Simon Says with protons and electrons

Poor man's way

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Though it feels great to calculate the change in the velocity vector relative to its magnitude along the path of the particles, it isn't strictly necessary. If the relation we seek is magical enough to guarantee identical trajectories in a monster field made of any number of coils, each of them pointing any which way, surely it is also good enough to promise the same in a uniform field. Therefore, let's consider an electron going into a region of uniform $B$ -field.

As usual, its trajectory is a circle of radius $\displaystyle \frac{q_eB}{m_ev_e} = \frac{q_eB}{p_e}$ . Similarly, we have $\displaystyle R_p = \frac{q_pB}{p_p}$ . Since $q_p=q_e$ and the field has the same magnitude in each instance, the condition for having the same radius is $p_e = p_p$ .

Now, we assume that all of the energy of the electron comes from accelerating at the potential difference $U_e$ . If so, we have $\displaystyle \frac{p_e^2}{2m_e} = q_eU_e \rightarrow p_e = \sqrt{2m_eq_e|U_e|}$ and similarly $\displaystyle p_p = \sqrt{2m_pq_p|U_p|}$ which implies $\displaystyle m_e|U_e|=m_p|U_p|$ . We have to remember that due to the opposite charge, we must run the proton through a gradient of opposite sign to the one used for an electron. The answer is then: $\displaystyle \boxed{U_p = \displaystyle -U_e \frac{m_e}{m_p} = \displaystyle -\frac{U_e}{6\pi^5}}$

Feel good way

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Briefly, we have

$\begin{aligned} ds &= \sqrt{dx^2+dy^2} \\ &= \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt \\ &= \sqrt{\vec{v}^2} dt \end{aligned}$

Therefore, $\displaystyle\frac{ds}{dt} = v$ .

Now,

$\begin{aligned} \frac{d}{ds}\frac{\vec{v}}{v} &= \frac{1}{v}\frac{dt}{ds}\frac{d}{dt}\vec{v} \\ &= \frac{1}{v^2}\left[\frac{q}{m}\vec{v}\times\vec{B}\left(\vec{r}\right)\right] \\ &= \frac{q}{p}\frac{\vec{v}}{v}\times\vec{B}\left(\vec{r}\right) \\ &= \frac{q}{p}\vec{i_s}\times\vec{B}\left(\vec{r}\right) \end{aligned}$

Where $\vec{i_s}$ is a vector of unit magnitude that points in the direction as the velocity vector.

Obviously, this quantity will be the same for any two particles that have the same value of $\frac{q}{p}$ . As the charge of the proton and the electron are equal in magnitude, the only factor to match is $\displaystyle p$ .

As this directional derivative quantity details the change in the direction of the particle velocity for every point along the path, it fully characterizes the trajectory of a path in space. If it is the same for two trajectories that have the same initial conditions, the trajectories are the same.

So, as before, the two particles have the same trajectory if they have the same momentum after being accelerated across the potential difference, which leads to the same conclusion as argued above.

The equation of motion for a particle of mass $m$ and charge $e$ moving in a magnetic field $\mathbf{B}$ is $m\dot{\mathbf{v}} \; = \; e\mathbf{v} \wedge \mathbf{B}$ This implies that $\mathbf{v}\cdot\dot{\mathbf{v}} \,=\, 0$ , and hence $v = |\mathbf{v}|$ is constant. Thus $\frac{\dot{\mathbf{v}}}{v} \; =\; \frac{d}{dt}\Big(\frac{\mathbf{v}}{v}\Big) \; = \; \frac{d}{ds}\Big(\frac{\mathbf{v}}{v}\Big)\times \frac{ds}{dt} \;= \;v\frac{d}{ds}\Big(\frac{\mathbf{v}}{v}\Big)$ and hence, if we define $\mathbf{j} \; =\; \frac{\mathbf{v}}{v}$ then $\frac{d\mathbf{j}}{ds} \; = \; \frac{1}{v^2}\dot{\mathbf{v}} \; = \; \frac{1}{mv^2}e\mathbf{v} \wedge \mathbf{B} \; = \; \frac{1}{mv}\mathbf{j} \wedge (e\mathbf{B})$ Thus, since $e\mathbf{B}$ is unchanged by reversing the currents and changing the sign of the particle, we deduce that the proton and the electron have the same trajectory provided that they have the same initial momentum. Since $\tfrac12m_ev_e^2 \; = \; |q|U_e \qquad \tfrac12m_pv_p^2 \; = \; qU_p$ where $q > 0$ is the charge on the proton, $m_e,m_p$ the masses of the electron and the proton and $v_e,v_p$ the speeds of the electron and the proton after the initial acceleration, we require that $m_eU_e = -m_pU_p$ , and so $U_p = -\frac{10^{-3}}{6\pi^5} = -5.45 \times 10^{-7}$ V.

Feels perfectly fine and efficient to me

First of all, $U_p$ will have to be negative to speed up the proton in the same direction as the electron.

Second, the trajectory's geometry will be the same but the kinetics different. So we reduce the problem to geometric quantities only. The important one is the radius of curvature, which is $r = \frac{mv}{qB}.$ Since $q$ and $B$ are the same as before (apart from a double change in sign), and we need $r$ to remain the same, we conclude that the proton's momentum $p = mv$ must be the same as the electron's.

This momentum is due to acceleration through a conservative force, so that $E_{pot} = \frac{p^2}{2m};$ the potential difference $U = E_{pot}/q$ is therefore proportional to $1/m$ . Thus $U_p = -U_e\cdot \frac{1}{m_p/m_e} = -1\times 10^{-3}\ \text{V} \cdot \frac{1}{6\pi^5} = \boxed{-5.45\times 10^{-7}\ \text{V}}.$

The above means,mv^2/r=qvB;means mv should be same for both.means (2mU)^(1/2) same.means mU same.of course minus sign for proton.