In the figure below, an electron with an initial kinetic energy of 4.10 keV ente
ID: 2059430 • Letter: I
Question
In the figure below, an electron with an initial kinetic energy of 4.10 keV enters region 1 at time t = 0. That region contains a uniform magnetic field directed into the screen, with magnitude 0.00510 T.
The electron goes through a half-circle and then exits region 1, headed toward region 2 across a gap of 27.0 cm. There is an electric potential difference V = 2000 V across the gap, with a polarity such that the electron's speed increases uniformly as it traverses the gap. Region 2 contains a uniform magnetic field directed out of the page, with magnitude 0.0204 T.
Explanation / Answer
The velocity of the electron in Region can be found given its initial energy
4.1 k2eV = 4100 eV. There are 1.602 X 10-19 J per eV so the electron has...
(4100)(1.602 X 10-19) = 6.57 X 10-16 J of energy
That energy is kinetic and will equal .5mv2
6.57 X 10-16 J = (.5)(9.11 X 10-31)(v2)
v = 3.80 X 107 m/s (save for later)
The formula to find the radius of the path in the magnetic field is
r = mv/qB
r = (9.11 X 10-31)(3.8 X 107)/(1.6 X 10-19)(.00510)
r = 4.24 X 10-2 m
Since the electron is traveling half of a circle, the distance it travels is half the circumference
D = r = (4.24 X 10-2) = .133 m
Using d = vt we can find the time the electron spent in Region 1
.133 = (3X107)t t = 4.44 X 10-9 s (save for later)
Then the electron goes into the gap where an electric field acts on it.
We know that it will gain KE from the acceleration such that qV = KE
qV = .5mvf2 - .5mvi2
(1.6 X 10-19)(2000) = (.5)(9.11 X 10-31)(vf2) - (.5)(9.11 X 10-31)(3.8 X 107)2
Solve for the new velocity
vf = 4.63 X 107 m/s
Now we can find the time through the gap.
Using d = (vf + Vi)t/2
.27 = (4.63 X 107 + 3.80 X 107)(t)/2
t = 6.41 X 10-9 s (save for later)
Now we know the speed the electron enters Region 2
Solve for that radius
r = mv.qB
r = (9.11 X 10-31)(4.63 X 107)/(1.6 X 10-19)(.0204)
r = 1.29 X 10-2 m
Again the distance traveled is half a circle
using d = vt, we can find that time
(1.29 X 10-2) = (4.63 X 107)(t)
t = 8.77 X 10-10 s
Finally, we can add up all three times
4.44 X 10-9 s + 6.41 X 10-9 s + 8.77 X 10-10 s
Total time equals 1.17 X 10-8 s which is 11.7 ns (nano seconds)