Learn about Magnetism 3

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 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 used this to show you that the units of a magnetic field, this is not updated so B—that the units of a magnetic field or the Tesla which is abbreviate with a capital T and that is equal to Newton seconds per column 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 t his up on a fly so I hope the numbers turn out. It’s inspired by a problem that I read in Barron's AP calculus book. So I want to draw a bunch of vectors or vector fields that popping out of the screen I could just do the top of those arrowheads I’ll draw them in magenta. So let’s say I have 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 to—so you get a sense that its field. It pervades the space. So these are 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 this .5 Tesla.

Let’s say I have some proton that come speeding along? I have some protons that comes speeding along and it’s speeding along at the velocity so the velocity of the proton is equal to 6 times 10 to the 7th meters per second. And that is actually about a 5th the velocity or a 5th of the speed of light. So this is a—we’re almost—we’re pretty much in the relativistic round but we won’t go too much into relativity because of the mass of the proton increase etcetera, etcetera. We just assume that the mass has increase significant at this point. So we this proton going at the 5th of the speed of light and it’s crossing through this magnetic field. Sp the first question is, what is the magnitude and direction of the force on this proton from this magnetic field.

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 the proton? Well, we don’t it right now but my calculator has it stored and if you have a TI graphing calculator you’re 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—other call it Q sub P times the magnitude of the velocity 6 times 10 to the 7th to meters per second. We’re using all the right units; it’s just the centimeters we’d probably want to convert to meters. 6 times 10 in the 7th meters per second and then times the magnitude of the magnetic field which is .5 Teslas, I don’t have write 0338 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, are going to have to do a little bit of three-dimensional visualization now? And this particle is moving in the plain of the field, what is the angle between them? As you visualize in the three-dimension they’re actually—they’re orthogonal to each other, they’re at right angles to each other. Because these vectors are popping out of the screen, they’re perpendicular to the plain that defines the screen while this proton is moving within this plain 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 radiance which is 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 data. 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 to that 90 degree angle, sine of 90 degrees well that’s just one.

So the magnitude of this—of the force is actually pretty easy to calculate if we know that charge and the proton and let’s see if we can figure out the charge on a proton. Let me get the TI 85 out and actually I—let me clear there just so you can appreciate TI 85—so if you press second and constant, that second and then the number of four, we have a little constant above it. You get their constant functions or their values and you said the built-in or accurately about the built in functions let me press F1. And they have a bunch of—Avogadro’s number and 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 off setting charges one is positive, one is negative. It’s just that a proton ism ore massive, that’s how they’re different and of course is positive. So let’s just confirm that that’s the charge of a electron, yes, that looks about right. But that’s also the charge of a proton and actually this positive values exact charge of a proton. They should have maybe put a negative number here but all we care about this is the value. So let’s use that again. The charge of electron and it’s positive so that’s simply a charge of a proton times 6 times 10 to the 7th—6E7 is just—you just press that EE button on your calculator times .5 Teslas make sure all your units are in Teslas, meters and columns and then your result will be in newtons and you get 4.8 times 10 the negative 12 newtons, let me write that down.

So the magnitude of this force--right that was a 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 or 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 will you take something just cross something, the first thing in the cross product is your index finger on your right hand so let’s see. And then the second thing is your middle finger pointed at the right angle with your index finger. So let’s see if I could 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 so let me see if I can pull that of.

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 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 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 plain 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 that. What happens when you have a force perpendicular to velocity? Well, think about it, if your force is here and the velocity is like that, if the particles—it will defected a little bit to the right. And then, since the force is always going to be perpendicular to the velocity of 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 it’s going to actually be like a centripetal force on the particle so that particles going to go onto a circle.

And then the next video once we figure out the radius of that circle. And just one thing I want to think—let you think about is it’s kind of I mean I don’t know it’s kind of weird of spooky to lead 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 your charged particles have 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 leave you at that. The next video will explore this magnetic phenomenon a little bit deeper. See you soon.