B Cross product evaluation (for the Lorentz Force).

1. Aug 11, 2015

rogerk8

Let's say we have:

$$\vec{E}=E_x\vec{i}_x+E_y\vec{i}_y+E_z\vec{i}_z$$

and

$$\vec{B}=B_x\vec{i}_x+B_y\vec{i}_y+B_z\vec{i}_z$$

and the Lorentz Force

$$0=q(\vec{E}+\vec{v}X\vec{B})$$

which due to

$$\vec{E}X\vec{B}=\vec{B}X(\vec{v}X\vec{B})=vB^2-B(\vec{v}\cdot \vec{B})$$

and transverse components only, gives

$$v_{gc}=\frac{\vec{E}X\vec{B}}{B^2}$$

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

$$\vec{C}=\vec{E}X\vec{B}$$

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. Aug 11, 2015

Staff: Mentor

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..

3. Aug 11, 2015

rogerk8

Simple, the equation comes from

$$\vec{E}X\vec{B}=-\vec{B}X\vec{E}$$

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

I don't like you.

/Roger

4. Aug 11, 2015

Staff: Mentor

You don't like being reminded of reality by a professional working physicist? Hmm

5. Aug 11, 2015

Staff: Mentor

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