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A charged particle moves in a region of uniform magnetic field along a helical path (radius 5,0 cm; pitch 12 cm; period 15.6 ms) What is the speed of this particle?...

Question

A charged particle moves in a region of uniform magnetic field along a helical path (radius 5,0 cm; pitch 12 cm; period 15.6 ms) What is the speed of this particle?B2rKPitch0 a, 26.54 @ b: 20,13 6 7.69d. 23.982. 21.55

A charged particle moves in a region of uniform magnetic field along a helical path (radius 5,0 cm; pitch 12 cm; period 15.6 ms) What is the speed of this particle? B 2r K Pitch 0 a, 26.54 @ b: 20,13 6 7.69 d. 23.98 2. 21.55



Answers

A magnetic field forces an electron to move in a circle with radial acceleration $3.0 \times 10^{14} \mathrm{m} / \mathrm{s}^{2}$ . (a) What is the speed of the electron if the radius of its circular path is 15 $\mathrm{cm} ?$ (b) What is the period
of the motion?

In this question we have positive on um That is uh there is traveling with the velocity that is 89 degrees from the magnetic field. So for example if the medic periods to the right, okay. The the velocity vector uh the positron is uh huh And angle 89 degrees on the monastic view. And so we want to find peter it. It's going to make a psychopath. Okay. You want to find a period, find a peach. And the radios of the path. Okay, so the path will look something like this. Yeah. And then the peach will be the distance. Yeah peach then. And every guess would be like this. Okay, It's something like this. Okay, so to solve this question for me to find a period you'll be using, he goes to two p.m. Or to be so he is too high times 9.11 time centered upon negative 31 mm two charge of one electron. And then the magic you is 0.1 Tesla. The period is 3.58 And send it out power negative 10 seconds. Mhm. And then be to find a pitch, pitch is equal to be for science data. Okay. Mhm. Okay. So we need to find V. Mhm. So using he goes to have and the square so we is two K. O. M. Spirits. So peach just be two K over em my roots for saint data times. T Okay. So just put in the numbers. Okay, 2000 electron wards cause I 89 degrees. Then multiply by T. 23.58 I have sent a negative 10. The answer is 1.16. Thanks handed. Huh? Found 94. He does then in Patsy fucking the yes our Okay. Yes it goes to M V. Or vi que mm. This is an over beat you James spirit to cave. And so we have graduate up to NK by beat. You keep putting the numbers. Mhm. So the answer is uh 1.51 times 10 to the power of negative three m. So this is the answer for this question, Patsy, and that's all.

In this question. Um They are given this situation so we have an electron and we have energy here. The energy on. Uh huh 1.20 E. Then it's moving in on uniform magnetic field. So yeah, he undergoes uniform circular motion. Radius is 25. Yeah. Yeah. We want to find the speed of the electron, the main menu two of the magnetic view frequency and the period of the circular motion. So to do a we find it speed. Yeah. Do you think Kay was to have and square. So yes. Two kids that I am scared. So just in the numbers. So um Kay E. V. Is 1.6 times 10 to the power like a big 16 drew and then area is never won one holiday. This is they take the square root. The answer is 0.5 times 10 to the power seven. He does hope. Okay, this is the answer for partying then. Part B Yeah. Community of the magnetic field. To do that. To be using. Our goal is to envy over BQ. Okay so B. S. Uh Maybe you are food using a speed that we calculated in A. So the answer is 4.67 and sent to the parliament before. So this is the unsellable pipi. And then you proxy. Want to find a frequency a circular motion probably using. Um And it was too one of the key wishes we have to buy our. Yeah they're putting their numbers. I took my hero five takes into account 17 divided by hi differently side. And the answer is 1231 and said with the power heaven. Okay so this is beyond several patsies then paddy period. Yeah using he was. So if you read yes the recipe girl frequency 765 and standard. How many it? Okay so this is the answer for you and that's all. All this passion.

OK, folks. So in this video, we're gonna be talking about this problem. We have an electron of kinetic energy 1.2 Uh que evey that that is, circling in a playing perpendicular to a uniform magnetic field. The orbit radius is 25 centimeters. We're gonna find four things we're gonna find first of all of the electrons speed, and then we're gonna find a beef yellow, the magnitude, and then we're gonna find a certain frequency. And then after that, we're gonna find the period of the motion. So basically, this probably is not necessarily conceptually speaking difficult. You just have to realize that they, uh that there's some equations that describe how the motion of the electron is gonna be at soon as a printer's a region where there's a uniform magnetic field. So basically, you have, you know, a uniform magnetic field that points into the page, right? And then and then there's an electron that sliced into this region. As soon as it slides into this region is gonna start whirling around in a circular motion, right? And when we're given is the quantities such as the getting energy, which which I'm just going to denote by t. Um, and that number in this case is gonna be 1.2 kilo electron votes. All right. And what were also given is the radius of the orbit. Does this radius is swirling around and that whirling has a radius, which is r equals 25 centimeters. Now, while I'm doing this problem while I'm excuse me while I'm doing this problem, I'm not gonna bother. I'm plugging in all of the numbers. I'm just gonna write everything in terms of symbols and at the end of the problem, our book plunk all of the numbers into the calculator and get all of the answers I came anyway. Um, So first of all, the way we're gonna do this is by recalling a Newton's second law which says force equals empty. And the acceleration in this problem because there's an electron is in a circular motion. The acceleration is just gonna be v squared over r. And we happen to know our That's a bonus that is equal to something else. And it's equal to the Lawrence Force, which is given by Q V. Crosbie. Usually there would be a cross spotted. You know, the cross me, but I'm not gonna do it here because, um, the velocity vector is perpendicular to the b field director here. This problem because it says here that this circles in a plane circles and appoint perpendicular to a uniform Byfield. Right? So? So whenever you have a, you know, velocity vector that is perpendicular to the Byfield? No, to the background. Byfield, you don't have to write that cross brought up because that cross brought up has a sign, you know, And that sign, it's just one. Okay. Okay. Anyway, I'm assuming that you're familiar with the cross product on everything. It's a Q V b. Usually there would be a scientist, but I'm not gonna write it. Um, okay, so let's simplify this expression of another bit. We have a beat groups. We have a V here, B squared here so we can cancel out one of the V's. Um and we're left with, um are our equals and the over Q b. So this doesn't really tell us a lot. Actually, this does tell us, tell us a lot because we know what our is and we know where and this and this is the mass of an electron, which is something that you can just go ahead. Google. So we know what, Agnes. We know what Q is. And what we don't know is we don't know. We don't know B and we don't know the So we want to find those two things. Okay, so what can we get out of the fact that we know about the kinetic energy of the left front? We know that the King of Energy could be expressed as 1/2 ab the squared right stack. That means V is gonna give you It is gonna be to tea over m the whole thing square bit. And we know what TV is. Redo. What is so we can figure out what V is. Okay, so now we have the velocity, The speed. Um, we can plug that in here, um, and give the be field magnitude. This means the be field magnitude is gonna be m B over Q, Are we know what? At this, we know what b is. We know what Q is, and we know what the radius is. So let's pluck the expression for V here. We have screwed up to t over m Q times are which is just gonna be squint, Whatever you are. Square root test some tea. AB, this right here is be okay on. So we know the speed. We know the magnitude of the be felled. And what also, what other things do we need to figure out? Oh, we need to figure out the circle in frequency. Well, um, a circular frequency and period of the motion. So let's figure out where the period of motion is first because, um, because frequency is just one of her tea, but t is easier to figure out in my opinion of you as a team, you know, what was the period was the total amount of time that it takes for the electron to finish circling around. How long does it take to finish circling around once, right. That is gonna be the total distance that this electron travels wishes just which is to the circumference of the circle, which is two pi times the radius of the radius we know and that divided by the speed, the speed we also know so we can pluck all of those in. So we have two pi r divided by Lee Ah, squared of two key over. M huh? I realize I'm using very Americans using them letters here. This tea is the period. This tea is the kinetic energy. They're not the same teeth. This is the kinetic energy. Okay, you're gonna have even have tea. Excuse me for using these confusing matters. Um, anyway, I ask that this is you go to okay, but that's just leaving like that. So we have the period. Was the circle in frequency. The circles, The circling frequency is one over the period, so it's going to be square root of two t over em over to pie, huh? So this is part see, Mrs Part D. And this is partly this is part B. Yeah. Let me use a different color just to make it out in a bit more clear. All right, so these four boxes are all that we need to find. Of course, I didn't plug in the numbers yet by SMS you plugging on of the numbers into your calculator and your calculator is gonna give you all of the new miracle answers. Which is that the clothes 2.5 times 10 to the seven as a unit is meeting over second part B. Um, this is equal to 4.67 times 10 to the negative four. Test less ad part See us is equal. But one point who 1.31 tenants tend to the seven hurts on the clear is 7.63 times 10 to the negative eight seconds. That's how long it takes for the electron to finish one cycle. Okay, folks. So, like I said, this is not particularly conceptually speaking that difficult to understand. Um, you have some relations and some equations. All you need to do is is to solve those equations. Okay, so let me rephrase that. All of it. Um, basically, you have a beefy onto this perpendicular to the direction of velocity of the electron. And the first thing you need to realize in order to do this problem is that this electron circles around. And when you realize that you could use Newton's second law to find a relationship between you know, the centripetal acceleration and they Lord force, and once you have that equation, you can figure out all of the other quantities, like the or period or frequency and blah, blah, blah. Okay, so that's basically a quick little suddenly, Um, if there's anything you need to understand a few free D'oh! Leave a comment down below. Okay, I'll see you in the next video.

In this problem. We have given a positron with Canada Energy, but two electron volt is projected into a uniformed magnetic field which is 0.100 Tesla with its velocity vector meeting an angle. So this is a velocity vector. And suppose this is the velocity vector and it is making I suppose this is magnetic field director. It is main 89 degree. With the magnetic field, we have to find the time period which and radius. So we know that if yeah, velocity is making some angle with magnetic field, there will be a joke on two component off velocity one is perpendicular and oneness parallel due to its parallel, confident and vertical confident. It will be moved like a Helling in a helical path like this, positron will move in a helical path and we have to find the time period so time period is equal toe. I am by Q V by putting the values off M, which is 9.11 to 10 to the power minus 31 g on the charge which is 1.6 into 10 to the power minus 19 into magnetic field with this which is 0.100 Tesla. So the time period will be equal. Toe 3.58 into 10 to the power minus stance again. No, here. If you see the diagram, it is moving in a helical path. So from this point toe, this point it is a quilt work which we know that it is moving in a circular path due to its vertical component. And it is moving ahead in this direction because off it parallel component of velocity. So for pitch first, we will find it's velocity. We know that cannot take an industry is equal to a half and we squared. So from here we will find velocity. We just on the route to K I am from here we will put the value off kinetic energy which is to elect on world toe into 10 to the power three electron world. Now we will hear We have given connecting energies to kill electron world kill direct on hold. So this is two into 10 to the power three. Now we will change it electron wall to Jewell So we will multiply with 1.6 into 10 to the power minus 19 Now the value off em yes is 9.11 to 10 to the power minus 31 kg. After solving this, we will get the the loss of the richest 2.65 and two 10 to the power 7 m per second. Now we know that it is the velocity and it has two components. One is spelled and one is come for political. So pitch will be called toe parallel component in two time period so fellow confident will be equal to V cost 89 in two time period. We put the value of V and time period. We already know after putting the value, we get the pages equal toe 1.66 into 10 to the power minus 3 m. Now we have to find the radius in part C. We know that our is equal to M the perpendicular divided by Q and T V. Again we will put the value off, which is 9.1 100 into the bar, one of 31. And but when dealer come from confident of velocity Yes, V sign 89 divided by charge. Q and we, after putting the value off Q V and V. We get the radius, which is equal to 1.51 into 10 to the power minus 3 m.


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