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Electron equations of motion through an uniform magnetic field

  1. Oct 31, 2009 #1
    1. The problem statement, all variables and given/known data

    An electron enters a zone of uniform magnetic field [tex]\vec B = 0,4T{\rm{ }}\hat j[/tex] with velocity [tex]{\vec V_0} = {10^5}m/s{\rm{ }}\hat i[/tex]. Find the differential equations that govern its motion through the field, and solve them to find the equations of motion. What happens to its kinetic energy?
    2. Relevant equations

    - Lorentz Force = [tex]q\vec V \otimes \vec B[/tex]
    - Newton's Second Law = [tex]\sum {\vec F} = m\frac{{{\partial ^2}\vec r}}{{\partial {t^2}}}[/tex]
    - Conservation of Kinetic Energy = [tex]\Delta {E_k} = {W_{all{\rm{ }}forces}}[/tex]


    3. The attempt at a solution

    I know that the answer should be that the electron's trajectory is a circle. But I can't get there throught the differential equations:

    If I don't take the electron's weight into account, I have that the only force acting upon it is the Lorentz Force. Using Newton's Second Law:

    [tex]\vec F = q\vec V \otimes \vec B = m\frac{{d\vec V}}{{dt}}[/tex]
    -[tex] 0 = m\frac{{d{V_x}}}{{dt}}[/tex]
    -[tex] 0 = m\frac{{d{V_y}}}{{dt}}[/tex]
    -[tex] q{V_x}B = m\frac{{d{V_z}}}{{dt}}[/tex]

    Then

    -[tex]{V_x} = {10^5}m/s[/tex]
    -[tex]{V_y} = 0[/tex]
    -[tex]\frac{{q{V_x}B}}{m}t = {V_z}[/tex]

    I know there's something wrong: since the only force acting upon the electron is the Lorentz Force, being a central force (perpendicular to the trajectory), it doesn't do any work, the kinetic energy conserves and therefore the module of V should be constant. Which doesn't happen if the solution I found is true (I know it's wrong).

    What's wrong with my resolution?

    Thanks.
     
  2. jcsd
  3. Nov 1, 2009 #2

    ehild

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    You can not assume that the x component of the velocity stays constant. Decompose the Lorentz force into all components, assuming non-zero components of velocity, and discuss what you got.

    ehild
     
  4. Nov 1, 2009 #3
    You say that instead of keeping the Lorentz Force this way:

    [tex]\vec F = q\vec V \otimes \vec B = q{V_x}B{\rm{ }}\hat i[/tex]

    I should rewrite the Lorentz Force this way?:

    [tex]\vec F = q\vec V \otimes \vec B = q\left| {\begin{array}{*{20}{c}}
    {\hat i} & {\hat j} & {\hat k} \\
    {{V_x}} & {{V_y}} & {{V_z}} \\
    0 & B & 0 \\
    \end{array}} \right| = - q{V_z}B{\rm{ }}\hat i + q{V_x}B{\rm{ }}\hat k[/tex]
     
  5. Nov 1, 2009 #4

    ehild

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    It is correct. Now you can set up the differential equations for Vx, Vy and Vz.

    ehild
     
  6. Nov 1, 2009 #5
    Could somebody correct me if I'm wrong?

    Given the new set of differential equations:

    [tex]\begin{array}{l}
    \frac{{d{V_x}}}{{dt}} = \frac{{q{V_z}B}}{m} \\
    \frac{{d{V_y}}}{{dt}} = 0 \\
    \frac{{d{V_z}}}{{dt}} = \frac{{q{V_x}B}}{m} \\
    \end{array}
    [/tex]

    I calculate:

    [tex]\begin{array}{l}
    {V_z} = \left( {\frac{{d{V_x}}}{{dt}}} \right)\frac{m}{{qB}} \\
    {V_y} = 0 \\
    \frac{{d{V_z}}}{{dt}} = \frac{{q{V_x}B}}{m} \\
    \end{array}
    [/tex]

    [tex]\begin{array}{l}
    {V_z} = \left( {\frac{{d{V_x}}}{{dt}}} \right)\frac{m}{{qB}} \\
    {V_y} = 0 \\
    \frac{{d\left( {\frac{{d{V_x}}}{{dt}}\frac{m}{{qB}}} \right)}}{{dt}} = \frac{{q{V_x}B}}{m} \\
    \end{array}
    [/tex]

    [tex]\begin{array}{l}
    {V_z} = \left( {\frac{{d{V_x}}}{{dt}}} \right)\frac{m}{{qB}} \\
    {V_y} = 0 \\
    \frac{{{d^2}{V_x}}}{{d{t^2}}}\frac{m}{{qB}} = \frac{{q{V_x}B}}{m} \\
    \end{array}
    [/tex]

    [tex]\begin{array}{l}
    {V_z} = \left( {\frac{{d{V_x}}}{{dt}}} \right)\frac{m}{{qB}} \\
    {V_y} = 0 \\
    \frac{{{d^2}{V_x}}}{{d{t^2}}} = \frac{{{q^2}{B^2}}}{{{m^2}}}{V_x} \\
    \end{array}
    [/tex]

    I propose the solution for Vx (should I include the 50 that way?):

    [tex]{V_x} = 50\cos \left( {\frac{{qB}}{m}t} \right)
    [/tex]

    Then I solve that:

    [tex]\begin{array}{l}
    {V_x} = 50\cos \left( {\frac{{qB}}{m}t} \right) \\
    {V_y} = 0 \\
    {V_z} = - 50\sin \left( {\frac{{qB}}{m}t} \right) \\
    \end{array}

    [/tex]

    I will get a similar expression for rx, ry and rz, which translates (I think) into the equation of a circle.

    Is this OK?
     
  7. Nov 1, 2009 #6

    ehild

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    Homework Helper
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    Just a small mistake:

    [tex]

    \frac{dV_x}{dt} = -\frac{qV_zB}{m}
    [/tex]

    ehild
     
  8. Nov 1, 2009 #7
    Thank you very much.
     
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