- #1
Opus_723
- 178
- 3
Homework Statement
The position of a proton at time t is given by the distance vector
[itex]\vec{r}(t) = \hat{i}x(t) + \hat{j}y(t) + \hat{k}z(t)[/itex]
A magnetic induction field along the z-axis, [itex]\vec{B} = \hat{k}B_{z}[/itex] exerts a force on the proton
[itex]\vec{F}[/itex] = e[itex]\vec{v}[/itex][itex]\times[/itex][itex]\vec{B}[/itex]
a.) For initial conditions:
x(0) = x[itex]_{0}[/itex]
[itex]\dot{x}[/itex](0) = 0
y(0) = 0
[itex]\dot{y}[/itex](0) = v[itex]_{y0}[/itex]
z(0) = 0
[itex]\dot{z}[/itex](0) = v[itex]_{z0}[/itex]
and using cartesian coordinates calculate the orbit of the proton.
b.) Rephrase this entire problem in circular cylindrical coordinates and solve in circular cylindrical coordinates
The Attempt at a Solution
I think I managed to solve it in Cartesian by solving a coupled system of linear differential equations. My result was:
x(t) = [itex]\frac{v_{y0}m}{eB_{z}}cos(\frac{eB_{z}}{m}t) + (x_{0}-\frac{v_{y0}m}{eB_{z}})[/itex]
y(t) = [itex]\frac{v_{y0}m}{eB_{z}}sin(\frac{eB_{z}}{m}t)[/itex]
z(y) = v[itex]_{z0}[/itex]t
But I don't see how to do this in cylindrical coordinates. When I set up the differential equations using Newton's Law, I get:
[itex]\ddot{\rho}-\rho\dot{\varphi}^{2} = \frac{eB_{z}}{m}\rho\dot{\varphi}[/itex]
[itex]\rho\ddot{\varphi}+2\dot{\rho}\dot{\varphi} = \frac{-eB_{z}}{m}\dot{\rho}[/itex]
[itex]\ddot{z} = 0[/itex]
And of course the initial conditions
[itex]\rho(0) = x_{0}[/itex]
[itex]\dot{\rho}(0) = 0[/itex]
[itex]\varphi(0) = 0[/itex]
[itex]\dot{\varphi}(0) = \frac{v_{y0}}{x_{0}}[/itex]
z(0) = 0
[itex]\dot{z}(0) = v_{z0}[/itex]
But those equations are nonlinear, and I don't see any way to do this problem, although I assumed the problem would be easier in cylindrical coordinates because of the symmetry involved. I can see how the solutions for [itex]\rho(t)[/itex] and [itex]\varphi(t)[/itex] could get really weird in the general case where the helix isn't centered on the z-axis. But I don't see how to simplify this particular problem.