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Analsis magnet, loretz force

  1. Jun 11, 2007 #1

    malawi_glenn

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    1. The problem statement, all variables and given/known data

    I have a source producing ions of He: He2+ and He+, they are accelerated towards a tandem accelerator, se atteched figure. They are accelerated with a voltage towards the first magnet, were a selection of the ions are made. We only want the He2+ ions the proceed to the tandem accelerator.

    Show how this is done in the first analysis magnet.

    answer according to my teacher:
    the He2+ will get higher energy and therefore less bent, larger radius, then the He+ ions.

    2. Relevant equations and 3. The attempt at a solution

    But according to this, I found that is the opposite, becase F= qvB and the bigger charge, the more force, and hence smaller radius.

    [tex]|F| = qvB \Rightarrow r = \frac{mv}{qB}[/tex]

    m and B is the (practically) same for all, I have He+ and He2+ ions. T is kinetical energy.

    [tex]|F| = qvB \Rightarrow r = \frac{mv}{qB}[/tex]


    [tex]r\propto \frac{v}{q} [/tex]


    [tex]U=T=qV = mc^{2} \Rightarrow v=c\sqrt{1-\left( \frac{mc^{2}}{qV+mc^{2}}\right) ^{2}}}
    [/tex]


    Ratio for radius of He2+ and He+ "r(2q)/r(q)"

    gives me that He2+ has smaller radius than He+ if they are accelerated with the same potential V and is bent i same magnetiv field.

    Non-relativistic gives me:

    [tex]r\propto \frac{v}{q} [/tex]


    [tex]v=\sqrt{\frac{2qV}{m}} [/tex]


    [tex]r\propto \frac{\sqrt{\frac{2qV}{m}}}{q} = \frac{constant}{\sqrt{q}}[/tex]

    LOL help =)
     
    Last edited: Jun 12, 2007
  2. jcsd
  3. Jun 11, 2007 #2

    malawi_glenn

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    And yeah, this last also show that the higher q, the smaller r, and the ion is more bent.

    And the attachemt is here too.
     

    Attached Files:

  4. Jun 11, 2007 #3

    malawi_glenn

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    lol

    should be:

    [tex]|F| = qvB \Rightarrow r = \frac{mv}{qB}

    r\propto \frac{v}{q} \\

    U=T=qV = mc^{2}(\gamma -1) \Rightarrow v=c\sqrt{1-\left( \frac{mc^{2}}{qV+mc^{2}}\right) ^{2}}}
    [/tex]
     
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