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Beam of particles focused by magnetic field

  1. Apr 15, 2009 #1
    1. The problem statement, all variables and given/known data

    A beam of particles of charge q and velocity v is emitted from a point source, roughly parallel with a magnetic field B, but with a small angular dispersion. Show that the effect of the field is to focus the beam to a point at a distance

    [tex]z=\frac{2\pi mv}{qB}[/tex]

    from the source

    2. The attempt at a solution

    All that I know for sure is that

    [tex]\vec{F}=q \vec{v} \wedge \vec{B}[/tex]

    I also think that, due to the angular dispersion, I'll have to somehow find an equation that will relate to all particles in the beam. We know that if v is initially perpendicular to B then the particle describes a circle with radius r = mv/qB. But what happens when the velocity is NOT perpendicular to B (nor parallel, in which case the particle will experience no force)?

    My understanding is that with "focus" they mean that particles will move in a spiral around the z-axis (set in the direction of B) from where they were emitted to where they are focused, i.e. if distance r between the particle and the z-axis is initially r, then the final distance at point z will be 0. Is this correct?

    So far, however, all attempts at a solution have been unsuccessful.

    Could someone please give me a hint?

    Thanks!
    phyz
     
    Last edited: Apr 15, 2009
  2. jcsd
  3. Apr 16, 2009 #2

    Redbelly98

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    For problems like these, it's usually best to consider two parts of the motion separately:

    • Motion along B
      Velocity = vz = constant

      and

    • Motion perpendicular to B
      The trajectory is circular
      Centripetal force & acceleration play a role

    Hope that helps.
     
  4. Apr 18, 2009 #3
    OK, I've got some time today so I'll try and work it again from this angle. Thanks Red!
     
  5. Apr 28, 2009 #4
    Hi Red!

    Well, I've been working on this problem in between the half a million other things I need to do and I believe I've made some progress. From the theory, we can easily determine that the force component along B (z-axis) will be zero, and the component perpendicular to B (z-axis) will result in the particle describing a circle of radius

    [tex]F=qvB[/tex]
    [tex]ma_r=qvB[/tex]
    [tex]m \frac{v^2}{r}=qvB[/tex]
    [tex]mv^2=rqvB[/tex]
    [tex]r=\frac{mv}{qB}[/tex]

    Now, it seems from the question (pls see OP) that this ultimately means that the distance z we're looking for is equal to 2 pi r. I don't believe it is coincidental that this distance z is therefore equal to the circumference of the circle described by the particle's motion, but now I can't seem to make the connection between the two.

    Is there some theorem/reasoning I can use to prove that z = s (circumference of the circle) or even that the linear velocity component of the particle in the xy plane will be equal to the velocity component in the z direction (which will lead to the same result)?
     
  6. Apr 28, 2009 #5

    Redbelly98

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    You're on the right track for the most part. However, note that the vB term should actually be a cross product,

    |v×B| = vBsinθ

    where θ is the angle that an individual particle makes with respect to the z-axis. vsinθ is simply the v-component perpendicular to B. I.e., when ignoring the z-component of motion, the speed is vsinθ

    Hint for next step: how long does it take a particle to complete one loop of the circle?
     
  7. May 12, 2009 #6
    Hi, I seem to be struggling with the same problem...

    (P.S. phyz, you're not perhaps from south africa and studying at unisa...)

    I have been searching all over the internet and in all the relevant textbooks i have, but can't seem to find a solution...

    I also got something similar to what phyz got, but i don't know where to go from there, i don't even know if i understand the problem correctly, because surely i must have found some hints as to what to do by now...

    Hope you can help Red, please!!!
     
  8. May 13, 2009 #7

    Redbelly98

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    Welcome to Physics Forums :smile:

    If you show how far you have gotten, we'll take it from there.
     
  9. May 17, 2009 #8
    Ok, what I was trying to do is to derive the equation for the position of the particle, which i found to be:

    [tex]\bar{r}[/tex]=[tex]\left\{v_{x0}\frac{m}{qB}sin\left(\frac{q}{m}Bt\right)-v_{y0}\left[\frac{m}{qB}cos\left(\frac{q}{m}Bt\right)-\frac{m}{qB}\right]\right\}[/tex][tex]\hat{i}[/tex] + [tex]\left\{v_{x0}\left[\frac{m}{qB}cos\left(\frac{q}{m}Bt\right)-\frac{m}{qB}\right]+v_{y0}\frac{m}{qB}sin\left(\frac{q}{m}Bt\right)\right\}[/tex][tex]\hat{j}[/tex] + [tex]\left(v_{z0}t\right)[/tex][tex]\hat{k}[/tex]

    Where [tex]v_{x0}[/tex], [tex]v_{y0}[/tex] and [tex]v_{z0}[/tex] are the seperate initial components (at time=0) of the velocity in the x, y and z directions.

    Now this seems to be a valid result, since I plotted the position equation with the help of Matlab and it gave the expected result, a helix which starts at x=0, y=0, z=0, since this was accepted as the initial starting position of the particle.

    Now to determine the position of focus, I was planning to work out at what time the x and y components of the position returns back to (0; 0), since then the particle will be in line with the origin of the beam (along the z axis) and thus focused, and then substitute this time into the equation for the z component of the position to determine how far this point is from the origin along the z-axis.

    Now I tried to do the above process, but couldn't get a reasonable result for the time and the z component of the position that looks anything like the result the expect and if I do get a result for the z component, it is dependent on only the z component of the initial velocity and not the whole initial velocity as they expect you to get, which can be seen from the fact that the z component of the position is equal to the z component of the initial velocity times time.

    I'm sure there must be a much shorter and easier way of doing this, which I'm just not thinking about, so please tell me what you think and if I'm just missing something.

    Thanks
     
  10. May 17, 2009 #9

    Redbelly98

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    You're almost there!

    Your expressions for the x and y components involve periodic functions. One period corresponds to the time it takes the argument of the sin or cos functions to change by 2π.

    Does that help?
     
  11. May 18, 2009 #10
    Ok, so you basically have to work out what the time will be when the arguments for the sine and cosine functions are equal to [tex]2\pi[/tex] and then substitute this value for time into the equation for the z component of the position, which will give you the right answer that they expect you to prove, except the answer is in terms of [tex]v_{z0}[/tex] instead of the complete [tex]v_{0}[/tex] (the size of the initial velocity), which the expect, but then i thought, they say that the beam is "roughly parallel" to the magnetic field with only "a small angular dispersion", which implies that [tex]v_{z0}>>v_{y0}[/tex] and [tex]v_{z0}>>v_{x0}[/tex] and thus [tex]v_{0} \approx v_{z0}[/tex], which will give you the correct answer as they expect it, right?
     
  12. May 18, 2009 #11

    Redbelly98

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  13. May 18, 2009 #12
    Well, I guess then it's solved :smile: Thanks for guiding my thoughts a bit.
     
  14. May 19, 2009 #13
    Hello hello!

    Hope you are both doing well and sorry for the slow response time but the last two weeks were spent on exams

    I think I finally have a complete answer here after combining both of your inputs into something that works for my way of reasoning. It's pretty straightforward so I'd appreciate it if you can check it for me one last time. Here goes:

    The component in the xy-plane (perpendicular to the z-axis and with theta = pi/2) will result in the particle describing a circle with radius r and angular velocity shown to be

    [tex]F=qvB\sin\theta[/tex]
    [tex]ma_r=qvB\sin(\frac{\pi}{2})[/tex]
    [tex]m \frac{v^2}{r}=qvB[/tex]
    [tex]\frac{v}{r}=\omega =\frac{qB}{m}[/tex]

    The time taken by the particle to complete one loop of the circle is therefore

    [tex]t=\frac{2\pi}{\omega}[/tex]

    [tex]t=\frac{2\pi m}{qB}[/tex]

    Applying the reasoning from your last couple of posts and substitution then gives us our final answer. Having got there, I still have some questions though. RaStevey, I know this is probably going to be a nightmare of LaTex for you, but could you show me how you got to:

    Please. I think I might be lacking some maths for this one...Finally, why does this approach yield the correct answer?

    How do we know that the beam is focused after one period and not two or three or whatever? What I'm trying to understand is why the time taken for the field to focus the beam is the same as the time taken for a particle to complete one revolution of the z-axis.

    Thank you both for your help! Would never have got this far on my own :smile:
     
  15. May 19, 2009 #14

    Redbelly98

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    The beam will be in focus after any integer multiple of a period.

    To be in focus just means that x=y=0 for all particles in the beam.
     
  16. May 20, 2009 #15
    Oh, ok cool. Thanks Red! You're a legend! :biggrin:
     
  17. May 24, 2009 #16
    Well you start with:

    [tex]\bar{F}=q\bar{v}\wedge\bar{B}[/tex]
    [tex]m\frac{d\bar{v}}{dt}=q\bar{v}\wedge\bar{B}[/tex]
    [tex]\frac{d\bar{v}}{dt}=-\frac{q}{m}\bar{B}\wedge\bar{v}[/tex]

    Now if you break this up into its x, y and z components you will get:

    [tex]\frac{dv_{x}}{dt}=\frac{q}{m}Bv_{y}[/tex]
    [tex]\frac{dv_{y}}{dt}=-\frac{q}{m}Bv_{x}[/tex]
    [tex]\frac{dv_{z}}{dt}=0[/tex]

    Now if you solve the differential equations for [tex]v_{x}[/tex] and [tex]v_{y}[/tex] you will get the velocity to be:

    [tex]\bar{v}=\left[v_{x0}cos\left(\frac{q}{m}Bt\right)+v_{y0}sin\left(\frac{q}{m}Bt\right)\right]\hat{i} + \left[-v_{x0}sin\left(\frac{q}{m}Bt\right)+v_{y0}cos\left(\frac{q}{m}Bt\right)\right]\hat{j} + \left(v_{z0}\right)\hat{k}[/tex]

    The displacement can now just be determined by integrating the velocity :smile:
     
  18. May 24, 2009 #17
    Legend! Thanks! :smile:
     
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