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Radius of Curvature of Ion in Magnetic Field

  1. Feb 14, 2009 #1
    1. The problem statement, all variables and given/known data

    A doubly ionized atom (charge = +2e) whose mass is 5.15E-26 kg is accelerated by a voltage of 3450 V and enters a region where a uniform magnetic field B = 0.100 T acts perpendicular to its motion. What is the radius of curvature of the path of the ion in the B-field?


    2. Relevant equations

    r=[tex]\frac{mv}{qB}[/tex]
    e = 1.602 x 10-19


    3. The attempt at a solution

    Given the equation above this is a very simple problem except I don't know how to calculate the velocity of the ion from voltage. The only idea I had was conservation of energy: Kf = Ui or [tex]\frac{1}{2}[/tex]mv2 = q[tex]\Delta[/tex]V. But this ended up wrong and in retrospect I'm not sure if it entirely makes sense to begin with...
     
  2. jcsd
  3. Feb 14, 2009 #2

    LowlyPion

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    Homework Helper

    Looks like you are on the right track. The Δ Voltage will result in the ΔPE which becomes KE.

    Looks to me like ½mv² gets related to the Lorentz force component and the centripetal acceleration as you've already written.

    I end up with an equation that looks like:

    R = 1/B*(2 * m * ΔV/ q)1/2
     
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