Potentials between regions. An electron with an initial kinetic energy of 4.0 ke
ID: 1389796 • Letter: P
Question
Potentials between regions.
An electron with an initial kinetic energy of 4.0 keV enters region 1 at time t = 0. That region contains a uniform B field as shown on 0.0 10 T. The electron goes through the half circle and then exits headed towards region 2 across a gap of 25 cm. There is an electric field with a potential difference of 2000V across the gap with a polarity such that the electron? s speed increases uniformly as it moves through the gap. Region 2 has another uniform B field as shown of magnitude 0.020T. Again the electron goes around a semi circle and exits. What is the time when it leaves region 2?Explanation / Answer
The speed of the particle entering region1 comes form its kinetic energy
the initial kinetic energy,K = 1/2*m*v^2 = 2000eV*1.60x10^-19J/ev = 3.2x10^-16J
so v = sqrt(2*3.2x10^-16 J/9.11x10^-31 kg) = 2.65x10^7 m/s
The period in the field T = 1/f = 2?/? = 2?*m/(q*B) (w=qB/m)
= 2*3.14*9.11x10^-31 kg/(1.60x10^-19 C*0.010 T) = 3.575x10^-9 s=3.575 ns
but since it only travels in a semi circle the time t1 = 3.575 ns/2 = 1.787 ns
Part two it will exit region 1 with a speed of 2.65x10^7 m/s
it will undergo a constant acceleration of F=m*a=>q*E=m*a
rearrange the term, a=E*q/m
= (?V/d)*q/m
=( 2000 V*1.60x10^-19 C)/(0.25 m*9.11x10^-31 kg) = 1.405x10^15 m/s^2
so the time to cross the gap comes from x = v*t + 1/2*a*t^2
re arrange the term,
(1.405x10^15 m/s^2/2)*t^2 + 2.65x10^7 m/s*t - 0.25 m = 0 => t2 = 5.38x10^-8s
Part three The period in the field T = 1/f = 2?/? = 2?*m/(q*B)
= 2*3.14*9.11x10^-31 kg/(1.60x10^-19 C*0.020 T) = 1.79x10^-9s
but since it only travels in a semi circle the time t3 = 1.79x10^-9 s/2 = 8.94x10^-10s
Therefore total time = 1.79x10^-9 + 5.38x10^-8 + 8.94x10^-10 = 5.6484x10^-8 s
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