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What speed would a proton need to orbit in an exact circle ?

  1. Jun 29, 2009 #1
    1. The problem statement, all variables and given/known data
    Assume the Earth’s magnetic dipole moment
    is aligned with the Earth’s rotational axis,
    and the Earth’s magnetic field is cylindrically
    symmetric (like an ideal bar magnetic).
    What speed would a proton need to orbit in
    an exact circle around the Earth at a height of
    476 km, where the Earth’s magnetic field has
    an intensity of 4.81 × 10−8 T? The mass of
    a proton is 1.67262 × 10−27 kg and the radius
    of Earth is 6.37 × 106 m.
    Answer in units of m/s.


    2. Relevant equations

    r= m*v/(q*B)


    3. The attempt at a solution

    i solved for the velocity but it is wrong i think i'm missing a step =S
     
  2. jcsd
  3. Jun 29, 2009 #2

    diazona

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    From my perspective you're missing more than a step :wink: What did you get and how did you get it?
     
  4. Jun 29, 2009 #3
    v= (r*q*B)/m
    v=(6.3*10^6)(1.6*10^-19)(4.81*10^-8)/(1.67*10^-27)
     
  5. Jun 30, 2009 #4

    alphysicist

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    I believe your value for r in this equation is incorrect. The value r is the radius of the particles path; do you see what it needs to be?
     
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