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B Cross product evaluation (for the Lorentz Force).

  1. Aug 11, 2015 #1
    Let's say we have:

    [tex]\vec{E}=E_x\vec{i}_x+E_y\vec{i}_y+E_z\vec{i}_z[/tex]

    and

    [tex]\vec{B}=B_x\vec{i}_x+B_y\vec{i}_y+B_z\vec{i}_z[/tex]

    and the Lorentz Force

    [tex]0=q(\vec{E}+\vec{v}X\vec{B})[/tex]

    which due to

    [tex]\vec{E}X\vec{B}=\vec{B}X(\vec{v}X\vec{B})=vB^2-B(\vec{v}\cdot \vec{B})[/tex]

    and transverse components only, gives

    [tex]v_{gc}=\frac{\vec{E}X\vec{B}}{B^2}[/tex]

    where Vgc is the guiding center drift of the charged particles in a magnetic field with an electric field.

    My question now is how to calculate

    [tex]\vec{C}=\vec{E}X\vec{B}[/tex]

    I could have chosen pure math for this but I'm tired of theory that is hard to see the practical use of.

    So what is C, with my definitions of E & B?

    And how do I calculate it (the manitude and resulting direction is easy but I whish to see it in math)?

    Does anyone want to help me refresh this knowledge?

    Best regards, Roger
     
  2. jcsd
  3. Aug 11, 2015 #2

    mfb

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    I don't understand where that equation comes from.
    With the standard way to express the cross-product via components. You can find the formula in literally every textbook, in all the wikipedia articles and in thousands of other websites related to the cross-product..
     
  4. Aug 11, 2015 #3
    Simple, the equation comes from

    [tex]\vec{E}X\vec{B}=-\vec{B}X\vec{E}[/tex]

    Please don't answer (or whatever you think you are trying to do) any more of my posts.

    I don't like you.

    /Roger
     
  5. Aug 11, 2015 #4

    berkeman

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    You don't like being reminded of reality by a professional working physicist? Hmm
     
  6. Aug 11, 2015 #5

    berkeman

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    Staff: Mentor

    Roger has left the building, so this thread is now closed.
     
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