Yes, it's always line segments. The important thing is that those line segments are not constrained to a single direction and can describe curved paths.
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That's it, now we're talking.
Electron is entering static magnetic field B from the left. We can see it in five successive...
I'm not sure if we completely understand each other. It can not be a line integral, it must not be limited to change in just one direction. It has to be arbitrary path or curve integral, so electrons may, if they are compelled to do so, move in lateral direction to their initial velocity vector.
I'm not making any assertions, please ask for reference if you are not familiar with something I said.
This is from Wikipedia:
http://en.wikipedia.org/wiki/Work_%28physics%29#Mathematical_calculation
That equation will produce correct results. You can use dx instead of ds only...
Because that second velocity vector is supposed to be in the direction of the force and/or wire displacement. Why would you calculate work done relative to the direction of the current when neither the force nor wire displacement is in that direction? That equation works for electric force, but...
Please provide some reference for that.
I drift direction X-axis: {1, 0, 0}
B field vector Y-axis: {0, 1, 0}
F force vector Z-axis: {0, 0, 1}
1. Lorentz force acts along Z-axis?
2. Lorentz force accelerate electrons along Z-axis?
It's not arbitrary, I chose Lorentz force and wire...
I don't think that's how vector addition works. Let electron move with 10m/s along x-axis, then add 5m/s along y-axis: {10,0,0} + {0,5,0} = {10,5,0}. Vector magnitude increases from 10m/s to 11.18m/s.
You didn't mark axis or specify any directions. Please point at which step do you disagree...
Where is the glitch?
1. Let a single free electron move along x-axis with constant speed Vx. It encounters magnetic field on its right side, so the Lorentz force accelerates it along y-axis and it gains speed Vy.
--- Glitch or pass?
2. The electron does not slow down in x direction when...
What is significance of F.dx = 0? Dx is not the displacement vector we are interested in. Only displacement in the direction of the force matters.
I don't know. What equation are you referring to and why would it have precedence over W= F*s?
Where did you get that .v at the end of the first line, what is it? That's extra. There is charge velocity vector, there is B field vector, and cross product between them, thus force vector is perpendicular to both.
It's just a geometrical rotation defined via unit vector, it doesn't change...
Work is force times displacement, speed is not a part of the equation. Anyway, do you think a free electron would not change its speed when passing next to a permanent magnet where this external B field maximum gradient density is not located perpendicularly to is trajectory, but this time more...
I don't see how is this any different from calculating Coulomb's attraction and work done by electric force, or work done by gravity on a falling object. There is a force vector, there is a displacement vector, and multiplied together they represent work done.
Take a single electron moving...
For the Lorentz force it's all the same whether those are a pair of parallel traveling electrons, or two parallel electron beams, or two current carrying wires, or a single electron moving next to a permanent magnet, or electron beam passing by permanent magnet, or current carrying wire placed...