Crossed Fields: Discovery of the Electron — Question 4

Question 4

A short velocity selector has a calculable finite acceptance band

In a selector with downward E and into-page B, positive particles enter with speed $v_0+\delta v$, where $v_0=E/B$ and $|\delta v|\ll v_0$. The field length is L and an exit slit admits transverse displacement $|y|\le a$. Assume $|q|BL/(mv_0)\ll1$, so transverse acceleration is nearly constant during transit.

1. Derive the leading exit displacement in terms of delta v.

2. Find the approximate allowed fractional speed error and state how it changes when L doubles.

Question 4 - Solution

A short velocity selector has a calculable finite acceptance band

(a) Method and calculation

$$F_y=q[B(v_0+\delta v)-E]=qB\delta v,\quad t\simeq L/v_0.$$$$y\simeq\frac{qB\delta v}{2m}\frac{L^2}{v_0^2}.$$

(b) Interpretation and further calculation

$$\frac{|\delta v|}{v_0}\lesssim\frac{2mav_0}{|q|BL^2}.$$$$L\prime=2L\Rightarrow\text{fractional acceptance}\prime=\tfrac14\text{(original)},\quad\text{within the short-transit regime}.$$

Physical check

This is a controlled short-transit approximation, not the exact trajectory through a long crossed-field region.