Ch 26: Magnetic Fields and ForcesSee all chapters

Sections | |||
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Magnets and Magnetic Fields | 22 mins | 0 completed | Learn |

Summary of Magnetism Problems | 10 mins | 0 completed | Learn |

Force on Moving Charges & Right Hand Rule | 31 mins | 0 completed | Learn Summary |

Circular Motion of Charges in Magnetic Fields | 18 mins | 0 completed | Learn Summary |

Mass Spectrometer | 34 mins | 0 completed | Learn Summary |

Magnetic Force on Current-Carrying Wire | 13 mins | 0 completed | Learn Summary |

Force and Torque on Current Loops | 18 mins | 0 completed | Learn |

Concept #1: Mass Spectrometers

**Transcript**

Hey guys. So, next we're going to talk about mass spectrometers which are instruments used to measure mass. Now, this can be overwhelming at first because there are three different parts and there are three different equations you can use but I will break it down really simply for you and hopefully by the others video you agree with me that this is totally manageable, let's check it out. Alright, so as I said mass spectrometers are instruments that measure mass of a known charge, meaning that experimentally you would know your q and be looking for m. Now, in a physics problem you may actually be given the m and asked for a q, right? There are four steps to a mass spectrometer that you should know, I merged the first two steps because they are pretty basic. So, steps one and two are ionization and acceleration, ionizing something means that you are adding or removing electrons so that it becomes either positively or negatively charged and the reason we want to do this is because particles need to have charge, an excess of electrons or protons, so that they actually feel a force. So that, so you do ionization so that the particle, the particle feel a magnetic force and an electric force, if you, if you don't have a charge you don't do those two forces, so right here you have something, that's called an ionizer. So, let's just draw that there and what it does it's got some, some particles chillin here and somehow it magically strips them of electrons so that they become positively charged it then sort of shoots them in this direction so that these bunch of positive charges are hanging out over here, okay? We're going to looks with positive charges but it could be with negative charges as well. Now, so that's ionization, we ionize them and then we threw them over here. Now, they're going to get, they're going to go through acceleration, they're going to be accelerated through a potential difference, meaning you are going to have a positive potential here and a negative potential here, electric fields are created whenever you have a situation like this, always going from high to low so this is the direction of an electric field. Remember, that positive charges feel a net force in the direction of the electric field. So, these guys will be pushed this way by an electric force, which will result in result in them accelerating this way, by the way, the initial velocity here, will be 0, okay? I'm dropping a ton information on you, but when we do an example you see that this is pretty straight forward. So, they are going to get over here, with some sort of final velocity, okay? So, between here and here, they're actually going to have a constant velocity, okay? Now, for this first part the only equation need to know is you have one equation per part, you remember that the work done on a charge as it moves a potential difference is q, Delta V, where Deta V is the potential difference between the two plates and that's typically given to you, or you're asked to calculate it. Remember also, that work is also the change in kinetic energy, okay? So, because these two are both work, I can just say that one equals the other, okay? So, I can say that q, Delta V equals Delta k. Now, remember Delta means change and Delta k is k final minus k initial but this is almost always, they're probably always going to be 0 the initial velocity. So, what you really end up with is the final kinetic energy which is going to be half m, V final squared, so this is the first equation that you have for this problem and there's two more, okay? Now, if you don't want to remember all these letters you can remember this part and sort of work your way there, cool? So, that's this first part, this thing is accelerated it moves with a constant speed and now it's going to go through what's called a velocity selector, so the second part is velocity selection through a velocity selector and the idea is you want to filter desired speeds. So, let's talk about that, the only way to calculate a mass on a mass spectrometer is if you basically know everything else, and I'll show you this later with the equations, you have to know everything else, meaning you have to know the velocity even to be able to calculate it. So, what's going to happen here is that these guys will have different masses, this is mass 1, this is mass 2, etcetera, because they have different masses they're going to have different speeds here and this is because of just basic F equals m, a, right? Acceleration is force over mass. So, if you have different masses you're going to have different accelerations, therefore you have different speeds, but remember, we have to control the speeds. So, what we're going to do is we're going to make sure that all these charges run through this device so that we can get rid of the ones that are not the right speed, and the way to do this is by running them through an electric field again, So, notice that this is going from positive to negative here. So, there's an electric field that's pointing down, these are all positive charges. So, when these charges are here, they're going to feel a force this way, force electric, okay? And they're going to get potentially deflected but you want the right ones, you want the right ones to go straight through, okay? So, you want the, you want the particles with the target desired ideal speed to make it, okay? You want them to make it all the way to the end and come out of a gate over here into this green area that we're going to talk about, but if you have an electron electric field here, they're going to pull down which means they're going to hit this wall and that's bad news, right? Slow ones are going to hit over here, fast ones are going to hit it over here. So, what you do is you want to try to cancel that electric force with another force and we're going to try to cancel with the magnetic force. So, over here, we're going to try to have a magnetic force that cancels to that electric force, so the next equation you can write is that Fe equals FB, okay? Now, let's expand this real quick, electric force is q, E and magnetic force is q, v, B, sine of theta, the angle here between V and B is going to be 90, we'll talk about that, so this is just going to go away, okay? I can cancel the Q's and I end up with this relationship here, which is that, I can simplify or not simplify, I can make this more standard looking by moving the v around. So, it's going to look like this, this is the second equation that you can use or that you will need to use in a lot of these questions v equals E over B, okay? So, let's talk about the direction of this thing. So, I have a positive charge. So, I'm going to use the right hand rule, right? With my right hand, and I want the force to be up, how do you get the force to be up? Well, my palm has to be going up because this is force. So, it's something like this, right? Something like this or something like this? Well, it's actually in this orientation because I want my velocity to be going to the right If you look at the diagram, right? So, velocity to the right, force up, means that my hand is going away from me and into the page, right? Don't just look at the video, do this yourself, away from you into the page, so that means that to accomplish this cancellation of forces my magnetic field needs to be into the page here, okay? So, that's the direction. Now, if you look at this equation, this equation ties E, B and V and the idea is that this only is going to cancel the charges that are going to have the target speed, because if you have a different speed, if you look at FB, FB equals q, v, B, if you have the wrong speed you're going to have the wrong force that isn't going to exactly cancel E. So, what you're going to end up doing is you're going to either hit over here, hit over here, right? So, this is sort of a filter and only the forces, only those particles with the right speed will make it over here, okay? Now, we're going to get to the last part, when you get here, the way that these devices are built is that not only there's an electric, a magnetic field here, inside of the velocity selector, there's also the same exact magnetic field out here, into the deflection area, okay? Now, what's going to happen it's a little bit different here is, when you come out the gate here, right? You still have the magnetic force because you still have a magnetic field, but you no longer have an electric force. So, I want you to write it and then scratch it out, you no longer have electric force because there's no longer an electric field, right? You only have that between the parallel plates. So, what happens is you're now going to move in a circular path like this, and that's a terrible semicircle, it's pretty good. So, remember, if you're going into, if you have a charge moving into a constant magnetic field you have circular motion, okay? And this circular motion is going to have a radius R and you may remember our circular motion equation, which is the third equation we're going to use, which is R equals m, v over q, B, we talked about this in the previous video, you have to memorize that equation, not just for those simpler problems but also because you're going to need it here, okay? One last thing that I want to mention about equations is that sometimes you were given or asked for, not the radius but this distance here, okay? This distance, I'm going to write over here, and hopefully you see that this distance it's just one radius, two radius, okay? So, I'm going to erase this so that it's not very messy, hopefully you believe me, distance is 2R, that's another one, it's not really an equation, it's just sort of like an accessory that you may need to use, cool? So, if you know the velocity, because you've filtered only the good particles with the right velocity and you know the magnetic field because you control the device. So, you can adjust the magnetic field and you know the charge, you know the charge because you know how much charge you gave these things over here on the ionizer, and you can measure the distance, right? You like take a picture of this and you see, oh crap they're hitting over here. Now, I can measure d and from d I can find R, right? Now, you know all three, all four variables, which means you're able to solve for mass. Now, if you want you can rewrite this and say you know mass equals q, B, R over v, so that it's a more straightforward equation but now you got to remember two equations, I don't think it's worth it, I wouldn't really do this, I would just leave it as the standard rotation equation, circular motion equation, cool? Alright, so there's a lot of crap but let's a problem here and I think you see that it's not that bad, this is quite a long problem because I want to hit up a lot of different things. So, here, and by the way, a lot of the text here is just describing how a mass spectrometer works so you get familiar with language. A charge of two. So, charge positive 2C, positive means we're going to use the right hand rule, not the left, it's accelerating through an x, accelerating the positive x-axis, which is to the right. So, we have a positive charge. So, if you want to draw, you don't have to draw, but I'm just going to draw here, you got your little q's, by the way there's a, so you got your q here and you are going to have a little negative there, right? It gets accelerated to a potential difference Delta V, so the potential difference here, Delta V. Delta is not given to us and if you look at question d, we're being asked for, what is Delta V. So, that's coming, we want to know what that is later, it then passes through horizontal plates. So, it gets accelerated here and then it goes here with a constant speed and then here's accelerated and it's going to pass through horizontal plates right here, horizontal plates that have electric field 3 Newtons per Coulomb. So, electric field over here, E equals 3, that points up, by the way, if it points up it means that you're positive here and negative here, always high to low, it has a magnetic field, magnetic field V equals 4 Tesla, that also exists between the plates. So, inside of here there's a magnetic field, I don't know the direction yet. So, let's not draw that, and remember here you have sort of the deflection zone because this guy will go straight through here, right? It's going to go straight through here and it's going to deflect either this way or that way, we don't know yet. So, let's not write that yet, and by the way, the same magnetic field 4 is also going to exist in this deflection zone over here, okay? That's what it says here, this magnetic field also exists outside of the plates and causes the charge to deflect with a circular arc of radius 5 centimeters. So, this means that the radius of this deflection is going to be in 0.05 meters, okay? Alright, question, what must the direction of the magnetic field be? So, the direction of the magnetic field you will only determine that by looking here, okay? So, you have a positive charge which means the electric force will be up because the electric force for a positive charge goes in the direction of the electric field, the backwards if you have a negative charge, therefore you want your magnetic charge to be going down, that's the step. Now, we're going to use our right hand rule to determine the direction that the magnetic field has to have so that you actually get a force down. So, force down means palm down and it could be this but actually my charge is supposed to be going to the right. So, you end up with something like this, okay? Actually, that's bad because now my palm is up. So, you actually have to do this, okay? So, your fingers are pointing towards your face, your thumb is to the right. please do this. Now, my force is down which is what we want, what this means is that my fingers are coming at my face which means are popping out of the page into me, which means that this is a direction of out of the page, which is given by a dot, so the direction of magnetic fields is little dots everywhere, right? Little dots everywhere, lots of little dots. So, it's going to look like this. So, what is the direction? The direction is out of the page and we got the first part done.

Sketch the deflection that the charge will experience there's two options here, and it's either going to be like this or like this, right? And that will depend on the direction of the force. So, the direction of the magnetic force, the magnetic force will be the same out here and because this is being pulled down is going to arc down, okay? So, that's the sketch, we got the sketch done they could have asked, is this going to arc up or down? It's arcing down, okay? And then now we're going to calculate the mass of the charge, right? So, now we're actually going to start calculating stuff, how do we do this? Well, we have three equations, you can think of it as sort of a menu and I'm going to write them here, one is q, Delta V equals the change in kinetic energy or you can write q, Delta V equals the final kinetic energy, right? Squared, that's equation one, the second equation is V equals and we wrote this up there, v equals e, b, e over B and the third equation is the radius one, m, v over q, B, so that's it, those are the three equations, that's all you got to play with. Now this is just an algebra problem and I'm going to get out of the way. So, we're looking for mass, where do you see mass?There's mass here and there's mass here. So, let's see, do I know Delta V, I don't know Delta V, do I know the velocity, we double the velocity either and I know q. So, this equation is kind of ugly, I have two unknowns and I'm looking for m, I have three unknowns, that's terrible, what about this equation do I know v, I don't know v but I know q, I know B, right? I know q, q is 2, I know B, B is 4, and I know R because the radius is 5 centimeters, so I know this as well, so this equation is still not totally ready but at least here I have three unknowns, which is bad news, and here I only have two. So, this equation is not as bad but before I can find m, I'm gonna have to find V? Well, there's a V here and there's a V here, right? And this is just problem-solving skills, this equation we already said was bad. So, let's try to get the V from here, do I know E and B and I do so the first thing we're going to do is actually do V equals E over B and E is 3, B is 4, so the answer is going to be 0.75 meters per second for this part. Now, I got V, that's good news. So, we can find m. So, I'm going to rearrange this equation m is going to be q, B, R over V, please do this carefully, move this stuff around carefully and q is 2, B is 4, R is 0.5, 0.5, and all of that divided by V, V is 0.75, okay? And if you do all of this, if you do all of this you get that the answer is 0.53 kilograms, okay? 0.53 kilograms, that is the mass. So, we're done with Part C. Now, we're just down to Part D, which is finding Delta V, Delta V shows up in this equation only. So, that's what we're going to go for, okay? Running out of space here, but we're going to write the Delta V equals m, v final squared divided by q, okay? The mass is 0.53, the velocity square is 0.75 squared, and the charge, we know is 2, right here. So, if you plug all this end you get 0.075 volts, that's the potential difference that this thing must have been accelerated through, okay? So, this was a long video and this was a long example but I wanted to move slowly through explaining the spectrometer so you understand all the characteristics, all the properties and I wanted to do a long example where we talked about a bunch of different things and I wanted to move through that slowly as well but hopefully you're seeing that all you got to do, no matter what the questions are asked of you, is just use a combination of these three things as well as think about directions with the right-hand rule, cool? That's it for this one, let's keep going.

Practice: A negative charge in a spectrometer is accelerated in the negative x-axis. It is later deflected and collides some distance ABOVE velocity selector. What are the orientations of the electric and magnetic fields, respectively, inside the selector?

(a) up and out of the page

(b) up and into the page

(c) down and out of the page

(d) down and into the page

Practice: A 2 kg, – 3 C charge is accelerated through a potential difference of 4 V. The velocity selector has an electric field of magnitude 5 N/C. How far from the velocity selector will the charge collide against the spectrometer “wall”?

Example #1: Find Mass-to-Charge Ratio in Spectrometer

**Transcript**

Hey guys. So, some questions will ask you to find the mass-to-charge ratio of particles going through a mass spectrometer and that's what we're gonna do here. So, let me show you. So, here it says, A mass spectrometer is a velocity selector electric field of magnitude 20. So, remember, you get accelerated and then you go through a velocity selector vs, that's going to have an electric field and we're telling you here that that is the magnitude of 20, it says, when a certain charge is accelerated to a constant 30, so it gets accelerated over here, it goes through the little hole and then it's now going to move here, with a constant 30, it collides 40 meters away from the velocity selector, what does that look like? Well, it means that it's going to curve this way and it's going to hit this wall here, at a distance d equals 40 meters. Remember, also the distance is twice the radius because you have radius and radius which means, if the distance is 40, this means that the radius is 20, and the reason we want to change it to radius is because our other equations or the one other equation we're going to need here is in terms of radius and not in terms of distance, I drew it down even though it could have gone up, we don't actually know, we are not being given enough information to figure out which way this is going, but it doesn't matter, I just picked one for the sake of illustration, okay? In this question we want to find the mass-to-charge ratio, okay? Mass to charge ratio, which is m over q, by the way, if I hadn't told you that we were looking for m over q you can just interpret that from the question, the mass to charge ratio, you would do this and say, hey, I'm looking for this, which means you're not looking for m or q, you want to leave this as a fixed unit and then solve it. So, how the heck are we going to do that? Well, we got these three equations and one of them, hopefully will work for us, okay?

So, R equals, actually let's star with the first one. Alright, just to keep it in order. So, q, q, Delta V equals 1/2 m, v squared, by the way, this is a potential difference or voltage and this is a velocity, two different things, the other equation is that V velocity equals E over B and the other equation is that R equals m, v over q, B, okay? And we're looking for the ratio m over q, luckily, this is actually really straightforward because if you look here you'll find one of these equations has and m and a q and it's right here, and not only do they have it they are already sitting next to each other which is awesome. So, all you got to do is move stuff around in such a way that the m and the q stay exactly where they are. So, we're going to move the B to the other side. So, I'm going to get the B, R over v. Do I have B? Yep, do I have B? No, I don't have B, got excited there for a second, do I have R? Yes, R is 20, do I have the velocity? The velocity is 30. So, we don't have B, we got to get it, can we get B? This equation here seems to be, seems like it's going to work. So, v equals E over B therefore B is E over v, the e is 20, electric field strength, and the velocity is 30. So, B is 0.67, okay? 0.67 Tesla. So, that's what's going to go over here, 0.67 Tesla, and if you do all of this, you're going to get that that ratio is 0.44, 4.44, the units are kilograms per Coulomb, okay? So, that's how you could do this, the second you got stuck here, because you didn't have v, you just go through a different equation and you get, alright? That's it for this one, let's keep going.

0 of 4 completed

A small particle with negative charge enters a region of uniform magnetic field and travels along the path shown in the sketch. The magnetic field in the region has direction
(a) into the page
(b) out of the page
(c) toward the top of the page
(d) toward the bottom of the page
(e) to the left
(f) to the right

In the rectangular region shown in the sketch there is a uniform magnetic field. A small particle with mass m = 4.0 x 10-12 kg and negative charge q = - 8.0 x 10 -6 C enters the field with velocity v = 5.0 x 104 m/s in the direction shown. The particle travels with constant speed along the semicircular path shown in the sketch. The radius of the path is R = 0.40 m. What are the magnitude and direction of the magnetic field in the region?

A small sphere with charge q enters a region of uniform magnetic field that is directed into the page. While it is in the field, the sphere moves along the path shown in the sketch and then exits the field. The charge on the sphere is
(a) positive
(b) negative

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