Magnetic force on a proton example (part 1)

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, and 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-- but 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 a vector field that's popping out of the screen, I could just do the top of the arrowheads. 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. 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 0.5 teslas. Let's say 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 seventh meters per second. And that is actually about 1/5 of the velocity or 1/5 of the speed of light. So 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 significantly at this point. So we have this proton going at a 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, 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. And 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 seventh 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 seventh meters per second. And then times the magnitude of the magnetic field, which is 0.5 soon. teslas-- I didn't have to write the units there, but I'll do it there-- times sine of the angle between them. I'll write that down right now. 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 orthogonal to each other. They're at right angles to each other. Because these vectors are popping out of the screen. They are perpendicular to the plane that defines the screen, 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 2? Either way, if you want to deal in radians. Well, it's just equal to 1. The whole-- 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 1. So this is just 1. So the magnitude of the force is actually pretty easy to calculate, if we know the charge on a proton. And let's see if we can figure out the charge on a proton. Let me get the trusty TI-85 out. Let me clear there, just so you can appreciate the TI-85 store. If you press second and constant-- that's second and then the number 4. They have a little constant above it. You get their constant functions. Or their values. And you say the 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, they have 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. Let's just confirm that that's the charge of an electron. 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 an electron-- and it is positive, so that's the same thing as the charge for a proton-- times 6 times 10 to the seventh-- 6 E 7, you just press that EE button on your calculator-- times 0.5 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 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 you is where we break out-- 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 when you take something crossed something, the first thing in the cross product is your index finger on your right hand. And then the second thing is your middle finger pointed at a right angle with your index finger. Let's see if I can do this. So I want my index finger on my right hand to point to the right. But I want my middle finger to point upwards. Let me see if I can pull that off. So my right hand is going to look something like this. And 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 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-- 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. 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. 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 that-- what happens when you have a force perpendicular to velocity? Well, think about it. If you have a force here and the velocity is 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. 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. And in the next video we'll actually figure out the radius of that circle. And just one thing I want to let you think about. 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 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, you know, how does that magnetic field know how fast you're moving? But anyway, I'll leave you with that. In the next video we'll explore this magnetic phenomenon a little bit deeper. See