The position of a particle with mass m traveling on a helical ...

Question

The position of a particle with mass m traveling on a helical path (see Fig. 11–48) is given by $\vec{r}$ = R cos (2πz/d) î + R sin (2πz/d) ĵ + zk̂ where R and d are the radius and pitch of the helix, respectively, and z has time dependence z = v_𝓏t where v_𝓏 is the (constant) component of velocity in the z direction. Determine the time-dependent angular momentum $\vec{L}$ of the particle about the origin.

Accepted Answer

Welcome back. Everyone. In this problem, an electron moves in a helical trajectory within a uniform magnetic field oriented along the y axis. The position of the electron is described by the vector R equals B plus two pi Y divided by C plus YJ plus B sine, two pi Y divided by CK where B represents the radius of the helical path. C denotes the pitch of the helix and Y varies as Y equals VT where VY is the constant component of velocity along the Y axis, determine the time dependent angular momentum L of the electron about the origin for our answer choices. A says that it's MVV multiplied by the expression two pi y divided by C two pi y divided by C minus sine, two pi Y divided by C minus two pi divided by CBJ plus two pi divided by C plus two pi divided by C sine, two pi Y divided by CK for B. It's similar to a first answer but no, the sine function comes first and then the cosine for our I component, the J component is the same and the sine and cosine functions, the sine comes first and the cosine comes second. For K. For C, the cosine term comes first and then the sine for I, there is no J component and for K, the cosine comes first and then the sine and for D it's MV multiplied by two pi B squared divided by CJ. Now, in our problem, as we said, we want to determine the time dependent angular momentum L of the electron about origin. What do we know about angular momentum? We recall that our angular momentum L is the cross product of the position vector R and our linear momentum vector P. But we can rewrite our linear momentum as the mass multiplied by the speed vector or the velocity vector rather. So we can write this as R crossed with M multiplied by V. And now we can take out our constant M to say that our angular momentum L equals M multiplied by the cross product of R and V. Now we might not have V readily available here. OK. But recall as well that V is the derivative of R with respect to T. So if we can express our term R in terms of T, then we will be able to differentiate with respect to T find a value for V and then find the cross product to find the angular momentum. So how can we do that? Well, this equation can help us Y equals VT notice we have Y terms in our vector for R. So we could replace them with Vyt and differentiate with respect to T. No. That means here two pi ye is going to be equal to two pi multiplied by V divided by C multiplied by T. And now you notice that we can take this term to make it simpler while we work and replace it with a common term here. Let's just call it alpha. So no, it's gonna be a bit easier for us to work. We won't have to use as much terms. Uh while we differentiate and find a cross product and so on. Now here, since we've given this value for alpha, then if we rewrite R, if we rewrite our vector R, that means R with this simplified notation is going to be B cos alpha T I, OK plus V I TJ because remember uh it was YJ plus B sign alpha T K. Now, since that's the case, we can differentiate R with respect to T to get V. Now, when we differentiate R with respect to T, then we should get it to be equal to negative alpha B sine alpha T, OK, I plus VYJ. OK. Plus alpha B multiplied by the cosine of alpha T. Now that we have an expression for our velocity, we can go ahead and find our angular momentum because no, this means our angular momentum L is going to be the cross product of R and V multiplied by M. So it's gonna be M multiplied by. Well, let me write all to our components here. We have IJ and K it's gonna be B cos alpha te multiplied by VT multiplied by B sine alpha T. OK. And for our next term, it's gonna be a negative alpha B for V. It's gonna be negative alpha B multiplied by the sine of alpha TV for J component. And alpha B costs alpha T for our K component. So now we can go ahead and find a cross product no, in inches of time. I'm not gonna find a cross product for all the terms. I'm just going to do it for one. And then I want you to try to do the rest. But essentially remember that for our cross product, our, it is going to be positive. Let me just uh write it here. OK? Our, it is gonna be positive. Our J term is gonna be negative and our K term is gonna be positive. And to find the cross product, we remove the row and the column essentially of the, the, the, the component we're trying to find and then multiply the remaining terms diagonally. So, so for example, for I, OK, if I were to write that here, let me put that in. Let me write it back in black here. So it's gonna be M multiplied by Vyt, OK? Multiplied by alpha BC alpha T minus the, the, the, the product of the other diagonal that is minus VY multiplied by B sine alpha T. OK. So that all of this, I mean, I should put this in a square bracket here. All of this is for or, or I component next for the J component, then you'd remove these and basically do the same thing. OK? I know when you do that, you're going to get the product of B CAS alpha T and alpha B CAS alpha T, which is going to be equal to my apologies, which is going to be equal to let me write this properly here. Alpha B squared sine squared, alpha T minus alpha B squared cas squared, alpha T, OK. And that's our J component. And finally, for our key component, it's going to be BVBVY Bvyk alpha T, OK, plus VT multiplied by alpha B sine alpha T K. OK. Now you realize that's, that's, we've written a lot of stuff here. OK? But no, essentially we finally found the cross product. Now that we have the cross product, we can make it a bit simpler by factoring OB and Vy notice that it's in most of our terms. OK. So when we do that, then our angular momentum L is going to be equal to MB multiplied by alpha T plus alpha T minus sine alpha T that's for our item. Well, OK. Yeah, let's write that here or, or item minus alpha B divided by VY J. That's for our J term plus OK. Plus alpha T multiplied by the sine of alpha T for our K term. OK. Or rather let me write that properly here. So cos alpha T rather plus alpha T multiplied by the sine of alpha T or over K term. OK. And no, no, then we can based on what we have here, know that we've simplified, then we can try to further simplify by putting in our values for alpha. Remember we said alpha was equal to two pi Y or two pi Vy divided by CT and we can replace Vyt with Y. Now again, in the interest of time, I'll allow you to do that. But when you do, let me bring this over a bit, when you do, you should find that you get L to be equal to mbvy multiplied by two pi YC divided by sorry two pi Y divided by C. And you can see that here because this would replace alpha T basically multiplied by the cosine of two pi Y divided by C minus the sine of two pi Y divided by C in minus two pi divided by CBJ. OK. OK. Plus, write that properly here plus the cosine of two pi Y divided by C plus two pi Y divided by C multiplied by the sine of two pi Y divided by CK. And now for all our terms, you can see where two pi Y divided by C replaced alpha, T where let me just make a note of that we know that alpha T would be equal to two pi Y divided by C. Thus, this is our angular momentum. And if we compare what we have here to our answer choices that tells us that A is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Angular Momentum

Helical Motion

Time Dependence in Motion

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