(adsbygoogle = window.adsbygoogle || []).push({}); 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 dependant 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 dont 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 dependant on z, is it? T is indpednant. However since the velocity is changing isnt 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 dependance 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?

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# Homework Help: Particle in a field

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