In the figure, an electron with an initial kinetic energy of 3.70 keV enters reg
ID: 1313521 • Letter: I
Question
In the figure, an electron with an initial kinetic energy of 3.70 keV enters region 1 at time t = 0. That region contains a uniform magnetic field directed into the page, with magnitude 0.00630 T. The electron goes through a half-circle and then exits region 1, headed toward region 2 across a gap of 23.0 cm. There is an electric potential difference ?V = 1900 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
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In the figure, an electron with an initial kinetic energy of 3.80keV enters region 1 at time t = 0. That region contains auniform magnetic field directed into the page, with magnitude0.0110 T. The electron goes through a half-circle and then exitsregion 1, headed toward region 2 across a gap of 28.0 cm. There isan electric potential difference ?V = 2000 V across the gap,with a polarity such that the electron's speed increases uniformlyas it traverses the gap. Region 2 contains a uniform magnetic fielddirected out of the page, with magnitude 0.0201 T. The electrongoes through a half-circle and then leaves region 2. At what timet does it leave?
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)
= 1.626 ns
Similarly, t2 = ? mv2 / q B2 v2
= ?* 9.11 x 10-31 kg / (1.60 x10-19 C * 0.0201 T)
= 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 * 3800 eV * 1.60 x 10-19 J / eV) / 9.11 x10-31 kg]
= 3.653 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 = 3800 eV * 1.6 x10-19 J / eV + (2000 V * 1.60 x10-19 C)
= 9.28 x 10-16 J
v2 = ?(2 * 9.28 x 10-16 J / 9.11 x10-31 kg)
= 4.154 x 107 m/s
Now throw all that back into the orange equation:
?t = 9.11 x 10-31 * 0.280 m * (4.154 -3.653 x 107 ) / 1.60 x10-19 * 2000 V
= 6.86 ns
Finally add t = t1 + ?t + t2
= 9.38 ns