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Magnetism question 2

  1. Apr 2, 2010 #1
    1. The problem statement, all variables and given/known data

    What speed would a proton need to achieve
    in order to circle Earth 1670 km above the
    magnetic equator, where the Earth’s mag-
    netic field is directed on a line between mag-
    netic north and south and has an intensity of
    4.1 × 10−8 T?
    The mass of a proton is 1.673 × 10−27 kg.
    Answer in units of m/s.

    2. Relevant equations

    f1=qvbsin(theta)
    f2=mv^2/(d+R)
    f3=Gm(proton)m(earth)/(d+R)^2 where d is 1670 km, and R is Earth distance


    3. The attempt at a solution

    I tried doing f1+f2=f3, b ut im not sure if this is right and i dont know how to solve for v
     
  2. jcsd
  3. Apr 2, 2010 #2

    rock.freak667

    User Avatar
    Homework Helper

    I think you'd want to equate f2 to f3 as the gravitational force of attraction will provide the centripetal force.
     
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