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Find distance proton moves between adjacent turns of helix

  1. Apr 20, 2012 #1
    1. The problem statement, all variables and given/known data
    A proton moves with a speed of 4300 m/s in a direction 76.0° above positive x axis. It enters a region where a magnetic filed of 25x10^-6 Tesla exists in the positive x direction. Find the radius of the helix formed by the protons path and the distance between adjacent turns of helix.

    To find the distance, I'm not sure. I know I need to find distance around and time it takes for a complete circle.


    2. Relevant equations

    Force of mag. field = qvsin∅B
    F=qvsin∅B=ma
    F= qvsin∅B = mv^2/r

    r = mvsin∅/ qB

    v = 2∏r/t

    t = 2∏r/v


    xf=xi + vt
    vyfinal = vyinitial + at



    3. The attempt at a solution

    r = (1.67*10^-27 * 4300sin76°) / (1.602*10^-19 * 25*10^-6)
    r = 1.74m


    t = 2∏r/v
    t = .002542s

    so, vyfinal = vyinitial + at

    xf = 0 + 4300m/s(.002542m)

    xf = 10.9, which is NOT the answer.
     
  2. jcsd
  3. Apr 20, 2012 #2

    gneill

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    Staff: Mentor

    Okay, you've used the velocity of the particle that is perpendicular to the field. Good. That should give you the correct radius.
    Something fishy here. Did you use the speed of the particle rather than its velocity component in the (moving) plane of the loop? Hint: The speed of the particle in the loop should be the same as that which was used in the calculation of the loop radius.
     
  4. Apr 21, 2012 #3
    so I have Δy=vsin∅t + at^2

    Δy= 2∏r

    but vsin∅final = vsin∅initial + at

    F = ma

    qvsin∅B = ma

    a = qvsin∅B/m

    a = (25*10^-6T)(4300sin76m/s)(1.602*10^-19C)/ (1.67*10^-27kg) = 1.0006*10^7
    t = vsin∅final/a = 4172.27/1.0006*10^7 = .000417s

    xfinal = vinitial*t = 4300m/s*.000417s =1.793m but it's wrong according to my professors answers. 8-(
     
    Last edited: Apr 21, 2012
  5. Apr 22, 2012 #4

    gneill

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    Staff: Mentor

    Can you add some explanation to the above formulas? What acceleration are you dealing with? As far as I can tell the only acceleration you need to worry about is that involved with the circular motion -- centripetal acceleration.
    I think you're missing a factor of ##2\pi## in the above calculation.
    The proton does not have the full speed of 4300m/s in the x-direction.
     
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