In the figure, an electron with an initial kinetic energy of 3.60 keV enters reg
ID: 1596968 • Letter: I
Question
In the figure, an electron with an initial kinetic energy of 3.60 keV enters region 1 at time t = 0. That region contains a uniform magnetic field directed into the page, with magnitude 0.0100 T. The electron goes through a half-circle and then exits region 1, headed toward region 2 across a gap of 29.0 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 0.0212 T. The electron goes through a half-circle and then leaves region 2. At what time t does it leave?
Explanation / Answer
Within the B field regions, theelectron's speed is constant but the direction is changing(circular motion). Between the regions, it is increasing inspeed.
Region 1: Time for ahalf circle: t = T / 2 = 2 r / 2 v1
t1 = r1 / v1
Region 2: Similarly we canderive t2 = r2 /v2
Betweenregions: t = (v2 - v1) / a
We can get the acceleration. Since we know the potentialdifference V and the gap length d, we can find the Efield, force, and hence acceleration:
a = F / m = q E / m = q V / md
So weget: t = m d (v2 - v1) / q V
That done, we know that in region 1, the mag forcesupplies centripetal force and the speed isconstant
F = q v B = m v2 / r
qB = m v / r
r1 = m v1 / qB1
Similarly r2 = mv2 / q B2
Plug back into the first yellow equation:
t1 = m v1 / q B1v1 interestingly, we don't need the speed
= * 9.11 x 10-31 kg / (1.60 x 10-19 C * 0.0110 T)
t1 = 1.626 ns
Similarly, t2 = mv2 / q B2 v2
= * 9.11 x 10-31 kg / (1.60 x10-19 C * 0.0201 T)
t2= 0.8899 ns
For the middle time t we'll need to know both v1and v2. Those we can get from the kinetic energies.
At first v1 = (2 * KE1 / m)
=[(2 * 3600 eV * 1.60 x 10-19 J / eV) / 9.11 x10-31 kg]
= 3.556 x 107 m/s
As for v2, well, we know it gains q V of kineticenergy as it accelerates through the voltage
So KE2 = KE1 + q V = (3600 eV * 1.6 x10-19 J / eV) + (2000 V * 1.60 x10-19 C)
= 8.96 x 10-16 J
v2 = (2 * 8.96 x 10-16 J / 9.11 x10-31 kg)
= 4.435 x 107 m/s
Now throw all that back into the orange equation:
t = 9.11 x 10-31 * 0.280 m * ((4.435 -3.556) x 107 ) / 1.60 x10-19 * 2000 V
= 7.0067 ns
Finally add t = t1 + t + t2
= 1.626 + 7.0067 + 0.8899
= 9.5226 ns