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Current time:0:00Total duration:10:49

Video transcript

- [Voiceover] In the last video, we learned, or at least I showed you. I don't know if you've learned it yet but we'll learn it in this video but we learned that the force on a moving charge, from a magnetic field, when it's a vector quantity, is equal to the charge on the moving charge times the cross product of the velocity of the charge and the magnetic field. And we use this to show you that the units of a magnetic field, this is not a beta, it's a B, the units of a magnetic field are the Tesla, which is abbreviated with a capital T, and that is equal to Newton seconds per coulomb meters. So let's see if we can apply that to an actual problem. So let's say that I have a magnetic field and let's say it's popping out of the screen. I'm making this up on the fly so I hope the numbers turn out. It's inspired by a problem that I read in Barron's AP Calculus book. So if I want to draw a bunch of vectors or vector field, that's popping out of the screen, I could just do the top of the arrow heads. I'll draw them in magenta. So let's say I have a vector field, so you can imagine a bunch of arrows popping out of the screen. I'll just draw a couple of them. Just so you get the sense that it's a field. It pervades the space. So these are a bunch of arrows popping out, and the field is popping out and the magnitude of the field, let's say it is, I don't know, let's say it is .5 Teslas. Let's say I have some proton that comes speeding along. I have some proton that comes speeding along and it's speeding along at a velocity, so the velocity of the proton is equal to 6 times 10 to the 7th meters per second, and that is actually about 1/5 the velocity, or 1/5 of the speed of light. So this is a, you know, we're almost, we're pretty much in the relativistic realm, but we won't go too much into relativity because then the mass of the proton increases, et cetera, et cetera. We just assume that the mass hasn't increased signficantly at this point. So we have this proton going at 1/5 of the speed of light and it's crossing through this magnetic field. So the first question is, what is the magnitude and direction of the force on this proton from this magnetic field? Well let's figure out the magnitude first. So how can we figure out the magnitude? Well, the cross product, well, first of all, what is the charge on a proton? Well, we don't know it right now but my calculator has that stored in it. If you have a TI graphing calculator, your calculator would also have it stored in it, so let's just write that down as a variable right now. So the magnitude of the force on the particle is going to be equal to the charge of a proton, I'll call it Q sub P, times the magnitude of the velocity, 6 times 10 to the 7th meters per second. We're using all the right units. If this was centimeters we'd probably want to convert it to meters. 6 times 10 to the 7th meters per second and then, times, times the magnitude of the magnetic field, which is .5 Teslas. I didn't have to write units there, but I'll do it there, times sine of the angle between them. I'll write that down right now. Sine of the angle between them. But let me ask you a question. If the magnetic field is pointing straight out of the screen and you're going to have to do a little bit of three-dimensional visualization now, and this particle is moving in the plane of the field, what is the angle between them? If you visualize it in three dimensions, they're actually, they're orthogonal to each other. They're at right angles to each other, right? Because these vectors are popping out of the screen. They are perpendicular to the plane that defines the screen, right? While this proton is moving within this plane, so the angle between them, if you can visualize it in three dimensions, is 90 degrees, or they're perfectly perpendicular, and when things are perfectly perpendicular, what is the sine of 90 degrees, or the sine of pi over two? Either way, if you want to deal in radians. Well, it's just equal to one. Right? Hopefully intuition you got about the cross product is we only want to multiply the components of the two vectors that are perpendicular to each other, and that's why we have the sine of theta, but if the entire vectors are perpendicular to each other, then we just multiply the magnitude of the vector, or if you even forget to do that, you say, oh, well, they're perpendicular, they're at 90 degree angles, sine of 90 degrees, well that's just one. So this is just one. So, the magnitude of this, of the force, is actually pretty easy to calculate if we know the charge on a proton. Now let's see if we can figure out the charge on a proton. Let me get the trusty TI-85 out. And actually, I, well let me clear the air just so you can appreciate that T-85 stores... if you press 2nd and constant, that's 2nd and then the number four, they have a little constant above it, you get their constant functions, or their values and you say they're built in, I care about the built-in functions so let me press "F1" and they have a bunch of, you know, this is Avogadro's number, and they have a bunch of, you know, a bunch of interesting, this is the charge of an electron, which is actually the same thing as the charge of a proton, so let's use that. Electrons, just remember, electrons and protons have offsetting charges. One's positive and one's negative. It's just that a proton is more massive. That's how they're different, and of course it's positive, so let's just confirm that that's the charge of an electron. Yup, that looks about right, but that's also the charge of a proton, and actually this positive value is the exact charge of a proton. They should have maybe put a negative number here, but all we care about is the value, so let's use that again. The charge of electron, and it is positive, so that's the same thing as the charge of a proton, times six times 10 to the seventh, 6 E 7 is just, you just press that EE button on your calculator, times .5 Teslas. Make sure all your units are in Teslas, meters, and coulombs, and then your result will be in newtons. And you get 4.8 times 10 to the negative 12 newtons. Let me write that down. So the magnitude of this force, right, now I know it's the magnitude, the magnitude of this force is equal to 4.8 times 10 to the minus 12 newtons, so that's the magnitude. Now what is the direction? What is the direction of this force? Well this is where we break out our, we put our pens down if we're right-handed and we use our right hand rule to figure out the direction. So what do we have to do? So we take something cross something. The first thing in the cross product is your index finger on your right hand, right? So let's see, so if I, and then the second thing is your middle finger pointed at a right angle with your index finger, so let's see if I can do this. So I want my, I want my index finger on my right hand to point to the right, but I want my middle finger to point upwards, so let me see if I can pull that off. Let me see if I can pull that off. So my right hand is going to look something like this. My hand is brown, so my right hand is going to look something like this. My index finger is pointing in the direction of the velocity vector while my middle finger is pointing in the direction of the magnetic field, so my index finger is going to point straight up, so all you see is the tip of it, and then my other fingers are just going to go like that, and then my thumb is going to do what? My thumb is going to, you know, this is the, heel of my thumb, and so my thumb is going to be at a right angle to both of them, so my thumb points down like this. This is often the hardest part, just making sure you get your hand visualization right with the cross product. So, just as a review, this is the direction of V. This is the direction of the magnetic field, right? It's popping out, and so if I arrange my right hand like that, my thumb points down, so this is the direction of the force. So as this particle moves to the right with some velocity, there's actually going to be a downward force, downward on this plane, so the force is going to move in this direction and so what's going to happen? Well what happens, if you remember a little bit about your circular motion and your centripetal acceleration, and all of that. What happens when you have a force perpendicular to velocity? Well think about it. If you have a force here and the velocity's like that, if the particles, it'll be deflected a little bit to the right, and then, since the force is always going to be perpendicular to the velocity vector, the force is going to charge like that, so the particle is actually going to go in a circle, right? As long as it's in the magnetic field, the force applied to the particle by the magnetic field is going to be perpendicular to the velocity of the particle, so the velocity of the particle, so it's going to actually be like a centripetal force on the particle, so the particle is going to go into a circle. In the next video, we'll actually figure out the radius of that circle. And just one thing I want to think, let you think ahout is it's kind of, I mean, I don't know, it's kind of weird or spooky to me that the force on a moving particle, it doesn't matter about the particle's mass, it doesn't, it just matters the particle's velocity and charge, so it's kind of a strange phenomenon, that the faster you move through a magnetic field, or at least if you're charged, if you're a charged particle, the faster you move through a magnetic field, the more force that magnetic field is going to apply to you. It seems a little bit, how does that magnetic field know how fast you're moving? But anyway, I'll let you at that, and next video, we'll explore this magnetic phenomenon a little bit deeper. See you soon.