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Movement of electron in an nonuniform E and B fields

  1. Dec 7, 2016 #1
    1. The problem statement, all variables and given/known data
    The system consists of a long cylindrical anode of radius [itex]a[/itex] and a coaxial cylindrical cathode with radius [itex]b[/itex] ([itex]b <a[/itex]). On the axis of the system has a thread with a heating current [itex]I[/itex],
    creates in the surrounding magnetic field.
    Find the smallest potential difference between the cathode and anode, whereby thermoelectrons with zeros initial velocity will reach the anode.

    2. Relevant equations
    Equations of motion
    $$\ddot{x} = \frac{e}{m}\frac{Ux}{r^2\ln{b/a}} - \frac{e}{mc}\dot{z} B_y$$
    $$\ddot{y} = \frac{e}{m}\frac{Uy}{r^2\ln{b/a}} + \frac{e}{mc}\dot{z} B_x$$
    $$\ddot{z} = \frac{e}{mc}(\dot{x} B_y - \dot{y} B_x)$$
    and
    $$\vec B = \frac{2I}{cr^2}(-y\hat e_x + x\hat e_y )$$

    3. The attempt at a solution
    From equation for [itex]\ddot z[/itex]I found
    $$\dot z = \frac{2eI}{mc^2}\ln{\frac{r}{b}}$$
    Near anode:
    $$\dot z(r=a) = \frac{2eI}{mc^2}\ln{\frac{a}{b}}$$

    From the Energy conservation law:
    $$\frac12 m v_a^2 = eU$$

    I get
    $$U = \frac{2eI^2}{mc^4}\ln^2{a/b}$$

    The solution in textbook is differ from above:
    $$U = 2\frac{e}{m} \frac{I^2}{c^2}\ln a/b$$
     
    Last edited: Dec 7, 2016
  2. jcsd
  3. Dec 7, 2016 #2

    haruspex

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    Hi know nothing about the subject, but I can see that your answer, with c4 in the denominator, is dimensionally wrong. You should be able to find an error by going through your working to find where a dimensional inconsistency crept in.
     
  4. Dec 7, 2016 #3
    Dimention of charge from [itex]ma = q^2/r^2[/itex]
    $$q^2 = \frac{g\cdot cm^3}{c^2}$$
    Dim of U is [itex]q/cm[/itex]
    Thus
    $$\frac{q}{cm} = \frac{q\cdot q^2}{c^2\cdot g} \frac{c^4}{cm^4} = \frac{q\cdot \frac{g\cdot cm^3}{c^2}}{c^2\cdot g} \frac{c^4}{cm^4}.$$

    It seems, with dimension everything ok.
     
  5. Dec 7, 2016 #4

    haruspex

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    You've lost me completely with that analysis. Can you express all of these variables in the standard M, L, T, Q dimensional notation?
    E.g. your ma=q2/r2 looks like it translates to MLT-2=Q2L-2, which is clearly not true.
    What is g?
     
  6. Dec 7, 2016 #5

    TSny

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    This looks correct.
    The units seem to check out (cgs system), although I don't follow your unit analysis in post #3.
     
  7. Dec 7, 2016 #6
    Ok. I use cgs units
    Unit of square of charge
    $$Q^2 = \frac{M\cdot L^3}{T^2}$$
    Dimension of U in cgs is [itex]QL^{-1}[/itex]
    Thus
    $$\frac{Q}{L} = \frac{Q\cdot Q^2}{T^2\cdot M} \frac{T^4}{L^4} = \frac{Q\cdot \frac{M\cdot L^3}{T^2}}{T^2\cdot M} \frac{T^4}{L^4}.$$
     
  8. Dec 7, 2016 #7

    haruspex

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    Ok, I am guessing that this is a quantum analysis trick that fixes some physical constant which, to most physicists, has dimension, as a dimensionless 1, thereby deriving a translation between dimensions which would conventionally be taken as independent. E.g. c=1, so L=T, or something like that.
     
  9. Dec 7, 2016 #8

    TSny

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    OK. One thing that confused me in post #3 was your use of ##c## to apparently denote seconds. You wrote ##q^2 = \frac{g \cdot cm^3}{c^2}##
     
  10. Dec 7, 2016 #9
    Yes, sorry :sorry:. [itex]c[/itex] is the seсond, but in cyrillic notation, I forgot.
     
  11. Dec 8, 2016 #10

    TSny

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    Ah. Interesting.
     
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