Teaching Modern Physics - QuarkNet - Fermilab

For path B…
The cathode ray interacts with both the electric field
and the magnetic field simultaneously.
FE = FB
qE = qvB
v = E/B
Using v = E/B and q/m = v/(Br)…
q/m = (E/B)/(Br) or
q/m = E/(B 2 r) since V = Ed
q/m = V/(B 2 rd)
(Thompson could determine the charge-to-mass
ratio of the “cathode ray” by measuring V, I
(magnetic field), radius of curvature and the
distance between the plates.)
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