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Find V of a partical in B field :O!

  1. Apr 11, 2009 #1
    1. The problem statement, all variables and given/known data
    A particle with a charge of 5.40nC is moving in a uniform magnetic field of B= -(1.20 )Tz^. The magnetic force on the particle is measured to be F=3.50×10−7N X^ + 7.60×10−7N Y^ .

    Calculate the y component of the velocity of the particle.

    2. Relevant equations
    F= qv x B
    F = qvBSin(theta)


    3. The attempt at a solution
    lost trying to use the magnitude to find also
    i tried
    F/qb=Vsin(theta) doesnt seem to yield results yet :/
     
  2. jcsd
  3. Apr 11, 2009 #2

    tiny-tim

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    Hi SoulofLoneWlf! :smile:
    Hint: call the velocity (Vx,Vy,Vz) …

    then what is the x-component of the force? :wink:
     
  4. Apr 11, 2009 #3
    so would it be possible for me to say v = Vx + Vy + Vz
    ...
    Fy=qVyBsin(thetha)?
     
  5. Apr 11, 2009 #4

    tiny-tim

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    erm :redface:

    right idea :smile:, wrong result :cry:
     
  6. Apr 11, 2009 #5
    hahaha felt that was wrong hmm let me try again i guess ><"
     
  7. Apr 11, 2009 #6
    F=x^(5.4Vy*1.2Z^T)-Y^(5.4Vx*1.2TZ^)
    then place value for F as 3.50x10^-7 etc and equal x to x and y to y??
     
  8. Apr 11, 2009 #7

    tiny-tim

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    Yup! :biggrin:
     
  9. Apr 11, 2009 #8
    ur amazing thank you ^-^
     
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