Magnetic force on a proton example (part 2)
Sal determines the radius of the circle traveled by the proton in the previous example by using the formula for centripetal acceleration. Created by Sal Khan.
Want to join the conversation?
- Why does the path of the force vector change? It is because the velocity vector is constantly changing too? If this is the case, why does the velocity vector change?(25 votes)
- At Marthasilvabio. You maybe know from dynamics that when we have a force on an object it changes its velocity over time. Changing velocity can mean that the object starts or stops moving or goes in different directions. Please remember the difference between velocity (vector ) and speed (scalar). If an charge particle is moving throw a magnetic field. The magnetic field will created a force on it. This force is always perpendicular to the velocity vector and the magnetic field vector. When the force is acting on the particle it brings it to move in direction of the force because of the newtons law . But force will try to stay perpendicular to the velocity so it move away from the velocity. Velocity is hunting force or force is pulling velocity. And that results in a circle. The same you see in Mechanics by Gyroscopic precession. https://www.youtube.com/watch?v=ty9QSiVC2g0(1 vote)
- the proton should have come out after executing half the circle.because by then it would have got out of the magnetic field.rite?(11 votes)
- I believe that at first, the particle would simply bend and continue traveling into the magnetic field. Eventually, it would complete a circle, but I guess that we are meant to assume that either the field was large enough to contain the circular path of the proton, or that the proton traveled far enough into the medium to enter a circular path.(3 votes)
- Sal did an awesome job on this video! However I do understand the centripetal acceleration formula but I do not understand why at2:42he put an M in front of the MA formula? I do know that he may have broken up the force in MA but where did the A go? Thank you! Also please explain why he made it equal the force after denoting the MA, thank you!(3 votes)
- because centripetal force=ma and a=v^2/r so F centripetal =(mv^2)/r
F magnetic=F centripetal
simplifying (mv)/r=QB(17 votes)
- i dont understand how the direction of the proton will still be perpendicular to the field lines after deflection. Please help..(3 votes)
- The force/acceleration is always perpendicular to the field and velocity. If the velocity is initially perpendicular to the field then the charge will move in a circle which is always perpendicular to the field.(5 votes)
- On 4.43,the calculator use to find the mass of the proton,is that only in TI-85,because I have TI-84 & I can't figure how to find that on my calculator(1 vote)
- Hi, I checked it out on my TI-84 as well. They do have these constants build in.
Go to "APPS", then to "Scitools", then "2:UNIT CONVERTER". If you press f1 or f2 now, you'll end up in the constants menu!!
Good luck(6 votes)
- Is this the same method they've been using to smash particles in search of the Higgs Boson? I know it's a little sidebar, but, I think it will help me grasp it a little more if this is the same process. Thanks.(3 votes)
- My Physics reference book states that the direction a proton moves in is anticlockwise, while the electron moves clockwise, which is opposite from what Sal said. Can anyone explain to me why, or my book has a mistake?(1 vote)
- It depends which way the magnetic field is going.
If you have a magnetic field into the page, a proton coming from the left will circulate counterclockwise. You can determine that with the right hand rule
An electron would go the opposite direction.
If you reverse the magnetic field, both particles will reverse, too.
Sal's field is coming out of the page.(4 votes)
- A question: what happens to the radius if the velocity increases? Does it stay the same? If velocity increases the force increases also. In the first yellow line both increases would cancel out and the radius in unaffected?(2 votes)
- From the combination of centripetal and magnetic force formulas (m*v^2/r, q*v*B*sin(a)) we can figure out the radius which is r=m*v/q*B. If the velocity increases then the radius increases too.(1 vote)
- At3:06, why is the centripetal force equal to only the MAGNITUDE of the magnetic force and not the vector force? I thought that the magnetic force WAS the centripetal force in this case, just like how gravity or tension or friction are sometimes the centripetal force.(2 votes)
- Good question! you are correct in noting that magnetic force should be a vector. In general, the force on a charge in a magnetic field is F = q v X B (Asterisks denote vectors here, and the X denotes a cross product.)
However, Sal decided to ignore the cross product, since v and B are perpendicular (B comes out of the screen, so anything in the plane of the screen is perpendicular to it), and we can thus simplify the magnitude of magnetic force to be q v B. Moreover, he uses only the magnitude of this force because we hardly ever use the vector form of centripetal force. If you look back at the circular motion videos, you'll notice that unit vectors hardly ever come up.(1 vote)
- At6:00, Sal says that the radius is 1.25 m. But since the proton is moving at 1/5th the speed of light, shouldn't we take relativity into account? The mass of the proton actually increases as per the formula
And the corrected radius is actually 1.28m. Isn't it??(1 vote)
- I didn't check your math but it looks right. He probably should have used a lower velocity. As you probably know, relativistic answers are always more correct than Newtonian ones, but the difference is usually too small to worry about and it makes it a lot harder to teach and learn basic physics. All those problems where we ignore friction are also "wrong", right?(3 votes)
In the last video we figured out that if we had a proton coming into 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 would be 4.8 times 10 to the negative 12 newtons. And then the direction, we used our right hand rule because it's a cross product. And we figured out that it would be perpendicular-- well, it has to be perpendicular to both, because 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. 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 because it's always staying in this plane. But the orientation does change within the plane. And because of that, because the orientation of the 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 let's see if we can break out our toolkit 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 radiuses of circles, et cetera? 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 the centripetal force? Well, it's the magnetic field and we've figured that out. 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 magnitude is 4.8 times 10 to the minus 12 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 1 over 4.8 times 10 to the minus 12. I could just figure out what that number is, but I won't worry about that now. Then we can multiply both sides 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 divided by the force from the magnetic field. The centripetal force. 4.8 times 10 to the minus 12 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 constant 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 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 6 times 10 to the seventh meters per second. So times 6 times 10 to the seventh meters per second squared. And 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. Divided by 4.8 E minus 12. Let's see. Hopefully we don't get something funky. There we go. That's actually a pretty neat number. 1.25 meters. That's actually a number that we can imagine. So if you have a proton going in this direction at 1/5 the speed of light through a-- what'd 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 because 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 can actually say has a direct application 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, because you're more likely to hurt yourself because there's a vacuum in there that can implode, and all 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 these 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 could hold it and guide it around in 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 actually use those electric fields. I don't claim to have any expertise in this, but then they can keep speeding it up using the same devices, because it keeps passing through the same part of the collider. And then once it collides, you've probably seen those pictures. You know, that you spend 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, 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 mass is changing, and all that. But the basic idea is what we just learned. They move in circles. They move in circles because they're going through a magnetic field. But their radiuses are different because their charges and their velocities are going to be different. And actually some will move to the left and some will move to the right. And that might be because they're positive or negative, and then the radius will be dependent on their masses. Anyway, I don't want to confuse you. But I just wanted to show you that we actually are touching on some physics that a physicist would actually care about. Now with that said, what would have happened if this wasn't a proton but if this was an electron moving at this velocity at 6 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 have still been safe. The magnitude of the force is the charge-- but it wouldn't have been the charge of a proton, it would have been the charge of an electron, times 6 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 used 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. But now, all of a sudden, if we reverse the charge, if we say we have a negative charge-- the same magnitude but it's negative, because it's an electron-- what happens? The force is now in this direction, using the right hand rule, but it is a 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 a question. Is that circle going to be a tighter circle or a wider circle? Well, the mass of an electron is 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 this mass is smaller and the radius is going to be smaller. So the electron's path would actually move up and it would be a smaller radius. Actually proportional to the difference in their radiuses 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.