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We will consider a very simple but beautiful problem that gives ussome insight into the behaviour of electrons in a two-dimensional gas.The applications are often to systems treated with quantummechanics thin semiconductor sheets, graphene layers etc. Thetreatment here however will be classical and restricted to theequations of motion of two electrons repelling each other. Thepresence of a uniform magnetic field means that the trajectories arecurved rather than linear and the question becomes one ofdetermining what shape these trajectories take they form througha combination of the Coulomb repulsion between the electrons andthe Lorentz motion in a magnetic field.- The geometry of the problem is illustrated in Figure 1.1.

The trajectories of the electrons are the red curves. The electrons are initiallyequally spaced, each at distance from the origin of our coordinate systemat O. The magnetic field points out of the paper. We will use plane polar coordinates (r,) to describe the motion.Since the motion paths of the two electrons are mirror images, weneed only solve for one of them to find the motion of both.In plane polar coordinates, the expressions for the velocity andacceleration are: \vec{V}=r \vec{e}_{r}+r \dot{\phi} \vec{e}_{\phi} In the equations above, e, represents a radial unit vector and e, atangential unit vector. \vec{a}=\left(\ddot{r}-r \dot{\phi}^{2}\right) \vec{e}_{r}+(r \ddot{\phi}+2 \dot{r} \dot{\phi}) \vec{e}_{\phi} Show that the equations of motion for one of our electrons are givenby: m\left(\ddot{r}-r \dot{\phi}^{2}\right)=\frac{e^{2}}{16 \pi \varepsilon_{r} r^{2}}-e B r \dot{\phi} m(r \ddot{\phi}+2 \dot{r} \dot{\phi})=e B \dot{r} In the equations above, e is a positive quantity equal to themagnitude of the electron charge. r(0)=r_{0} \quad \dot{r}(0)=0 \quad \phi(0)=0 \quad \dot{\phi}(0)=0

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