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Electron Velocity

  1. Jun 16, 2009 #1
    1. The problem statement, all variables and given/known data

    An electron gun (applied voltage of 3000 volts) is emitting electrons. THe electrons enter a region of constant magnetic field (B=.025 Tesla.) The magnetic field is perpendicular to the velocity of the beam.

    2. Relevant equations

    W=Q(Delta V) = F*d

    V=F/eB

    3. The attempt at a solution

    -1.6X10^-19 C (3000 (N*m/C)) = F*d
    -4.8 x 10^-16 = F*d

    I'm lost as to how to eliminate d. I'm almost positive it doesn't drop out due to the dot product, but I need to get rid of it to find F and no distance between the gun and the field is given. There's two more parts to this problem that I know how to do, my problem is I need the velocity for both those parts. So if I can be nudged in the proper direction of how to eliminate d so I can find force, that'd be great.
     
  2. jcsd
  3. Jun 16, 2009 #2

    berkeman

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    Staff: Mentor

    What are you asked to find? The radius of the circular path? Write the equation for the Lorentz force (which you have in your work above), and relate that to the centripital acceleration of uniform circular motion.
     
  4. Jun 16, 2009 #3
    I need to find the force. Once I find that everything else isn't difficult.

    Part I)
    Find the speed of the electron:
    v=F/(eB)
    W=q(Delta V) = F*d, where W is -4.8*10^-16 N*m

    Part II)
    Find the distance between the been entering and exiting the field.
    r=mv/(eB) where r is the radius so the distance is 2r

    Part III)
    Obtain the acceleration of the electrons in the magnetic field
    F=veB=ma

    Like I said, I'm having difficulty finding out how to eliminate distance in W=F*d since no distance is given.
     
  5. Jun 16, 2009 #4

    berkeman

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    The speed is derived from the electron energy. All of the energy is kinetic energy. You don't need force * distance.
     
  6. Jun 16, 2009 #5
    -4.8*10^-16 = .5*(9.109*10^-31)*(v^2)

    So v = 3.246 * 10^7 m/s?
     
  7. Jun 16, 2009 #6

    berkeman

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    From google:

    1 electron volt = 1.60217646 × 10-19 joules

    What is on the lefthand side (LHS) of your first equation? Is that supposed to be 3keV converted to Joules? It helps in calculations like these to show units for each quantity, to be sure that you are doing the unit conversions correctly, and ending up with compatible units.
     
    Last edited: Jun 16, 2009
  8. Jun 16, 2009 #7
    W=q(Delta V) = Change in kinetic energy, which is (-1.6*10^-19 C)(3000V) = -4.8*10^-16 N*m

    -4.8*10^-16 N*m = .5*(9.109*10^-31 kg)*(v^2)

    Since a newton is kg*m/s^2, the kg's cancel and im left with m^2/s^2

    Sorry for the confusion.
     
  9. Jun 16, 2009 #8

    berkeman

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    Okay, that looks right then. The answer you got for the electron velocity is about 1/10 of c, which sounds about right for a 3keV beam. So does the rest of the question ask you to figure out the radius of the circular motion of the electrons based on the Lorentz force?
     
  10. Jun 16, 2009 #9

    ideasrule

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    Homework Helper

    For Part I, you wrote:

    v=F/(eB)

    Where did that come from? If you're trying to find out how much velocity the electric field added, the magnetic field doesn't matter; it can't affect the answer in any way.
     
  11. Jun 16, 2009 #10

    berkeman

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    It came from the Lorentz force. The OP isn't confused about some electric field. The e in that equation is the charge on an electron.
     
  12. Jun 17, 2009 #11

    ideasrule

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    The equation is valid, but the OP (apparently that term isn't unique to Wikipedia) was trying to find the electron's speed in Part I. The magnetic force has nothing to do with finding the speed, which depends solely on the kinetic energy added to the electron.
     
  13. Jun 17, 2009 #12

    berkeman

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    That's why s/he correctly converted the 3keV energy to a velocity. Where is the electric field term that you allude to?
     
  14. Jun 17, 2009 #13

    ideasrule

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    The electric field is the field that applied the 3000V potential difference. The OP did indeed convert 3 keV to velocity correctly, without using v=F/eB; that's why I said v=F/eB is not relevant here.
     
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