In the figure, an electron with an initial kinetic energy of 4.10 keV enters reg
ID: 1434330 • 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 0.0120 T. The electron goes through a half-circle and then exits region 1, headed toward region 2 across a gap of 26.0 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 0.0150 T. The electron goes through a half-circle and then leaves region 2. At what time t does it leave?
Explanation / Answer
We need the three individual times t1, t2 & t3
the speed of the particle entering region1 comes form its kinetic energy
K = 1/2*m*v^2 = 4100eV*1.60x10^-19J/ev = 6.56 x10^-16 J
so v = sqrt(2*6.56x10^-16/9.11x10^-31) = 3.8 x10^7m/s
The period in the field T = 1/f = 2/ = 2*m/(q*B) = 2*9.11x10^-31/(1.60x10^-19*0.0120) = 2.98 x10^-9s
but since it only travels in a semi circle the time t1 = 2.98 x10^-9s/2 = 1.49x10^-9s
Part two it will exit region 1 with a speed of 3.8 x10^7m/s
it will undergo a constant acceleration of E*q/m = (V/d)*q/m
= 2100*1.60x10^-19/(0.26*9.11x10^-31) = 1.42 x10^15m/s^2
so the time to cross the gap comes from x = vx0*t + 1/2*a*t^2
or (0.5*1.42x10^15)*t^2 + 3.8 x10^7*t - 0.26 = 0 or t2 = 6.14 x10^-9 s
Part three The period in the field T = 1/f = 2/ = 2*m/(q*B)
T = 2*9.11x10^-31/(1.60x10^-19*0.015) = 2.38 x10^-9 s
but since it only travels in a semi circle the time t3 = 2.38 x10^-9s/2 = 1.19 x10^-9 s
Therefore total time = 1.49x10^-9 + 6.14 x10^-9 + 1.19x10^-9 = 8.82*10^-9 s