Magnetism - Part 4
The magnetic force as a centripetal force: calculating the radius of the path traveled by a proton. Applications: cathode ray tube (CRT) monitors, cyclotrons in particle accelerators. Created by Sal Khan.
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- Two things:
1. It would probably be better not to use a calculator in example problems since we cannot use a calculator on the MCAT.
2. Do these four videos on Magnetism cover all of the content concerning magnetism that the MCAT will test on? What about the magnetic fields caused by current carrying wires and how magnetic fields affect each other?
Keep up the good work!(36 votes)
- These videos have been up for years. They have been nicely organized to cover the MCAT content, but not all of them were specifically created for the purpose of the MCAT. You should check the AAMC website for an exact breakdown for the content you are required to know.
Make sure you do lots of practice on your own.. MCAT math is not simple math. The numbers are complex and ugly- this is done on purpose..it is supposed to be hard. Try pausing the video and challenging yourself to get the answer on your own without a calculator. Be sure to do lots and lots and lots of practice.(24 votes)
- Did Sal ever make a video about how TVs work?
That would be really interesting.(13 votes)
- If you look at the top left corner at7:48you see Sal's most visited sites and it says "America's next top model" lOOOOOOol
I guess he is a fashion guru as well(11 votes)
- Starting at about9:27, I get a little confused as to how we would come up with the direction of the force on the electron using the right hand rule. I understood how we got the direction for the proton, and I understand that the direction of the force will be opposite of that of the proton since it's negative. But I don't know if I fully understand how we would find it if we did not already know the direction of the proton.
When using the right hand rule, we are using our pointer finger for the direction of the velocity, and we are using our middle finger for the direction of the magnetic field. Since those will still be the same, we will still have our thumb pointing downward, right?
So, it's only the fact that our force is negative that we change the direction of the force to the opposite direction? Is this correct? So, anytime we end up with a negative force, we automatically change the direction obtained from from the right hand rule to the opposite direction? Or is this only if we have a negative charge? In any case of having a negative force, couldn't we just use our left hand instead and get the right direction without having to change it?
Sorry if my explanations are long and confusing, I just want to know if I am understanding this part correctly. :)(3 votes)
- 1. apply the same concept with the left hand
2. use the right hand and take the opposite of the direction of the force you find.(4 votes)
- is there ever going to be a section dedicated to AP Physics C? I see that there is an AP Physics 1 and 2, and so i was wondering if there is going to be an AP Physics C in the future(4 votes)
- Hey khan academy, please stop using calculators when we can't even use them on the MCAT..(0 votes)
- As someone else has pointed out, this video was not created exclusively for the MCAT but was included in this playlist to help viewers gain a deeper understanding of the concepts.(13 votes)
- no calc on MCAT...(3 votes)
- Is this partly why the planets in the solar system orbit the sun is the sun's magnetic field?(1 vote)
- Good thought, but the intensity of the magnetic force of the planet/sun system is not strong enough to cause the planets to orbit the sun. Instead, the orbital motion that we observe is due to the mutual force of gravity between the planet and the sun. You may also suspect that other planets perturb each other's paths, but the major effect is due to the sun/planet mutual gravitation. I hope that was clear.(2 votes)
- what do you mean by changing electric field? Does it create a magnetic field? If a positron moves perpendicular to the changing electric field , then would there be any magnetic force on the particle? Please help(1 vote)
- When do you account accelerating voltage when finding the radius?(1 vote)
- In the last video we figured out that if we had a proton coming in to the right, at a velocity of 6 times 10 to the seventh meters per second, so the magnitude of the velocity is 1/5 the speed of light, and if it were to cross this magnetic field, we used this formula to figure out that the magnitude of the force on this proton, not protein, would be 4.8 times 10 to the negative 12 newtons, and then the direction, we used our right hand rule, because this was a cross product, and we figured out that it would be perpendicular, it would have to be perpendicular to both 'cause we're taking the cross product, and right when it enters, the net force will be downwards, but then think about what happens if you have a downward force right there, then the particle will be deflected downward a little bit so its velocity vector will then look something like that. But it's still in the magnetic field, right? And not only is it still in the magnetic field, but since the particle is still moving within the plane of your video screen, it's still completely perpendicular to the magnetic field, and so the magnitude of the force, on the moving particle won't change, just the direction will. Right? Because if we do the right hand rule here, but if we just move our hand down a little bit, if we tilt it down, then our thumb's going to be pointing in this direction. And that just keeps happening. It gets deflected that way a little bit. So the magnitude of the velocity doesn't ever change, it always stays perpendicular to the magnetic field 'cause it's always staying in this plane, but the orientation does change within the plane. And because of that, because the orientation of velocity changes, the orientation of the force changes. So when the velocity is here, the force is perpendicular. So it acts as kind of a centripetal force. And so the particle will start moving in a circle. So lets see if we can break out our tool kit from what we've learned before in classical mechanics, and figure out what the radius of that circle is. And that might seem more daunting than it really is. Well, what do we know about centripetal forces and radius-es of circles, et cetera. Let me make some space, first. Let's see, I probably don't need this. I don't even need the first few of these field lines. Well I'll leave everything else, I'll get rid of that field line, that should give us enough space. I'll try to write small, but legibly. So what is the formula for centripetal force? And we proved it many, many videos ago, early in the physics playlist, well, centripetal acceleration is the magnitude of the velocity vector, squared, over the radius of the circle, and since this is acceleration, if we want to know the centripetal force, it's just the mass times acceleration. So it's the mass of the particle or the object in question, times the magnitude of its velocity squared, divided by the radius of the circle, in this case, this is the radius of the circle. And that's what we're going to try to solve for. And what do we know about the centripetal force? What is causing this centripetal force? Well, it's the magnetic field. And we figured that out, right? This is going to be equal to this, which we figured out, is going to be equal to at least the magnitudes, the magnitude of this is equal to the magnitude of this, and that, the magnitude is 4.8... times 10 to the minus 12th newtons, and so the radius is going to be, let's see, if we flip both sides of this equation, we get radius over mass, velocity squared, is equal to one over 4.8 times 10 to the minus 12. I could just figure out what that number is, but I won't worry about it right now. Then we could multiply both sides times this, times this MV squared, and we get that the radius of the circle is going to be equal to the mass of the proton, times the magnitude of its velocity squared, the magnitude of its velocity squared, divided by the force from the magnetic field. The centripetal force, 4.8 times 10 to the minus 12th newtons. And the radius should be in meters, since everything is kind of in the standard SI units, and let's see if we can figure this out. Get our calculator. And this is where that constants function is useful again, because what is the mass of a proton? Well, that's something that I personally don't have memorized, but if we go into the built-in constants on the TI-85, let's see, more... mass of a proton. This is mass of an elect... let me scoot it over a little bit. This is mass of an electron, this is mass of a proton. So mass of a proton, that's what we care about, times the magnitude of the velocity squared. What was the velocity? It was six times 10 to the seventh meters per second. So times... six times 10 to the seventh meters per second, squared, right, 'cause I have to take the... And then all of that, divided by, all of that divided by the magnitude of the centripetal force, which is the force that's being generated by the magnetic field. That's 4.8 times 10 to the negative 12. So divided by 4.8, E minus 12. Now let's see. Hopefully we don't get something funky. There we go. Now that's actually a pretty neat number. 1.25 meters. That's actually kind of a number that we can imagine. So if you have a proton, going through, going in this direction, at 1/5 the speed of light, through a, what did I say it was, it was a 0.5 tesla magnetic field, where the vectors are pointing out of the video, we have just shown that this proton will go in a circle of radius 1.25 meters. Which is neat, 'cause it's a number that I can actually visualize. And so this whole business of magnetic fields making charged particles go into circles, this is one of the few times that I could actually say has a direct application into some things, into things that you've seen. Namely, your TV, or at least the old-school TVs, the non-plasma or LCD TVs, your cathode-ray TVs, take advantage of this, where you essentially have a beam of not protons, but electrons, and a magnet, if you take apart a TV, which I don't think you should do, 'cause you're more likely to hurt yourself 'cause there's a vacuum in there that can implode and all of that, but essentially, you have a magnet that deflects this electron beam, and does it really fast, so it scans your entire screen of different intensities, and that's what forms the image. I won't go into that detail. Maybe one day, I'll do a whole video on how TVs work. So that's one application of a magnetic field causing a beam of charged particles to curve. And then the other application, and this is actually one where it's actually useful to make the particle go in a circle, is the cyclotrons that you read about, where they take these protons, and they make them go in circles really, really fast, and then they smash them together. Well, have you ever wondered, how do they even make a proton go in a circle? It's not like you can hold it, and, you know, guide it around a circle. Well, that's what they do. They pass it through an appropriate, strength magnetic field, and it curves the path of the proton so that it can keep going through the same field over and over again, and then they can use those, then they can actually use... electric fields, I don't claim to have any expertise in this, but then they can keep speeding it up, using the same devices, 'cause it keeps passing through the same part of the... collider. And then once it collides, you've probably seen those pictures. Those pictures, right? You know, that you've spent billions of dollars on supercolliders, and you end up with these pictures, and somehow these physicists are able to take these pictures and say, oh, this is some new particle, because of the way it moved. Well, actually, what they're actually talking about is these are moving at relativistic speeds, and since they're at relativistic speeds, as they move at different velocities, their masses change and all of that, but the basic idea is what we just learned. They move in circles. They move in circles, 'cause they're going through a magnetic field, but their radius-es are different, because their charges and their velocities are going to be different, and actually some will move to the left, some will move to the right, and that might be because they're positive or negative, or you know, and then the radius will be dependent on their masses. Anyway, I don't want to confuse you, but I just want to show you that we actually are touching on some physics that a physicist would actually care about. Now, with that said, what would've happened, what would've happened if this wasn't a proton, but if this was an electron moving at this velocity at six times 10 to the seventh meters per second, through a 0.5 tesla magnetic field, popping out of this video? What would have happened? Well, this formula would've still been the same, the magnitude of the force is the charge, but it wouldn't be the charge of the proton, it'd be the charge of an electron, times six times 10 to the seventh meters per second, times 0.5 teslas. So what's the difference between the charge of a proton and the charge of an electron? Well, the charge of an electron is negative. So if this was an electron, then the net force would actually end up being a negative number. So what does that mean? Well, when we use the right hand rule with the proton example, we said that the, at least when the proton is moving in this direction, that the net force would be downwards. Right? But now all of a sudden, if we reverse the charge, if we say we have a negative charge, this is the same magnitude, but it's negative, 'cause it's an electron, what happens? The force is now in this direction, using the right hand rule, but it is negative. So really, it's going to be a positive force of the same magnitude, in this direction. So if we have a proton, it'll go in a circle in this direction, it'll go like this... but if we have an electron, it'll go in a circle of the other direction. Now, let me ask you a question. Is that circle going to be a tighter circle, or a wider circle? Well, the mass of an electron's a lot smaller than the mass of a proton, and we had the radius is equal to the mass times the velocity squared, divided by the centripetal force. So if this mass is smaller, the radius is going to be smaller. So the electron's path would actually, it would move up, and it would be a smaller radius. Actually proportional to the difference, the differences in the radius-es is the difference in their masses, actually. But that would be the path of the electron. Anyway. I thought you'd be interested in that as well. I have run out of time. I will see you in the next video.