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Find the magnitude of the maximum magnetic force

  1. Jul 18, 2006 #1
    In a television set, electrons are accelerated from rest through a potential difference of 25 kV. The electrons then pass through a 0.26 T magnetic field that deflects them to the appropriate spot on the screen. Find the magnitude of the maximum magnetic force that an electron can experience.

    I know to use F=qvBsin(theta) but the v is lacking. What do I do to find v?
     
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  3. Jul 18, 2006 #2

    quasar987

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    The answer lies in "electrons are accelerated from rest through a potential difference of 25 kV"

    What is the relation btw potential difference and the variation in kinetic energy of a particle passing through it.
     
  4. Jul 18, 2006 #3

    berkeman

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    Or put another way, the 25keV is a measure of the electron's kinetic energy, all due to its velocity. Convert the keV units to Joules, look up the mass of the electron, and.....
     
  5. Jul 18, 2006 #4
    I know it lies within that statement, but I cannot find a relationship between the two.

    I know P=IV, but I lack I. If thats even the right approach to begin with. I figure the energy in the potential difference (Volts) can be converted into a E=mA, but can the E in that equation be expressed in Volts?
     
  6. Jul 18, 2006 #5
    ok, I see where you're coming from and I converted it into Joules.

    I used 1eV=1.6e-19 joules

    x 25kV = 4e-15J

    My answer was wrong however, and I believe it is because I converted it with eV instead of V. eV = how many V again?
     
  7. Jul 18, 2006 #6

    quasar987

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    [tex]q\Delta V = \Delta U[/tex]

    Where U is the potential energy. And how are variations of potential energy related to variations of kinetic energy?
     
  8. Jul 18, 2006 #7
    This is what I've tried

    E=mA = 1.6x10^-19 (25000)

    A = E/A = (1.6x10^-19)(25000) / 9.11x10^-31 = 4.39 x 10^15

    F = (1.6 x 10^-19) (.26T) (sin 89.9) (4.39 x 10 ^15)

    I've also tried without the q because maybe it already was alrady included with the eV.
    I've also tried with different thetas, assuming 90* is the biggest sign, but not plausible (although tried), and 89.9* since maybe its plausible and also 45*.

    Nothing has been right. ???
     
  9. Jul 18, 2006 #8

    Office_Shredder

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    This question is meaningless. eV is a measure of energy, V is a measure of energy / culoumb.

    It's confusing at first, but you get used to it
     
  10. Jul 19, 2006 #9

    berkeman

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    [tex]KE = \frac{1}{2} m v^2[/tex] in the mks system of units, [tex][J] = [kg] [m^2 / s^2][/tex]
     
  11. Jul 20, 2006 #10
    From rest to pd of 25kV, the electrons move with kinetic energy equal to 0.5ev^2 = 25ekV. That gives you the v.
     
  12. Jul 20, 2006 #11

    berkeman

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    Say what? Why are you taking 0.5eV and squaring it? and what's 25ekV?


    EDIT -- Oh, I see what you were trying to type. You should fix the typos on both sides of that equation so it doesn't confuse the OP.
     
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