![]() The direction of the velocity vector, while my middle finger Your middle finger pointed at a right angle with your So what do we have to do? So when you take somethingĬrossed something, the first thing in the cross product is We use our right hand rule to figure out the direction. Out- we put our pens down if we're right handed, and Of this force? Well, this you is where we break Now what is the direction? What is the direction Is equal to 4.8 times 10 to the minus 12 newtons. ![]() In teslas, meters, and coulombs, and then your result Just press that EE button on your calculator- times It is positive, so that's the same thing as the charge forĪ proton- times 6 times 10 to the seventh- 6 E 7, you Negative number here, but all we care about is the value. ![]() Electrons- just remember-Įlectrons and protons have offsetting charges. Interesting- this is the charge of an electron. Know, this is Avogadro's number, they have a bunch of I care about the built-in functions, so let me press F1. If you press second andĬonstant- that's second and then the number 4. Just so you can appreciate the TI-85 store. Is actually pretty easy to calculate, if we know theĬharge on a proton. That, you say, oh well, they're perpendicular. Perpendicular to each other, then we just multiply the Multiply the components of the two vectors that are Intuition you got about the cross product is we only want to Perpendicular, what is the sine of 90 degrees? Or the sine of pi over 2? Either way, if you want You can visualize it in three dimensions, is 90 degrees. Plane that defines the screen, while this proton is moving Of the field, what is the angle between them? If you visualize it in threeĭimensions, they're actually orthogonal to each other. Now- and this particle is moving in the plane Have to do a little bit of three-dimensional visualization Pointing straight out of the screen- and you're going to Have to write the units there, but I'll do it there. We'd probably want to convert it to meters. Velocity, 6 times 10 to the seventh meters per second. Proton- I'll call it Q sub p- times the magnitude of the The particle is going to be equal to the charge of a So let's just write that downĪs a variable right now. And if you have a TI graphingĬalculator, your calculator would also have it Now, but my calculator has that stored in it. The charge on a proton? Well, we don't know it right So how can we figure out the magnitude? Well, first of all, what is The magnitude and direction of the force on this protonįrom this magnetic field? Well, let's figure out the So we have this proton going atĪ 1/5 of the speed of light and it's crossing through Hasn't increased significantly at this point. Relativity because then the mass of the proton increases,Įt cetera, et cetera. Relativistic realm, but we won't go too much into Of the velocity or 1/5 of the speed of light. Velocity- so the velocity of the proton is equal toĦ times 10 to the seventh meters per second. Just so you get the sense that it's a field. So let's say I haveĪ bunch of arrows popping out of the screen. Screen, I could just do the top of the arrowheads. Vectors or a vector field that's popping out of the Magnetic field, and let's say it's popping outįly, so I hope the numbers turn out. Is equal to newton seconds per coulomb meters. Which is abbreviated with a capital T- and that The units of a magnetic field- this is not a beta, it'sĪ B- but the units of a magnetic field are the tesla. Vector quantity, is equal to the charge- on the movingĬharge- times the cross product of the velocity of theĬharge and the magnetic field. On a moving charge from a magnetic field, and it's a Or at least I showed you, I don't know if you've learned it You don't have to learn them all, you will be fine if you remember one of them and with which charge it works. However, just like the other rule, we just have to invert the sense of the vector to find out if the charge is positive. As you may have guessed, this is true only if the charge is negative. ![]() If we use this rule we will find out that the force is pointing north, but that is not true. If it is negative, we must invert the sense of the vector. This is true only if the charge is positive. Sal concludes in the video that the force is pointed south. The diference between them (and the reason because there are more than one) is because the sense of the vector-force depends on the charge of the particle, whether it is positive or negative.
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