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Radius of helical motion in a magetic field

  1. Apr 4, 2014 #1
    1. The problem statement, all variables and given/known data
    When the charge enters the magnetic feld at an angle x , how the radius can be expressed as?


    2. Relevant equations
    Force of charge in a magentic field = qvbsinx
    centripetal force = mv^2/r

    3. The attempt at a solution
    when i combined the two equations above, i got r=mv/qbsinx,
    but the question i did said the radius is directly proportional to sinx. What i have done wrong?Hope someone can help me :)
     
  2. jcsd
  3. Apr 4, 2014 #2

    TSny

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    Seems to me your answer is correct. What if the angle x = 0? What does your answer give for r for this case? Does it make sense?

    What would r be for x = 0 if r were proportional to sinx? Would that make sense?
     
  4. Apr 4, 2014 #3

    lightgrav

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    comment: 2) Force applied to the moving charge is q v B sinθ ... Force "of" the charge sounds like the charge is applying that Force.

    If the Force is zero, the curvature is zero ... curvature is proportional to sinθ.
    I hope you mis-read the question's "discussion".
     
  5. Apr 4, 2014 #4
    but the angle x is a θ less then 90 degrees but bigger than 0 degree...the force applied is never 0
     
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