If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content
Current time:0:00Total duration:10:47

Video transcript

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 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 this is not a beta it's a B that the units of a magnetic field are the Tesla which is 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 an 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 can 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 these and 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 point five Tesla's let's say I have some proton that comes speeding along I have some proton that comes feeding along and it's beating along at a velocity so the velocity of the proton is equal to six times ten to the seventh meters per second and that is actually about a fifth the the velocity or a fifth of speed of light so this is a you know are 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 etc etc we just assume that where the mass hasn't increased significantly 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 could 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 it stored in and if you have a TI graphing calculator your calculator will 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 six 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 six times ten to the seventh meters per second and then x times the magnitude of the magnetic field which is 0.5 Tesla's I didn't have to write the unit's there but I'll do it there times sine of the angle between them right 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 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 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 right with the hole the hole 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 90 degree angle sine of 90 degrees well that's just 1 so this is just 1 so the magnitude of this of the force is actually pretty easy to calculate if we know that the charge on a proton let's see if we can figure out the charge on a proton let me get the trusty ti-85 out and actually I let me clear there just so you can appreciate that 85 Stewart so if you press second and constant that's second and then the number for their little constant above it you get their constant functions or their values and you say the built in oh I care about the built in function so let me press f1 and they have a bunch of you know this is abogados 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 right electrons just remember electrons and protons have offsetting charges once positive ones 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 a charge of a electron yep that looks about right but that's also the charge of a proton and actually this positive values the exact charge of a proton they should have maybe put a negative number here but all we care about is the value let's use that again the charge of electron and it is positive so that's the same thing as a charge of a proton times 6 times 10 to the seventh 6 e 7 is just you just press that EE button on your calculator times 0.5 Tesla's times 0.5 Tesla's make sure all your units are in Tesla's meters and coulombs and then your result will be in you and you get 4.8 times 10 to the negative 12 Newtons let me write that down so the magnitude of this force alright I know it's a magnitude the magnitude of this force is equal to four point eight times ten to the minus twelve Newton so that's the magnitude now what is the direction what is the direction of this force well this is where we break out are 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 cross something the first thing in the cross product is your index finger on your right hand right so let's see if I and then the second thing is your your middle finger pointed at a right angle with your index finger so let's see if I could 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 and my hand is brown so so my right hand is going to look something like this my index finger is pointing in the vault 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 to have you know this is the heel of my thumb and so my thumb is going to be the 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 visually 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 it's been down word 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 all of that what happens when you have a force perpendicular to velocity well think about if you have force here and the velocities 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 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 think let you think about it it's kind of I mean I don't know it's it's kind of weird or spooky to me that that the force on a moving particle it doesn't matter about the particles mass it doesn't it just matters the particles 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 charged particle the faster you move through 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 the next video we'll explore this this magnetic phenomenon a little bit deeper see you soon