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Given magnetic field, particle charge, and force (vector) Calculate velocity?

  1. Apr 6, 2006 #1
    A particle with charge -5 nC is moving in a uniform magnetic field B = -(1.2 T)k. The magnetic force on the particle is measured to be F = -(3.6*10^(-7) N)i + (7.6*10^-7)j. Calculate the x and y components of the particle's velocity.

    F = q(V X B)
    (force equals charge multiplied by the cross product of V and B)

    Calculate the scalar product (dot product) v*F
    vx*Fx + vy*Fy + vz*Fz?

    What is the angle between v and F (in degrees)?
    F = qvB sin theta
    theta = arcsin [F/(qvB)]?
     
  2. jcsd
  3. Apr 6, 2006 #2
    Fz is zero, right? So do you need to worry about Vz, or the angle between V and B? Or do you just need to worry about the component of V that is perpendicular to B? (Which would be in the xy plane only).
     
  4. Apr 7, 2006 #3
    I believe that I just need to worry about the component of V that is perpendicular to B (x and y components of V) since force exists in the x and y directions only.
     
  5. Apr 7, 2006 #4
    I found equations for each of the x and y components:

    V_x = (-F_y)/(q*B)
    V_x = -(-7.6E-7)/(-5E-6*-1.20)
    V_x = -0.127 m/s

    V_y = (-F_x)/(q*B)
    V_y = -(-3.6E-7)/(-5E-6*-1.2)
    V_y = 0.06 m/s

    But those answers aren't right... What am I doing wrong? Are the equations not right? Am I using incorrect values in correct equations? I'm so confused!
     
  6. Apr 7, 2006 #5
    Kay, nevermind... :uhh:
     
  7. Apr 7, 2006 #6
    Just calculate the cross product q(vxB), assuming a certain variable for each component of the velocity, and equate the result to the force.

    Btw, vxF is zero and it doesn't help much.
     
  8. Apr 7, 2006 #7
    Thanks. I found out my problem. I'm so used to converting micro 10^-6 that when I came across "nano," I used micro's conversion value. :uhh:
     
  9. Apr 27, 2010 #8
    v*F being zero tells you that the angle between them is 90 degrees, so it does actually help
     
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