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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 \begin{equation} L=\frac{1}{2} m v^{2}+\frac{Q}{c} \vec{v} \cdot \vec{A} \end{equation} the momentum in $x$-direction should be conserved ($A$ doesn't depend on the $x$-coordinat), so \begin{equation} v_x=const. \end{equation} When the particle enters the magnetic field, the Lorentz-force bands its trajectory and it gains velocity in the $y$-direction, so \begin{equation} v_y\neq0 \end{equation} 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 \begin{equation} v_{ges}=\sqrt{v_{x}^2+v_{y}^2}=const. \end{equation} 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.

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    $\begingroup$ 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 $\frac{\partial L}{\partial y}$. $\endgroup$ – user196574 Apr 11 '20 at 21:17
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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 $\mathbf {F}\propto \nabla (\mathbf{v\cdot A}) -(\mathbf{v}\cdot \nabla) \mathbf{A}$.

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  • $\begingroup$ 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. $\endgroup$ – NicAG Apr 11 '20 at 21:35
  • $\begingroup$ 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. $\endgroup$ – NicAG Apr 11 '20 at 21:42
  • $\begingroup$ You had a bogus paradox, which is now settled. You are then reverting to a "hidden" homework problem that should have been the original problem, with the full homework-problems tag. Once you write down the full, all time!, equations of motion in the magnetic field, you could fit the solution of the circular trajectory to your ark. Your thinking in this comment is unsound. $\endgroup$ – Cosmas Zachos Apr 11 '20 at 21:42
  • $\begingroup$ Sorry, but anyway thank you very much. I got it now. $\endgroup$ – NicAG Apr 11 '20 at 22:20

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