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Motion in a Magnetic Field 2

  1. Feb 23, 2013 #1
    1. The problem statement, all variables and given/known data
    A proton (q = 1.6 X 10-19 C, m = 1.67 X 10-27 kg) moving with constant velocity enters a region containing a constant magnetic field that is directed along the z-axis at (x,y) = (0,0) as shown. The magnetic field extends for a distance D = 0.64 m in the x-direction. The proton leaves the field having a velocity vector (vx, vy) = (2.2 X e5 m/s, 0.9 X e5 m/s).

    What is R, the radius of curvature of the motion of the proton while it is in the region containing the magnetic field?

    2. Relevant equations
    F=qv*b
    m*(v^2)/R=F

    3. The attempt at a solution

    Well, I first wanted to try to find theta. So i did arctan(.9e5/2.2e5). Then using that angle i did .64 (my dimension D) divided by sin of that angle. Which gives me the correct radius. Yet, when i look at the trig behind this is does not make sense to me. I would think i should use cos of the angle, but that is incorrect.

    https://www.smartphysics.com/Content/Media/Images/EM/12/h12_bendtheta.png [Broken]
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Feb 23, 2013 #2

    TSny

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    Hi Gee Wiz.

    You probably new this was coming :smile:
    Please explain why you think you should use cosine?
     
  4. Feb 23, 2013 #3
    alright well, because cosine is adjacent over hypotenuse. I assumed (perhaps incorrectly) that the angle i found from the velocity would the the angle that the radius and D would create. So, therefore D is adjacent to the angle.
     
  5. Feb 23, 2013 #4

    TSny

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    The radius that goes through the point where the particle exits the square does not make an angle with respect to the horizontal that is equal to the angle that the velocity makes to the horizontal. Did you construct a diagram to check your assumption?
     
  6. Feb 23, 2013 #5
    I tried to draw a diagram, but it didn't really look right to me. Since the, i guess the vertical direction is not a straight line, but rather a curve.
     
  7. Feb 23, 2013 #6

    TSny

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    I'm not following your comment. Let's first make sure we're clear on some basics. Can you describe the general shape of the path of the particle as it moves in the magnetic field?
     
  8. Feb 23, 2013 #7
    Well, a particle in a magnetic field would follow a circular path.
     
  9. Feb 23, 2013 #8

    TSny

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    Ok. How is the direction of the velocity vector of the particle related to the circular path?
     
  10. Feb 23, 2013 #9
    The velocity vector is tangential to the path and perpendicular to the force vector
     
  11. Feb 23, 2013 #10

    TSny

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    Good. Can you tell roughly where the center of the circle will be? Will it be somewhere on the y-axis? The x-axis? Neither?
     
  12. Feb 23, 2013 #11
    I believe the center would be at the origin?
     
  13. Feb 23, 2013 #12

    TSny

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    Well, the particle passes through the origin. So, are you saying that the particle passes through the center of the circle that it travels on?
     
  14. Feb 23, 2013 #13
    Ahh, okay so perhaps that's my issue. I was kind of imagining the square was containing a fourth of a circle centered at the origin. But, now i'm kind of thinking that the center of this circle would be (.32,.32)
     
  15. Feb 23, 2013 #14

    TSny

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    Let's see. Consider the instant when the particle is at the origin and moving horizontally to the right. We know the velocity is tangent to the circle. So, that means the velocity is perpendicular to the radius. So, if the velocity is horizontal at the origin, what would be the direction of the radius of the circle at that point?
     
  16. Feb 23, 2013 #15
    In the positive y direction
     
  17. Feb 23, 2013 #16

    TSny

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    Yes. So, does that help to see roughly where the center of the circle would be?
     
  18. Feb 23, 2013 #17
    Above (direction of the positive y axis) the square. Would it be centered at x=0, and some y value..1.69(since that's the radius i magically stumbled upon)?
     
  19. Feb 23, 2013 #18

    TSny

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    Right, the center will be somewhere on the y-axis. To get a better idea of the location of the center on the y-axis, go to the point where the particle leaves the blue square and the velocity is tilted up at the angle θ. Can you draw the direction of the radius at this point?
     
  20. Feb 23, 2013 #19
    I think i would go from the point where the particle leaves the blue square and go up at some angle to the y-axis. The x-direction or component would be d, would this then make the angle that the y-axis and the radius to the point make the same angle as the one at the velocity vector? (yes, i realize that is wordy. my apologies)
     
  21. Feb 23, 2013 #20

    TSny

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    Yes, that's right. The radius is always perpendicular to the velocity. So, if the velocity makes angle θ to the horizontal, then the radius will make the same angle θ with respect to the vertical.
     
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