Solve Equation of Motion: Particle in Field

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SUMMARY

The discussion focuses on solving the equation of motion for a particle with mass m and charge q in a constant electric field E0 and magnetic field B0. The equation m ˙v - q E_{0} + (q/C) v × B_{0} is analyzed to determine the particle's velocity v(t) and trajectory r(t). Participants explore the implications of the chosen coordinate system, particularly the z-axis along B0 and the y-axis perpendicular to the E0 and B0 plane, leading to specific expressions for the electric field components and their time dependence. The necessity of considering separate Lagrangians for different directions and the potential's dependence on z and t are also debated.

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Solve the equation of motion for a particle of mass m and charge q moving in a constant electric and magnetic field E0 and B0 tat is solve the equation
[tex]m \dot{v} - q E_{0} + \frac{q}{C} v \times B_{0}[/tex]
to determine the velocity v(t) and thence the trajctory r(t) of the particle. Discuss the hsape of the trajectory.

Hint: Choose the z axis along B0 and the y-axis perpendicular to the plane of E0 and B0


Wel accoridng to the hint Bx and By would be zero
since Y is perpendicular ot the plane of E an B then
then
[tex]E_{x} = E_{0} \cos(t)[/tex]
[tex]E_{z} = E_{0} \sin(t)[/tex]
Ey = 0
does v hasve to be dependent on some angle??

so does this mean i have to have three separate lagrangians?
would phi depend on t since the velocity will change ??
force in the Z direction is
[tex]m \dot{v_{z}} = q E_{0} \sin(t) + \frac{q}{c} v_{?} \times B_{0}[/tex]
not sure about the direction of v. I don't think its possible... is it? After i find the force do i have to find the potential V(z)? But nothing in that equation is dependent on z, is it? T is indpednant. However since the velocity is changing isn't v depdnant on z??

for hte Y direction velocity in the X direction yes?
[tex]m \dot{v_{x}} = q E_{0} \cos(t) + \frac{q}{C} v_{x} \times B_{0}[/tex]
again... what about the potnetial and its dependence on t or y??

for the Y direction
[tex]m \dot_{y} = \frac{q}{c} (-v_{x}) \times B_{0}[/tex]
are there right so far?
 
Last edited:
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If the fields are constant, why did you put a time dependence for the electric field strength...?

Daniel.
 
i was meaning to correct that ...
it should just be some angle, phi, shouldn't it?
 

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