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The electron beam emerging from a certain high-energy electron accelerator has a circular cross section of radius 1.80 mm. (a) The beam current is 8.10 µA. Find the current density in the beam assuming it is uniform throughout. (b) The speed of the electrons is so close to the speed of light that their speed can be taken as 300 Mm/s with negligible error. Find the electron density in the beam. (c) Over what time interval does Avogadro's number of electrons emerge from the accelerator? Part 1 of 5 - Conceptualize For a small current in a metal wire, the current density can be several amps per square meter. In such a metal wire, the electron density is high and the electron speed is low. In the high-energy electron beam described in this problem, the speed is high and the electron density is very low. A long time will be required for Avogadro's number of electrons to emerge. Part 2 of 5 - Categorize We will use the usual equations for current density and current flow to find the time interval. Part 3 of 5 - Analyze (a) The current density is given by the following equation, where we know the radius of the circular cross- section of the beam. I J = A 8.10 x 10-6 A) 8.1 T|1.80 1.8 x 10-3 0.79 0.796 A/m2 Part 4 of 5 - Analyze (b) From the equation relating current density to drift speed va, we have J = nqvd' In the equation above, n is the density of charge carriers per volume and q is the charge on each carrier. In this case, |q| = e is the magnitude of the electron charge. Solving for the charge density, we have ev | A/m²) 1.60 x 10-19 x 108 m/s) |× 1010 m-3

The electron beam emerging from a certain high-energy electron accelerator has a circular cross section of radius 1.80 mm. (a) The beam current is 8.10 µA. Find the current density in the beam assuming it is uniform throughout. (b) The speed of the electrons is so close to the speed of light that their speed can be taken as 300 Mm/s with negligible error. Find the electron density in the beam. (c) Over what time interval does Avogadro's number of electrons emerge from the accelerator? Part 1 of 5 - Conceptualize For a small current in a metal wire, the current density can be several amps per square meter. In such a metal wire, the electron density is high and the electron speed is low. In the high-energy electron beam described in this problem, the speed is high and the electron density is very low. A long time will be required for Avogadro's number of electrons to emerge. Part 2 of 5 - Categorize We will use the usual equations for current density and current flow to find the time interval. Part 3 of 5 - Analyze (a) The current density is given by the following equation, where we know the radius of the circular cross- section of the beam. I J = A 8.10 x 10-6 A) 8.1 T|1.80 1.8 x 10-3 0.79 0.796 A/m2 Part 4 of 5 - Analyze (b) From the equation relating current density to drift speed va, we have J = nqvd' In the equation above, n is the density of charge carriers per volume and q is the charge on each carrier. In this case, |q| = e is the magnitude of the electron charge. Solving for the charge density, we have ev | A/m²) 1.60 x 10-19 x 108 m/s) |× 1010 m-3

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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The electron beam emerging from a certain high-energy electron accelerator has a circular cross section of radius 1.80 mm. (a) The beam current is 8.10 µA. Find the current density in the beam assuming it is uniform throughout. (b) The speed of the electrons is so close to the speed of light that their speed can be taken as 300 Mm/s with negligible error. Find the electron density in the beam. (c) Over what time interval does Avogadro's number of electrons emerge from the accelerator? Part 1 of 5 - Conceptualize For a small current in a metal wire, the current density can be several amps per square meter. In such a metal wire, the electron density is high and the electron speed is low. In the high-energy electron beam described in this problem, the speed is high and the electron density is very low. A long time will be required for Avogadro's number of electrons to emerge. Part 2 of 5 - Categorize We will use the usual equations for current density and current flow to find the time interval. Part 3 of 5 - Analyze (a) The current density is given by the following equation, where we know the radius of the circular cross- section of the beam. I J = A 8.10 × 10-6 8.1 m)³ T1.80 1.8 x 10-3 = 0.79 V 0.796 A/m2 Part 4 of 5 - Analyze (b) From the equation relating current density to drift speed va, we have J = ngvd' In the equation above, n is the density of charge carriers per volume and q is the charge on each carrier. In this case, Ig| = e is the magnitude of the electron charge. Solving for the charge density, we have n = ev d | A/m²) 1.60 x 10-19 |× 108 m/s) × 1010 m
Transcribed Image Text:The electron beam emerging from a certain high-energy electron accelerator has a circular cross section of radius 1.80 mm. (a) The beam current is 8.10 µA. Find the current density in the beam assuming it is uniform throughout. (b) The speed of the electrons is so close to the speed of light that their speed can be taken as 300 Mm/s with negligible error. Find the electron density in the beam. (c) Over what time interval does Avogadro's number of electrons emerge from the accelerator? Part 1 of 5 - Conceptualize For a small current in a metal wire, the current density can be several amps per square meter. In such a metal wire, the electron density is high and the electron speed is low. In the high-energy electron beam described in this problem, the speed is high and the electron density is very low. A long time will be required for Avogadro's number of electrons to emerge. Part 2 of 5 - Categorize We will use the usual equations for current density and current flow to find the time interval. Part 3 of 5 - Analyze (a) The current density is given by the following equation, where we know the radius of the circular cross- section of the beam. I J = A 8.10 × 10-6 8.1 m)³ T1.80 1.8 x 10-3 = 0.79 V 0.796 A/m2 Part 4 of 5 - Analyze (b) From the equation relating current density to drift speed va, we have J = ngvd' In the equation above, n is the density of charge carriers per volume and q is the charge on each carrier. In this case, Ig| = e is the magnitude of the electron charge. Solving for the charge density, we have n = ev d | A/m²) 1.60 x 10-19 |× 108 m/s) × 1010 m
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