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Another Magnetic Force Problem

  1. Sep 24, 2009 #1
    1. The problem statement, all variables and given/known data
    A singly charged positive ion has a mass of 2.51 X 10-26 kg. After being accelerated through a potential difference of 264 V, the ion enters a magnetic field of 0.570 T, in a direction perpendicular to the field. Calculate the radius of the path of the ion in the field


    2. Relevant equations
    r=mv/qB


    3. The attempt at a solution
    Im not sure how to get the value of v from the information in the problem.
     
  2. jcsd
  3. Sep 24, 2009 #2

    kuruman

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    What is the kinetic energy of a singly charged ion that is accelerated (from rest we assume) through a potential difference of 264 Volts?
     
  4. Sep 27, 2009 #3
    1/2mvf^2?
     
  5. Sep 27, 2009 #4
    That's definitely a true expression; but the question is, what is its value equal to?
     
  6. Sep 27, 2009 #5
    the potential energy?
     
  7. Sep 27, 2009 #6
    Strictly, no. It is equivalent to the change in potential energy of the system magnitude-wise. (recall conservation of energy)
     
  8. Sep 27, 2009 #7
    1/2mvf^2+1/2mvi^2= mghi+mghf
     
  9. Sep 27, 2009 #8
    Err..no gravitational potential energy is involved here; the particle is accelerated through a (electrical) potential difference. What is the change in electrical potential energy of the particle upon passing through the electric field?
     
  10. Sep 27, 2009 #9
    would it be Change In P.E.=(q)(potential difference)
    (1.6X10^-19)(264)= 4.224X10^-17
     
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