In 1897, J. J. Thomson "discovered" the electron by measuring the charge-to-mass ratio of "cathode rays" (actually, streams of electrons, with charge $q $ and mass $m $) as follows:
(a) First he passed the beam through uniform crossed electric and magnetic fields $\mathbf{E} $ and $\mathbf{B} $ (mutually perpendicular, and both of them perpendicular to the beam), and adjusted the electric field until he got zero deflection. What, then, was the speed of the particles (in terms of $E $ and $B $)?
(b) Then he turned off the electric field, and measured the radius of curvature, $R $, of the beam, as deflected by the magnetic field alone. In terms of $E $, $B $, and $R $, what is the charge-to-mass ratio $(q/m) $ of the particles?
We want zero deflection. That means zero force:
\[\mathbf{F}=q(\mathbf{E}+\mathbf{v}\times \mathbf{B})=\mathbf{0} \]We can cancel the factor of $q $, and we are left with:
\[\mathbf{E}=-\mathbf{v}\times \mathbf{F} =\mathbf{F}\times \mathbf{v}\]With magnitudes
\[E=vB \]So
\[v=\frac{E}{B} \]We can use the already studied cyclotron motion (see Example 5.1). Circular uniform motion is sustained when
\[QvB=m\frac{v^2 }{R} \tag{Eq. 5.3, page 211}\]We can solve for the ratio $q/m $:
\begin{align*} \frac{q}{m} &= \frac{v}{BR} \\ &= \frac{E/B}{BR}\\ &= \frac{E}{B^2 R} \end{align*}