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Radius of ion's path in field

  1. Mar 29, 2009 #1
    1. A singly charged positive ion has a mass = 1.08e-26 kg
    After being accelerated through a potential difference of 240 V the ion enters a magnetic field of 0.595 T in a direction perpendicular to the field.
    The charge on the ion is 1.602 e-19 C.
    Find the radius of the ion's path in the field.
    Answer in units of cm.



    2. m= 1.08e-26 kg
    potential difference= 240 V
    B= 0.595 T
    q= 1.602e-19
    sin theta= 1 (enters in direction perpendicular to the field)
    radius= mv/ qB




    3. I am confused on exactly how to calculate velocity and the relevance of the potential difference. But, I tried:
    F=qvBsin theta= ma
    F=qvB= ma
    1.602e-19(v) (0.595T)= 1.08e-26(-9.81m/s^2)
    v= 1.112e-6 m/s
    I then plugged this into r= mv/ qB
    r= 1.08e-26kg (1.112e-6 m/s )/ 1.602e-19 (0.595T) and calculated r= 1.8997 e-11 cm, but this incorrect. What am I missing?
     
  2. jcsd
  3. Mar 29, 2009 #2

    rock.freak667

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    Homework Helper

    Try using the fact that the potential difference*charge of the ion gives the kinetic energy of it. (you can now find v)
     
  4. Mar 29, 2009 #3
    Thank you! I just needed that missing piece!
     
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