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The magnetron

  1. Oct 22, 2009 #1
    1. The problem statement, all variables and given/known data
    A magnetron is a microwave source in which a "bunch" of electrons travel on a circular orbit in a uniform magnetic field. As the electrons pass the electrodes, a high-frequency alternating voltage difference occurs.

    Suppose the resonant frequency is f = 1.91×1010 s−1; that is, the electron period of revolution is T = 5.24×10-11 s. What is the corresponding strength of the magnetic field?
    Set the centripetal force equal to the magnetic force. The electron charge is -1.6E-19 C and the electron mass is 9.1E-31 kg.


    2. Relevant equations
    mv^2/r= qv*b


    3. The attempt at a solution

    none i can't find away to find the radius or speed
     
  2. jcsd
  3. Oct 22, 2009 #2

    rock.freak667

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    So in

    [tex]\frac{mv^2}{r}=BQv[/tex]

    does anything happen to cancel out?


    Do you know a relation between some sort of velocity and period?
     
  4. Oct 22, 2009 #3
    just one V is i am solving for b and the only thing i know is what is at the begging
     
  5. Oct 22, 2009 #4

    rock.freak667

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    good good. So now we have

    [tex]\frac{mv}{r}=Bq[/tex]


    Right?


    so do you know any other formula where 'something' is equal to v/r ?
     
  6. Oct 22, 2009 #5
    kinetic engergy

    1/2mv
    ______
    r
     
  7. Oct 22, 2009 #6

    rock.freak667

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    no no, I went a bit over your head there.

    Do you know of angular velocity?
     
  8. Oct 22, 2009 #7
    ω = θ / t (2a)

    where

    ω= angular velocity (rad/s)

    θ = angular displacement (rad)

    t = time (s)
     
  9. Oct 22, 2009 #8

    rock.freak667

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    so if θ=s/r (s= arc length)

    then ω=(s/r)/t = (s/t)/r

    Since s is a distance, what does distance/time give?
     
  10. Oct 22, 2009 #9
    velocity, but how do i find that out from what i know
     
  11. Oct 22, 2009 #10

    rock.freak667

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    Now if s/t=v

    then ω=v/r right?


    Now what is ω with what you have?

    If ω is the angular displacement in one revolution/time taken to make one revolution, what is ω equal to?
     
  12. Oct 22, 2009 #11
    so s=360 T= 5.24×10-11 s
     
  13. Oct 22, 2009 #12

    rock.freak667

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    ω=θ/T

    Right right, we're reaching somewhere. In one revolution, how many radians does it rotate?
     
  14. Oct 22, 2009 #13
  15. Oct 22, 2009 #14

    rock.freak667

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    Nice! (2pi=2π)

    so now ω=2π/T


    so can we find ω now?


    Can we replace v/r by ω in this formula?

    [tex]\frac{mv}{r}=BQ[/tex]
     
  16. Oct 22, 2009 #15
    m*(2pi/t)/r= Bq
     
  17. Oct 22, 2009 #16
    but how di i find r
     
  18. Oct 22, 2009 #17

    rock.freak667

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    no no

    [tex]\omega= \frac{v}{r}=\frac{2\pi}{T}[/tex]

    so your formula changes to?
     
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