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Radius of ions path in a 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 30, 2009 #2

    lanedance

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    Homework Helper

    Hi rinarez7


    the acceleration to use in this problem is not gravity!!! in fact you can likely neglect gravity... check at the end that your calculated accerlation is much larger than g.

    the ion will undergo circular motion due to the magnetic field, and i think you have used this to find r, however v is unknown as you don't yet know a...

    to find v, think about the energy of the ion due to being accelerated by the potential difference and how it relates to velocity
     
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