In the figure, an electron with an initial kinetic energy of 4.10 keV enters reg

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 30.0 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 0.0209 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 = 2000eV*1.60x10^-19J/ev = 3.2x10^-16J

so v = sqrt(2*3.2x10^-16/9.11x10^-31) = 2.65x10^7m/s

The period in the field T = 1/f = 2/ = 2*m/(q*B) = 2*9.11x10^-31/(1.60x10^-19*0.012) = 2.979x10^-9s

but since it only travels in a semi circle the time t1 = 2.979x10^-9s/2 = 1.48x10^-9s

Part two it will exit region 1 with a speed of 2.65x10^7m/s

it will undergo a constant acceleration of E*q/m = (V/d)*q/m = 2000*1.60x10^-19/(0.30*9.11x10^-31 = 1.17x10^15m/s^2

so the time to cross the gap comes from x = vx0*t + 1/2*a*t^2

or (1.17x10^15/2)*t^2 + 2.65x10^7*t - 0.30 = 0 or t2 = 9.38x10^-9s

Part three The period in the field T = 1/f = 2/ = 2*m/(q*B) = 2*9.11x10^-31/(1.60x10^-31*0.0209) = 1.71x10^-9s

but since it only travels in a semi circle the time t3 = 1.71x10^-9s/2 = 8.55x10^-10s

Therefore total time = 1.17x10^-9 + 9.38x10^-9 + 8.55x10^-10 = 1.91x10^-8s

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