# Momentum Conservation when particle enters magnetic field

Suppose a particle flies in $$x$$-direction (with velocity $$v_{ges}=v_x$$ ) into a constant magnetic field in $$z$$-direction. Looking at the Lagrangian $$$$L=\frac{1}{2} m v^{2}+\frac{Q}{c} \vec{v} \cdot \vec{A}$$$$ the momentum in $$x$$-direction should be conserved ($$A$$ doesn't depend on the $$x$$-coordinat), so $$$$v_x=const.$$$$ When the particle enters the magnetic field, the Lorentz-force bands its trajectory and it gains velocity in the $$y$$-direction, so $$$$v_y\neq0$$$$ The Lorenz-force doesn't do work so the kinetic energy has to be conserved (here I am not entirely sure, maybe the particle gains some potential energy) leading to $$$$v_{ges}=\sqrt{v_{x}^2+v_{y}^2}=const.$$$$ But that's a contradiction to $$v_x=const.$$ or $$v_y\neq0$$. I think the solution is very simple but I am kind of stuck at the moment.

• I think a good exercise would be to write out the vector potential and the Lagrangian explicitly in terms of the coordinates. There's freedom in the choice of vector potential, see feynmanlectures.caltech.edu/II_14.html, but you'll typically get at least one of $x$ dependence or $v_x$ coupling, both of which cause $v_x$ to change in time. From your post, it seems you might be using the vector potential of $A_x = -y B_0$ - I recommend writing the equations of motion explicitly and paying close attention to $\frac{\partial L}{\partial \dot{x}}$ and $\frac{\partial L}{\partial y}$. Apr 11 '20 at 21:17

Yes, the answer is simple. Indeed, at the very beginning, only, $$\dot v_x=0; \qquad \dot v_y \propto v_x,$$ so $$\dot T=0$$. But as soon as $$v_y$$ departs from zero, you get an opposing force in the x direction slowing $$v_x$$ down. When you write the full solution, you find the cyclotron trajectory.
Recall $$\mathbf {F}\propto \nabla (\mathbf{v\cdot A}) -(\mathbf{v}\cdot \nabla) \mathbf{A}$$.
• So $v_x^2=v_{x,inital}^2-(qv_{x,inital}Bt/m)^2$. But how do I get the time $t$ the particle stays in the magnet (if one supposes that it has the length $l$). The initial question is, if a particle with charge $e$ is deflected by $s$ (in the $y$ direction) and we know the length $l$ of the magnet and $B$ and the initial value of $v$. What's the mass of the particle? I used $s=1/2at^2$ wit $a=qv_{x,inital}B/m$ and $t=l/v_{x,inital}$ but the last equation is wrong because $v_x$ changes. Apr 11 '20 at 21:35
• If I use $t=l/v_x$ with the first equation in the last commend(for $v_x$) and solve for $t$ the solution is very ugly. I thought a question like this has a simple answer. Apr 11 '20 at 21:42