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Circular Motion in a Simple Mass Spectrometer

  1. Mar 21, 2016 #1
    1. The problem statement, all variables and given/known data
    In the simple mass spectrometer shown in the figure below, positive ions are generated in the ion source. They are released, traveling at very low speed, into the region between two accelerating plates between which there is a potential difference ΔV. In the shaded region there is a uniform magnetic field B rightarrowhead.gif ; outside this region there is negligible magnetic field. The semicircle traces the path of one singly charged positive ion of mass M, which travels through the accelerating plates into the magnetic field region, and hits the ion detector as shown.

    Determine the appropriate magnitude and direction of the magnetic field B rightarrowhead.gif , in terms of the known quantities shown in the figure below (in addition to M and q, where q is the charge on an ion).

    Magnitude B = ?
    direction = ?

    14f06f7925.png

    2. Relevant equations
    Fmag = dp/dtmag = qvB
    dp/dtmag = p(v/R) = p(omega), p = ymv
    omega = q_mag * b / (ym)

    3. The attempt at a solution
    deltaV (I don't know what to do with electrical potential)
    M (mass)
    q (charge)
    d/2 = R (radius)
    v << c, y = 1, p = mv (approximation)

    Fmag = dp/dtmag
    qvB = p(v/R)
    qvB = p(v/(d/2))
    qB = p/(d/2))
    B = p/(d/2))/q
    B = 2p/dq, p = mv (approx.)
    B = 2Mv/dq

    This is wrong, probably because I'm not given velocity. However, I don't know how to get magnetic field without velocity? I think the problem is I don't know what to do with electric potential.
     
    Last edited by a moderator: Apr 16, 2017
  2. jcsd
  3. Mar 21, 2016 #2

    gneill

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    Staff: Mentor

    Your ion with charge q is accelerating through the ΔV between the plates. Look for an equation relating the energy acquired by an electric charge accelerating through a potential difference.
     
  4. Mar 21, 2016 #3
    1/2 mv^2 = q deltaV

    This kinetic energy equation?

    mv^2 = 2q deltaV
    v^2 = 2q deltaV / m
    v = sqr rt (2q delta V / m)
     
  5. Mar 21, 2016 #4
    deltaV = Ed

    But I'm missing both E and d? d is the separation between the plates but the variable d in the image seems to be a length.
     
  6. Mar 21, 2016 #5

    gneill

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    Staff: Mentor

    There's another equation that involves the charge.
     
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