Part 1 In class you saw a demo where beam of electrons in an Crookes tube was de

ID: 2092750 • Letter: P

Question

In class you saw a demo where beam of electrons in an Crookes tube was deflected by a magnet. Imagine that the tube was oriented on the table so that the electrons moved horizontally from east to west and have an energy of 16 keV. In Boulder, the vertical component of the Earth's magnetic field has a magnitude of 49 uT and points down. In which direction does the force on the electrons act (enter N for north, S for South, E for East, or W for West)?

What is the magnitude of the acceleration of an electron in the tube due to the vertical component of the Earth's magnetic field?

By the right hand rule, the direction of the force is North (N)

To find the acceleration, apply F = ma and F = qvB

so a = qvB/m

To find the velocity, use KE = .5mv^2

(16000)(1.6 X 10^-19) = (.5)(9.11 X 10^-31)(v)^2

v = 7.50 X 10^7 m/s

Then a = (1.6 X 10^-19)(7.50 X 10^7)(49 X 10^-6)/(9.11 X 10^-31)

a = 6.45 X 10^14 m/s^2