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An electron has an energy of 100 eV as it enters a magnetic field of 3.50 X 10-2 T. Find the radius of the orbit....

Question

An electron has an energy of 100 eV as it enters a magnetic field of 3.50 X 10-2 T. Find the radius of the orbit.

An electron has an energy of 100 eV as it enters a magnetic field of 3.50 X 10-2 T. Find the radius of the orbit.



Answers

Find the radius of the orbit when (a) an electron or (b) a proton moves perpendicular to a magnetic field of 0.86 T with a speed of $6.47 \times 10^{5} \mathrm{m} / \mathrm{s}$.

Hi for this question. We have the particles accelerated from arrest three potential difference of equal votes And then it moves on the magnitude of 30 G. So we have the magnetic force and difficult Qv be charged times the velocity time stick like magnetic field and centripetal forces. Mv squared of our mass of the particles B squared velocity squared went right by the radius of the path. So we're going to quit this together. So Q V. P. You go to M V squared over R. V consumes here. So we have we left. So making already subject are equal to that M. V over a QB. Yeah. Now we need to find velocity and what's going to do that? We use equity. The electric potential energy which is going to take you to the kinetic energy at the final point. So the electra production energy allows it to gain kinetic energy And to make it make him feel the subject textile view. We call to skirt of two VQ divided by M Yeah, no substitute this into this. So our culture and our Q. B. Screwed Children like you. All right. Yeah. And simplifying this will be go to one will be Squirtle to VM virtue. So now we can get the problem doesn't question given that the potential difference vis is 800 votes And my 90 field be did you go to 30 goes and one glasses culture. Okay 10 to the power of -4 Tesla speak 30. Mhm That's why I'm running for kessler and m muslim electron would be 9.1 Door. And I understand that file Minour taxes one killer grammars charging an electron minus Which is the culture -1.6% power -19. Colombia subsidies into the form of R is equal to 1/30 times 10. -4. The spirit of two 800 terms nine times. Okay. One point I'm Jeremiah Physical My Life Statue one. All right. Bye -1.6 sounds 10 12 -19. And we can actually remove the minus here. This uh, this should be culture herb equal to Their points 32 m And it should be called to 1st part. I want to find the optimal frequency. Mm and we can first find the angle of velocity. So this is going to be over her and from this equation can figure it out subject and just be cool too. Mhm Q B. I am culture -1.6% power For the next 19 terms taxi terms turn to the problem with us for right. Right 9.1 during their 2% of all my statue one. Yeah. Yeah. Yes. But this will be in recounts. My second means cover these revolutions per second was planned by Yeah, supplier divided by super intense. What's going to be one position. And then did you don't have a negative sign? Of course I guess that's just direction of the force. So this there's also against the bill Richie. Approximately 84. So I'm started part of six. It's revolutions per second and this week tends bar of sixties mega. So this is it's a four like our heads traditional party. This is for heartbeat, physical part.

OK, folks. So in this video, we're gonna be talking about this problem we have. We're gonna find the frequency of the revolution of an electron with an energy of 100 C B in a uniform magnetic field of magnitude 35.0 Mute Tesla's. Um so that's question number one Part B of the question is calculated ingredient of the path of this electron if its velocity is perpendicular to the magnetic field, Um, so, so far apart, A. What we're gonna do is we're gonna find the frequency, right? So an equation that's gonna be a very helpful for us is this relation which is f equal, this one over the period, right? Abso there's an inverse relationship between the frequency of the period and period of something very easy to calculate because period of revolution is just the it's just gonna be the distance that you travel in a circle divided by the speed of your particle s o. The distance that you travel in a circle is obviously just the circumference of the circle, which is two pi r, right, and in the speed, let's just say that it's vey right, and so we have this period s o The frequency is just gonna be the inverse of this period, which is be over to pie. Okay, so that means as long as as soon as we find the and are we're done for partner. How do we find B and R? Well, we think we can take a look at what other information were given in this problem. For example, we know that we have a uniform Byfield with a constant magnitude, which is 35.0 Mute testaments were So we have a constant Byfield magnitude everywhere. So we know from Newton's second Law that an A, which in this case becomes M v squared over R. It's just Q times the B. This right here is the magnetic force. I'm assuming that the velocity vector and the magnetic director is perpendicular to each other so we can ignore that scientific term. So let's simplify this a little bit. We have a V here, and then we have a V squared over here. We can cross out one of the V's so we have V over r equals Q B divided by the mass. Okay, so so we have this This relation between V over R and Q B o Gram, If you look at this carefully, you see that all of the terms on the right hand side are constant. For example, this example this this Q is just the charge of the electron, right? And that is a constant something that you can just go ahead and Google be in system attitude of the Byfield, which is given in this problem, um, em It's just the mass of the electron, which is also a constant that you can Google. So So basically we have the answer to part a already this if you planted If you plan V over R two here, you will find the frequency of revolution. So let me just do that. So we have one over to pie times V over R. But the over are we know to be Q B over Mass. So this is the answer to partner, right? This is the frequency of revolution of the electron traveling in a circle. Um um, so that was part eh? Let's go ahead and do part B. We're going to calculate the radios of the past. Now, in order to do this, we have to use one of the other pieces of information that were given to this problem, which is the total energy, the total energy, which is which. In this case, it's just It's just the kinetic energy because there's no potential energy in this case, the magnetic field. That's not have a potential energy. Um, so we can write this at K, which stands for kinetic energy. K is equal to 1/2 and the square that means the is equal to two K over M square root. We're assuming level off these negative. I mean positive because, you know, obviously it has to be positive, but anyway is square root of two K over him. Um, what does it do for us? What does knowing the velocity due for us? Well, if we know the velocity, we can pluck this velocity into here and find the radius. So let's do that. We have radius equals, um, the, um over Q B, right? If you just do your algebra, you know, you move your terms from one side of the equation to the other side of the equation, you will find that Are you close? The Times Remember Q B and let me plug in V. Here's we have the equal squared of to kinetic energy over m times M over Q B. This is an equation for the radius of revolution. Okay, um, if you I'm not gonna bother plugging in all of the numbers because we don't have enough time in this video, you can do that yourself. After you plugging the numbers into their calculator. Your calculator is gonna give you the answer. Which, her party. The answer's gonna be F Nichols. 0.978 megahertz. Um, and for part B bar is going to be 0.964 meters. Okay, so So this is so. Here we are. We have Ah, we have the answers to part a part B and we're done. Okay, I will see you in the next video.

Here. The radius of the circular path is gonna be equaling the mass of the particle times V the velocity developed by the charge multiplied by the magnetic field. This is gonna be equaling the mass of an electron 9.11 times 10 to the negative 31st kilograms multiplied by the velocity of 7.50 times 10 to the sixth meters per second. Divided by the charge on an electron. We know to be 1.60 times 10 to the negative 19 cool arms multiplied by be the magnitude of the magnetic field 1.0 times 10 to the negative fifth Tesla's. And we can then say that here the radius of the circular path for the electron would be 4.27 meters. This would be our final answer. That is the end of the solution. Thank you for

All right. So we got a cause of great photons. It's in deep space. Uh, and it's got a certain energy was our inner cellar. There are because of a proton interstellar space making this really huge, huge orbit. Um, that is like energy here. That's equal to 10 kill electron bulls. All right, it executes this circular orbit. It's actually crazy enough. This circular orbit right here is actually the same radius as, Ah, Mercury's orbit around the sun is insane. It's ah, five point beat zero times, 10 t 10. I'm years. Wow. So we want to figure out what is the magnetic field. Now we obviously know there's a magnetic field going through here. What is it though? We figure it out from, So let's give it a shot. First things first. You've been using this quite a bit here, um, physics. And that's pranic energy. We know that's 1/2 times mass times velocity. Well, the proton master, the Portland velocity, the proton. Swear I'm gonna go ahead and solve this for velocity. When you isolate that, you get two times the tank energy that's divided by a mass of the proton. Cool and you can also use a different equation here for the radius. Use this quite a bit as well. This is massive proton times velocity. It's divided by charge of the proton times that my nephew now what I want to you is I want to take this V. I have here almost substituted in this guy. All right? And then I want Oh, go ahead and solve. Um, I'm gonna do that. Almost saw for the first very quickly here, So b equals massive her tone, times velocity and by charge of four times more specific. Now times. Yes, I start magnetic field. We're gonna be using a second. I just rolled down and seemingly more space here were really filling this up here one second. All right, Some fresh real estate here. So we had a couple important equations after we had a radius. Just reiterate this charge and that I feel that we had our the value that we solved for from this. When we did that, we had mass times, velocity times, charge. Andres and I could put the little sub scripts massive proton, proton. But we know I'm telling you guys, it's it's pretty clear That's problem is discussing the proton. So we know what value supplies in there. We'll cross that bridge. All right, so let's go ahead and take our velocity that we found earlier. I just refresh our memory. That's this guy right here. We're gonna fool in that in, just like I only should earlier into that velocity. You not actually this guy in here. Oh, yeah. So we'll do that. E equals mass by the by in charge times the radius where you're gonna have that square root from that velocity Two times athletic energy divided by mass we can rearrange is a little bit more We'll get B equals two times kinetic energy times the mass of the proton And down here we got charged you all multiplied by the radius R That would be the magnetic field that would be experiencing that this program will be experiencing out here and deep space. All we gotta do is plug in America values to get actually numerical solution. So it's going to times like energy, which we know 1.6 one to the last 12 0.7 times, 10 to the negative 27. All right, When there's gonna be that routes over this guy here. And then we have the denominator. All right. The denominator we gots Our charge is that 1.6 I was 10 to the minus 19 who longs when we have our radius Just 5.8 times 10 to the 10 meters. That's that same radius that mercury experiences for its orbit. Look, I'll let it. And we're looking at a Byfield value of 7.88 I was 10 to the negative. 12 Tesla's This magnetic field indeed, face on these magnetic fields in deep space have quite a big influence on part of his face and just in general, a lot of weight phenomenon that since able to do these magnetic fields in deep space, it's very, very interesting stuff.


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