rogerk8
- 288
- 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?
Roger
[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?
Roger