Can Lorentz force be expressed as a function of z in a static electromagnetic field?

AI Thread Summary
The discussion revolves around expressing the Lorentz force as a function of z in a static electromagnetic field, where the electric field E and magnetic field B depend on z. The participant proposes an equation for the Lorentz force but faces confusion regarding the inclusion of the term 1/v. They attempt to relate the force to the particle's motion along the z-axis, acknowledging that their initial velocity has only one component. However, there is disagreement about the implications of the particle's motion, specifically regarding the derivatives of x and y, which are not zero. The conversation highlights the complexities of transforming differential equations into a function of z while grappling with algebraic manipulations.
1Keenan
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Hi,

I have a particle moving in a static electromagnetic field in which E and B have the following components:
E=(Ex, 0, 0)
B=(-Bx, 0, 0)
and both depend on z, namely Ex(z) and Bx(z).
The particle is moving along z with constant velocity v=(0, 0, vz).

If I want to express Lorentz force as a function of z, is it correct to write:

F(z)=q [(E *cross (1/v)) *cross (1/v) + B *cross (1/v)]?

I get this equation considering F=q(E+v *cross B) as a system of 6 differential equations:

dx/dt=0
dy/dt=0
dz/dt=vz
dvx/dt= q*Ex/m
dvy/dt=q*Bx*vz/m
dvz/dt=0

and expressing them as a function of z
 
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Hi 1Keenan! :smile:
1Keenan said:
dx/dt=0
dy/dt=0

No. :confused:
If I want to express Lorentz force as a function of z, is it correct to write:

F(z)=q [(E *cross (1/v)) *cross (1/v) + B *cross (1/v)]?

(write "x" not "*cross" :wink:)

Where do the 1/v come from? :confused:
 
tiny-tim said:
No. :confused:

Why?
My particle is moving along z and v=[0,0,vz]
of course it will have a displacement along x and y but the intial velocity has only one component.
Could you please explain your point?

tiny-tim said:
Where do the 1/v come from? :confused:



it comes from the differential equation:
dvx/dt=qEx/m
dvy/dt=qBxvz/m

I change the variable t in z... it is a bit of algebra I can write you everything if you want so you can double check my manipulation.
 
1Keenan said:
… the intial velocity has only one component.

but that doesn't mean that dx/dt = dy/dt = 0, not even initially :redface:
it comes from the differential equation:
dvx/dt=qEx/m
dvy/dt=qBxvz/m

I change the variable t in z... it is a bit of algebra I can write you everything if you want so you can double check my manipulation.

i still don't get it :redface:
 
tiny-tim said:
but that doesn't mean that dx/dt = dy/dt = 0, not even initially :redface:

What does it mean? :confused:

tiny-tim said:
i still don't get it :redface:

How do you write it down?
I don't understand what is tricky for you...
 
I was thinking, and actually I'm doing something stupid, but I'm really interested in expressing those differential equation as function of z and I'm lost in papers full of my wrong formulas...
at the moment I'm not able to calculate 1+1... :(
 
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