1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.

  4. Jan 5, 2005 #3


    User Avatar
    Science Advisor
    Homework Helper

    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...?

  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


    User Avatar
    Science Advisor
    Homework Helper

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

  7. Jan 5, 2005 #6


    User Avatar
    Homework Helper

    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

    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


    User Avatar
    Science Advisor
    Homework Helper

    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]


  10. Jan 6, 2005 #9


    User Avatar
    Homework Helper

    Taking your first equation

    if I solve for v I get

    Then I plug it into the KE equation



    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 ;)
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Kinetic energy proof question