Q.An electron gun with its collector at a potential of $100\ \text{V}$ fires out electrons in a spherical bulb containing hydrogen gas at low pressure ($\sim 10^{-2}\ \text{mm of Hg}$). A magnetic field of $2.83 \times 10^{-4}\ \text{T}$ curves the path of the electrons in a circular orbit of radius $12.0\ \text{cm}$. (The path can be viewed because the gas ions in the path focus the beam by attracting electrons, and emitting light by electron capture; this method is known as the 'fine beam tube' method.) Determine e/m from the data.

Combining the energy gained from the accelerating potential with the radius of the resulting circular path in the magnetic field lets the electron's mass and charge appear only in the ratio $e/m$, which can then be solved for directly.

Step 1 — Two independent relations for the same electron.

Acceleration through potential $V$ gives it speed $v$ via

The magnetic field then curves this same electron into a circle of radius $r$:

Step 2 — Eliminate $v$ to solve for e/m.

Squaring (ii): $v^2 = \left(\dfrac{e}{m}\right)^2 B^2r^2$. Setting this equal to (i):

Step 3 — Substitute the numbers.

$V=100\ \text{V}$, $B=2.83\times10^{-4}\ \text{T}$, $r=0.12\ \text{m}$:

This is close to the accepted value of $1.76\times10^{11}\ \text{C/kg}$, the small difference being ordinary experimental rounding in the given data.