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Homework Help: Magnetic field: the Banebridge velocity selector

  1. Feb 6, 2010 #1
    1. The problem statement, all variables and given/known data

    The electric field intensity between the plates of the Banebridge velocity selector (I am not a native English speaker …) is 120 [Volt/cm] and the magnetic field magnitude of both fields is 0.6 [Weber/m2].
    A singly charged Neon ion beam makes a radius of 7.28 [cm]. What is it's mass number.
    The result must be 21, 10 times smaller than mine.

    2. Relevant equations

    Velocity of beam: V=[itex]\frac{E}{B}[/itex]
    radius of the beam: [tex]R[m]=\frac{mV}{qB}[/tex]
    mass number=mass/proton mass: [tex]N=\frac{m}{m_{p}}[/tex]

    3. The attempt at a solution

    120[Volt/cm]=12,000[Volt/m]

    [tex]V=\frac{E}{B}=\frac{12,000}{0.6}=20,000[m/sec][/tex]

    [tex]R=0.728=\frac{mV}{qB}=\frac{m*20,000}{1.602\ \times\ 10^{-19}*0.6}\Rightarrow m=3.5\ \times\ 10^{-25}[/tex]

    [tex]N=\frac{m}{m_{p}}=\frac{3.5\ \times\ 10^{-25}}{1.672\ \times\ 10^{-27}}=209[/tex]
     

    Attached Files:

  2. jcsd
  3. Feb 6, 2010 #2
    what is your questions exactly?
     
  4. Feb 6, 2010 #3
    My result is 10 times bigger, according to the book and the periodic table of elements, it should come out 20.9
     
  5. Feb 6, 2010 #4
    I cant see whats wrong with your solution , it seems okay ..
     
  6. Feb 6, 2010 #5
    Thanks for your effort, Ihope I'll get another answer before I decide there was a mistake in the book. By the way, it's taken from a translation to Sears & Zemansky, 1964...
     
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