In the figure, an electron with an initial kinetic energy of 4.10 keV enters reg
ID: 778023 • Letter: I
Question
In the figure, 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 page, with magnitude 0.00970 T. The electron goes through a half-circle and then exits region 1, headed toward region 2 across a gap of 22.0 cm. There is an electric potential difference V = 2100 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 0.0233 T. The electron goes through a half-circle and then leaves region 2. At what time t does it leave?
Explanation / Answer
The velocity of the electron in Region can be found given its initial energy
Region: 1
Initial kinetic energy = 4.10 keV = 4100 eV.
There are 1.602 X 10-19 J per eV
Total Kinetic energy = (4100)(1.602 X 10-19) = 6.57 X 10-16 J of energy
That kinetic energy is equal to (1/2)mv2
6.57 X 10-16 J = (.5)(9.11 X 10-31)(v2)
v = 3.80 X 107 m/s
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)(.00970)
r = 2.23 X 10-2 m
Since the electron is traveling half of a circle, the distance it travels is half the circumference
D = r = (2.23 X 10-2) =0.07 m
Using d = vt we can find the time the electron spent in Region 1
0.07 = (3.8 X 107)t t = 1.842 X 10-9 s ---------------(1)
Between the regions:
Then the electron goes into the gap where an electric field acts on it.
We know that electron will gain KE from the acceleration such that qV = KE
qV = (1/2)mvf2 – (1/2)mvi2
(1.6 X 10-19)(2100) = (.5)(9.11 X 10-31)(vf2) - (.5)(9.11 X 10-31)(3.8 X 107)2
Solve for the new velocity
vf = 4.67 X 107 m/s
Now we can find the time through the gap.
Using d = (vf + Vi)t/2
.22 = (4.67 X 107 + 3.80 X 107)(t)/2
t = 5.194 X 10-9 s -------------- (2)
Region: 2
Solve for the radius
r = mv/qB
r = (9.11 X 10-31)(4.67 X 107)/(1.6 X 10-19)(.0233)
r = 1.141 X 10-2 m
Again the distance traveled is half a circle
using d = vt, we can find that time
(1.41 X 10-2) = (4.67 X 107)(t)
t = 0.948 X 10-9 s -------------------(3)
Finally, we can add up all three times (equations)
T = 1.842 X 10-9 s + 5.194 X 10-9 s + 0.948 X 10-9 s
T = 7.984 *10-9 sec = 7.984 nsec