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Homework Help: Kinetic energy proof question

  1. Jan 5, 2005 #1
    this should probably go in the HS section since it is review of some material covered there but.... I'm guessing the same peole brose the two forums so I might as well post it here since it came up in college.

    now the questions seems quite straight forward:

    for a particle of mass m and charge q, moving in a circular path in a magnetic field B, show that its kinetic energe is proportional to the square of the radius of curvature of its path.

    so we have:

    r = (mv^2)/qvB
    r = mv^2/F

    F = ma
    F = m((v^2)/r)

    r = (mv^2)/(m((v^2)/r))

    now.... it all evens out and I'm stuck with nothing.... could someon tell me where I'm going wrong with this
     
  2. jcsd
  3. Jan 5, 2005 #2
    When you get a result like 1 = 1 or r = r, it means you're substituting an equation into itself, which should naturally take you nowhere.

    --J
     
  4. Jan 5, 2005 #3

    dextercioby

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    1.Do you know the expression of KE in terms of velocity??

    2.Can u compute/find a relation between the radius of trajectory and the velocity the particle has...?

    Daniel.
     
  5. Jan 5, 2005 #4
    KE = 1/2 mv^2

    yeah I know that

    still don't see how I can go from r = (mv^2)/F to r^2 = 1/2 mv^2
     
  6. Jan 5, 2005 #5

    dextercioby

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    Can u prove that:
    [tex] r_{traj.}=\frac{mv}{qB} [/tex]

    Daniel.
     
  7. Jan 5, 2005 #6

    learningphysics

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    This is all you need. As well as KE=1/2 mv^2

    Use the two equations you've got:
    r=(mv^2)/qvB and
    KE=1/2mv^2

    to get a relationship between KE and r.
     
  8. Jan 5, 2005 #7
    Forgive my thickness but I can't figure out how to go from those two to proving that (r^2) is proportional to (1/2 mv^2)


    well yeah that's just the v at the bottom is cancelled out by v^2 and the square is thus also gone, but where do I go from there
     
  9. Jan 6, 2005 #8

    dextercioby

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    If u can prove that,it means u can prove it is valid even when it's squared:
    [tex] r_{traj.}^{2}=\frac{m^{2}v^{2}}{q^{2}B^{2}}=\frac{2m}{q^{2}B^{2}}\frac{mv^{2}}{2}=C\cdot (KE) [/tex]

    VoilĂ .

    Daniel.
     
  10. Jan 6, 2005 #9

    learningphysics

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    Taking your first equation
    r=(mv^2)/qvB

    if I solve for v I get
    v=rqB/m

    Then I plug it into the KE equation
    KE=(1/2)mv^2

    [tex]KE=\frac{1}{2}m(\frac{rqB}{m})^2[/tex]

    [tex]KE=\frac{q^2B^2}{2m}r^2[/tex]

    since [tex]\frac{q^2B^2}{2m}[/tex] is a constant, this proves KE is proportional to r^2.

    You needed an expression for KE in terms of r, and constants (v is not a constant here as it changes with r), so you needed to get rid of that v... so solve for v in the first equation and plug it in to the second... so all that is left is constants and r's.
     
    Last edited: Jan 6, 2005
  11. Jan 6, 2005 #10
    ah okay I finally get it thnx to both of you, great help, really appreciate it ;)
     
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