Newton's 2nd law and Lorentz Force

[Dvsdvs]

ok, the lorentz force equation is F=q(E+v x B) and Newton's 2nd law is F=ma. I need to combine the two to show that ma.v=qE.V
I don't really know what to do first here...
i can see the simple substitution so that ma=qE+q(v x B) but beyond that...do u change the a to dv/dt. Any type of help to get me started will work.

 

[Doc Al]

i can see the simple substitution so that ma=qE+q(v x B) but beyond that...

Good. Now just multiply each side by v (scalar multiplication, aka dot product).

 

[Dvsdvs]

nice. thank you very much. How do i show that v.(qE+(v x B)=qE.v in other words why is it that v.(V x b)=0. i know this is probably an elementary question but I am not too good at working with vectors.

 

[diazona]

The cross product of two vectors is always perpendicular to those vectors. So \vec{v}\times\vec{B} is perpendicular to v (and to B). What does that tell you about the dot product \vec{v}\cdot(\vec{v}\times\vec{B}) (and why)?

 

[Dvsdvs]

oh wooow i feel so stupid really. yeah so dot product = 0 b/c they are perpendicular...Also the overall question for the exercise was to show that if speed is constant. then show that E does now work along the path of the particle.

For this it means that //v(t)// is constant which is to say that
v(t).a=0 by a proposition I proved in a previous prob. Does this mean that
ma.v=0=qE.v?? and if it does this shows that E is perpendicular to v. Being perpendicular to direction of motion, is it sound to say that it does no useful work along path of particle? This is the last part of this whole thing thank you very much for help so far!

 

[diazona]

Well, if you've shown that v.a = 0, and that v.a is proportional to E.v, then that would mean that E.v = 0 ... that's just simple math ;-)

 

[Doc Al]

For this it means that //v(t)// is constant which is to say that
v(t).a=0 by a proposition I proved in a previous prob. Does this mean that
ma.v=0=qE.v??

Yep.

and if it does this shows that E is perpendicular to v.

Yep.

Being perpendicular to direction of motion, is it sound to say that it does no useful work along path of particle?

Yep again.