapterTopicVideoID":21530,"CourseChapterTopicPlaylistID":99484,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.800","Text":"Hello. In this lesson,"},{"Start":"00:01.800 ","End":"00:04.470","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.470 ","End":"00:09.825","Text":"We have a proton which is entering a uniform magnetic field."},{"Start":"00:09.825 ","End":"00:11.715","Text":"Let\u0027s draw that over here."},{"Start":"00:11.715 ","End":"00:18.570","Text":"We\u0027re told that the magnetic field has a magnitude of 0.15 Teslas"},{"Start":"00:18.570 ","End":"00:27.090","Text":"and the proton is traveling at an angle of 30 degrees to the field."},{"Start":"00:27.090 ","End":"00:31.233","Text":"This angle over here is 30 degrees,"},{"Start":"00:31.233 ","End":"00:32.760","Text":"this is our protons,"},{"Start":"00:32.760 ","End":"00:35.325","Text":"so it has this positive charge,"},{"Start":"00:35.325 ","End":"00:42.875","Text":"and we\u0027re told the magnitude of its velocity is 10^6 meters per second."},{"Start":"00:42.875 ","End":"00:44.390","Text":"We\u0027re given in the question,"},{"Start":"00:44.390 ","End":"00:46.640","Text":"the values for the proton\u0027s charge and mass."},{"Start":"00:46.640 ","End":"00:52.740","Text":"We\u0027re being asked to find the radius of rotation of the proton."},{"Start":"00:52.820 ","End":"01:03.770","Text":"First of all, let\u0027s remember that a charged particle which is moving in a magnetic field,"},{"Start":"01:03.770 ","End":"01:08.600","Text":"where its velocity is perpendicular to the magnetic field,"},{"Start":"01:08.600 ","End":"01:17.510","Text":"the charged particle will move in a circular motion and it will have some radius,"},{"Start":"01:17.510 ","End":"01:21.350","Text":"R, over here, which is what we\u0027re trying to find."},{"Start":"01:21.350 ","End":"01:27.740","Text":"First of all, we can see that we\u0027re given some velocity over here."},{"Start":"01:27.740 ","End":"01:32.195","Text":"We can see that it\u0027s at 30 degrees to the magnetic field,"},{"Start":"01:32.195 ","End":"01:35.855","Text":"which means that it has a perpendicular component."},{"Start":"01:35.855 ","End":"01:38.120","Text":"That\u0027s this over here,"},{"Start":"01:38.120 ","End":"01:39.950","Text":"the perpendicular component,"},{"Start":"01:39.950 ","End":"01:45.825","Text":"and it has over here the parallel component of the magnetic field."},{"Start":"01:45.825 ","End":"01:47.750","Text":"In previous lessons, we saw that"},{"Start":"01:47.750 ","End":"01:52.490","Text":"the parallel component towards the magnetic field doesn\u0027t"},{"Start":"01:52.490 ","End":"02:01.190","Text":"change anything in the question when dealing with the force on this particle."},{"Start":"02:01.190 ","End":"02:06.780","Text":"However, the perpendicular component of the velocity does."},{"Start":"02:07.190 ","End":"02:13.445","Text":"We saw that the force on such a particle is equal to,"},{"Start":"02:13.445 ","End":"02:16.760","Text":"in general, it\u0027s equal to the charge multiplied by"},{"Start":"02:16.760 ","End":"02:20.765","Text":"the velocity cross product with the magnetic fields."},{"Start":"02:20.765 ","End":"02:22.685","Text":"Or in other words,"},{"Start":"02:22.685 ","End":"02:25.865","Text":"it\u0027s equal to the charge multiplied by"},{"Start":"02:25.865 ","End":"02:32.910","Text":"the perpendicular component of the velocity multiplied by the magnetic field."},{"Start":"02:32.910 ","End":"02:35.780","Text":"Then this will exhibit circular motion."},{"Start":"02:35.780 ","End":"02:38.720","Text":"If this is exhibiting circular motion,"},{"Start":"02:38.720 ","End":"02:41.570","Text":"then it has to abide by the equation for the force on"},{"Start":"02:41.570 ","End":"02:45.460","Text":"a particle in circular motion, which is mv^2."},{"Start":"02:45.460 ","End":"02:48.545","Text":"Of course, here it\u0027s the perpendicular velocity as well,"},{"Start":"02:48.545 ","End":"02:53.605","Text":"divided by the radius over here mapped out."},{"Start":"02:53.605 ","End":"02:56.720","Text":"We saw that if we isolate out our R,"},{"Start":"02:56.720 ","End":"02:58.010","Text":"because that\u0027s what we\u0027re trying to find,"},{"Start":"02:58.010 ","End":"03:00.170","Text":"the radius of rotation,"},{"Start":"03:00.170 ","End":"03:05.760","Text":"we can divide both sides by v perpendicular and what we\u0027re left"},{"Start":"03:05.760 ","End":"03:12.210","Text":"with is mv perpendicular divided by qB."},{"Start":"03:13.520 ","End":"03:18.200","Text":"Now all we have to do is we have to plug in our values."},{"Start":"03:18.200 ","End":"03:21.200","Text":"The mass of the proton is this."},{"Start":"03:21.200 ","End":"03:27.590","Text":"It\u0027s 1.67 times 10^negative 27 kilograms"},{"Start":"03:27.590 ","End":"03:32.455","Text":"multiplied by the perpendicular component of the velocity."},{"Start":"03:32.455 ","End":"03:34.785","Text":"Let\u0027s just take a look."},{"Start":"03:34.785 ","End":"03:38.555","Text":"The perpendicular component of the velocity is"},{"Start":"03:38.555 ","End":"03:42.935","Text":"equal to the magnitude of the velocity that we\u0027re given over here,"},{"Start":"03:42.935 ","End":"03:51.350","Text":"multiplied by sine of the angle between the magnetic field and the velocity vector,"},{"Start":"03:51.350 ","End":"03:54.610","Text":"which we were told is equal to 30 degrees."},{"Start":"03:54.610 ","End":"04:03.110","Text":"What we\u0027re going to get is that this is simply equal to 10^6 multiplied by,"},{"Start":"04:03.110 ","End":"04:06.590","Text":"so sine of 30 degrees is 0.5."},{"Start":"04:06.590 ","End":"04:12.870","Text":"This is multiplied by 10^6, multiplied by 0.5."},{"Start":"04:12.870 ","End":"04:18.095","Text":"This is v perpendicular and this is divided by the charge of a proton,"},{"Start":"04:18.095 ","End":"04:20.990","Text":"which we\u0027re given in the question as 1.6,"},{"Start":"04:20.990 ","End":"04:28.880","Text":"times 10^negative 19 coulombs multiplied by the magnitude of the magnetic field,"},{"Start":"04:28.880 ","End":"04:36.140","Text":"which we\u0027re being told is equal to 0.15 Teslas."},{"Start":"04:36.140 ","End":"04:39.075","Text":"Once we plug all of this in,"},{"Start":"04:39.075 ","End":"04:43.400","Text":"we\u0027ll get that the radius is approximately equal to"},{"Start":"04:43.400 ","End":"04:51.480","Text":"3.48 times 10^negative 2 meters."},{"Start":"04:52.090 ","End":"04:56.240","Text":"This is the radius of the circle mapped out by"},{"Start":"04:56.240 ","End":"05:01.075","Text":"the circular motion of this particle traveling in this magnetic field."},{"Start":"05:01.075 ","End":"05:03.920","Text":"That is the end of this lesson."}],"ID":22385},{"Watched":false,"Name":"Exercise 7","Duration":"26m 55s","ChapterTopicVideoID":21531,"CourseChapterTopicPlaylistID":99484,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.830","Text":"Hello. In this lesson,"},{"Start":"00:01.830 ","End":"00:04.425","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.425 ","End":"00:08.970","Text":"A suspended square loop with a side length of L and"},{"Start":"00:08.970 ","End":"00:15.015","Text":"mass m is suspended from the x-axis and is free to rotate around it."},{"Start":"00:15.015 ","End":"00:18.000","Text":"This side of the square loop is attached to"},{"Start":"00:18.000 ","End":"00:23.550","Text":"the x-axis and it can just rotate around this axis."},{"Start":"00:23.550 ","End":"00:28.650","Text":"The loop carries a current of I and the current is flowing"},{"Start":"00:28.650 ","End":"00:34.325","Text":"through the side along the x-axis in the direction of the x-axis."},{"Start":"00:34.325 ","End":"00:38.345","Text":"This is the positive direction of current over here."},{"Start":"00:38.345 ","End":"00:44.730","Text":"Then it flows down over here and then around like so."},{"Start":"00:46.220 ","End":"00:51.290","Text":"This is the 3-dimensional view and in 2-dimensions."},{"Start":"00:51.290 ","End":"00:56.005","Text":"If we\u0027re looking down the x-axis,"},{"Start":"00:56.005 ","End":"01:00.875","Text":"what we\u0027ll see is just this side of the square loop,"},{"Start":"01:00.875 ","End":"01:02.450","Text":"which is this over here."},{"Start":"01:02.450 ","End":"01:10.470","Text":"Which means that over here the current is flowing in this downwards diagonal direction."},{"Start":"01:10.820 ","End":"01:17.720","Text":"Question number 1 is to find the magnitude of the magnetic fields required to act in"},{"Start":"01:17.720 ","End":"01:21.080","Text":"the z-axis direction to bring the loop to"},{"Start":"01:21.080 ","End":"01:26.370","Text":"rest at an angle of Theta relative to the z-axis."},{"Start":"01:27.200 ","End":"01:30.275","Text":"What does this question mean?"},{"Start":"01:30.275 ","End":"01:40.100","Text":"We\u0027re being asked what magnetic field B would have to be acting in the z-direction such"},{"Start":"01:40.100 ","End":"01:46.430","Text":"that our square loop will be at rest at this angle of Theta so that"},{"Start":"01:46.430 ","End":"01:53.855","Text":"our square loop will stay at this angle Theta and won\u0027t carry on rotating around."},{"Start":"01:53.855 ","End":"01:58.330","Text":"We can say that our B field is equal"},{"Start":"01:58.330 ","End":"02:04.530","Text":"to some value B naught and it\u0027s acting in the z-direction."},{"Start":"02:04.530 ","End":"02:08.335","Text":"What we want to do is we want to find what this B naught is,"},{"Start":"02:08.335 ","End":"02:15.430","Text":"such that our loop will stay in this position and we\u0027ll keep on rotating."},{"Start":"02:16.820 ","End":"02:19.000","Text":"Let\u0027s for 1 moment,"},{"Start":"02:19.000 ","End":"02:22.810","Text":"imagine that we don\u0027t have this magnetic field."},{"Start":"02:22.810 ","End":"02:32.215","Text":"What we will have is our gravitational force acting on this wire over here."},{"Start":"02:32.215 ","End":"02:36.265","Text":"We\u0027ll have mg acting in this direction over here."},{"Start":"02:36.265 ","End":"02:42.935","Text":"This force mg is going to produce a torque or a moment,"},{"Start":"02:42.935 ","End":"02:49.005","Text":"which is going to try and rotate the loop in this direction, like so."},{"Start":"02:49.005 ","End":"02:50.570","Text":"Just as a reminder,"},{"Start":"02:50.570 ","End":"02:53.120","Text":"the equation for torque,"},{"Start":"02:53.120 ","End":"02:55.070","Text":"which we learned in mechanics,"},{"Start":"02:55.070 ","End":"03:03.225","Text":"is equal to the radius cross-product with the force."},{"Start":"03:03.225 ","End":"03:07.025","Text":"So mg, the force is going to cause a torque"},{"Start":"03:07.025 ","End":"03:11.255","Text":"which is trying to rotate our loop in this direction."},{"Start":"03:11.255 ","End":"03:16.325","Text":"What we\u0027re trying to find is what magnetic field is going to produce"},{"Start":"03:16.325 ","End":"03:22.475","Text":"a magnetic force that will cancel out this rotation."},{"Start":"03:22.475 ","End":"03:28.380","Text":"The rotation from the magnetic fields will have to be equal and opposite."},{"Start":"03:28.380 ","End":"03:31.185","Text":"They will cancel each other out."},{"Start":"03:31.185 ","End":"03:35.600","Text":"Then our loop will be at rest because there won\u0027t be"},{"Start":"03:35.600 ","End":"03:40.680","Text":"any net torque rotating it in any direction."},{"Start":"03:40.680 ","End":"03:47.660","Text":"What we can say is that the sum of all of our torques is going to be equal to"},{"Start":"03:47.660 ","End":"03:55.760","Text":"the torque due to our magnetic field plus the torque due to our gravitational force."},{"Start":"03:55.760 ","End":"03:59.255","Text":"All of this is going to have to be equal to 0."},{"Start":"03:59.255 ","End":"04:00.755","Text":"You will have to cancel out,"},{"Start":"04:00.755 ","End":"04:04.500","Text":"and then our loop will remain static."},{"Start":"04:04.500 ","End":"04:11.615","Text":"Let\u0027s first calculate the torque for our force due to gravity."},{"Start":"04:11.615 ","End":"04:15.230","Text":"Tau, mg. Now, first of all,"},{"Start":"04:15.230 ","End":"04:20.720","Text":"we know that the torque always acts from the center of gravity."},{"Start":"04:20.720 ","End":"04:22.235","Text":"If we look here,"},{"Start":"04:22.235 ","End":"04:24.530","Text":"this is where our force, mg,"},{"Start":"04:24.530 ","End":"04:26.870","Text":"is going to be acting from"},{"Start":"04:26.870 ","End":"04:32.090","Text":"the center of our square loop because this is a center of gravity."},{"Start":"04:33.790 ","End":"04:41.630","Text":"This is going to be equal to the size of our vector,"},{"Start":"04:41.630 ","End":"04:44.495","Text":"which is the distance from the axis of rotation,"},{"Start":"04:44.495 ","End":"04:48.800","Text":"multiplied by the size of our force."},{"Start":"04:48.800 ","End":"04:55.830","Text":"Then multiplied by sine of the angle between the 2."},{"Start":"04:56.920 ","End":"05:02.300","Text":"What is our distance from the axis of rotation?"},{"Start":"05:02.300 ","End":"05:05.450","Text":"We\u0027re looking at this distance over here."},{"Start":"05:05.450 ","End":"05:10.535","Text":"The distance from the center of gravity to the axes of rotation."},{"Start":"05:10.535 ","End":"05:13.025","Text":"This distance, we can see is 1/2."},{"Start":"05:13.025 ","End":"05:16.370","Text":"It\u0027s equal to 1/2 a side of the square loop."},{"Start":"05:16.370 ","End":"05:19.685","Text":"We\u0027re being told that our loop has sides of length"},{"Start":"05:19.685 ","End":"05:27.820","Text":"L. That means that the magnitude of r vector is equal to half the side of the loop,"},{"Start":"05:27.820 ","End":"05:30.205","Text":"which is L divided by 2."},{"Start":"05:30.205 ","End":"05:34.000","Text":"Then we\u0027re multiplying by the magnitude of the force."},{"Start":"05:34.000 ","End":"05:35.950","Text":"What is the magnitude of the force?"},{"Start":"05:35.950 ","End":"05:45.075","Text":"That\u0027s of course mg. Then we\u0027re looking at sine of the angle between the 2."},{"Start":"05:45.075 ","End":"05:49.250","Text":"What exactly is the angle between the 2?"},{"Start":"05:50.100 ","End":"05:52.765","Text":"Let\u0027s draw that."},{"Start":"05:52.765 ","End":"05:56.365","Text":"First of all, we have this distance r,"},{"Start":"05:56.365 ","End":"05:58.640","Text":"which is going like so,"},{"Start":"05:58.640 ","End":"06:01.260","Text":"this is our vector."},{"Start":"06:01.260 ","End":"06:04.280","Text":"Then we have pointing down,"},{"Start":"06:04.280 ","End":"06:07.740","Text":"we have our force like so."},{"Start":"06:07.740 ","End":"06:13.610","Text":"This is the direction of a force mg. From our equation for torque,"},{"Start":"06:13.610 ","End":"06:17.792","Text":"we can see that we have I cross our force."},{"Start":"06:17.792 ","End":"06:23.880","Text":"That means that we\u0027re rotating in this direction,"},{"Start":"06:23.880 ","End":"06:25.230","Text":"so r cross force."},{"Start":"06:25.230 ","End":"06:30.200","Text":"We go from the r to the direction of the force."},{"Start":"06:30.200 ","End":"06:34.250","Text":"Which, if we use our right-hand rule,"},{"Start":"06:34.250 ","End":"06:44.850","Text":"we will see that we will get that the torque is therefore going into the page."},{"Start":"06:45.200 ","End":"06:51.670","Text":"If you hold your 4 fingers in the direction of r and then you rotate them inwards to"},{"Start":"06:51.670 ","End":"06:55.150","Text":"the direction of the force you\u0027ll see that"},{"Start":"06:55.150 ","End":"07:00.130","Text":"your thumb will be therefore pointing into the page."},{"Start":"07:00.130 ","End":"07:04.945","Text":"We can see that our x-axis is pointing out of the page."},{"Start":"07:04.945 ","End":"07:07.150","Text":"If this is pointing into the page,"},{"Start":"07:07.150 ","End":"07:14.620","Text":"that means that this is in the direction of the negative x-axes. Like so."},{"Start":"07:15.410 ","End":"07:25.450","Text":"Of course, we know that this angle over here between the 2 forces is of course Theta."},{"Start":"07:25.450 ","End":"07:28.970","Text":"Sorry, I skipped that stage. Why is that?"},{"Start":"07:28.970 ","End":"07:31.860","Text":"If this angle is Theta over here,"},{"Start":"07:31.860 ","End":"07:35.010","Text":"we can see that our mg,"},{"Start":"07:35.010 ","End":"07:37.310","Text":"our force due to gravity,"},{"Start":"07:37.310 ","End":"07:41.270","Text":"is parallel to our z-axes over here."},{"Start":"07:41.270 ","End":"07:47.480","Text":"Then we have this over here, these parallel lines,"},{"Start":"07:47.480 ","End":"07:50.090","Text":"which means that if this angle over here is Theta,"},{"Start":"07:50.090 ","End":"07:54.060","Text":"then this angle over here is Theta as well."},{"Start":"07:55.070 ","End":"07:57.410","Text":"This is of course,"},{"Start":"07:57.410 ","End":"08:01.920","Text":"due to corresponding angles."},{"Start":"08:03.410 ","End":"08:11.870","Text":"Now, I want to calculate the moment or the torque due to my magnetic field."},{"Start":"08:11.870 ","End":"08:17.030","Text":"The equation that I\u0027m going to be using first is dF,"},{"Start":"08:17.030 ","End":"08:18.965","Text":"the force on the wire,"},{"Start":"08:18.965 ","End":"08:25.775","Text":"which is equal to the current multiplied by the length of the wire,"},{"Start":"08:25.775 ","End":"08:30.120","Text":"cross-product with my magnetic field."},{"Start":"08:30.500 ","End":"08:39.065","Text":"First of all, let\u0027s look at the force on this side of my loop."},{"Start":"08:39.065 ","End":"08:42.440","Text":"Now, of course, we\u0027re going to have a force over here."},{"Start":"08:42.440 ","End":"08:47.060","Text":"However, when I use the force on this side of"},{"Start":"08:47.060 ","End":"08:52.385","Text":"the wire and I plug it into this equation over here for r cross F,"},{"Start":"08:52.385 ","End":"08:56.615","Text":"r of course being the distance from the axis of rotation."},{"Start":"08:56.615 ","End":"09:02.015","Text":"Because this side of my wire is on the axis of rotation,"},{"Start":"09:02.015 ","End":"09:04.985","Text":"that means that my r vector is equal to 0."},{"Start":"09:04.985 ","End":"09:10.280","Text":"Then I have a 0 vector crossed with y force on this side of the loop."},{"Start":"09:10.280 ","End":"09:12.760","Text":"That is of course equal to 0."},{"Start":"09:12.760 ","End":"09:16.080","Text":"There\u0027s no torque acting on this side."},{"Start":"09:16.080 ","End":"09:21.764","Text":"Then, with respect to the torque on these 2 sides."},{"Start":"09:21.764 ","End":"09:26.840","Text":"We\u0027ll see that the torque on this side cancels out with a torque on this side."},{"Start":"09:26.840 ","End":"09:28.805","Text":"Why is that?"},{"Start":"09:28.805 ","End":"09:33.530","Text":"You can use this equation or the right-hand rule or other."},{"Start":"09:33.530 ","End":"09:36.845","Text":"You\u0027ll see that on this side of the loop,"},{"Start":"09:36.845 ","End":"09:41.715","Text":"force due to the magnetic field is in this direction and on"},{"Start":"09:41.715 ","End":"09:46.470","Text":"this side of the loop of force is in the opposite direction."},{"Start":"09:46.470 ","End":"09:49.659","Text":"Of course, these forces are equal,"},{"Start":"09:49.659 ","End":"09:53.810","Text":"because we have the same current and each side is of the same length."},{"Start":"09:53.810 ","End":"09:56.780","Text":"That means that these forces are equal and opposite."},{"Start":"09:56.780 ","End":"09:59.270","Text":"Which means that when we add up their torques,"},{"Start":"09:59.270 ","End":"10:01.400","Text":"their torques are going to be equal and opposite"},{"Start":"10:01.400 ","End":"10:04.370","Text":"and therefore they will cancel each other out."},{"Start":"10:04.370 ","End":"10:11.350","Text":"All we\u0027re left with is this side of the loop over here."},{"Start":"10:11.350 ","End":"10:14.750","Text":"What we want to to do is we want to work out the force on"},{"Start":"10:14.750 ","End":"10:19.010","Text":"this side and then plug that in to calculate the torque."},{"Start":"10:19.010 ","End":"10:23.015","Text":"We can see that it\u0027s only the side of the loop which is going to have"},{"Start":"10:23.015 ","End":"10:29.550","Text":"any sort of effect on the total torque."},{"Start":"10:31.710 ","End":"10:36.655","Text":"We know that our B,"},{"Start":"10:36.655 ","End":"10:44.755","Text":"our magnetic field is uniform and we know that our I is also uniform, so, therefore,"},{"Start":"10:44.755 ","End":"10:50.140","Text":"we can say that our force on this side of the loop"},{"Start":"10:50.140 ","End":"10:55.825","Text":"is simply equal to the magnitude of the magnetic field multiplied by the current,"},{"Start":"10:55.825 ","End":"10:59.364","Text":"multiplied by the side length,"},{"Start":"10:59.364 ","End":"11:01.090","Text":"which we know is L,"},{"Start":"11:01.090 ","End":"11:10.520","Text":"and then multiplied by sine of the angle between the current and the magnetic field."},{"Start":"11:10.520 ","End":"11:13.385","Text":"Let\u0027s look over here."},{"Start":"11:13.385 ","End":"11:18.280","Text":"We know that this side is this over here."},{"Start":"11:18.280 ","End":"11:20.395","Text":"Now current is going down,"},{"Start":"11:20.395 ","End":"11:23.950","Text":"and then this side when we\u0027re looking down"},{"Start":"11:23.950 ","End":"11:31.240","Text":"the x-axis it\u0027s just going to look like a dot because it goes into the page like so."},{"Start":"11:31.240 ","End":"11:33.550","Text":"Of course, it\u0027s going into the page,"},{"Start":"11:33.550 ","End":"11:36.235","Text":"so let\u0027s just redraw that."},{"Start":"11:36.235 ","End":"11:41.950","Text":"It\u0027s just going into the page like this and we of course"},{"Start":"11:41.950 ","End":"11:47.950","Text":"know that our B field is in this direction and the z axes,"},{"Start":"11:47.950 ","End":"11:51.110","Text":"that\u0027s what we\u0027re being asked in the question."},{"Start":"11:51.570 ","End":"11:59.120","Text":"From this equation, we\u0027re going to point our thumb into the page,"},{"Start":"12:06.390 ","End":"12:11.050","Text":"point out forefinger up in"},{"Start":"12:11.050 ","End":"12:19.300","Text":"this z direction and then our middle finger will point in the direction of the force."},{"Start":"12:19.300 ","End":"12:24.580","Text":"In this case, our middle finger is going to be pointing in"},{"Start":"12:24.580 ","End":"12:30.595","Text":"this rightwards direction and this is our force due to the magnetic field."},{"Start":"12:30.595 ","End":"12:36.565","Text":"Thumb in the direction of the current, so inwards,"},{"Start":"12:36.565 ","End":"12:39.460","Text":"our forefinger in the direction of the B field,"},{"Start":"12:39.460 ","End":"12:44.125","Text":"so upwards and then our middle finger points in the direction of the force."},{"Start":"12:44.125 ","End":"12:52.970","Text":"The force due to the magnetic field is pointing like so in the y direction."},{"Start":"12:55.770 ","End":"12:59.560","Text":"Now we can just write out our force,"},{"Start":"12:59.560 ","End":"13:01.480","Text":"which is of course a vector."},{"Start":"13:01.480 ","End":"13:03.160","Text":"What we have is our B field,"},{"Start":"13:03.160 ","End":"13:05.350","Text":"which we said is some kind of B Naught,"},{"Start":"13:05.350 ","End":"13:08.485","Text":"which we\u0027re trying to find multiplied by the current,"},{"Start":"13:08.485 ","End":"13:10.480","Text":"multiplied by the side length,"},{"Start":"13:10.480 ","End":"13:14.910","Text":"and then sine of the angle between the current and"},{"Start":"13:14.910 ","End":"13:20.740","Text":"the magnetic field so our angle is this over here,"},{"Start":"13:20.740 ","End":"13:23.770","Text":"which we can see is 90 degrees."},{"Start":"13:23.770 ","End":"13:26.920","Text":"Our B field is in the z direction and our force is in"},{"Start":"13:26.920 ","End":"13:31.144","Text":"the y-direction so they are perpendicular to one another."},{"Start":"13:31.144 ","End":"13:38.230","Text":"Of course, we know that sine of 90 is equal to 1 so we multiply this by 1."},{"Start":"13:38.230 ","End":"13:39.340","Text":"As we\u0027ve already said,"},{"Start":"13:39.340 ","End":"13:41.770","Text":"this is in the y direction."},{"Start":"13:41.770 ","End":"13:48.355","Text":"Now we\u0027re ready to calculate the torque due to the magnetic field."},{"Start":"13:48.355 ","End":"13:54.235","Text":"This is of course going to be equal to the magnitude of"},{"Start":"13:54.235 ","End":"14:00.460","Text":"the distance from the axis of rotation so that\u0027s this distance over here,"},{"Start":"14:00.460 ","End":"14:04.975","Text":"which as we can see is equal to the side length so that\u0027s L"},{"Start":"14:04.975 ","End":"14:10.420","Text":"multiplied by the magnitude of the force,"},{"Start":"14:10.420 ","End":"14:16.970","Text":"which is this multiplied by B Naught IL."},{"Start":"14:17.940 ","End":"14:24.550","Text":"Then, this is multiplied by sine of the angle"},{"Start":"14:24.550 ","End":"14:32.725","Text":"between the radius and the force."},{"Start":"14:32.725 ","End":"14:40.690","Text":"What we can see is we can draw the force in this direction like so."},{"Start":"14:40.690 ","End":"14:42.865","Text":"Then our radius, as we saw,"},{"Start":"14:42.865 ","End":"14:51.150","Text":"is going in this direction like so where if this is the z-axis,"},{"Start":"14:51.150 ","End":"14:53.610","Text":"so we know that this angle over here."},{"Start":"14:53.610 ","End":"14:57.285","Text":"Of course, the radius is pointing along this line over here."},{"Start":"14:57.285 ","End":"15:00.180","Text":"This angle here is Theta and of course,"},{"Start":"15:00.180 ","End":"15:03.825","Text":"the angle between our force and the z-axis"},{"Start":"15:03.825 ","End":"15:07.965","Text":"is 90 degrees because our force is in the y-direction,"},{"Start":"15:07.965 ","End":"15:15.080","Text":"so that means that this angle over here is equal to 90 minus Theta."},{"Start":"15:15.080 ","End":"15:20.125","Text":"We can put that this is equal to sine of 90 minus Theta,"},{"Start":"15:20.125 ","End":"15:26.400","Text":"which is also equal to cosine of Theta."},{"Start":"15:26.400 ","End":"15:29.190","Text":"We can rewrite this as"},{"Start":"15:29.190 ","End":"15:38.090","Text":"L^2 B naught I multiplied it by cosine of Theta."},{"Start":"15:38.270 ","End":"15:40.830","Text":"Of course, torque is a vector,"},{"Start":"15:40.830 ","End":"15:46.230","Text":"so it also has direction so we saw that the torque is equal to r cross F."},{"Start":"15:46.230 ","End":"15:53.455","Text":"That means that we have to rotate our fingers from r to F,"},{"Start":"15:53.455 ","End":"15:57.505","Text":"which means that our thumb is going to be pointing"},{"Start":"15:57.505 ","End":"16:01.660","Text":"out of the screen so this is the direction and out of the screen,"},{"Start":"16:01.660 ","End":"16:04.165","Text":"as we saw, is the x-direction."},{"Start":"16:04.165 ","End":"16:11.210","Text":"This is all going to be in the positive x direction."},{"Start":"16:11.610 ","End":"16:15.790","Text":"Now we\u0027ve calculated the torque of our force"},{"Start":"16:15.790 ","End":"16:19.795","Text":"due to gravity and our torque due to the magnetic field."},{"Start":"16:19.795 ","End":"16:24.535","Text":"What we\u0027re going to do is we\u0027re going to plug that back into this equation over here."},{"Start":"16:24.535 ","End":"16:28.105","Text":"Let\u0027s just scroll down a little bit."},{"Start":"16:28.105 ","End":"16:33.415","Text":"What we have is that the sum of all of the torques is equal to,"},{"Start":"16:33.415 ","End":"16:43.090","Text":"so we have L^2 B naught I cosine of Theta in"},{"Start":"16:43.090 ","End":"16:51.580","Text":"the x-direction minus L"},{"Start":"16:51.580 ","End":"16:59.665","Text":"divided by 2mg sine of Theta also in the x-direction."},{"Start":"16:59.665 ","End":"17:01.510","Text":"I just added this,"},{"Start":"17:01.510 ","End":"17:04.240","Text":"but it has a minus coefficient."},{"Start":"17:04.240 ","End":"17:08.695","Text":"In other words, I can just move this over,"},{"Start":"17:08.695 ","End":"17:10.705","Text":"and of course, this is equal to 0."},{"Start":"17:10.705 ","End":"17:19.330","Text":"What I can do is I can just write that L^2 B naught I"},{"Start":"17:19.330 ","End":"17:29.380","Text":"cosine of Theta is equal to L divided by 2 mg sine of Theta."},{"Start":"17:29.380 ","End":"17:33.370","Text":"Now I can divide both sides by L and all I want to do is I want"},{"Start":"17:33.370 ","End":"17:37.299","Text":"to isolate out my B naught because that\u0027s what I\u0027m trying to find,"},{"Start":"17:37.299 ","End":"17:40.030","Text":"the magnitude of the magnetic field."},{"Start":"17:40.030 ","End":"17:50.590","Text":"B naught is simply equal to mg sine of Theta divided by"},{"Start":"17:50.590 ","End":"18:00.250","Text":"2IL cosine of Theta and sine Theta divided by"},{"Start":"18:00.250 ","End":"18:05.320","Text":"cosine Theta is tan Theta so what we\u0027re left with is mg divided"},{"Start":"18:05.320 ","End":"18:13.220","Text":"by 2IL multiplied by tan of Theta."},{"Start":"18:13.740 ","End":"18:18.880","Text":"This is our B naught and we saw that our B field is"},{"Start":"18:18.880 ","End":"18:24.130","Text":"equal to this constant B naught and in the z-direction."},{"Start":"18:24.130 ","End":"18:30.430","Text":"To finish up, we know that B is equal to mg divided by"},{"Start":"18:30.430 ","End":"18:37.990","Text":"2IL tan of Theta in the z-direction."},{"Start":"18:37.990 ","End":"18:41.785","Text":"This is the answer to question number 1,"},{"Start":"18:41.785 ","End":"18:46.135","Text":"which was to find the magnitude of the magnetic field"},{"Start":"18:46.135 ","End":"18:51.175","Text":"required to keep this loop stationary at this angle Theta."},{"Start":"18:51.175 ","End":"18:54.475","Text":"Now, let\u0027s take a look at question number 2."},{"Start":"18:54.475 ","End":"18:57.340","Text":"Now, question number 2 is the exact same thing."},{"Start":"18:57.340 ","End":"19:02.710","Text":"Just this time, our B field is in the y direction."},{"Start":"19:02.710 ","End":"19:08.074","Text":"If we draw our B field like so."},{"Start":"19:08.074 ","End":"19:13.720","Text":"Again, we\u0027re trying to find this B naught just in the y-direction this time"},{"Start":"19:13.720 ","End":"19:21.110","Text":"that will keep our square loop stationary at this angle of Theta."},{"Start":"19:27.810 ","End":"19:32.155","Text":"Will write the changes in the equation."},{"Start":"19:32.155 ","End":"19:40.075","Text":"Now B is 3 equal to B naught but in the y-direction."},{"Start":"19:40.075 ","End":"19:43.075","Text":"We\u0027re again going to use this equation for"},{"Start":"19:43.075 ","End":"19:47.274","Text":"torque and that the sum of all the torques is equal to 0."},{"Start":"19:47.274 ","End":"19:51.085","Text":"Because that means that this will be stationary."},{"Start":"19:51.085 ","End":"19:58.570","Text":"The torque due to the gravitational field is going to remain the same."},{"Start":"19:58.570 ","End":"19:59.980","Text":"This doesn\u0027t change."},{"Start":"19:59.980 ","End":"20:06.805","Text":"But now let\u0027s take a look at what will happen with our torque due to the magnetic field."},{"Start":"20:06.805 ","End":"20:12.865","Text":"First, let\u0027s calculate the size of the force on the wire."},{"Start":"20:12.865 ","End":"20:16.780","Text":"Again, we\u0027re using this equation which becomes this,"},{"Start":"20:16.780 ","End":"20:18.520","Text":"so just the magnitude."},{"Start":"20:18.520 ","End":"20:22.000","Text":"We have a uniform magnetic field multiplied by the current,"},{"Start":"20:22.000 ","End":"20:23.740","Text":"multiplied by the side length,"},{"Start":"20:23.740 ","End":"20:28.633","Text":"multiplied by the angle between the magnetic field and the current,"},{"Start":"20:28.633 ","End":"20:33.835","Text":"and the angle between the magnetic field and the current is still 90 degrees."},{"Start":"20:33.835 ","End":"20:36.175","Text":"There\u0027s still perpendicular to 1 another."},{"Start":"20:36.175 ","End":"20:43.790","Text":"Which means that the magnitude of a force or so over here is equal to BIL."},{"Start":"20:44.100 ","End":"20:47.875","Text":"Now let\u0027s work out the direction."},{"Start":"20:47.875 ","End":"20:50.469","Text":"First of all, just like before,"},{"Start":"20:50.469 ","End":"20:54.595","Text":"we saw that the current is traveling like so."},{"Start":"20:54.595 ","End":"21:00.130","Text":"And it goes down and then it\u0027s going in this direction in 3-dimensional view,"},{"Start":"21:00.130 ","End":"21:03.280","Text":"which means that in 2-dimensional view, just like before,"},{"Start":"21:03.280 ","End":"21:05.485","Text":"it\u0027s going in to the page,"},{"Start":"21:05.485 ","End":"21:08.410","Text":"okay, if we\u0027re looking down the x-axis."},{"Start":"21:08.410 ","End":"21:10.930","Text":"The current is going into the page,"},{"Start":"21:10.930 ","End":"21:15.500","Text":"but the B field is in the rightwards direction."},{"Start":"21:19.650 ","End":"21:23.605","Text":"If this is the current and this is the B field,"},{"Start":"21:23.605 ","End":"21:26.020","Text":"using the right-hand rule we\u0027ll get"},{"Start":"21:26.020 ","End":"21:32.770","Text":"that our force due to the magnetic field is pointing downwards."},{"Start":"21:32.770 ","End":"21:36.445","Text":"You can either point your thumb into the page,"},{"Start":"21:36.445 ","End":"21:38.995","Text":"your forefinger in the right direction,"},{"Start":"21:38.995 ","End":"21:42.160","Text":"and then your middle finger representing the size of"},{"Start":"21:42.160 ","End":"21:45.955","Text":"the force or the direction of the force will be pointing downwards."},{"Start":"21:45.955 ","End":"21:50.410","Text":"Or alternatively, you can point all 4 fingers in"},{"Start":"21:50.410 ","End":"21:55.270","Text":"the direction of the current and then curl them to meet the B field."},{"Start":"21:55.270 ","End":"21:57.910","Text":"And then you\u0027ll see that your thumb is pointing"},{"Start":"21:57.910 ","End":"22:01.225","Text":"downwards representing the direction of the force."},{"Start":"22:01.225 ","End":"22:06.320","Text":"Either way, you\u0027ll get that the force is pointing downwards."},{"Start":"22:06.320 ","End":"22:11.635","Text":"In that case, our force is pointing in the negative z direction."},{"Start":"22:11.635 ","End":"22:16.225","Text":"But we know that our force due to gravity"},{"Start":"22:16.225 ","End":"22:21.745","Text":"mg is also pointing in the negative z direction."},{"Start":"22:21.745 ","End":"22:24.910","Text":"That means that both our force due to"},{"Start":"22:24.910 ","End":"22:30.190","Text":"the magnetic field and a force due to gravity are pointing in the same direction."},{"Start":"22:30.190 ","End":"22:37.060","Text":"Which means that the sum of the torques is never going to be equal to 0."},{"Start":"22:37.060 ","End":"22:40.750","Text":"Here we have a problem that we can solve this question."},{"Start":"22:40.750 ","End":"22:44.050","Text":"How then can we solve the question?"},{"Start":"22:44.050 ","End":"22:48.625","Text":"I\u0027ll just draw arrows to make this a bit clearer."},{"Start":"22:48.625 ","End":"22:55.240","Text":"We have this arrow for mg and in the same direction,"},{"Start":"22:55.240 ","End":"23:00.355","Text":"we have this arrow for torque due to the magnetic field."},{"Start":"23:00.355 ","End":"23:02.035","Text":"How can I solve this question?"},{"Start":"23:02.035 ","End":"23:07.644","Text":"I can solve this if I say that my magnetic field is in the opposite direction,"},{"Start":"23:07.644 ","End":"23:10.330","Text":"in the negative y-direction."},{"Start":"23:10.330 ","End":"23:13.960","Text":"If my magnetic field is in the negative y-direction,"},{"Start":"23:13.960 ","End":"23:18.505","Text":"then I can draw it like so."},{"Start":"23:18.505 ","End":"23:22.555","Text":"Alright, so that means that over here,"},{"Start":"23:22.555 ","End":"23:26.395","Text":"my B field also changes direction."},{"Start":"23:26.395 ","End":"23:31.915","Text":"In that case, my force due to gravity will also be in the opposite direction,"},{"Start":"23:31.915 ","End":"23:34.660","Text":"sorry, my force due to the magnetic field."},{"Start":"23:34.660 ","End":"23:37.344","Text":"FB will be pointing upwards."},{"Start":"23:37.344 ","End":"23:42.830","Text":"Then, this will be like so."},{"Start":"23:43.170 ","End":"23:49.735","Text":"Then the sum of the torques can cancel out and we can solve this question."},{"Start":"23:49.735 ","End":"23:55.160","Text":"Now we know that our B field is in the negative y direction."},{"Start":"23:55.680 ","End":"24:01.705","Text":"Our force due to the magnetic field is in the z direction,"},{"Start":"24:01.705 ","End":"24:04.375","Text":"so we can add that in."},{"Start":"24:04.375 ","End":"24:06.505","Text":"Now let\u0027s calculate our torque."},{"Start":"24:06.505 ","End":"24:08.785","Text":"We\u0027re looking at this over here."},{"Start":"24:08.785 ","End":"24:12.325","Text":"We have our torque due to the magnetic field."},{"Start":"24:12.325 ","End":"24:14.545","Text":"All of this is the same."},{"Start":"24:14.545 ","End":"24:17.920","Text":"However, now if we wanted to look at the angle,"},{"Start":"24:17.920 ","End":"24:19.840","Text":"so first of all,"},{"Start":"24:19.840 ","End":"24:21.670","Text":"if we complete this,"},{"Start":"24:21.670 ","End":"24:30.825","Text":"so we\u0027re looking at this angle over here between our loop and our force."},{"Start":"24:30.825 ","End":"24:35.700","Text":"What we see is that we have this z angle here or alternate angles."},{"Start":"24:35.700 ","End":"24:40.905","Text":"That means that this angle over here is also Theta."},{"Start":"24:40.905 ","End":"24:45.675","Text":"Instead of sine of 90 minus Theta,"},{"Start":"24:45.675 ","End":"24:50.660","Text":"this just remains sine of Theta."},{"Start":"24:50.660 ","End":"24:55.840","Text":"Then, therefore, what we get, let\u0027s write this out."},{"Start":"24:55.840 ","End":"25:01.300","Text":"Our torque due to the magnetic field is equal to"},{"Start":"25:01.300 ","End":"25:09.860","Text":"L^2 B _0 I multiplied by sine of Theta."},{"Start":"25:10.110 ","End":"25:15.265","Text":"And this is still in the x-direction."},{"Start":"25:15.265 ","End":"25:18.550","Text":"Now when we calculate out our sum of forces,"},{"Start":"25:18.550 ","End":"25:24.715","Text":"so here we\u0027re going to have sine of Theta instead of cosine of Theta."},{"Start":"25:24.715 ","End":"25:28.300","Text":"Let\u0027s just rewrite this over here."},{"Start":"25:28.300 ","End":"25:30.070","Text":"I\u0027ll stay in green."},{"Start":"25:30.070 ","End":"25:32.590","Text":"We\u0027re answering question number 2."},{"Start":"25:32.590 ","End":"25:36.910","Text":"We have the sum of forces is equal to"},{"Start":"25:36.910 ","End":"25:43.180","Text":"L^2 B_0I multiplied by sine of Theta."},{"Start":"25:43.180 ","End":"25:45.745","Text":"All of this is in the x-direction, so I\u0027ll leave it."},{"Start":"25:45.745 ","End":"25:47.320","Text":"This is equal to,"},{"Start":"25:47.320 ","End":"25:50.199","Text":"so moving this over to this side of the equation,"},{"Start":"25:50.199 ","End":"25:57.610","Text":"L divided by 2mg sine of Theta."},{"Start":"25:57.610 ","End":"26:00.760","Text":"This L and this L squared will cancel out,"},{"Start":"26:00.760 ","End":"26:04.780","Text":"sine of Theta and sine of Theta will cancel out."},{"Start":"26:04.780 ","End":"26:09.460","Text":"And therefore we will be left with B_0 being equal"},{"Start":"26:09.460 ","End":"26:16.405","Text":"to mg divided by LB_0I."},{"Start":"26:16.405 ","End":"26:20.275","Text":"And then here we have a 2, so 2L,"},{"Start":"26:20.275 ","End":"26:26.630","Text":"sorry, there\u0027s B_0 obviously stays over here because we\u0027re isolating out 2LI."},{"Start":"26:28.800 ","End":"26:32.275","Text":"Then our answer to question number 2,"},{"Start":"26:32.275 ","End":"26:36.610","Text":"because we remember that our magnetic field is in the negative y-direction."},{"Start":"26:36.610 ","End":"26:39.580","Text":"Our answer for the magnetic field is equal to"},{"Start":"26:39.580 ","End":"26:45.805","Text":"negative mg divided by 2LI in the y direction."},{"Start":"26:45.805 ","End":"26:49.360","Text":"We can see that there\u0027s no dependence on"},{"Start":"26:49.360 ","End":"26:52.990","Text":"this angle when the B field is in the y-direction."},{"Start":"26:52.990 ","End":"26:56.390","Text":"That is the end of this lesson."}],"ID":22386},{"Watched":false,"Name":"Exercise 8","Duration":"23m 3s","ChapterTopicVideoID":21532,"CourseChapterTopicPlaylistID":99484,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.710","Text":"Hello. In this lesson,"},{"Start":"00:01.710 ","End":"00:04.050","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.050 ","End":"00:12.330","Text":"A proton is accelerated with ME capacitor subject to a voltage of 10^5 volts."},{"Start":"00:12.330 ","End":"00:17.369","Text":"It then passes through a uniform magnetic field before striking"},{"Start":"00:17.369 ","End":"00:22.770","Text":"a screen located a distance of 15 centimeters from the capacitor."},{"Start":"00:22.770 ","End":"00:28.545","Text":"The magnetic field is of intensity 0.2 Teslas."},{"Start":"00:28.545 ","End":"00:31.326","Text":"Question Number 1 is,"},{"Start":"00:31.326 ","End":"00:38.140","Text":"find the horizontal distance traversed by the proton before striking the screen."},{"Start":"00:38.380 ","End":"00:42.710","Text":"Our proton reaches this point over here."},{"Start":"00:42.710 ","End":"00:49.445","Text":"It\u0027s exited the capacitor and then it has some velocity like so."},{"Start":"00:49.445 ","End":"00:52.354","Text":"Now, as we\u0027ve seen in previous lessons,"},{"Start":"00:52.354 ","End":"00:54.980","Text":"due to it having a velocity going in"},{"Start":"00:54.980 ","End":"00:59.840","Text":"this direction and the B field coming out of the screen,"},{"Start":"00:59.840 ","End":"01:05.660","Text":"we can see that the velocity and the B field are perpendicular to one another."},{"Start":"01:05.660 ","End":"01:09.980","Text":"We saw that if the velocity and the B field are perpendicular to one another,"},{"Start":"01:09.980 ","End":"01:13.010","Text":"the charged particle is going to move in"},{"Start":"01:13.010 ","End":"01:18.650","Text":"circular motion until it eventually lands at this point over here."},{"Start":"01:18.650 ","End":"01:22.715","Text":"We want to calculate where this point exactly is."},{"Start":"01:22.715 ","End":"01:26.030","Text":"The first thing that we\u0027re going to do in order to find that is we\u0027re going to"},{"Start":"01:26.030 ","End":"01:29.970","Text":"try and calculate this velocity over here."},{"Start":"01:30.710 ","End":"01:34.865","Text":"This is very similar to what we saw in the last lesson."},{"Start":"01:34.865 ","End":"01:39.215","Text":"If this is the voltage between the 2 capacitor plates,"},{"Start":"01:39.215 ","End":"01:44.750","Text":"then we know that voltage is the same as potential difference."},{"Start":"01:44.750 ","End":"01:48.530","Text":"The potential difference between the 2 plates is this,"},{"Start":"01:48.530 ","End":"01:53.270","Text":"which means that we can say that 1 plate has a potential of"},{"Start":"01:53.270 ","End":"01:59.150","Text":"10^5 volts and the other plate has a potential of 0."},{"Start":"01:59.150 ","End":"02:04.390","Text":"Then the potential difference between the plates will be this."},{"Start":"02:06.320 ","End":"02:12.695","Text":"Because in the question we\u0027re being told that it\u0027s a proton that is traveling from"},{"Start":"02:12.695 ","End":"02:18.745","Text":"this plate over here and then it\u0027s accelerated to this plate over here,"},{"Start":"02:18.745 ","End":"02:21.020","Text":"we know that this is going to be"},{"Start":"02:21.020 ","End":"02:25.835","Text":"the positive plate and this is going to be the negative plate."},{"Start":"02:25.835 ","End":"02:28.235","Text":"In that case, if this is the positive plate,"},{"Start":"02:28.235 ","End":"02:31.610","Text":"we can say that the potential over here is equal to"},{"Start":"02:31.610 ","End":"02:39.380","Text":"10^5 volts and that the potential over here is equal to 0,"},{"Start":"02:39.380 ","End":"02:43.260","Text":"just like what we saw in the previous question."},{"Start":"02:44.120 ","End":"02:48.160","Text":"Now, because we\u0027re dealing with electric force,"},{"Start":"02:48.160 ","End":"02:49.840","Text":"we know that electric force is"},{"Start":"02:49.840 ","End":"02:53.995","Text":"a conservative force or the electric field is a conservative field."},{"Start":"02:53.995 ","End":"02:57.805","Text":"Which means that we can use the idea of energy conservation."},{"Start":"02:57.805 ","End":"03:01.765","Text":"We can say that the initial energy is equal to"},{"Start":"03:01.765 ","End":"03:06.900","Text":"the kinetic energy plus the potential energy."},{"Start":"03:06.900 ","End":"03:10.990","Text":"It starts from rest,"},{"Start":"03:10.990 ","End":"03:15.475","Text":"so the kinetic energy is going to be equal to 0 and"},{"Start":"03:15.475 ","End":"03:20.500","Text":"the potential energy is equal to the charge of the proton."},{"Start":"03:20.500 ","End":"03:26.370","Text":"Let\u0027s do charge of the proton multiplied by the potential at that point,"},{"Start":"03:26.370 ","End":"03:30.784","Text":"which we already saw was equal to 10^5 volts."},{"Start":"03:30.784 ","End":"03:35.760","Text":"The final energy is going to also be equal to the kinetic energy,"},{"Start":"03:35.760 ","End":"03:38.855","Text":"so here we know we have some kind of voltage."},{"Start":"03:38.855 ","End":"03:43.970","Text":"Here we have some velocity."},{"Start":"03:43.970 ","End":"03:50.520","Text":"1\\2mv squared is the kinetic energy plus our potential energy,"},{"Start":"03:50.520 ","End":"03:55.100","Text":"so it\u0027s our charge of the proton multiplied by the potential at this point,"},{"Start":"03:55.100 ","End":"03:57.250","Text":"which is equal to 0."},{"Start":"03:57.250 ","End":"04:05.714","Text":"Now we get that the charge of the proton multiplied by 10^5 volts,"},{"Start":"04:05.714 ","End":"04:08.550","Text":"let\u0027s just add that in,"},{"Start":"04:08.980 ","End":"04:16.590","Text":"is equal to 1\\2 mv squared."},{"Start":"04:16.620 ","End":"04:20.920","Text":"What I\u0027m going to do is I\u0027m actually going to erase this."},{"Start":"04:20.920 ","End":"04:22.910","Text":"I\u0027m not going to put in the values."},{"Start":"04:22.910 ","End":"04:28.115","Text":"I\u0027ll substitute them in a bit later."},{"Start":"04:28.115 ","End":"04:30.924","Text":"We have the charge multiplied by the voltage,"},{"Start":"04:30.924 ","End":"04:33.475","Text":"we\u0027ll soon add in what the voltage is."},{"Start":"04:33.475 ","End":"04:35.290","Text":"Here it\u0027s the same thing."},{"Start":"04:35.290 ","End":"04:39.560","Text":"I just don\u0027t want to substitute in values just yet."},{"Start":"04:39.560 ","End":"04:45.775","Text":"We saw that the equation for the radius of this circle is"},{"Start":"04:45.775 ","End":"04:52.040","Text":"equal to mv divided by qB."},{"Start":"04:52.040 ","End":"04:55.054","Text":"We saw how we got to this equation."},{"Start":"04:55.054 ","End":"04:56.870","Text":"In previous lessons,"},{"Start":"04:56.870 ","End":"05:01.730","Text":"we saw that the force by the magnetic field on the particle is equal to"},{"Start":"05:01.730 ","End":"05:07.700","Text":"Q velocity cross product with the magnetic field."},{"Start":"05:07.700 ","End":"05:13.475","Text":"Then when we took that the velocity is perpendicular to the magnetic fields,"},{"Start":"05:13.475 ","End":"05:17.983","Text":"we got that this was equal to simply qvB."},{"Start":"05:17.983 ","End":"05:21.620","Text":"We know that when the velocity is perpendicular to"},{"Start":"05:21.620 ","End":"05:27.326","Text":"the magnetic field the particle will move in circular motions,"},{"Start":"05:27.326 ","End":"05:32.570","Text":"so that means that qvB is equal to mv squared divided by R,"},{"Start":"05:32.570 ","End":"05:35.660","Text":"where of course this v is the velocity."},{"Start":"05:35.660 ","End":"05:39.359","Text":"Then we just isolated that R to get this equation."},{"Start":"05:40.360 ","End":"05:49.175","Text":"From this equation we get that the velocity squared is equal to"},{"Start":"05:49.175 ","End":"05:53.945","Text":"2 times the charge of the proton multiplied by"},{"Start":"05:53.945 ","End":"05:59.090","Text":"the voltage divided by the mass of the proton."},{"Start":"05:59.090 ","End":"06:05.730","Text":"Now we can plug this in to our equation for R. We"},{"Start":"06:05.730 ","End":"06:12.840","Text":"get that m divided by qB multiplied by v,"},{"Start":"06:12.840 ","End":"06:14.240","Text":"so we take the square root of this,"},{"Start":"06:14.240 ","End":"06:19.355","Text":"multiply the square root of 2 charge of the proton also here,"},{"Start":"06:19.355 ","End":"06:24.570","Text":"multiplied by the voltage divided by the mass."},{"Start":"06:25.010 ","End":"06:27.590","Text":"We can put everything into"},{"Start":"06:27.590 ","End":"06:31.385","Text":"the square root sign so that it\u0027s a bit easier to see what\u0027s happening."},{"Start":"06:31.385 ","End":"06:41.635","Text":"We have 2 mv divided by qB squared and we take the square root of all of that."},{"Start":"06:41.635 ","End":"06:45.150","Text":"Now let\u0027s plug in our values."},{"Start":"06:45.150 ","End":"06:50.630","Text":"We\u0027re taking the square root of 2 multiplied by the mass of a proton,"},{"Start":"06:50.630 ","End":"06:58.145","Text":"which is equal to 1.67 times 10^negative 27 kilograms,"},{"Start":"06:58.145 ","End":"07:04.175","Text":"multiplied by 10^negative 5 volts,"},{"Start":"07:04.175 ","End":"07:08.870","Text":"divided by the charge of the proton,"},{"Start":"07:08.870 ","End":"07:16.370","Text":"which is equal to 1.602 times 10^negative 19"},{"Start":"07:16.370 ","End":"07:24.635","Text":"coulombs multiplied by the intensity of the magnetic field squared,"},{"Start":"07:24.635 ","End":"07:29.905","Text":"which is equal to 0.2 squared."},{"Start":"07:29.905 ","End":"07:33.110","Text":"Once we plug this into our calculator,"},{"Start":"07:33.110 ","End":"07:42.420","Text":"we\u0027ll get that the radius of the circle is equal to 0.228 meters."},{"Start":"07:42.830 ","End":"07:50.915","Text":"What we\u0027re trying to find now is this horizontal distance over here."},{"Start":"07:50.915 ","End":"07:53.180","Text":"Let\u0027s call this delta x."},{"Start":"07:53.180 ","End":"07:57.390","Text":"What we\u0027re going to do is we\u0027re going to use geometry."},{"Start":"07:57.530 ","End":"08:02.945","Text":"We know that this arc over here is just part of the circle."},{"Start":"08:02.945 ","End":"08:07.350","Text":"We don\u0027t know what part if it\u0027s a 1\\2 or the top corner,"},{"Start":"08:07.350 ","End":"08:13.325","Text":"but what we do know is that the center is at some point over here and"},{"Start":"08:13.325 ","End":"08:20.090","Text":"that this distance over here is the radius which we just calculated."},{"Start":"08:20.090 ","End":"08:26.435","Text":"Similarly, that this distance over here is also the radius."},{"Start":"08:26.435 ","End":"08:33.365","Text":"Then we\u0027re going to draw a line straight down from here like so,"},{"Start":"08:33.365 ","End":"08:37.235","Text":"such that this angle over here is a 90 degree angle."},{"Start":"08:37.235 ","End":"08:40.010","Text":"This line, the length of it,"},{"Start":"08:40.010 ","End":"08:42.560","Text":"we\u0027re going to call D just for now,"},{"Start":"08:42.560 ","End":"08:45.905","Text":"but we know that it\u0027s equal to 15 centimeters."},{"Start":"08:45.905 ","End":"08:50.070","Text":"It\u0027s this total length over here."},{"Start":"08:50.150 ","End":"08:54.680","Text":"Then we know that this distance over here is"},{"Start":"08:54.680 ","End":"09:02.070","Text":"delta x and this distance over here is also delta x because they\u0027re parallel."},{"Start":"09:02.960 ","End":"09:08.825","Text":"Now let\u0027s call this angle over here Theta."},{"Start":"09:08.825 ","End":"09:17.720","Text":"We know that this full length over here is R. It\u0027s this blue line over here."},{"Start":"09:17.720 ","End":"09:18.800","Text":"I just shifted it down."},{"Start":"09:18.800 ","End":"09:24.380","Text":"This is R. What I want to know is what this length is over here,"},{"Start":"09:24.380 ","End":"09:29.480","Text":"this length from my rule of sines and cosines."},{"Start":"09:29.480 ","End":"09:34.610","Text":"I know that this length is equal to R multiplied by cosine of this angle"},{"Start":"09:34.610 ","End":"09:40.370","Text":"Theta because I have the hypotenuse and they have the adjacent side."},{"Start":"09:40.370 ","End":"09:42.755","Text":"So that corresponds to cosine."},{"Start":"09:42.755 ","End":"09:46.850","Text":"This length that\u0027s left over here,"},{"Start":"09:46.850 ","End":"09:53.005","Text":"this is my length delta x that I showed over here."},{"Start":"09:53.005 ","End":"10:02.195","Text":"I can see that my delta x is simply equal to this total length of R minus this over here."},{"Start":"10:02.195 ","End":"10:04.200","Text":"What is this over here?"},{"Start":"10:04.200 ","End":"10:07.545","Text":"This is R cosine of Theta."},{"Start":"10:07.545 ","End":"10:11.785","Text":"I take the total length and I subtract this and I\u0027m left with delta x."},{"Start":"10:11.785 ","End":"10:14.210","Text":"Now I can take out my R as a common multiple,"},{"Start":"10:14.210 ","End":"10:17.533","Text":"so I have 1 minus cosine of Theta."},{"Start":"10:17.533 ","End":"10:22.380","Text":"Now I just want to know what cosine of Theta is because they don\u0027t know what Theta is."},{"Start":"10:22.550 ","End":"10:27.980","Text":"What I do know is that sine of Theta is equal"},{"Start":"10:27.980 ","End":"10:34.200","Text":"to opposite over hypotenuse."},{"Start":"10:34.200 ","End":"10:40.970","Text":"My opposite side is this d and my hypotenuse is this side of"},{"Start":"10:40.970 ","End":"10:48.810","Text":"length R. My d I know is equal to 15 centimeters."},{"Start":"10:49.150 ","End":"10:52.520","Text":"I want to convert it to meters,"},{"Start":"10:52.520 ","End":"10:58.400","Text":"so it\u0027s equal to 0.15 meters divided"},{"Start":"10:58.400 ","End":"11:06.340","Text":"by my radius which is equal to 0.228 meters."},{"Start":"11:06.340 ","End":"11:10.565","Text":"Then once we plug this into a calculator we get that this"},{"Start":"11:10.565 ","End":"11:18.833","Text":"is approximately equal to 0.658."},{"Start":"11:18.833 ","End":"11:22.810","Text":"Now, we know that there\u0027s"},{"Start":"11:22.810 ","End":"11:27.970","Text":"an equation that links cosine of Theta and sine of Theta and that"},{"Start":"11:27.970 ","End":"11:38.960","Text":"is that cosine of Theta is equal to square root of 1 minus sine squared of Theta."},{"Start":"11:39.000 ","End":"11:41.080","Text":"Where does this come from?"},{"Start":"11:41.080 ","End":"11:50.200","Text":"We know that cosine squared Theta plus sine squared Theta is equal to 1,"},{"Start":"11:50.200 ","End":"11:53.575","Text":"and then you just rearrange it to get this equation."},{"Start":"11:53.575 ","End":"12:02.990","Text":"It\u0027s very simple, so that means that we\u0027re taking the square root of 1 minus 0.658."},{"Start":"12:03.780 ","End":"12:08.120","Text":"What does that equal to?"},{"Start":"12:08.190 ","End":"12:14.450","Text":"This is approximately equal to 0.585."},{"Start":"12:15.330 ","End":"12:18.520","Text":"Now, we know what a cosine of Theta is."},{"Start":"12:18.520 ","End":"12:21.040","Text":"Now, we can plug it back into this equation."},{"Start":"12:21.040 ","End":"12:29.410","Text":"We get that Delta x is equal to R multiplied by 1 minus cosine of Theta,"},{"Start":"12:29.410 ","End":"12:32.720","Text":"so 1 minus 0.585."},{"Start":"12:35.520 ","End":"12:41.035","Text":"Where of course, R is equal to 0.228 meters."},{"Start":"12:41.035 ","End":"12:50.360","Text":"What we get is that Delta x is equal to 0.095 meters."},{"Start":"12:50.430 ","End":"12:53.400","Text":"Of course, this is 2,"},{"Start":"12:53.400 ","End":"12:56.275","Text":"3 decimal places, so it\u0027s an approximation."},{"Start":"12:56.275 ","End":"12:59.155","Text":"This is the answer to question Number 1."},{"Start":"12:59.155 ","End":"13:01.780","Text":"Now, let\u0027s move on to question Number 2."},{"Start":"13:01.780 ","End":"13:08.330","Text":"Find the time that elapsed until the screen was stricken."},{"Start":"13:09.420 ","End":"13:11.680","Text":"To answer this question,"},{"Start":"13:11.680 ","End":"13:15.189","Text":"the first thing that we need to know is that because"},{"Start":"13:15.189 ","End":"13:20.395","Text":"our velocity is perpendicular to our magnetic field,"},{"Start":"13:20.395 ","End":"13:25.405","Text":"that means that the magnitude of the velocity isn\u0027t affected,"},{"Start":"13:25.405 ","End":"13:26.950","Text":"the direction is,"},{"Start":"13:26.950 ","End":"13:30.415","Text":"the particle starts moving in a circular motion."},{"Start":"13:30.415 ","End":"13:36.145","Text":"However, the magnitude of it is not changed."},{"Start":"13:36.145 ","End":"13:38.140","Text":"That\u0027s very important."},{"Start":"13:38.140 ","End":"13:43.465","Text":"That means that the velocity or the magnitude of the velocity is a constant."},{"Start":"13:43.465 ","End":"13:46.840","Text":"What we want to know is the time taken for"},{"Start":"13:46.840 ","End":"13:51.110","Text":"a particle to go from this point to this point."},{"Start":"13:51.360 ","End":"13:58.510","Text":"The equation that we need to use is we need to take the length of the path S,"},{"Start":"13:58.510 ","End":"14:02.545","Text":"which is this shape over here S. Which is"},{"Start":"14:02.545 ","End":"14:08.330","Text":"equal to the velocity multiplied by the time taken."},{"Start":"14:20.070 ","End":"14:24.370","Text":"The magnitude of the velocity."},{"Start":"14:24.370 ","End":"14:28.180","Text":"Just saw here voltage and I accidentally said."},{"Start":"14:28.180 ","End":"14:30.475","Text":"This is velocity."},{"Start":"14:30.475 ","End":"14:33.460","Text":"The magnitude of the velocity is,"},{"Start":"14:33.460 ","End":"14:35.830","Text":"from here we take the square root,"},{"Start":"14:35.830 ","End":"14:40.930","Text":"and then we have 2 times the charge of the proton,"},{"Start":"14:40.930 ","End":"14:47.124","Text":"which is equal to 1.602 times"},{"Start":"14:47.124 ","End":"14:54.355","Text":"10 to the negative 19 coulombs multiplied by the voltage."},{"Start":"14:54.355 ","End":"15:03.715","Text":"The voltage is 10 to the power of 5 volts divided by the mass of the proton,"},{"Start":"15:03.715 ","End":"15:14.095","Text":"which is equal to 1.672 times 10 to the negative 27 kilograms."},{"Start":"15:14.095 ","End":"15:17.890","Text":"Once we plug all of this into the calculator,"},{"Start":"15:17.890 ","End":"15:19.810","Text":"we get that the velocity,"},{"Start":"15:19.810 ","End":"15:22.180","Text":"the magnitude of the velocity,"},{"Start":"15:22.180 ","End":"15:32.210","Text":"is equal to 0.0457 meters per second."},{"Start":"15:33.030 ","End":"15:37.120","Text":"Now, let\u0027s scroll down to have some more space."},{"Start":"15:37.120 ","End":"15:42.055","Text":"An equation for the arc length is equal to"},{"Start":"15:42.055 ","End":"15:47.850","Text":"the radius multiplied by the angle Theta, over here."},{"Start":"15:47.850 ","End":"15:51.630","Text":"This will give us this path length over here."},{"Start":"15:51.630 ","End":"16:02.080","Text":"Our radius is equal to 0.228 and our angle Theta is,"},{"Start":"16:02.080 ","End":"16:05.170","Text":"so cosine of Theta is equal to this."},{"Start":"16:05.170 ","End":"16:09.175","Text":"If we arc cos this will be left with Theta."},{"Start":"16:09.175 ","End":"16:14.800","Text":"Theta is equal to arc cosine or cosine to the negative 1."},{"Start":"16:14.800 ","End":"16:16.285","Text":"Let\u0027s write that, rather."},{"Start":"16:16.285 ","End":"16:23.150","Text":"We have cosine to the negative 1 of 0.585."},{"Start":"16:26.850 ","End":"16:32.740","Text":"Then what we get once we plug all of this into our calculator,"},{"Start":"16:32.740 ","End":"16:37.190","Text":"we get that our arc length is equal to 0.216."},{"Start":"16:38.490 ","End":"16:48.910","Text":"Therefore, our time t is equal to our arc length s divided by our velocity."},{"Start":"16:48.910 ","End":"16:53.530","Text":"We have 0.216. Of course,"},{"Start":"16:53.530 ","End":"16:57.400","Text":"this is in meters divided by our velocity,"},{"Start":"16:57.400 ","End":"17:00.320","Text":"which is equal to 0.0457."},{"Start":"17:04.650 ","End":"17:15.620","Text":"All of this is equal to 4.719 seconds."},{"Start":"17:16.890 ","End":"17:19.885","Text":"This is the answer to question Number 2."},{"Start":"17:19.885 ","End":"17:22.390","Text":"Now, let\u0027s answer question Number 3."},{"Start":"17:22.390 ","End":"17:29.960","Text":"What is the minimum voltage required for the proton to strike the screen?"},{"Start":"17:30.390 ","End":"17:34.990","Text":"We know that the proton is doing some circular motion."},{"Start":"17:34.990 ","End":"17:37.930","Text":"At some point, if the screen is close enough,"},{"Start":"17:37.930 ","End":"17:41.230","Text":"at some point in this motion,"},{"Start":"17:41.230 ","End":"17:43.510","Text":"the proton will hit the screen."},{"Start":"17:43.510 ","End":"17:47.395","Text":"However, if we move this screen back a little bit,"},{"Start":"17:47.395 ","End":"17:52.825","Text":"then the proton will carry on moving."},{"Start":"17:52.825 ","End":"17:55.360","Text":"Depending on its radius,"},{"Start":"17:55.360 ","End":"17:58.780","Text":"the proton will just carry on doing circular motion."},{"Start":"17:58.780 ","End":"18:01.930","Text":"If the screen has moved back,"},{"Start":"18:01.930 ","End":"18:04.360","Text":"then the proton won\u0027t reach the screen,"},{"Start":"18:04.360 ","End":"18:10.255","Text":"and it will just carry on in a circle and come back to this point over here."},{"Start":"18:10.255 ","End":"18:13.060","Text":"That\u0027s what happens if we move the screen."},{"Start":"18:13.060 ","End":"18:16.675","Text":"However, they\u0027re asking about the minimum voltage."},{"Start":"18:16.675 ","End":"18:22.540","Text":"It\u0027s the same thing, but now we have some radius r. We know"},{"Start":"18:22.540 ","End":"18:29.440","Text":"that the voltage source impacts this radius r,"},{"Start":"18:29.440 ","End":"18:31.525","Text":"as we saw over here,"},{"Start":"18:31.525 ","End":"18:35.515","Text":"our radius r is dependent on the voltage."},{"Start":"18:35.515 ","End":"18:38.065","Text":"If we change the voltage,"},{"Start":"18:38.065 ","End":"18:41.215","Text":"our radius will change as well."},{"Start":"18:41.215 ","End":"18:43.690","Text":"If our voltage is low enough,"},{"Start":"18:43.690 ","End":"18:45.640","Text":"our radius will be low enough,"},{"Start":"18:45.640 ","End":"18:49.345","Text":"such that even without moving the screen,"},{"Start":"18:49.345 ","End":"18:51.175","Text":"the screen will be at the same point,"},{"Start":"18:51.175 ","End":"18:54.700","Text":"but our proton will have a smaller radius,"},{"Start":"18:54.700 ","End":"18:56.260","Text":"and then it won\u0027t reach the screen,"},{"Start":"18:56.260 ","End":"18:58.330","Text":"and it will come back."},{"Start":"18:58.330 ","End":"19:04.660","Text":"What we want to do is we want to find the minimum voltage such that this radius is still"},{"Start":"19:04.660 ","End":"19:11.840","Text":"enough so that at some point in the screen our proton will hit it."},{"Start":"19:12.270 ","End":"19:16.990","Text":"What we\u0027re going to do is we\u0027re going to use this equation."},{"Start":"19:16.990 ","End":"19:22.735","Text":"I\u0027m just going to scroll down and copy it out over here."},{"Start":"19:22.735 ","End":"19:27.220","Text":"We saw that the radius is equal to the square root."},{"Start":"19:27.220 ","End":"19:32.155","Text":"I\u0027m just copying what was written on top of 2 multiplied by the mass,"},{"Start":"19:32.155 ","End":"19:34.405","Text":"multiplied by the voltage,"},{"Start":"19:34.405 ","End":"19:37.600","Text":"divided by the charge,"},{"Start":"19:37.600 ","End":"19:39.880","Text":"multiplied by B squared."},{"Start":"19:39.880 ","End":"19:43.940","Text":"We take the square root of all of this."},{"Start":"19:44.010 ","End":"19:47.770","Text":"Now, let\u0027s write out R squared."},{"Start":"19:47.770 ","End":"19:51.760","Text":"We\u0027ll get that R squared is equal to"},{"Start":"19:51.760 ","End":"19:59.660","Text":"2 times mass multiplied by the voltage divided by QB squared."},{"Start":"19:59.660 ","End":"20:02.440","Text":"Now, I want to find the minimum voltage."},{"Start":"20:02.440 ","End":"20:07.630","Text":"I\u0027m going to isolate out my v. I get that my voltage is equal to"},{"Start":"20:07.630 ","End":"20:16.280","Text":"QB squared multiplied by R squared divided by twice the mass."},{"Start":"20:16.950 ","End":"20:22.855","Text":"We\u0027re trying to find the minimum voltage where our proton will hit the screen."},{"Start":"20:22.855 ","End":"20:27.850","Text":"We can see that if our radius is equal to d,"},{"Start":"20:27.850 ","End":"20:33.204","Text":"is equal to this distance between the capacitor plates and the screen,"},{"Start":"20:33.204 ","End":"20:39.760","Text":"then that is the boundary at which our particle will hit the screen."},{"Start":"20:39.760 ","End":"20:43.405","Text":"If this is our radius r,"},{"Start":"20:43.405 ","End":"20:46.540","Text":"and r is equal to d. Here,"},{"Start":"20:46.540 ","End":"20:48.640","Text":"it\u0027s equal to 15 centimeters,"},{"Start":"20:48.640 ","End":"20:54.160","Text":"then our particle will hit the screen over here, and that\u0027s it."},{"Start":"20:54.160 ","End":"20:58.540","Text":"If our radius is smaller than d,"},{"Start":"20:58.540 ","End":"21:01.960","Text":"then the particle will just carry on moving in circular motion,"},{"Start":"21:01.960 ","End":"21:03.385","Text":"won\u0027t hit the screen."},{"Start":"21:03.385 ","End":"21:08.665","Text":"Of course, a smaller radius corresponds to a smaller voltage."},{"Start":"21:08.665 ","End":"21:11.365","Text":"What we\u0027re trying to find is this minimum voltage."},{"Start":"21:11.365 ","End":"21:13.705","Text":"We saw that the minimum voltage,"},{"Start":"21:13.705 ","End":"21:15.820","Text":"so that the particle will hit the screen,"},{"Start":"21:15.820 ","End":"21:19.300","Text":"is if a radius is equal to d."},{"Start":"21:19.300 ","End":"21:24.320","Text":"What we\u0027re going to do now is we\u0027re going to plug in the values."},{"Start":"21:24.530 ","End":"21:35.885","Text":"The charge of the proton is 1.602 times 10 to the negative 19."},{"Start":"21:35.885 ","End":"21:39.880","Text":"Our magnetic field has a magnitude of 0.2,"},{"Start":"21:39.880 ","End":"21:43.060","Text":"so we have 0.2 squared,"},{"Start":"21:43.060 ","End":"21:45.850","Text":"and we\u0027re multiplying by our radius squared,"},{"Start":"21:45.850 ","End":"21:52.615","Text":"where our radius has to be equal to our d. That\u0027s equal to 0.15 meters,"},{"Start":"21:52.615 ","End":"21:59.125","Text":"convert to meters, multiplied by 0.15 meters squared."},{"Start":"21:59.125 ","End":"22:05.530","Text":"Then divided by 2 times the mass of the proton,"},{"Start":"22:05.530 ","End":"22:16.220","Text":"which is equal to 1.67 times 10 to the negative 27 kilograms."},{"Start":"22:16.230 ","End":"22:20.305","Text":"Once we plug this all into the calculator,"},{"Start":"22:20.305 ","End":"22:22.780","Text":"we get that the value for the voltage,"},{"Start":"22:22.780 ","End":"22:27.400","Text":"the minimum value that we can have such that the particle will still hit the screen,"},{"Start":"22:27.400 ","End":"22:33.850","Text":"is equal to 4.312 times 10 to the power 4 volts."},{"Start":"22:33.850 ","End":"22:37.210","Text":"This is the answer to question Number 3,"},{"Start":"22:37.210 ","End":"22:39.895","Text":"and we see that we have"},{"Start":"22:39.895 ","End":"22:48.215","Text":"approximately just under half of the voltage that we\u0027re given originally in the question."},{"Start":"22:48.215 ","End":"22:51.470","Text":"We can have this voltage even more."},{"Start":"22:51.470 ","End":"22:56.945","Text":"Our particle will still hit the screen but if we take any lower voltage,"},{"Start":"22:56.945 ","End":"23:01.800","Text":"then our particle will just carry on in a circle without touching the screen."},{"Start":"23:01.800 ","End":"23:04.730","Text":"That\u0027s the end of this lesson."}],"ID":22387},{"Watched":false,"Name":"Exercise 9","Duration":"21m 18s","ChapterTopicVideoID":24812,"CourseChapterTopicPlaylistID":99484,"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":"00:04.560","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.560 ","End":"00:08.925","Text":"An electric field is present where x is smaller than 0."},{"Start":"00:08.925 ","End":"00:11.265","Text":"If this is the origin over here,"},{"Start":"00:11.265 ","End":"00:14.760","Text":"so this is the region where x is less than 0,"},{"Start":"00:14.760 ","End":"00:19.155","Text":"such that above the x-axis or y is greater than 0,"},{"Start":"00:19.155 ","End":"00:23.725","Text":"we have an E field in the negative y-direction."},{"Start":"00:23.725 ","End":"00:28.580","Text":"Below the x-axis, where y is smaller than 0,"},{"Start":"00:28.580 ","End":"00:35.660","Text":"we have an E field of the same magnitude but in the positive y direction."},{"Start":"00:35.660 ","End":"00:38.465","Text":"A uniform magnetic field of"},{"Start":"00:38.465 ","End":"00:42.875","Text":"unknown magnitude and direction is also present in the entire space."},{"Start":"00:42.875 ","End":"00:48.640","Text":"That means also in the region where x is greater than 0."},{"Start":"00:48.640 ","End":"00:56.760","Text":"A particle of mass m and charge q arrives from x is equal to negative infinity,"},{"Start":"00:56.760 ","End":"00:58.750","Text":"so it arrives from somewhere here,"},{"Start":"00:58.750 ","End":"01:02.780","Text":"and moves in a straight line at a constant velocity."},{"Start":"01:02.780 ","End":"01:08.525","Text":"The height of the particle\u0027s trajectory is y is equal to negative R."},{"Start":"01:08.525 ","End":"01:17.630","Text":"This over here is at negative R. When the particle crosses the y-axis,"},{"Start":"01:17.630 ","End":"01:21.150","Text":"so it crosses this line over here."},{"Start":"01:21.470 ","End":"01:27.454","Text":"It travels along the perimeter of a quarter circle of radius"},{"Start":"01:27.454 ","End":"01:36.150","Text":"R. Question number 1 is to sketch the rest of the particle\u0027s trajectory."},{"Start":"01:36.950 ","End":"01:42.170","Text":"In order to answer this question, first of all,"},{"Start":"01:42.170 ","End":"01:44.480","Text":"we know that in this entire region,"},{"Start":"01:44.480 ","End":"01:47.240","Text":"we have a uniform magnetic field."},{"Start":"01:47.240 ","End":"01:52.160","Text":"Now, we know that the force on a particle in"},{"Start":"01:52.160 ","End":"01:58.370","Text":"a magnetic field is going to cause the particle to move in circular motion,"},{"Start":"01:58.370 ","End":"02:00.125","Text":"to move in a circle."},{"Start":"02:00.125 ","End":"02:04.400","Text":"We can see that in this region where x is greater than 0,"},{"Start":"02:04.400 ","End":"02:07.970","Text":"the particle is only subject to the magnetic field."},{"Start":"02:07.970 ","End":"02:13.750","Text":"The only force acting on it is going to be a force due to the magnetic field,"},{"Start":"02:13.750 ","End":"02:16.625","Text":"so that means that the bicycle in this whole region"},{"Start":"02:16.625 ","End":"02:20.005","Text":"is going to carry on moving in circular motion."},{"Start":"02:20.005 ","End":"02:25.010","Text":"It\u0027s going to carry on moving over here another quarter circle,"},{"Start":"02:25.010 ","End":"02:28.480","Text":"so in total, we have a half circle over here."},{"Start":"02:28.480 ","End":"02:32.615","Text":"If a particle is traveling in circular motion over here,"},{"Start":"02:32.615 ","End":"02:35.870","Text":"it\u0027s of course, due to the magnetic force."},{"Start":"02:35.870 ","End":"02:42.930","Text":"The magnetic force is always pointing to the center of the circle,"},{"Start":"02:42.930 ","End":"02:49.260","Text":"so it\u0027s always going to be pointing like so,"},{"Start":"02:49.260 ","End":"02:54.120","Text":"and of course, or so here and here and so on."},{"Start":"02:54.890 ","End":"02:58.790","Text":"What we can see over here on the border"},{"Start":"02:58.790 ","End":"03:06.500","Text":"between a magnetic field and E field acting and chest and magnetic field acting,"},{"Start":"03:06.500 ","End":"03:10.625","Text":"our magnetic force is pointing directly upwards."},{"Start":"03:10.625 ","End":"03:13.805","Text":"That means that also over here,"},{"Start":"03:13.805 ","End":"03:16.895","Text":"where the E field is acting,"},{"Start":"03:16.895 ","End":"03:22.610","Text":"the magnetic force also has to be pointing in this direction."},{"Start":"03:22.610 ","End":"03:30.150","Text":"This we can also get from the right-hand rule if this is the direction of travel,"},{"Start":"03:30.170 ","End":"03:36.325","Text":"and this over here is the force."},{"Start":"03:36.325 ","End":"03:41.015","Text":"Over here, the force must also be acting in the same direction."},{"Start":"03:41.015 ","End":"03:45.540","Text":"Here we can see we have our F_B."},{"Start":"03:46.010 ","End":"03:50.240","Text":"For a similar reason over here where we see the magnetic force is"},{"Start":"03:50.240 ","End":"03:53.825","Text":"pointing downwards, also over here,"},{"Start":"03:53.825 ","End":"04:02.075","Text":"the magnetic force will be pointing downwards in this same direction."},{"Start":"04:02.075 ","End":"04:05.690","Text":"Again, due to the right-hand rule because here,"},{"Start":"04:05.690 ","End":"04:08.000","Text":"the velocity is going in the rightwards direction,"},{"Start":"04:08.000 ","End":"04:12.665","Text":"then the direction of the velocity changes around the half circle,"},{"Start":"04:12.665 ","End":"04:17.850","Text":"and then the direction of the velocity is in the leftward direction."},{"Start":"04:19.420 ","End":"04:24.630","Text":"In this region, we also have the E field."},{"Start":"04:25.520 ","End":"04:29.525","Text":"Because we\u0027re being told that over here in this region,"},{"Start":"04:29.525 ","End":"04:36.560","Text":"our particle is traveling in a straight line and the velocity is constant,"},{"Start":"04:36.560 ","End":"04:41.400","Text":"that means that there is no acceleration."},{"Start":"04:42.140 ","End":"04:45.305","Text":"If there is no acceleration,"},{"Start":"04:45.305 ","End":"04:49.250","Text":"that means that the sum of all of the forces,"},{"Start":"04:49.250 ","End":"04:52.625","Text":"specifically here, this is along the y-axis,"},{"Start":"04:52.625 ","End":"04:56.590","Text":"is equal to 0."},{"Start":"04:58.160 ","End":"05:03.095","Text":"What forces are acting on the particle in this region over here,"},{"Start":"05:03.095 ","End":"05:06.800","Text":"in the region where x is smaller than 0,"},{"Start":"05:06.800 ","End":"05:15.580","Text":"we have our force due to the magnetic field and we have our force due to the E field,"},{"Start":"05:15.580 ","End":"05:18.983","Text":"so this is the sum of all of the forces and this has to be equal to 0."},{"Start":"05:18.983 ","End":"05:27.780","Text":"Therefore, our magnetic force is equal to our electric force."},{"Start":"05:28.630 ","End":"05:33.280","Text":"Over here, let\u0027s draw it in green,"},{"Start":"05:33.280 ","End":"05:37.340","Text":"an equal and opposite force over here,"},{"Start":"05:37.340 ","End":"05:41.765","Text":"F_E, which will balance this out,"},{"Start":"05:41.765 ","End":"05:45.350","Text":"which is why the particle travels in a straight line."},{"Start":"05:45.350 ","End":"05:47.750","Text":"Similarly, over here,"},{"Start":"05:47.750 ","End":"05:54.580","Text":"the force will be equal and opposite as well."},{"Start":"05:57.710 ","End":"06:02.670","Text":"Now we can see that in this region,"},{"Start":"06:02.670 ","End":"06:08.710","Text":"the particle is going to carry on traveling in the straight line"},{"Start":"06:08.710 ","End":"06:15.700","Text":"just like in this region but the velocity is this time going to be in this direction."},{"Start":"06:15.700 ","End":"06:22.335","Text":"That is the particle\u0027s trajectory throughout the region."},{"Start":"06:22.335 ","End":"06:24.615","Text":"That\u0027s the onset of question number 1."},{"Start":"06:24.615 ","End":"06:26.970","Text":"Now let\u0027s answer question number 2,"},{"Start":"06:26.970 ","End":"06:30.580","Text":"what is the charge\u0027s polarity?"},{"Start":"06:30.650 ","End":"06:33.205","Text":"In order to know the sign,"},{"Start":"06:33.205 ","End":"06:38.635","Text":"we\u0027re actually going to look at the force due to our E field."},{"Start":"06:38.635 ","End":"06:42.450","Text":"We know that the electric force,"},{"Start":"06:42.450 ","End":"06:49.310","Text":"let\u0027s write F_E is equal to the charge multiplied by the E field."},{"Start":"06:49.310 ","End":"06:51.770","Text":"From that, we can see that the force from"},{"Start":"06:51.770 ","End":"06:56.105","Text":"the electric field is going to be parallel to the E field."},{"Start":"06:56.105 ","End":"07:01.505","Text":"However, we have to see if it\u0027s in the direction of the E field or not."},{"Start":"07:01.505 ","End":"07:08.030","Text":"For instance, if our q has a negative charge,"},{"Start":"07:08.030 ","End":"07:17.659","Text":"then our F_E will be in the opposite direction to the E field."},{"Start":"07:17.659 ","End":"07:24.650","Text":"Our F_E will be in the opposite direction to the E field,"},{"Start":"07:24.650 ","End":"07:28.145","Text":"and if our charge is positive,"},{"Start":"07:28.145 ","End":"07:34.080","Text":"then our force will be in the same direction."},{"Start":"07:34.310 ","End":"07:36.920","Text":"What we can see over here,"},{"Start":"07:36.920 ","End":"07:41.255","Text":"let\u0027s say is that our E field is in the positive y-direction."},{"Start":"07:41.255 ","End":"07:43.700","Text":"However, our F_E,"},{"Start":"07:43.700 ","End":"07:46.430","Text":"our electric force is in the negative y direction,"},{"Start":"07:46.430 ","End":"07:50.060","Text":"and that we saw because our B field is"},{"Start":"07:50.060 ","End":"07:54.400","Text":"over here at this point in the positive y direction,"},{"Start":"07:54.400 ","End":"07:56.720","Text":"and our forces over here,"},{"Start":"07:56.720 ","End":"08:01.865","Text":"how to balance out in order to maintain this constant velocity in this region."},{"Start":"08:01.865 ","End":"08:05.525","Text":"That meant that if this is"},{"Start":"08:05.525 ","End":"08:12.110","Text":"our magnetic force then this had to be our electric force pointing down,"},{"Start":"08:12.110 ","End":"08:16.760","Text":"and similarly here, if our magnetic force is pointing in this direction,"},{"Start":"08:16.760 ","End":"08:19.490","Text":"then our electric force has to be pointing in"},{"Start":"08:19.490 ","End":"08:22.670","Text":"the opposite direction to maintain the constant velocity,"},{"Start":"08:22.670 ","End":"08:24.755","Text":"to maintain the straight line."},{"Start":"08:24.755 ","End":"08:28.849","Text":"We can see that our electric force is in the opposite direction,"},{"Start":"08:28.849 ","End":"08:34.730","Text":"and similarly over here in the positive y region,"},{"Start":"08:34.730 ","End":"08:39.785","Text":"we can see that our E field is in the negative direction however,"},{"Start":"08:39.785 ","End":"08:43.925","Text":"our electric force is in the positive y-direction."},{"Start":"08:43.925 ","End":"08:48.710","Text":"Therefore, we can see that"},{"Start":"08:48.710 ","End":"08:56.690","Text":"the sign of our charge q is equal to a negative charge,"},{"Start":"08:56.690 ","End":"08:59.490","Text":"so the sign is negative 1."},{"Start":"09:00.130 ","End":"09:03.425","Text":"We\u0027re dealing with a negatively charged charge,"},{"Start":"09:03.425 ","End":"09:05.585","Text":"and that\u0027s the answer to question 2."},{"Start":"09:05.585 ","End":"09:08.120","Text":"Now let\u0027s answer question 3."},{"Start":"09:08.120 ","End":"09:13.020","Text":"Determine the charge velocity and magnetic field."},{"Start":"09:14.660 ","End":"09:20.180","Text":"First of all, we know that the magnitude of"},{"Start":"09:20.180 ","End":"09:29.265","Text":"our magnetic force is equal to the magnitude of our electric force."},{"Start":"09:29.265 ","End":"09:33.015","Text":"That\u0027s what we saw in question number 1."},{"Start":"09:33.015 ","End":"09:37.380","Text":"First of all, let\u0027s see what the magnitude of our electric force is equal to."},{"Start":"09:37.380 ","End":"09:41.935","Text":"It\u0027s equal to the magnitude of our charge,"},{"Start":"09:41.935 ","End":"09:45.930","Text":"multiplied by the magnitude of our E field."},{"Start":"09:45.930 ","End":"09:48.950","Text":"We\u0027re being told that our E field for"},{"Start":"09:48.950 ","End":"09:53.660","Text":"both regions of y is greater than and y is less than 0,"},{"Start":"09:53.660 ","End":"09:55.910","Text":"the magnitude is E_naught,"},{"Start":"09:55.910 ","End":"09:57.890","Text":"and then the direction just changes,"},{"Start":"09:57.890 ","End":"10:01.385","Text":"but the magnitude is E_naught."},{"Start":"10:01.385 ","End":"10:06.430","Text":"Then as for our magnetic force,"},{"Start":"10:06.430 ","End":"10:11.020","Text":"so F_B as we know that is equal to the charge"},{"Start":"10:11.020 ","End":"10:16.150","Text":"multiplied by the velocity cross product with the B field."},{"Start":"10:16.150 ","End":"10:24.160","Text":"If we\u0027re trying to find the magnitude of this magnetic force is equal to"},{"Start":"10:24.160 ","End":"10:28.750","Text":"the magnitude of our charge multiplied by"},{"Start":"10:28.750 ","End":"10:34.450","Text":"the magnitude of the velocity cross product with the magnetic field."},{"Start":"10:34.450 ","End":"10:38.290","Text":"Now we know in this question that the velocity"},{"Start":"10:38.290 ","End":"10:42.040","Text":"is always perpendicular to the magnetic field,"},{"Start":"10:42.040 ","End":"10:46.160","Text":"and we can just see that from the trajectory."},{"Start":"10:47.190 ","End":"10:52.270","Text":"That means that we can simply write this as"},{"Start":"10:52.270 ","End":"10:58.300","Text":"the magnitude of q if v and B are always perpendicular,"},{"Start":"10:58.300 ","End":"11:02.095","Text":"so multiplied by the magnitude of the velocity,"},{"Start":"11:02.095 ","End":"11:07.120","Text":"and then multiplied by the magnitude of the B field."},{"Start":"11:07.120 ","End":"11:12.535","Text":"Now we can equate these 2 because we already saw that they\u0027re equal."},{"Start":"11:12.535 ","End":"11:17.590","Text":"We have that the magnitude of the charge multiplied by E_naught"},{"Start":"11:17.590 ","End":"11:23.770","Text":"is equal to the magnitude of the charge multiplied by the magnitude of the velocity,"},{"Start":"11:23.770 ","End":"11:27.265","Text":"multiplied by the magnitude of the B field,"},{"Start":"11:27.265 ","End":"11:30.535","Text":"can divide both sides by the magnitude of q."},{"Start":"11:30.535 ","End":"11:34.360","Text":"What we get is that E_naught is equal to"},{"Start":"11:34.360 ","End":"11:38.005","Text":"the magnitude of v multiplied by the magnitude of B,"},{"Start":"11:38.005 ","End":"11:41.695","Text":"where of course our v and our B are unknown."},{"Start":"11:41.695 ","End":"11:46.040","Text":"We need another equation in order to solve this."},{"Start":"11:46.890 ","End":"11:53.523","Text":"What we can see is that we have this circular motion going on over here,"},{"Start":"11:53.523 ","End":"12:00.325","Text":"and we remember that an equation for the radius of circular motion is equal to"},{"Start":"12:00.325 ","End":"12:02.860","Text":"the mass of the particle multiplied by"},{"Start":"12:02.860 ","End":"12:08.930","Text":"the velocity divided by q multiplied by the B field."},{"Start":"12:09.630 ","End":"12:13.975","Text":"What we can do is we can isolate out our v,"},{"Start":"12:13.975 ","End":"12:16.165","Text":"and so we see that our v is equal to"},{"Start":"12:16.165 ","End":"12:24.480","Text":"qBR divided by m. Now we can plug this into this equation."},{"Start":"12:24.480 ","End":"12:29.145","Text":"We\u0027ll get that E_naught is equal to v,"},{"Start":"12:29.145 ","End":"12:36.510","Text":"so that\u0027s qBR divided by m multiplied by B,"},{"Start":"12:36.510 ","End":"12:39.170","Text":"so B^2 over here."},{"Start":"12:39.690 ","End":"12:44.515","Text":"Here, everything is given to us in the question aside from B."},{"Start":"12:44.515 ","End":"12:47.628","Text":"Therefore we can isolate out our B,"},{"Start":"12:47.628 ","End":"12:53.530","Text":"and we get that B is equal to the square root of E_naught multiplied by"},{"Start":"12:53.530 ","End":"13:01.893","Text":"m divided by qR."},{"Start":"13:01.893 ","End":"13:05.710","Text":"These are all values that were given in the question."},{"Start":"13:05.710 ","End":"13:15.670","Text":"Now, we can just substitute B in for v. What we\u0027ll get is that v is equal to,"},{"Start":"13:15.670 ","End":"13:22.550","Text":"so we saw it\u0027s equal to qR divided by m multiplied by B,"},{"Start":"13:22.680 ","End":"13:29.185","Text":"where B is equal to the square root of E_naught m divided by qR."},{"Start":"13:29.185 ","End":"13:34.312","Text":"Now, if we put everything inside the square root sign,"},{"Start":"13:34.312 ","End":"13:42.130","Text":"we\u0027ll get that v is equal to the square root of qR E_naught divided"},{"Start":"13:42.130 ","End":"13:50.710","Text":"by m. This is what the magnetic field is equal to,"},{"Start":"13:50.710 ","End":"13:53.905","Text":"and we can see all of these values are given in the question,"},{"Start":"13:53.905 ","End":"13:57.475","Text":"and this is what the velocity is equal to, and of course,"},{"Start":"13:57.475 ","End":"14:02.149","Text":"all of these values are also given to us in the question."},{"Start":"14:02.610 ","End":"14:07.075","Text":"Now, this is of course the magnitude of the B field,"},{"Start":"14:07.075 ","End":"14:13.090","Text":"but we know that the B field is a vector which means that it also has direction."},{"Start":"14:13.090 ","End":"14:17.485","Text":"What we want we do is we want to calculate the direction."},{"Start":"14:17.485 ","End":"14:25.669","Text":"Let\u0027s imagine that our z-axis is going out of the page."},{"Start":"14:25.890 ","End":"14:28.345","Text":"Let\u0027s draw it like this,"},{"Start":"14:28.345 ","End":"14:32.065","Text":"this is our z-axis, so y is this."},{"Start":"14:32.065 ","End":"14:34.428","Text":"If this is our x-axis,"},{"Start":"14:34.428 ","End":"14:36.124","Text":"and this is our y-axis,"},{"Start":"14:36.124 ","End":"14:38.870","Text":"so when we do x cross y,"},{"Start":"14:38.870 ","End":"14:40.840","Text":"we move like so,"},{"Start":"14:40.840 ","End":"14:45.760","Text":"which means that the z-axis has to be coming out over the page."},{"Start":"14:45.760 ","End":"14:54.640","Text":"You can use the same right-hand rule that we usually use when using the cross-product."},{"Start":"14:54.640 ","End":"14:56.965","Text":"Now let\u0027s take a look,"},{"Start":"14:56.965 ","End":"15:01.015","Text":"and let\u0027s first work out v cross B."},{"Start":"15:01.015 ","End":"15:04.524","Text":"Let\u0027s choose some random points along the semicircle,"},{"Start":"15:04.524 ","End":"15:12.250","Text":"and so of course we put our thumb in the direction of the velocity."},{"Start":"15:12.250 ","End":"15:17.920","Text":"This is our thumb in the direction of the velocity,"},{"Start":"15:17.920 ","End":"15:19.510","Text":"so at this point over here,"},{"Start":"15:19.510 ","End":"15:21.475","Text":"the velocity is traveling like so."},{"Start":"15:21.475 ","End":"15:26.050","Text":"Then we have our forefinger which represents our B field."},{"Start":"15:26.050 ","End":"15:28.315","Text":"Let\u0027s just leave that for a second."},{"Start":"15:28.315 ","End":"15:32.500","Text":"Then we have our middle finger which represents the onset,"},{"Start":"15:32.500 ","End":"15:35.450","Text":"which is the direction of the force."},{"Start":"15:35.490 ","End":"15:42.535","Text":"We know that the force is pointing in this direction into this circle."},{"Start":"15:42.535 ","End":"15:45.280","Text":"It\u0027s going to the center of the circle."},{"Start":"15:45.280 ","End":"15:51.745","Text":"That means that our middle finger is pointing like so."},{"Start":"15:51.745 ","End":"15:56.770","Text":"I\u0027ll just move the picture of the z-axis here not to confuse."},{"Start":"15:56.770 ","End":"16:00.865","Text":"Our middle axis which represents F_B is pointing here,"},{"Start":"16:00.865 ","End":"16:05.600","Text":"and our thumb which represents v is pointing in this direction."},{"Start":"16:06.060 ","End":"16:14.890","Text":"Now we have to see where our forefinger is pointing to get our B."},{"Start":"16:14.890 ","End":"16:19.300","Text":"Let\u0027s say that our forefinger would point into the page."},{"Start":"16:19.300 ","End":"16:22.750","Text":"Then what we can see is that our right-hand rule works out."},{"Start":"16:22.750 ","End":"16:27.970","Text":"However, we saw from question number 2 that our charge has"},{"Start":"16:27.970 ","End":"16:33.423","Text":"a negative sign which means that here we have to multiply by a minus."},{"Start":"16:33.423 ","End":"16:37.900","Text":"Therefore, if our forefinger was pointing into the page,"},{"Start":"16:37.900 ","End":"16:40.135","Text":"if our B was into the page,"},{"Start":"16:40.135 ","End":"16:45.535","Text":"then that means that our force would have to be pointing like so,"},{"Start":"16:45.535 ","End":"16:46.900","Text":"which we know isn\u0027t the case,"},{"Start":"16:46.900 ","End":"16:49.880","Text":"it\u0027s pointing in this direction."},{"Start":"16:50.670 ","End":"16:58.060","Text":"In that case, we can rub out this because we have to multiply by a minus,"},{"Start":"16:58.060 ","End":"17:03.100","Text":"and let\u0027s say that our forefinger is pointing out."},{"Start":"17:03.100 ","End":"17:06.505","Text":"Then we have v cross B,"},{"Start":"17:06.505 ","End":"17:09.085","Text":"so this represents our B."},{"Start":"17:09.085 ","End":"17:11.890","Text":"We get v cross B,"},{"Start":"17:11.890 ","End":"17:14.650","Text":"and this represents our forefinger."},{"Start":"17:14.650 ","End":"17:20.740","Text":"We get v cross B which will give us an F_B pointing like so."},{"Start":"17:20.740 ","End":"17:27.085","Text":"However, then we multiply it by q which has a negative on it,"},{"Start":"17:27.085 ","End":"17:30.610","Text":"which will then flip this over in this direction,"},{"Start":"17:30.610 ","End":"17:36.190","Text":"and the F_B will be in the right direction for what we\u0027re dealing with."},{"Start":"17:36.190 ","End":"17:39.760","Text":"Therefore we can see that our forefinger is"},{"Start":"17:39.760 ","End":"17:44.303","Text":"pointing out of the page which is in the direction of the z-axis,"},{"Start":"17:44.303 ","End":"17:49.460","Text":"so we can see that our B field is in the z-direction."},{"Start":"17:50.460 ","End":"17:56.095","Text":"Now let\u0027s move on to question number 4."},{"Start":"17:56.095 ","End":"18:02.260","Text":"Given the same conditions but a magnetic field 3 times as strong,"},{"Start":"18:02.260 ","End":"18:08.575","Text":"determine the particle\u0027s required mass for the same trajectory to be followed."},{"Start":"18:08.575 ","End":"18:11.800","Text":"If the same trajectory is to be followed,"},{"Start":"18:11.800 ","End":"18:20.150","Text":"then that means that this equation and this equation have to remain the same."},{"Start":"18:21.090 ","End":"18:24.715","Text":"E_naught still has to be equal to v B,"},{"Start":"18:24.715 ","End":"18:29.200","Text":"and of course, R has to be equal to this."},{"Start":"18:29.200 ","End":"18:39.190","Text":"Therefore, what we\u0027ll get is that E_naught is equal to,"},{"Start":"18:39.190 ","End":"18:47.110","Text":"so we got q multiplied by R divided by m multiplied by B^2,"},{"Start":"18:47.110 ","End":"18:50.410","Text":"this equation over here."},{"Start":"18:50.410 ","End":"18:55.450","Text":"Here, we have our B_1 and our m_1."},{"Start":"18:55.450 ","End":"19:02.125","Text":"This is our first magnetic field and our first mass where this equation is true,"},{"Start":"19:02.125 ","End":"19:10.510","Text":"and now we want this equation to be true for qR divided by,"},{"Start":"19:10.510 ","End":"19:12.910","Text":"so m_2, our second mass,"},{"Start":"19:12.910 ","End":"19:14.110","Text":"which is unknown,"},{"Start":"19:14.110 ","End":"19:17.770","Text":"we\u0027re trying to find this, multiplied by B_2^2."},{"Start":"19:17.770 ","End":"19:22.300","Text":"Our second magnetic field where of course,"},{"Start":"19:22.300 ","End":"19:26.205","Text":"we know that B_2 is equal to,"},{"Start":"19:26.205 ","End":"19:30.390","Text":"so our magnetic field is 3 times as strong,"},{"Start":"19:30.390 ","End":"19:34.540","Text":"so it\u0027s equal to 3 times B_1."},{"Start":"19:35.410 ","End":"19:42.550","Text":"Therefore, so let\u0027s plug this in,"},{"Start":"19:42.550 ","End":"19:47.740","Text":"so what we\u0027ll get is that E_naught has to be equal to qR divided"},{"Start":"19:47.740 ","End":"19:53.230","Text":"by m_2 which is unknown, multiplied by B_2^2."},{"Start":"19:53.230 ","End":"19:57.477","Text":"What is B_2? B_2 is 3 times B_1."},{"Start":"19:57.477 ","End":"19:59.500","Text":"All of this is squared,"},{"Start":"19:59.500 ","End":"20:03.700","Text":"which is simply equal to 9 multiplied by"},{"Start":"20:03.700 ","End":"20:11.125","Text":"qR divided by m_2 multiplied by B_1^2."},{"Start":"20:11.125 ","End":"20:13.137","Text":"Here we have a 9,"},{"Start":"20:13.137 ","End":"20:16.010","Text":"which means therefore that for"},{"Start":"20:16.010 ","End":"20:23.200","Text":"this equation to be equal to E_naught or equal to this equation,"},{"Start":"20:23.200 ","End":"20:31.040","Text":"we need our m_2 to be equal to 9 times m_1."},{"Start":"20:31.050 ","End":"20:34.516","Text":"Then what we\u0027ll get if so,"},{"Start":"20:34.516 ","End":"20:42.865","Text":"is we\u0027ll get qR divided by m_1 multiplied by B_1^2,"},{"Start":"20:42.865 ","End":"20:49.780","Text":"is equal to 9 times qR divided by"},{"Start":"20:49.780 ","End":"20:59.490","Text":"9m_1"},{"Start":"20:59.490 ","End":"21:00.300","Text":"multiplied"},{"Start":"21:00.300 ","End":"21:02.483","Text":"by B_1^2."},{"Start":"21:02.483 ","End":"21:06.830","Text":"Then the 9 over here and the 9 over here cancel out,"},{"Start":"21:06.830 ","End":"21:10.130","Text":"and we get the exact same equation."},{"Start":"21:10.130 ","End":"21:15.605","Text":"Therefore, this is the answer to question number 4."},{"Start":"21:15.605 ","End":"21:18.780","Text":"That is the end of this lesson.+"}],"ID":25725},{"Watched":false,"Name":"Exercise 10","Duration":"19m 7s","ChapterTopicVideoID":21318,"CourseChapterTopicPlaylistID":99484,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.680","Text":"Hello. In this lesson,"},{"Start":"00:01.680 ","End":"00:04.845","Text":"we\u0027re going to be answering questions which deal with"},{"Start":"00:04.845 ","End":"00:10.440","Text":"closed current loops subject to uniform magnetic fields."},{"Start":"00:10.440 ","End":"00:16.110","Text":"The first question is to show that the magnetic force acting on"},{"Start":"00:16.110 ","End":"00:22.890","Text":"a square current loop and a uniform field perpendicular to the loops plane cancels out."},{"Start":"00:22.890 ","End":"00:28.425","Text":"We have a square loop where we have a current flowing through."},{"Start":"00:28.425 ","End":"00:32.790","Text":"I\u0027ve chosen an arbitrary direction for the current to flow in"},{"Start":"00:32.790 ","End":"00:38.055","Text":"the clockwise direction and we drew this on the plane of the page."},{"Start":"00:38.055 ","End":"00:43.610","Text":"Now, we have a uniform magnetic field perpendicular to the loops plane."},{"Start":"00:43.610 ","End":"00:49.160","Text":"Which means that the magnetic fields can either be coming in or out of the page."},{"Start":"00:49.160 ","End":"00:55.570","Text":"Let\u0027s just say that the magnetic field is going into the page."},{"Start":"00:57.170 ","End":"01:02.090","Text":"What I want to do in this question is I want to show that the magnetic force"},{"Start":"01:02.090 ","End":"01:07.540","Text":"on this entire loop cancels, it equals 0."},{"Start":"01:07.850 ","End":"01:10.940","Text":"Let\u0027s just do the last thing."},{"Start":"01:10.940 ","End":"01:15.040","Text":"Let\u0027s say that each side length is a."},{"Start":"01:15.040 ","End":"01:21.815","Text":"What I\u0027m going to do is I\u0027m going to work out the force on each side of this square."},{"Start":"01:21.815 ","End":"01:25.475","Text":"Then I\u0027m going to add all of the forces up and we\u0027re going to see"},{"Start":"01:25.475 ","End":"01:31.010","Text":"whether the total force on the loop is equal to 0 or not."},{"Start":"01:31.010 ","End":"01:35.180","Text":"Let\u0027s work out the force on the upper edge,"},{"Start":"01:35.180 ","End":"01:38.030","Text":"so that\u0027s the force over here."},{"Start":"01:38.030 ","End":"01:42.907","Text":"Because our magnetic field we\u0027re being told is uniform,"},{"Start":"01:42.907 ","End":"01:50.690","Text":"we can use the equation for the force that it is equal to BI multiplied by L,"},{"Start":"01:50.690 ","End":"01:56.690","Text":"multiplied by sine of the angle between the magnetic field and the current."},{"Start":"01:56.690 ","End":"02:00.600","Text":"Our B is equal to B current is I,"},{"Start":"02:00.600 ","End":"02:04.110","Text":"our side length we said is a,"},{"Start":"02:04.110 ","End":"02:09.405","Text":"and the angle between the current and the magnetic field is 90 degrees."},{"Start":"02:09.405 ","End":"02:11.765","Text":"They\u0027re perpendicular to one another,"},{"Start":"02:11.765 ","End":"02:17.510","Text":"which means that sine of 90 degrees is equal to 1. We just have this."},{"Start":"02:17.510 ","End":"02:21.590","Text":"Now, what we want to do is we want to see the direction."},{"Start":"02:21.590 ","End":"02:23.930","Text":"If we use the right-hand rule,"},{"Start":"02:23.930 ","End":"02:31.416","Text":"so our thumb is in the direction of the current our pointing finger,"},{"Start":"02:31.416 ","End":"02:33.995","Text":"our forefinger is into the page,"},{"Start":"02:33.995 ","End":"02:35.930","Text":"which means that our middle finger,"},{"Start":"02:35.930 ","End":"02:40.880","Text":"which represents the direction of the force is going to be pointing upwards."},{"Start":"02:40.880 ","End":"02:42.680","Text":"I\u0027ll draw in green."},{"Start":"02:42.680 ","End":"02:48.090","Text":"Our force is in this direction."},{"Start":"02:48.290 ","End":"02:54.020","Text":"Let\u0027s just imagine that this is the positive y direction."},{"Start":"02:54.020 ","End":"02:58.710","Text":"We can say that the force is in the positive y direction."},{"Start":"02:58.820 ","End":"03:02.500","Text":"Now, let\u0027s look on the other sides."},{"Start":"03:02.500 ","End":"03:08.491","Text":"If we look at the force on this side over here at the bottom,"},{"Start":"03:08.491 ","End":"03:11.165","Text":"we can see that now the,"},{"Start":"03:11.165 ","End":"03:16.015","Text":"if on the top of the current was flowing in this rightwards direction."},{"Start":"03:16.015 ","End":"03:20.035","Text":"In the bottom, because the current is flowing in the clockwise direction,"},{"Start":"03:20.035 ","End":"03:23.980","Text":"on the bottom, the current is flowing in the leftward\u0027s direction."},{"Start":"03:23.980 ","End":"03:27.460","Text":"If the direction of the current has changed,"},{"Start":"03:27.460 ","End":"03:33.805","Text":"so that means that the force on the bottom side is going to be pointing down."},{"Start":"03:33.805 ","End":"03:39.435","Text":"Then we can see that over here we have equal forces."},{"Start":"03:39.435 ","End":"03:43.640","Text":"We still have BAL sine of Alpha, but they\u0027re opposite."},{"Start":"03:43.640 ","End":"03:46.430","Text":"The force over here is in the positive y-direction,"},{"Start":"03:46.430 ","End":"03:51.695","Text":"and the force over here pointing downwards is in the negative y-direction."},{"Start":"03:51.695 ","End":"03:58.460","Text":"These forces cancel out and if we do the exact same thing on the right and left side,"},{"Start":"03:58.460 ","End":"04:00.950","Text":"we\u0027ll get that the force on the right side is pointing in"},{"Start":"04:00.950 ","End":"04:06.290","Text":"this direction and the force on the left side is pointing in this direction."},{"Start":"04:06.290 ","End":"04:09.815","Text":"Of course, it\u0027s still the same force, it\u0027s still BIa."},{"Start":"04:09.815 ","End":"04:11.810","Text":"But this time it\u0027s in"},{"Start":"04:11.810 ","End":"04:15.905","Text":"the positive x direction and this time it\u0027s in the negative x-direction,"},{"Start":"04:15.905 ","End":"04:19.160","Text":"so they\u0027re equal and opposite and they cancel out."},{"Start":"04:19.160 ","End":"04:27.095","Text":"Therefore, the total force due to the magnetic field is equal to 0"},{"Start":"04:27.095 ","End":"04:30.140","Text":"for a square current loop subject to"},{"Start":"04:30.140 ","End":"04:35.405","Text":"a uniform magnetic field which is perpendicular to its plane."},{"Start":"04:35.405 ","End":"04:37.850","Text":"Now, let\u0027s look at question number 2."},{"Start":"04:37.850 ","End":"04:41.960","Text":"Again, show that the magnetic force acting on a square current loop in"},{"Start":"04:41.960 ","End":"04:48.270","Text":"a uniform field parallel to the loop\u0027s plane cancels out."},{"Start":"04:48.880 ","End":"04:53.240","Text":"Again, we have our square current loop,"},{"Start":"04:53.240 ","End":"04:57.065","Text":"and we\u0027ll choose an arbitrary direction for the current,"},{"Start":"04:57.065 ","End":"04:59.494","Text":"again in the clockwise direction."},{"Start":"04:59.494 ","End":"05:03.380","Text":"This time we\u0027re told that the B field is still uniform."},{"Start":"05:03.380 ","End":"05:06.215","Text":"However, it\u0027s parallel to the loops plane."},{"Start":"05:06.215 ","End":"05:11.212","Text":"Let\u0027s say this is the loops plane on our screens."},{"Start":"05:11.212 ","End":"05:13.910","Text":"Parallel would mean that the B field is either"},{"Start":"05:13.910 ","End":"05:16.924","Text":"pointing in the right direction or in the left direction."},{"Start":"05:16.924 ","End":"05:20.610","Text":"Let\u0027s point it in the right direction,"},{"Start":"05:20.610 ","End":"05:26.190","Text":"so it\u0027s a uniform and it\u0027s pointing like so."},{"Start":"05:26.650 ","End":"05:30.650","Text":"Again, we have the same side length,"},{"Start":"05:30.650 ","End":"05:39.080","Text":"each side over here is of length a and we want to show that our force cancels out."},{"Start":"05:39.080 ","End":"05:42.590","Text":"First of all, the force on these two sides,"},{"Start":"05:42.590 ","End":"05:44.015","Text":"the top and the bottom."},{"Start":"05:44.015 ","End":"05:48.045","Text":"Let\u0027s write this force on the upper side."},{"Start":"05:48.045 ","End":"05:50.370","Text":"Again, it\u0027s equal to BIL."},{"Start":"05:50.370 ","End":"05:54.215","Text":"We have BI and the side length is a,"},{"Start":"05:54.215 ","End":"05:57.920","Text":"and then multiplied by sine of the angle between the two."},{"Start":"05:57.920 ","End":"06:01.850","Text":"The angle between the current and our magnetic field"},{"Start":"06:01.850 ","End":"06:04.920","Text":"is equal to 0 because they\u0027re parallel."},{"Start":"06:04.920 ","End":"06:08.930","Text":"The angle is 0 and sine of 0 is 0."},{"Start":"06:08.930 ","End":"06:10.685","Text":"This is multiplied by 0,"},{"Start":"06:10.685 ","End":"06:12.550","Text":"which is equal to 0."},{"Start":"06:12.550 ","End":"06:14.223","Text":"This is of course,"},{"Start":"06:14.223 ","End":"06:20.870","Text":"also equal to the force over here on the bottom side."},{"Start":"06:20.870 ","End":"06:23.300","Text":"Again, for the same reason,"},{"Start":"06:23.300 ","End":"06:27.380","Text":"they\u0027re parallel so that angle is 0."},{"Start":"06:27.380 ","End":"06:30.330","Text":"Or in this case, because the current is in the opposite direction,"},{"Start":"06:30.330 ","End":"06:32.045","Text":"so the angle is 180."},{"Start":"06:32.045 ","End":"06:35.575","Text":"But of course, sine of 180 is 0."},{"Start":"06:35.575 ","End":"06:42.620","Text":"That means that all I have to do is I have to work out the force on the sides over here,"},{"Start":"06:42.620 ","End":"06:44.330","Text":"the right side and the left side,"},{"Start":"06:44.330 ","End":"06:47.300","Text":"and I want to see if those canceled out."},{"Start":"06:47.300 ","End":"06:48.620","Text":"Let\u0027s work this out,"},{"Start":"06:48.620 ","End":"06:50.720","Text":"the force on the right-hand side."},{"Start":"06:50.720 ","End":"06:54.200","Text":"Again, I have BI multiplied by"},{"Start":"06:54.200 ","End":"06:59.230","Text":"the side length a multiplied by sine of the angle between the two."},{"Start":"06:59.230 ","End":"07:06.020","Text":"Of course, we can see that the angle between the two is 90 degrees and sine of 90 is 1."},{"Start":"07:06.020 ","End":"07:11.145","Text":"Now, we want to know the direction."},{"Start":"07:11.145 ","End":"07:13.130","Text":"Through the right-hand rule,"},{"Start":"07:13.130 ","End":"07:16.645","Text":"our thumb points down in the direction of the current,"},{"Start":"07:16.645 ","End":"07:19.820","Text":"our forefinger points in the right direction,"},{"Start":"07:19.820 ","End":"07:21.650","Text":"in the direction of our B field,"},{"Start":"07:21.650 ","End":"07:25.640","Text":"and our middle finger representing the direction of the force will"},{"Start":"07:25.640 ","End":"07:31.425","Text":"then point outside of the page."},{"Start":"07:31.425 ","End":"07:33.170","Text":"Let\u0027s draw this in green."},{"Start":"07:33.170 ","End":"07:38.190","Text":"Our force over here is coming out of the page."},{"Start":"07:38.190 ","End":"07:42.630","Text":"Then what about the force on this side over here?"},{"Start":"07:46.700 ","End":"07:53.180","Text":"Let\u0027s just for the sake of it say that the z direction is coming out of the page."},{"Start":"07:53.180 ","End":"07:56.165","Text":"This is in the positive z direction."},{"Start":"07:56.165 ","End":"07:58.205","Text":"Then for this side,"},{"Start":"07:58.205 ","End":"08:00.680","Text":"so what we\u0027re going to have is our thumb"},{"Start":"08:00.680 ","End":"08:05.000","Text":"pointing up in the direction of the current over here."},{"Start":"08:05.000 ","End":"08:10.565","Text":"Our forefinger pointing right in the direction of the B field."},{"Start":"08:10.565 ","End":"08:12.890","Text":"Therefore, our middle finger representing"},{"Start":"08:12.890 ","End":"08:16.835","Text":"the direction of the force is pointing into the page."},{"Start":"08:16.835 ","End":"08:21.580","Text":"Here, our force is pointing like so."},{"Start":"08:21.580 ","End":"08:23.563","Text":"We can see that on these sides,"},{"Start":"08:23.563 ","End":"08:25.055","Text":"the forces are equal."},{"Start":"08:25.055 ","End":"08:31.225","Text":"However, they\u0027re in the opposite direction because this is in the negative z direction."},{"Start":"08:31.225 ","End":"08:37.685","Text":"Or of course, this is equal to negative the force on the left-hand side,"},{"Start":"08:37.685 ","End":"08:46.868","Text":"and therefore will get that the total force is equal to 0 plus 0,"},{"Start":"08:46.868 ","End":"08:50.520","Text":"and then plus BIa minus BIa,"},{"Start":"08:50.520 ","End":"08:53.805","Text":"which is equal to 0."},{"Start":"08:53.805 ","End":"08:58.340","Text":"The magnetic force cancels out."},{"Start":"08:58.340 ","End":"09:01.670","Text":"Now, let\u0027s take a look at question number 3,"},{"Start":"09:01.670 ","End":"09:05.420","Text":"show that the magnetic force acting on a square current loop in"},{"Start":"09:05.420 ","End":"09:09.980","Text":"a uniform field cancels out always."},{"Start":"09:09.980 ","End":"09:14.180","Text":"Now, just a quick note before we move on to 3,"},{"Start":"09:14.180 ","End":"09:19.445","Text":"what we can see is that our loop over here"},{"Start":"09:19.445 ","End":"09:22.100","Text":"is going to be rotating because here"},{"Start":"09:22.100 ","End":"09:25.175","Text":"the force is pushing it out and here it\u0027s pushing it in."},{"Start":"09:25.175 ","End":"09:27.755","Text":"The loop is going to rotate however,"},{"Start":"09:27.755 ","End":"09:29.795","Text":"the loop\u0027s center of mass,"},{"Start":"09:29.795 ","End":"09:32.035","Text":"which is over here, CM,"},{"Start":"09:32.035 ","End":"09:34.905","Text":"is going to remain stationary."},{"Start":"09:34.905 ","End":"09:37.050","Text":"It isn\u0027t going to move."},{"Start":"09:37.050 ","End":"09:42.185","Text":"That is, as a result of the sum of the forces being equal to 0."},{"Start":"09:42.185 ","End":"09:45.905","Text":"The sum of the torques is not equal to 0 because this is rotating,"},{"Start":"09:45.905 ","End":"09:49.820","Text":"but the sum of the forces on this body is equal to 0."},{"Start":"09:49.820 ","End":"09:54.740","Text":"That\u0027s why the center of mass isn\u0027t going to move or travel anywhere."},{"Start":"09:55.130 ","End":"09:57.960","Text":"Back to question number 3."},{"Start":"09:57.960 ","End":"10:02.570","Text":"Again, we\u0027re dealing with our square current loop and again,"},{"Start":"10:02.570 ","End":"10:06.800","Text":"we\u0027ll say that our current is traveling in this clockwise direction."},{"Start":"10:06.800 ","End":"10:10.055","Text":"Of course, it\u0027s correct if the current is reversed,"},{"Start":"10:10.055 ","End":"10:13.054","Text":"just the forces will also be reversed."},{"Start":"10:13.054 ","End":"10:18.785","Text":"Now, we\u0027re talking about a uniform magnetic field in any direction."},{"Start":"10:18.785 ","End":"10:20.750","Text":"Let\u0027s draw it."},{"Start":"10:20.750 ","End":"10:27.620","Text":"I\u0027m going to say that my magnetic field is in this direction like so."},{"Start":"10:27.620 ","End":"10:32.315","Text":"Now, of course, the magnetic field is also"},{"Start":"10:32.315 ","End":"10:38.375","Text":"pointing in the diagonal with respect to the plane to the screen."},{"Start":"10:38.375 ","End":"10:45.545","Text":"But it\u0027s also at some angle in the z-direction or in the y-direction."},{"Start":"10:45.545 ","End":"10:52.550","Text":"It\u0027s going inwards a little bit in this direction,"},{"Start":"10:52.550 ","End":"10:55.885","Text":"and in some direction like so."},{"Start":"10:55.885 ","End":"11:01.700","Text":"Because these arrows representing the B field are in three dimensions."},{"Start":"11:01.700 ","End":"11:05.790","Text":"I\u0027m going to write 3D, just remember this."},{"Start":"11:06.500 ","End":"11:09.390","Text":"The b field is in 3 dimensions,"},{"Start":"11:09.390 ","End":"11:15.750","Text":"which means that it has some angle Theta with respect to the plane of the page and"},{"Start":"11:15.750 ","End":"11:24.405","Text":"some angle Alpha going in or pointing out of the page however you want to look at it."},{"Start":"11:24.405 ","End":"11:32.670","Text":"What does that mean? That means that we can split this b field into components."},{"Start":"11:32.670 ","End":"11:33.959","Text":"What are the components?"},{"Start":"11:33.959 ","End":"11:38.490","Text":"We have b field that is perpendicular to the plane,"},{"Start":"11:38.490 ","End":"11:44.100","Text":"so that\u0027s going to be this little arrow over here going inwards."},{"Start":"11:44.100 ","End":"11:48.765","Text":"The b field, which is parallel to the plane."},{"Start":"11:48.765 ","End":"11:52.710","Text":"Which is this arrow going to the side."},{"Start":"11:52.710 ","End":"11:55.200","Text":"We have 2 components."},{"Start":"11:55.200 ","End":"11:56.880","Text":"From question 1 and 2,"},{"Start":"11:56.880 ","End":"12:02.625","Text":"we saw that if the b field is perpendicular or parallel to the plane,"},{"Start":"12:02.625 ","End":"12:07.410","Text":"the magnetic force acting on the loop cancels out."},{"Start":"12:07.410 ","End":"12:11.610","Text":"That\u0027s what we saw. If we have these 2 components,"},{"Start":"12:11.610 ","End":"12:13.230","Text":"we know that for each component,"},{"Start":"12:13.230 ","End":"12:15.390","Text":"if we calculate the force on the loop,"},{"Start":"12:15.390 ","End":"12:16.965","Text":"the force will cancel out."},{"Start":"12:16.965 ","End":"12:21.375","Text":"The f total parallel is equal to 0 as we saw,"},{"Start":"12:21.375 ","End":"12:28.120","Text":"and the f total perpendicular is equal to 0 as we saw."},{"Start":"12:29.540 ","End":"12:33.660","Text":"This we\u0027ve just proven in the previous questions."},{"Start":"12:33.660 ","End":"12:35.790","Text":"This is of course, correct,"},{"Start":"12:35.790 ","End":"12:40.395","Text":"whichever direction the b field is pointing in so long as it\u0027s uniform"},{"Start":"12:40.395 ","End":"12:46.125","Text":"because you can split it up into these 2 components and only these 2 components."},{"Start":"12:46.125 ","End":"12:51.420","Text":"Question number 4 is asking an even more generalized question."},{"Start":"12:51.420 ","End":"12:55.890","Text":"Here, we want to show that the magnetic force acting on"},{"Start":"12:55.890 ","End":"13:01.710","Text":"a closed current loop of any shape in a uniform field cancels out,"},{"Start":"13:01.710 ","End":"13:03.270","Text":"so in any shape."},{"Start":"13:03.270 ","End":"13:08.400","Text":"Here we\u0027ve been looking at a closed current loop in the shape of a square."},{"Start":"13:08.400 ","End":"13:11.025","Text":"What we want to show is any shape,"},{"Start":"13:11.025 ","End":"13:13.290","Text":"be it a square,"},{"Start":"13:13.290 ","End":"13:17.910","Text":"a circle, some kind of shape like this,"},{"Start":"13:17.910 ","End":"13:19.605","Text":"any shape that it may be,"},{"Start":"13:19.605 ","End":"13:23.910","Text":"the magnetic force will be equal to 0 if there is"},{"Start":"13:23.910 ","End":"13:30.010","Text":"a uniform magnetic field in this region pointing in any direction."},{"Start":"13:30.440 ","End":"13:35.605","Text":"I have some closed current loop."},{"Start":"13:35.605 ","End":"13:38.270","Text":"It\u0027s in this shape."},{"Start":"13:38.270 ","End":"13:45.720","Text":"Again, I\u0027m saying that my current is flowing in this clockwise direction like so."},{"Start":"13:45.720 ","End":"13:55.155","Text":"I know that I have a b field or so which is uniform, so uniform."},{"Start":"13:55.155 ","End":"13:59.100","Text":"But I don\u0027t know the direction of the b field."},{"Start":"13:59.100 ","End":"14:02.010","Text":"I don\u0027t know its orientation."},{"Start":"14:02.010 ","End":"14:05.130","Text":"What I\u0027m going to do now is I\u0027m going to take"},{"Start":"14:05.130 ","End":"14:09.015","Text":"some kind of slice from this closed current loop."},{"Start":"14:09.015 ","End":"14:13.755","Text":"The length of the slice is dl."},{"Start":"14:13.755 ","End":"14:20.370","Text":"Of course, the direction is in the direction of the slice,"},{"Start":"14:20.370 ","End":"14:23.085","Text":"which means it\u0027s in the direction of the current."},{"Start":"14:23.085 ","End":"14:25.214","Text":"Over here specifically,"},{"Start":"14:25.214 ","End":"14:28.330","Text":"it\u0027s pointing in this direction."},{"Start":"14:29.540 ","End":"14:38.595","Text":"We already know that the equation for the force on this small piece over here."},{"Start":"14:38.595 ","End":"14:46.150","Text":"That is going to be equal to Idl cross B."},{"Start":"14:46.640 ","End":"14:52.695","Text":"Now if I want to know the total force on this entire loop,"},{"Start":"14:52.695 ","End":"14:57.390","Text":"all I have to do is integrate along df,"},{"Start":"14:57.390 ","End":"15:05.200","Text":"which means that I\u0027m integrating along Idl cross B."},{"Start":"15:05.450 ","End":"15:13.950","Text":"Now an easier way to look at it is if I switch around my dl and my B."},{"Start":"15:13.950 ","End":"15:18.510","Text":"If I switch the order when I\u0027m doing a cross product,"},{"Start":"15:18.510 ","End":"15:23.175","Text":"then I have to multiply my equation by negative 1."},{"Start":"15:23.175 ","End":"15:32.205","Text":"What I have is negative the integral of IB cross dl."},{"Start":"15:32.205 ","End":"15:36.120","Text":"It\u0027s just going to be easier for us to work with it like this."},{"Start":"15:36.120 ","End":"15:37.410","Text":"You\u0027ll soon see why."},{"Start":"15:37.410 ","End":"15:41.175","Text":"Because I switched my dl with my B,"},{"Start":"15:41.175 ","End":"15:44.085","Text":"I have to multiply this by negative 1."},{"Start":"15:44.085 ","End":"15:48.075","Text":"Of course, my eye is constant throughout."},{"Start":"15:48.075 ","End":"15:50.130","Text":"If I\u0027m looking at this piece over here,"},{"Start":"15:50.130 ","End":"15:53.580","Text":"the current flowing through this piece over here is going to be"},{"Start":"15:53.580 ","End":"15:58.635","Text":"the same current flowing through this piece over here and so on."},{"Start":"15:58.635 ","End":"16:00.900","Text":"My current is constant."},{"Start":"16:00.900 ","End":"16:11.500","Text":"I can say that this is equal to negative I multiplied by the integral of B cross dl."},{"Start":"16:12.400 ","End":"16:18.475","Text":"The next thing that I know is that I know that my b is uniform."},{"Start":"16:18.475 ","End":"16:20.339","Text":"That\u0027s in the question."},{"Start":"16:20.339 ","End":"16:22.260","Text":"My B is uniform,"},{"Start":"16:22.260 ","End":"16:24.135","Text":"which means that it\u0027s a constant."},{"Start":"16:24.135 ","End":"16:26.160","Text":"Which means that if it\u0027s a constant,"},{"Start":"16:26.160 ","End":"16:29.835","Text":"I can take it out of the integral."},{"Start":"16:29.835 ","End":"16:32.805","Text":"Let\u0027s carry on over here to give us a bit more space."},{"Start":"16:32.805 ","End":"16:36.750","Text":"I have negative IB,"},{"Start":"16:36.750 ","End":"16:40.275","Text":"and then I still have this cross-product over here."},{"Start":"16:40.275 ","End":"16:47.160","Text":"I write the cross out here and then I keep my integral of dI like so."},{"Start":"16:47.160 ","End":"16:51.315","Text":"If you have a cross-product in the integral sign and you need to"},{"Start":"16:51.315 ","End":"16:56.655","Text":"split this up to get the first item in the cross-product out,"},{"Start":"16:56.655 ","End":"16:58.170","Text":"so you will take it out."},{"Start":"16:58.170 ","End":"16:59.340","Text":"Here it\u0027s the B,"},{"Start":"16:59.340 ","End":"17:01.545","Text":"you take the cross-product out as well."},{"Start":"17:01.545 ","End":"17:08.080","Text":"Then what you are in fact doing is the cross-product on the integral of dl."},{"Start":"17:09.440 ","End":"17:13.110","Text":"What I\u0027m doing now is I\u0027m summing up all of"},{"Start":"17:13.110 ","End":"17:18.420","Text":"my dl vectors and then I do this cross-product."},{"Start":"17:18.420 ","End":"17:20.535","Text":"What I\u0027m doing is,"},{"Start":"17:20.535 ","End":"17:22.965","Text":"so here\u0027s my first dl vector."},{"Start":"17:22.965 ","End":"17:27.360","Text":"I add that onto this over here to this over here,"},{"Start":"17:27.360 ","End":"17:30.405","Text":"and so on and so forth."},{"Start":"17:30.405 ","End":"17:34.300","Text":"I\u0027m summing up all my dl\u0027s around."},{"Start":"17:34.370 ","End":"17:39.285","Text":"Eventually, I\u0027ll get to my starting point."},{"Start":"17:39.285 ","End":"17:41.190","Text":"What happens with vectors,"},{"Start":"17:41.190 ","End":"17:44.730","Text":"if this was my starting point over here,"},{"Start":"17:44.730 ","End":"17:47.249","Text":"as we know with vectors,"},{"Start":"17:47.249 ","End":"17:49.410","Text":"I started my starting point."},{"Start":"17:49.410 ","End":"17:51.855","Text":"If I ended my starting point,"},{"Start":"17:51.855 ","End":"17:53.760","Text":"when I add all of these vectors up,"},{"Start":"17:53.760 ","End":"17:57.490","Text":"it\u0027s equal to 0."},{"Start":"17:58.070 ","End":"18:02.175","Text":"What we\u0027re doing over here and this integral along dl,"},{"Start":"18:02.175 ","End":"18:09.705","Text":"we\u0027re taking the displacement from point a all the way to point B."},{"Start":"18:09.705 ","End":"18:13.095","Text":"But point a and point b are the same place,"},{"Start":"18:13.095 ","End":"18:19.170","Text":"which means that the displacement between a and b is equal to 0."},{"Start":"18:19.170 ","End":"18:22.125","Text":"That means that when we sum up all of these vectors,"},{"Start":"18:22.125 ","End":"18:25.710","Text":"it\u0027s equal over here to 0."},{"Start":"18:25.710 ","End":"18:30.285","Text":"Then when we do this cross-product with 0,"},{"Start":"18:30.285 ","End":"18:32.730","Text":"this is equal to 0."},{"Start":"18:32.730 ","End":"18:41.410","Text":"Therefore we get that the total force is equal to 0."},{"Start":"18:42.530 ","End":"18:47.400","Text":"Now we can see that whatever shape to loop we have,"},{"Start":"18:47.400 ","End":"18:51.738","Text":"as long as it\u0027s a closed loop and there\u0027s a current flowing through it,"},{"Start":"18:51.738 ","End":"18:55.335","Text":"if this loop is subject to a uniform magnetic field,"},{"Start":"18:55.335 ","End":"18:58.905","Text":"no matter the orientation of the magnetic field,"},{"Start":"18:58.905 ","End":"19:04.830","Text":"the sum of all of the forces on the loop will be equal to 0."},{"Start":"19:04.830 ","End":"19:08.020","Text":"That is the end of this lesson."}],"ID":21410},{"Watched":false,"Name":"Exercise 11","Duration":"10m 7s","ChapterTopicVideoID":24814,"CourseChapterTopicPlaylistID":99484,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.380","Text":"Hello. In this lesson we\u0027re going to answer in the following question."},{"Start":"00:04.380 ","End":"00:08.535","Text":"A particle beam with a mass of m and charge q"},{"Start":"00:08.535 ","End":"00:13.799","Text":"encounters a region where a uniform magnetic field is present."},{"Start":"00:13.799 ","End":"00:19.590","Text":"The magnetic field is perpendicular to the plane of the page, and going inwards."},{"Start":"00:19.590 ","End":"00:23.025","Text":"The particles possess kinetic energy E_k,"},{"Start":"00:23.025 ","End":"00:30.390","Text":"and they enter the magnetized zone at an angle of Theta as shown in the diagram."},{"Start":"00:30.390 ","End":"00:37.200","Text":"Question number 1 is to determine the vertical distance y traversed by"},{"Start":"00:37.200 ","End":"00:45.480","Text":"the particles from their entry point into the magnetic zone and until they exit."},{"Start":"00:45.830 ","End":"00:51.470","Text":"As we know, when a particle travels through a magnetic field,"},{"Start":"00:51.470 ","End":"00:54.980","Text":"it moves in circular motion due to"},{"Start":"00:54.980 ","End":"00:59.975","Text":"the velocity being perpendicular to the magnetic field."},{"Start":"00:59.975 ","End":"01:06.230","Text":"Here we can see that the velocity is on the plane of the page and the magnetic field is"},{"Start":"01:06.230 ","End":"01:09.200","Text":"perpendicular to the plane of the page which means"},{"Start":"01:09.200 ","End":"01:13.070","Text":"that the velocity is perpendicular to the magnetic field."},{"Start":"01:13.070 ","End":"01:16.215","Text":"That means that our particle moves in"},{"Start":"01:16.215 ","End":"01:20.675","Text":"circular motion from its entry point to its exit point."},{"Start":"01:20.675 ","End":"01:26.795","Text":"Now let\u0027s just draw the rest of the circle, like so."},{"Start":"01:26.795 ","End":"01:32.735","Text":"Let\u0027s say that this is the center of the circle somewhere over here."},{"Start":"01:32.735 ","End":"01:36.305","Text":"We know that from the exit point to this point,"},{"Start":"01:36.305 ","End":"01:43.490","Text":"this is a radius of R. Similarly from the entrance point to the center of the circle,"},{"Start":"01:43.490 ","End":"01:49.190","Text":"this is also radius R. The next thing that we know is that here we have"},{"Start":"01:49.190 ","End":"01:55.195","Text":"a velocity and the velocity passes through into this region,"},{"Start":"01:55.195 ","End":"01:57.955","Text":"this point over here, the entrance point,"},{"Start":"01:57.955 ","End":"02:05.655","Text":"which means that over here the velocity has to be at 90 degrees to the radius."},{"Start":"02:05.655 ","End":"02:08.590","Text":"Remember, because if we\u0027re looking at a circle,"},{"Start":"02:08.590 ","End":"02:11.755","Text":"the velocity is tangent to the circle."},{"Start":"02:11.755 ","End":"02:14.140","Text":"Similarly here at the exit points,"},{"Start":"02:14.140 ","End":"02:19.030","Text":"this is also 90 degrees because then we have velocity over here,"},{"Start":"02:19.030 ","End":"02:23.570","Text":"which is also tangent to the circle."},{"Start":"02:23.600 ","End":"02:26.620","Text":"The diagram isn\u0027t exactly accurate,"},{"Start":"02:26.620 ","End":"02:27.820","Text":"but you can imagine."},{"Start":"02:27.820 ","End":"02:33.505","Text":"Remember the velocity is always tangent or at 90 degrees to the circle."},{"Start":"02:33.505 ","End":"02:42.704","Text":"That means if this angle over here is Theta and here we also have 90 degrees,"},{"Start":"02:42.704 ","End":"02:48.155","Text":"let me draw this in gray so that you can see it."},{"Start":"02:48.155 ","End":"02:52.069","Text":"Here we have 90 degrees between the v and the radius."},{"Start":"02:52.069 ","End":"02:55.940","Text":"But also here we have 90 degrees,"},{"Start":"02:55.940 ","End":"03:06.270","Text":"which means that this angle over here is 90 minus Theta."},{"Start":"03:06.270 ","End":"03:16.390","Text":"Then that means that this angle over here has to therefore be Theta as well."},{"Start":"03:17.480 ","End":"03:22.010","Text":"Now what I\u0027m going to do is I\u0027m going to draw a line"},{"Start":"03:22.010 ","End":"03:27.710","Text":"connecting the center of the circle to over here."},{"Start":"03:27.710 ","End":"03:30.830","Text":"Then we can see that we have an angle of"},{"Start":"03:30.830 ","End":"03:37.340","Text":"90 degrees over here between the dotted line and this gray line over here."},{"Start":"03:37.340 ","End":"03:40.085","Text":"I\u0027ll just draw it in black to make it clearer."},{"Start":"03:40.085 ","End":"03:42.110","Text":"We draw this line."},{"Start":"03:42.110 ","End":"03:44.900","Text":"Then between here and here,"},{"Start":"03:44.900 ","End":"03:48.185","Text":"we also have a 90 degree angle."},{"Start":"03:48.185 ","End":"03:55.775","Text":"If this angle over here is Theta between this gray line and our radius,"},{"Start":"03:55.775 ","End":"03:57.410","Text":"then as we can see,"},{"Start":"03:57.410 ","End":"03:59.555","Text":"we have the z shape over here,"},{"Start":"03:59.555 ","End":"04:02.240","Text":"which means we have alternate angles."},{"Start":"04:02.240 ","End":"04:06.890","Text":"Remember, if you have 2 parallel lines,"},{"Start":"04:06.890 ","End":"04:12.755","Text":"then this angle over here will be equal to this angle over here,"},{"Start":"04:12.755 ","End":"04:14.330","Text":"which is exactly what we have."},{"Start":"04:14.330 ","End":"04:16.700","Text":"This angle over here, Theta,"},{"Start":"04:16.700 ","End":"04:20.805","Text":"is equal to this angle over here."},{"Start":"04:20.805 ","End":"04:25.200","Text":"This angle over here is also Theta."},{"Start":"04:27.770 ","End":"04:31.335","Text":"Then we can draw this line over here."},{"Start":"04:31.335 ","End":"04:36.145","Text":"All in all we have this triangle down the radius,"},{"Start":"04:36.145 ","End":"04:38.020","Text":"up this length over here,"},{"Start":"04:38.020 ","End":"04:40.375","Text":"which is the length of 1/2y."},{"Start":"04:40.375 ","End":"04:42.820","Text":"The total length over here is y,"},{"Start":"04:42.820 ","End":"04:46.090","Text":"if you remember, it\u0027s written under this blue dot."},{"Start":"04:46.090 ","End":"04:50.155","Text":"This 1/2 line over here is 1/2y,"},{"Start":"04:50.155 ","End":"04:56.160","Text":"and then here the dotted line makes it full y and then over here."},{"Start":"04:56.160 ","End":"05:04.370","Text":"This is our triangle where this length over here is R. Here we have 90 degrees."},{"Start":"05:04.370 ","End":"05:12.110","Text":"What we have over here is a length R. Here an angle of Theta,"},{"Start":"05:12.110 ","End":"05:16.860","Text":"and this length over here is 1/2y."},{"Start":"05:17.950 ","End":"05:25.070","Text":"We know that the radius of this circle from"},{"Start":"05:25.070 ","End":"05:27.680","Text":"the equations we\u0027ve seen in previous lessons is equal to"},{"Start":"05:27.680 ","End":"05:32.460","Text":"mass times the velocity divided by qB."},{"Start":"05:35.090 ","End":"05:37.940","Text":"Now from trigonometry we have,"},{"Start":"05:37.940 ","End":"05:40.040","Text":"here\u0027s our angle Theta,"},{"Start":"05:40.040 ","End":"05:46.860","Text":"which means that the opposite side is equal to 1/2y."},{"Start":"05:47.200 ","End":"05:57.705","Text":"The hypotenuse is equal to R. We know R and 1/2y."},{"Start":"05:57.705 ","End":"05:59.705","Text":"What we\u0027re trying to find out is y,"},{"Start":"05:59.705 ","End":"06:02.005","Text":"we want to isolate this out."},{"Start":"06:02.005 ","End":"06:06.995","Text":"What trig identity links the opposite side and the hypotenuse?"},{"Start":"06:06.995 ","End":"06:08.870","Text":"It\u0027s sine."},{"Start":"06:08.870 ","End":"06:10.160","Text":"We know"},{"Start":"06:10.160 ","End":"06:20.044","Text":"that 1/2"},{"Start":"06:20.044 ","End":"06:24.370","Text":"of y is equal to r multiplied by sine of that angle Theta over here."},{"Start":"06:24.410 ","End":"06:31.500","Text":"Because we\u0027re using sine is equal to opposite over hypotenuse."},{"Start":"06:32.840 ","End":"06:35.670","Text":"Let\u0027s substitute this in."},{"Start":"06:35.670 ","End":"06:38.510","Text":"We have y is equal to,"},{"Start":"06:38.510 ","End":"06:40.520","Text":"we\u0027re multiplying both sides by 2."},{"Start":"06:40.520 ","End":"06:42.470","Text":"2 multiplied by R,"},{"Start":"06:42.470 ","End":"06:50.715","Text":"which is equal to mv divided by qB multiplied by sine of Theta."},{"Start":"06:50.715 ","End":"06:53.285","Text":"Of course, Theta is given to us in the question."},{"Start":"06:53.285 ","End":"06:55.040","Text":"Now, v isn\u0027t."},{"Start":"06:55.040 ","End":"06:58.805","Text":"But we\u0027re given in the question, the kinetic energy."},{"Start":"06:58.805 ","End":"07:05.205","Text":"As we know, E_k is equal to 1/2mv^2."},{"Start":"07:05.205 ","End":"07:10.400","Text":"Therefore we can isolate our v and we get that our velocity is equal to"},{"Start":"07:10.400 ","End":"07:18.115","Text":"the square root of 2 times the kinetic energy divided by the mass."},{"Start":"07:18.115 ","End":"07:20.940","Text":"Now we can substitute that in."},{"Start":"07:20.940 ","End":"07:28.520","Text":"We get that y is equal to 2m divided by qB multiplied by v,"},{"Start":"07:28.520 ","End":"07:32.210","Text":"which is a square root of 2E_k divided by"},{"Start":"07:32.210 ","End":"07:37.565","Text":"m. Then all of this is multiplied by sine of Theta."},{"Start":"07:37.565 ","End":"07:42.140","Text":"Now we can put this into the square root sign."},{"Start":"07:42.140 ","End":"07:47.495","Text":"What we\u0027ll get is we\u0027ll have 2^2"},{"Start":"07:47.495 ","End":"07:52.550","Text":"in the square root sign multiplied by 2 so we\u0027ll have 8 and then m"},{"Start":"07:52.550 ","End":"07:57.065","Text":"squared divided by m so we have m in the numerator multiplied by"},{"Start":"07:57.065 ","End":"08:04.865","Text":"E_k divided by q^2B."},{"Start":"08:04.865 ","End":"08:12.420","Text":"All of this square root multiplied by sine of Theta."},{"Start":"08:13.120 ","End":"08:19.040","Text":"Or alternatively, let\u0027s not put this in the square root sign,"},{"Start":"08:19.040 ","End":"08:22.370","Text":"so we just have the numerator and the square root sign,"},{"Start":"08:22.370 ","End":"08:24.155","Text":"and then divided by qB."},{"Start":"08:24.155 ","End":"08:27.120","Text":"There\u0027s no need to put that in."},{"Start":"08:28.220 ","End":"08:31.690","Text":"Now let\u0027s answer Question number 2,"},{"Start":"08:31.690 ","End":"08:35.890","Text":"determine the exit angle Theta tag."},{"Start":"08:35.890 ","End":"08:40.470","Text":"What we can see is that Theta tag is equal to Theta."},{"Start":"08:40.470 ","End":"08:47.415","Text":"This is exactly for the reason that we saw that the velocity is tangent to the circle."},{"Start":"08:47.415 ","End":"08:51.530","Text":"Of course, when a particle travels through a magnetic field,"},{"Start":"08:51.530 ","End":"08:55.055","Text":"the magnitude of the velocity doesn\u0027t change,"},{"Start":"08:55.055 ","End":"08:58.400","Text":"but the direction will change."},{"Start":"08:58.400 ","End":"08:59.825","Text":"At this exit point,"},{"Start":"08:59.825 ","End":"09:03.335","Text":"it\u0027s still going to be tangent to the circle."},{"Start":"09:03.335 ","End":"09:07.895","Text":"It\u0027s going to be at the same angle as when it was entering."},{"Start":"09:07.895 ","End":"09:12.750","Text":"Let\u0027s just draw another circle over here just to highlight this point."},{"Start":"09:13.130 ","End":"09:19.010","Text":"Here\u0027s the circle which represents the trajectory over here."},{"Start":"09:19.010 ","End":"09:21.470","Text":"From here to here,"},{"Start":"09:21.470 ","End":"09:27.770","Text":"it represents the trajectory of a particle traveling through this magnetic fields."},{"Start":"09:27.770 ","End":"09:36.080","Text":"Then what we can see is that if we draw the velocity over here in the entrance,"},{"Start":"09:36.080 ","End":"09:40.630","Text":"so it\u0027s going to be like this tangent to the circle."},{"Start":"09:40.630 ","End":"09:44.960","Text":"Then when it exits through the exit point,"},{"Start":"09:44.960 ","End":"09:49.265","Text":"it\u0027s going to be like so also tangent to the circle."},{"Start":"09:49.265 ","End":"09:51.965","Text":"We can see that this angle over here,"},{"Start":"09:51.965 ","End":"09:57.020","Text":"Theta is equal to this angle over here."},{"Start":"09:57.020 ","End":"10:02.030","Text":"Therefore we can see that Theta is equal to Theta tag."},{"Start":"10:02.030 ","End":"10:04.730","Text":"That\u0027s the answer to Question number 2."},{"Start":"10:04.730 ","End":"10:07.830","Text":"This is the end of the lesson."}],"ID":25727},{"Watched":false,"Name":"Exercise 12","Duration":"15m 27s","ChapterTopicVideoID":24811,"CourseChapterTopicPlaylistID":99484,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.920","Text":"Hello. In this lesson,"},{"Start":"00:01.920 ","End":"00:04.574","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.574 ","End":"00:12.030","Text":"A uniform magnetic field B is present in a rectangular area, a times b."},{"Start":"00:12.030 ","End":"00:15.225","Text":"Its direction is out of the page."},{"Start":"00:15.225 ","End":"00:17.010","Text":"Outside of this region,"},{"Start":"00:17.010 ","End":"00:18.715","Text":"the field is 0."},{"Start":"00:18.715 ","End":"00:24.200","Text":"Charge q enters the middle of the region with the velocity of"},{"Start":"00:24.200 ","End":"00:30.800","Text":"v and direction that is perpendicular to the side of the rectangle."},{"Start":"00:30.800 ","End":"00:36.685","Text":"The charge then exits the rectangle via its upper edge."},{"Start":"00:36.685 ","End":"00:38.635","Text":"Question Number 1 is,"},{"Start":"00:38.635 ","End":"00:41.055","Text":"what is the charge\u0027s polarity,"},{"Start":"00:41.055 ","End":"00:47.610","Text":"and what is the magnitude of its velocity as it exits the rectangle?"},{"Start":"00:49.040 ","End":"00:52.995","Text":"Our charge enters over here, the center."},{"Start":"00:52.995 ","End":"00:57.920","Text":"We know that when a charge is traveling through a magnetic field,"},{"Start":"00:57.920 ","End":"00:59.870","Text":"we can see that the magnetic field is"},{"Start":"00:59.870 ","End":"01:03.935","Text":"perpendicular to the direction of travel of the charge."},{"Start":"01:03.935 ","End":"01:06.620","Text":"Then we know that the charge is going to travel in"},{"Start":"01:06.620 ","End":"01:13.670","Text":"circular motion until it reaches this point over here along the upper edge,"},{"Start":"01:13.670 ","End":"01:17.510","Text":"where it will exit the region of the magnetic field,"},{"Start":"01:17.510 ","End":"01:20.660","Text":"and then it will travel outwards."},{"Start":"01:20.660 ","End":"01:27.420","Text":"What we\u0027re going to do now is we\u0027re going to show this circular motion."},{"Start":"01:27.420 ","End":"01:29.674","Text":"Let\u0027s draw the circle."},{"Start":"01:29.674 ","End":"01:32.790","Text":"So it\u0027s traveling like so,"},{"Start":"01:32.790 ","End":"01:35.460","Text":"it exits the edge."},{"Start":"01:35.460 ","End":"01:38.960","Text":"Then we can just carry on because it makes it easier to"},{"Start":"01:38.960 ","End":"01:42.890","Text":"understand the shape of the circle,"},{"Start":"01:42.890 ","End":"01:48.000","Text":"the motion that it would follow if this whole region was in the magnetic field."},{"Start":"01:48.140 ","End":"01:54.965","Text":"What we have is that our charge q travels like so in a straight line,"},{"Start":"01:54.965 ","End":"01:57.590","Text":"then it enters the region of the magnetic fields."},{"Start":"01:57.590 ","End":"01:59.525","Text":"It travels in a circular motion,"},{"Start":"01:59.525 ","End":"02:01.760","Text":"and then when it exits the magnetic fields,"},{"Start":"02:01.760 ","End":"02:07.385","Text":"it carries on like so moving in a straight line in this direction."},{"Start":"02:07.385 ","End":"02:09.335","Text":"That\u0027s important to know."},{"Start":"02:09.335 ","End":"02:14.960","Text":"Of course, its velocity is always going to be tangent here when it enters,"},{"Start":"02:14.960 ","End":"02:18.665","Text":"it\u0027s going to be tangent to the circle that we drew."},{"Start":"02:18.665 ","End":"02:20.750","Text":"Also over here at the exit,"},{"Start":"02:20.750 ","End":"02:25.525","Text":"the direction is tangent to the circle at this point."},{"Start":"02:25.525 ","End":"02:29.370","Text":"Now let\u0027s draw the radius of the circle."},{"Start":"02:29.370 ","End":"02:34.205","Text":"As we know, the radius is always going to be perpendicular to the tangent."},{"Start":"02:34.205 ","End":"02:37.145","Text":"If this over here is the exit point,"},{"Start":"02:37.145 ","End":"02:42.765","Text":"so we\u0027re going to draw a line which is perpendicular to this exit point."},{"Start":"02:42.765 ","End":"02:45.840","Text":"Here we have a 90-degree angle."},{"Start":"02:45.840 ","End":"02:49.370","Text":"That would mean that around about here is"},{"Start":"02:49.370 ","End":"02:54.335","Text":"the center of the circle and similarly over here."},{"Start":"02:54.335 ","End":"02:59.959","Text":"Here, of course, this radius over here is tangent to this line over here,"},{"Start":"02:59.959 ","End":"03:03.780","Text":"so this is also a 90-degree angle."},{"Start":"03:04.910 ","End":"03:08.190","Text":"This is the radius,"},{"Start":"03:08.190 ","End":"03:10.140","Text":"and this is the radius."},{"Start":"03:10.140 ","End":"03:16.505","Text":"What we\u0027ve done here will help us a bit in this question also in the next few questions."},{"Start":"03:16.505 ","End":"03:18.680","Text":"Let\u0027s go back to our Question Number 1,"},{"Start":"03:18.680 ","End":"03:22.385","Text":"we\u0027re trying to find the charge\u0027s polarity."},{"Start":"03:22.385 ","End":"03:25.040","Text":"The first thing that we\u0027re going to do is we\u0027re going to"},{"Start":"03:25.040 ","End":"03:28.430","Text":"remember our equation for the magnetic force,"},{"Start":"03:28.430 ","End":"03:30.350","Text":"and that is equal to q,"},{"Start":"03:30.350 ","End":"03:37.330","Text":"the charge multiplied by v cross B."},{"Start":"03:37.610 ","End":"03:41.510","Text":"What we\u0027re going to do is we\u0027re going to use the right-hand rule."},{"Start":"03:41.510 ","End":"03:44.675","Text":"Now, remember, we can either use the thumb,"},{"Start":"03:44.675 ","End":"03:47.195","Text":"the forefinger, and the middle finger,"},{"Start":"03:47.195 ","End":"03:50.525","Text":"or we can use our thumb,"},{"Start":"03:50.525 ","End":"03:54.890","Text":"our forefingers to point in the direction of the B."},{"Start":"03:54.890 ","End":"04:00.870","Text":"Then the curl in the direction of the force."},{"Start":"04:01.310 ","End":"04:03.770","Text":"We can use either one,"},{"Start":"04:03.770 ","End":"04:06.140","Text":"but let\u0027s use our thumb,"},{"Start":"04:06.140 ","End":"04:09.800","Text":"forefinger, or pointing finger and our middle finger."},{"Start":"04:09.800 ","End":"04:13.230","Text":"We have our thumb representing the velocity."},{"Start":"04:13.230 ","End":"04:18.390","Text":"So our thumb is going in this direction,"},{"Start":"04:18.390 ","End":"04:20.765","Text":"this is our thumb."},{"Start":"04:20.765 ","End":"04:24.395","Text":"Then we have our forefinger or a pointing finger"},{"Start":"04:24.395 ","End":"04:30.485","Text":"pointing out of the page in the direction of the B field."},{"Start":"04:30.485 ","End":"04:35.460","Text":"Let\u0027s write this out of the page in the direction of the B field."},{"Start":"04:35.920 ","End":"04:39.745","Text":"Here we have the rest."},{"Start":"04:39.745 ","End":"04:43.350","Text":"Now we have to draw our middle finger."},{"Start":"04:43.350 ","End":"04:45.410","Text":"When we do v cross B,"},{"Start":"04:45.410 ","End":"04:52.440","Text":"we can see that our middle finger is going to be pointing down in this direction."},{"Start":"04:52.440 ","End":"04:57.920","Text":"Then here are the other fingers, something like this."},{"Start":"04:57.920 ","End":"05:01.280","Text":"This is the middle finger and this is going to be pointing"},{"Start":"05:01.280 ","End":"05:05.045","Text":"down in the direction of our B,"},{"Start":"05:05.045 ","End":"05:07.890","Text":"of our magnetic force."},{"Start":"05:08.110 ","End":"05:12.740","Text":"However, let\u0027s take a look at our diagram."},{"Start":"05:12.740 ","End":"05:16.715","Text":"We can see in our diagram that our charge enters like this,"},{"Start":"05:16.715 ","End":"05:20.480","Text":"and then it travels upwards in the circular motion."},{"Start":"05:20.480 ","End":"05:24.035","Text":"It goes upwards, up until it reaches this point."},{"Start":"05:24.035 ","End":"05:25.820","Text":"If it\u0027s moving upwards,"},{"Start":"05:25.820 ","End":"05:28.160","Text":"that means that there\u0027s a force over here,"},{"Start":"05:28.160 ","End":"05:33.839","Text":"our magnetic force in this direction like so."},{"Start":"05:33.839 ","End":"05:40.100","Text":"Which means that let\u0027s say if we\u0027re looking at this area over here,"},{"Start":"05:40.100 ","End":"05:42.830","Text":"our F_B is pointing upwards."},{"Start":"05:42.830 ","End":"05:50.045","Text":"In reality, our middle finger should be pointing in this direction, not like so."},{"Start":"05:50.045 ","End":"05:55.340","Text":"How can we work with this equation and our hand movement?"},{"Start":"05:55.340 ","End":"05:59.975","Text":"We can see that if our charge is negatively charged,"},{"Start":"05:59.975 ","End":"06:04.190","Text":"then the direction of v cross B will be the opposite."},{"Start":"06:04.190 ","End":"06:06.304","Text":"If our charge is negative,"},{"Start":"06:06.304 ","End":"06:09.560","Text":"then this whole expression over here,"},{"Start":"06:09.560 ","End":"06:12.710","Text":"v cross B would be pointing in this upwards direction,"},{"Start":"06:12.710 ","End":"06:14.000","Text":"which from the diagram,"},{"Start":"06:14.000 ","End":"06:17.760","Text":"we see that is the correct direction."},{"Start":"06:17.960 ","End":"06:25.630","Text":"Therefore, we can say that our charge\u0027s polarity is negative."},{"Start":"06:26.360 ","End":"06:30.350","Text":"If the direction of v cross B over"},{"Start":"06:30.350 ","End":"06:34.294","Text":"here is opposite to the direction of the force in the diagram,"},{"Start":"06:34.294 ","End":"06:36.020","Text":"so here we got v cross B,"},{"Start":"06:36.020 ","End":"06:37.970","Text":"we would have a force pointing downwards,"},{"Start":"06:37.970 ","End":"06:39.965","Text":"but our actual force is pointing upwards."},{"Start":"06:39.965 ","End":"06:46.220","Text":"That means that the charge has to be a negative charge,"},{"Start":"06:46.220 ","End":"06:49.410","Text":"or the charge is less than 0."},{"Start":"06:49.450 ","End":"06:56.600","Text":"Now as for the magnitude of the velocity as it exits the rectangle."},{"Start":"06:56.600 ","End":"06:58.190","Text":"As we know,"},{"Start":"06:58.190 ","End":"07:01.474","Text":"also from our right-hand rule and also from the equation,"},{"Start":"07:01.474 ","End":"07:08.285","Text":"our force is always perpendicular to our velocity and our magnetic field."},{"Start":"07:08.285 ","End":"07:10.750","Text":"This is always true."},{"Start":"07:10.750 ","End":"07:15.200","Text":"Our velocity is also always perpendicular to the force of course,"},{"Start":"07:15.200 ","End":"07:17.720","Text":"and to the magnetic field."},{"Start":"07:17.720 ","End":"07:20.960","Text":"That means that our force and"},{"Start":"07:20.960 ","End":"07:26.810","Text":"the magnetic field will only affect the direction of the velocity,"},{"Start":"07:26.810 ","End":"07:29.800","Text":"but not its magnitude."},{"Start":"07:29.800 ","End":"07:34.370","Text":"Our magnetic force is perpendicular to the velocity,"},{"Start":"07:34.370 ","End":"07:38.270","Text":"which is also perpendicular to the magnetic field."},{"Start":"07:38.270 ","End":"07:42.185","Text":"Therefore, because they\u0027re all perpendicular to one another,"},{"Start":"07:42.185 ","End":"07:46.370","Text":"the magnetic force and the magnetic field will never"},{"Start":"07:46.370 ","End":"07:50.950","Text":"affect the size of the velocity, only its direction."},{"Start":"07:50.950 ","End":"07:57.140","Text":"That means that the magnitude of the initial velocity v in will be equal"},{"Start":"07:57.140 ","End":"08:04.725","Text":"to the magnitude of the velocity as the charge exits, so v out."},{"Start":"08:04.725 ","End":"08:08.860","Text":"This is the answer to Question 1."},{"Start":"08:08.860 ","End":"08:13.210","Text":"Now let\u0027s take a look at Question 2."},{"Start":"08:13.210 ","End":"08:15.865","Text":"What is the distance x between"},{"Start":"08:15.865 ","End":"08:23.830","Text":"the upper left corner of the rectangle and the charge\u0027s exit point along the upper edge?"},{"Start":"08:23.830 ","End":"08:31.225","Text":"We\u0027re looking at the distance between these two points."},{"Start":"08:31.225 ","End":"08:36.350","Text":"This distance over here is x."},{"Start":"08:37.170 ","End":"08:39.835","Text":"The first thing is,"},{"Start":"08:39.835 ","End":"08:43.100","Text":"we know what our radius is."},{"Start":"08:43.560 ","End":"08:48.940","Text":"We looked at this equation that we derive from the fact that our particle is moving"},{"Start":"08:48.940 ","End":"08:53.688","Text":"in circular motion and for our equation for force."},{"Start":"08:53.688 ","End":"09:00.235","Text":"We got that our radius is equal to mv divided by qB,"},{"Start":"09:00.235 ","End":"09:07.070","Text":"where all of these values are given to us in the question."},{"Start":"09:07.530 ","End":"09:12.100","Text":"Let\u0027s imagine that we were told the mass in the question,"},{"Start":"09:12.100 ","End":"09:13.825","Text":"it just isn\u0027t written in."},{"Start":"09:13.825 ","End":"09:15.520","Text":"So we know the radius."},{"Start":"09:15.520 ","End":"09:22.180","Text":"The next thing that we know is the distance from here to here;"},{"Start":"09:22.180 ","End":"09:24.745","Text":"so this vertical distance."},{"Start":"09:24.745 ","End":"09:32.180","Text":"In the question, we\u0027re told that charge q enters the middle of the region."},{"Start":"09:32.370 ","End":"09:37.015","Text":"That means it\u0027s entering the middle of this side over here,"},{"Start":"09:37.015 ","End":"09:39.820","Text":"where we know that it\u0027s of side length b."},{"Start":"09:39.820 ","End":"09:43.015","Text":"So we\u0027re entering right in the middle of side of length b,"},{"Start":"09:43.015 ","End":"09:50.600","Text":"which means that this length over here is half of b or b divided by 2."},{"Start":"09:51.300 ","End":"09:57.160","Text":"Now we want to know what this little length over here is equal to."},{"Start":"09:57.160 ","End":"10:00.040","Text":"Let\u0027s call this length y."},{"Start":"10:00.040 ","End":"10:05.725","Text":"So y is simply going to be equal to this total length over here,"},{"Start":"10:05.725 ","End":"10:08.950","Text":"which is b divided by 2,"},{"Start":"10:08.950 ","End":"10:11.560","Text":"minus this length over here,"},{"Start":"10:11.560 ","End":"10:14.230","Text":"which we of course know is R,"},{"Start":"10:14.230 ","End":"10:20.780","Text":"so minus R. Then we\u0027re left with this small length here."},{"Start":"10:22.350 ","End":"10:29.425","Text":"Now what we want to do is we want to look at this triangle over here."},{"Start":"10:29.425 ","End":"10:31.960","Text":"We know our length R,"},{"Start":"10:31.960 ","End":"10:33.535","Text":"it\u0027s this over here."},{"Start":"10:33.535 ","End":"10:35.260","Text":"We know our length y."},{"Start":"10:35.260 ","End":"10:37.990","Text":"It\u0027s made of these values over here,"},{"Start":"10:37.990 ","End":"10:42.925","Text":"which of course we\u0027re given in the question and we\u0027re trying to find x."},{"Start":"10:42.925 ","End":"10:45.850","Text":"We\u0027re just going to use Pythagoras."},{"Start":"10:45.850 ","End":"10:48.370","Text":"We know that the hypotenuse,"},{"Start":"10:48.370 ","End":"10:50.395","Text":"which is R^2,"},{"Start":"10:50.395 ","End":"10:56.125","Text":"is equal to x^2 plus y^2."},{"Start":"10:56.125 ","End":"10:59.650","Text":"What we have,"},{"Start":"10:59.650 ","End":"11:02.464","Text":"therefore, we can isolate out x^2."},{"Start":"11:02.464 ","End":"11:08.905","Text":"So we have a x^2 is equal to R^2 minus y^2,"},{"Start":"11:08.905 ","End":"11:11.470","Text":"and this is what we\u0027re trying to isolate."},{"Start":"11:11.470 ","End":"11:13.870","Text":"I\u0027m not going to substitute in my R,"},{"Start":"11:13.870 ","End":"11:16.990","Text":"so I\u0027ll just leave it as R^2, minus,"},{"Start":"11:16.990 ","End":"11:25.495","Text":"and then my y^2 is b divided by 2 minus R^2."},{"Start":"11:25.495 ","End":"11:26.714","Text":"Now let\u0027s do this here."},{"Start":"11:26.714 ","End":"11:29.140","Text":"We have that x^2 is equal to,"},{"Start":"11:29.140 ","End":"11:31.060","Text":"so we\u0027ve R^2,"},{"Start":"11:31.060 ","End":"11:33.190","Text":"and then we have minus."},{"Start":"11:33.190 ","End":"11:43.750","Text":"Then what we\u0027ll have is b^2 divided by 4 minus bR."},{"Start":"11:43.750 ","End":"11:49.150","Text":"We have minus b divided by 2R minus b divided by 2R,"},{"Start":"11:49.150 ","End":"11:55.690","Text":"so we\u0027re just left with minus b divided by R and then plus R^2."},{"Start":"11:55.690 ","End":"12:00.055","Text":"Then this will give us R^2"},{"Start":"12:00.055 ","End":"12:08.365","Text":"minus b^2 divided by 4 plus bR minus R^2."},{"Start":"12:08.365 ","End":"12:13.210","Text":"This R^2 cancels out with this R^2 and therefore,"},{"Start":"12:13.210 ","End":"12:22.880","Text":"we\u0027re left with x^2 being equal to bR minus b^2 divided by 4."},{"Start":"12:23.910 ","End":"12:31.300","Text":"Now what we can do is we can put in our common multiples,"},{"Start":"12:31.300 ","End":"12:38.350","Text":"so what we\u0027ll have is b multiplied by R minus b divided by 4."},{"Start":"12:38.350 ","End":"12:41.980","Text":"Then of course, we\u0027re trying to find x."},{"Start":"12:41.980 ","End":"12:49.570","Text":"So x is just the square root of bR minus b divided by 4."},{"Start":"12:49.570 ","End":"12:52.315","Text":"That\u0027s the answer to Question 2."},{"Start":"12:52.315 ","End":"12:53.620","Text":"Of course, if you want,"},{"Start":"12:53.620 ","End":"12:57.670","Text":"you can substitute in R over here."},{"Start":"12:57.670 ","End":"13:00.985","Text":"Now, let\u0027s answer Question 3."},{"Start":"13:00.985 ","End":"13:05.245","Text":"What is the angle Theta of the charges exiting"},{"Start":"13:05.245 ","End":"13:10.165","Text":"velocity relative to the upper edge of the rectangle?"},{"Start":"13:10.165 ","End":"13:14.065","Text":"Basically, we\u0027re just trying to find this angle Theta,"},{"Start":"13:14.065 ","End":"13:21.440","Text":"which will give us the direction of our exiting velocity over here."},{"Start":"13:22.860 ","End":"13:26.575","Text":"First of all, we remember that here,"},{"Start":"13:26.575 ","End":"13:31.090","Text":"we have this 90-degree angle because our radius over"},{"Start":"13:31.090 ","End":"13:37.075","Text":"here is of course perpendicular to the velocity over here."},{"Start":"13:37.075 ","End":"13:43.010","Text":"As we said, the velocity is tangent to the circle."},{"Start":"13:43.770 ","End":"13:47.770","Text":"We can say that this angle over here,"},{"Start":"13:47.770 ","End":"13:50.540","Text":"inside the pink triangle,"},{"Start":"13:51.420 ","End":"13:56.335","Text":"the angle of the pink triangle over here,"},{"Start":"13:56.335 ","End":"14:00.220","Text":"we know that this is 90 degrees and that this is Theta."},{"Start":"14:00.220 ","End":"14:04.360","Text":"So this angle over here in this pink triangle is going to be"},{"Start":"14:04.360 ","End":"14:09.440","Text":"equal to 90 degrees minus Theta."},{"Start":"14:10.530 ","End":"14:14.650","Text":"Now, what we can do is is we can do sine of this,"},{"Start":"14:14.650 ","End":"14:20.470","Text":"so we have that sine of 90 degrees minus Theta,"},{"Start":"14:20.470 ","End":"14:23.695","Text":"which is the same as saying cosine of Theta,"},{"Start":"14:23.695 ","End":"14:25.600","Text":"but let\u0027s just leave it as sine."},{"Start":"14:25.600 ","End":"14:28.330","Text":"Sine from our SOHCAHTOA,"},{"Start":"14:28.330 ","End":"14:30.070","Text":"we know that sine, SOH,"},{"Start":"14:30.070 ","End":"14:33.160","Text":"is opposite over hypotenuse."},{"Start":"14:33.160 ","End":"14:37.269","Text":"The opposite side to this angle over here is y,"},{"Start":"14:37.269 ","End":"14:42.055","Text":"and the hypotenuse of this triangle is R,"},{"Start":"14:42.055 ","End":"14:46.060","Text":"and of course we know that our y is equal to b divided by"},{"Start":"14:46.060 ","End":"14:51.680","Text":"2 minus R and then divided by R."},{"Start":"14:52.410 ","End":"14:59.575","Text":"Now we can say that cosine of Theta is equal to"},{"Start":"14:59.575 ","End":"15:07.210","Text":"b divided by 2R minus R divided by R, so minus 1."},{"Start":"15:07.210 ","End":"15:12.850","Text":"Therefore, we can say that Theta is equal to arc cosine or cosine"},{"Start":"15:12.850 ","End":"15:19.400","Text":"to the negative 1 of b divided by 2R minus 1."},{"Start":"15:20.430 ","End":"15:23.836","Text":"This is the answer to Question 3."},{"Start":"15:23.836 ","End":"15:27.260","Text":"Now we have finished this lesson."}],"ID":25724}],"Thumbnail":null,"ID":99484}]

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