Crossed Fields: Discovery of the Electron — Question 6

Question 6

An exact crossed-field trajectory shows why one exit slit is not a perfect speed test

Positive particles enter uniform $\vec E=-E\hat y$ and $\vec B=-B\hat z$ with initial velocity $(v_0+\delta)\hat x$, where $v_0=E/B$ and $\omega=qB/m$. Their initial position is the origin. No short-transit approximation is made.

1. Solve for vx, vy, x and y as functions of time.

2. Find position and velocity after $T=2\pi/\omega$ and explain how an off-speed particle can return to the undeflected line.

Question 6 - Solution

An exact crossed-field trajectory shows why one exit slit is not a perfect speed test

(a) Method and calculation

$$\dot v_x=-\omega v_y,\quad\dot v_y=\omega(v_x-v_0)\Rightarrow v_x=v_0+\delta\cos\omega t,\ v_y=\delta\sin\omega t.$$$$x=v_0t+\frac\delta\omega\sin\omega t,\quad y=\frac\delta\omega(1-\cos\omega t).$$

(b) Interpretation and further calculation

$$x(T)=v_0T,\quad y(T)=0,\quad\vec v(T)=(v_0+\delta)\hat x.$$$$y(t)\not\equiv0\quad\text{when }\delta\ne0.$$A slit at this return point admits the particle although its path between entrance and exit is curved.

Physical check

Exact force balance implies a straight path throughout; a single zero exit displacement is a weaker condition.