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stunner5000pt
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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
m \dot{v} - q E_{0} + \frac{q}{C} v \times B_{0}
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
E_{x} = E_{0} \cos(t)
E_{z} = E_{0} \sin(t)
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
m \dot{v_{z}} = q E_{0} \sin(t) + \frac{q}{c} v_{?} \times B_{0}
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 dependant 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?
m \dot{v_{x}} = q E_{0} \cos(t) + \frac{q}{C} v_{x} \times B_{0}
again... what about the potnetial and its dependence on t or y??
for the Y direction
m \dot_{y} = \frac{q}{c} (-v_{x}) \times B_{0}
are there right so far?
m \dot{v} - q E_{0} + \frac{q}{C} v \times B_{0}
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
E_{x} = E_{0} \cos(t)
E_{z} = E_{0} \sin(t)
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
m \dot{v_{z}} = q E_{0} \sin(t) + \frac{q}{c} v_{?} \times B_{0}
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 dependant 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?
m \dot{v_{x}} = q E_{0} \cos(t) + \frac{q}{C} v_{x} \times B_{0}
again... what about the potnetial and its dependence on t or y??
for the Y direction
m \dot_{y} = \frac{q}{c} (-v_{x}) \times B_{0}
are there right so far?
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