An electron in a cathode-ray tube is accelerated through a potential difference of 10 kV, then
passes through the d 2.2-cm-wide region of uniform magnetic field in the figure(Figure 1). Part
A What field strength will deflect the electron by 14° Express your answer to two significant
figures and include the appropriate units BValue Units Figure 1 of1 Submit Request Answer
Provide Feedback 0 V 10 kV
Solution
in a magnetic field
magnetic force = centripetal force
qvB = mv^2/r
B = mv/qr
if an electron is accelerated in a potential difference
then 1/2mv^2 = q*V
v = sqrt(2qV/m)
v = sqrt(2*1.6*10^-19*10000/9.1*10^-31)
v = 5.93*10^7 m/s
from the figure
sin(theta) = d/r
r = 2.2*10^-2/sin14
r = 0.091 m
B = 9.1*10^-31*5.93*10^7/1.6*10^-19*0.091
B = 3.71*10^-3 T