The diagram at right shows the path of an electron (mass m = 9.11 times 10^-31 k
ID: 1455368 • Letter: T
Question
The diagram at right shows the path of an electron (mass m = 9.11 times 10^-31 kg, charge q = 1.60 times 10^-19 C) in a region with several electric and magnetic fields. The electric potentials at points A, B, C, and D are V_A = 0 V, V_B = V_c = 1 kV, and V_D = V_E = 3 kV, respectively. The electron is released from rest at point A. The magnetic field out of the page has magnitude of 0.02 T, and the field into the page has a magnitude of 0.05 T. 25) What is the speed of the electron at the instant it enters the magnetic field at point B? 26) What force acts on the electron at the instant it enters the magnetic field at point B? 27) What is the distance from point B to point C? 28) What is the speed of the electron at the instant it enters the magnetic field at point D? What force acts on the electron at the instant it enters the magnetic field at point D? What is the distance between point C and point E? Between point B and point E? 31) If the electron enters the magnetic field at point B at time t = 0, t what time t does it leave the magnetic field at point C? 32) Repeat question #31 assuming that V_B = V_C = 2kV.Explanation / Answer
25.) The final kinetic energy is equal to the change in electrical potential energy which is:
KE = PE = qV = (1.6x10^-19 C)(1.0x10^3 V) = 1.6x10^-16 J
Use the formula for kinetic energy to find the speed:
KE = (1/2)mv^2
v = sqrt[2*KE/m] = sqrt[2(1.6x10^-16 J)/(9.11x10^-31 kg)] = 1.87x10^7 m/s
26.)Force = QvB
F = 1.6*10^-19*1.87*10^7*0.02 = 1.5*10^-13 N
27.) Mv^2 / R = F
R = 9.1*10^-31*(1.87*10^7)^2 / (1.5*10^-13)
R = 2.13* 10^-3 m
distance b/w B and C = 2R = 4.26*110^-3 m = 4.26 mm
28.) potential difference = 3Kv - 1Kv = 2Kv
The final kinetic energy is equal to the change in electrical potential energy which is:
KE = PE = initial K.E + qV = 1.6*10^-16 + (1.6x10^-19 C)(2.0x10^3 V) = 4.8x10^-16 J
Use the formula for kinetic energy to find the speed:
KE = (1/2)mv^2
v = sqrt[2*KE/m] = sqrt[2(4.8x10^-16 J)/(9.11x10^-31 kg)] = 3.24x10^7 m/s
29.)Force = QvB
F = 1.6*10^-19*3.24*10^7*0.05 = 2.59*10^-13 N