This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Chapter 25 Electric Current and DirectCurrent Circuits 34 An accelerator produces a beam of protons with a circular cross section that is 2.0 mm in diameter and has a current of 1.0 mA. The current density is uniformly distributed through the beam. The kinetic energy of each proton is 20 MeV. The beam strikes a metal target and is absorbed by the target. ( a ) What is the number density of the protons in the beam? ( b ) How many protons strike the target each minute? ( c ) What is the magnitude of the current density in this beam? Picture the Problem We can use neAv I = to relate the number n of protons per unit volume in the beam to current I . We can find the speed of the particles in the beam from their kinetic energy. In Part ( b ) we can express the number of protons N striking the target per unit time as the product of the number of protons per unit volume n in the beam and the volume of the cylinder containing those protons that will strike the target in an elapsed time t and solve for N. Finally, we can use the definition of current to express the charge arriving at the target as a function of time. ( a ) Use the relation between current and drift velocity (Equation 253) to relate I and n : neAv I = eAv I n = The kinetic energy of the protons is given by: 2 p 2 1 v m K = p 2 m K v = Relate the crosssectional area A of the beam to its diameter D : 2 4 1 D A = Substitute for v and A and simplify to obtain: K m eD I m K eD I n 2 4 2 p 2 p 2 4 1 = = Substitute numerical values and evaluate n : ( 29 ( 29 ( 29 ( 29 ( 29 3 13 3 13 19 27 2 19 m 10 2 . 3 m 10 21 . 3 J/eV 10 602 . 1 MeV 20 2 kg 10 673 . 1 mm 2 C 10 602 . 1 mA . 1 4 = = = n ( b ) Express the number of protons N striking the target per unit time as the product of the number n of protons per unit volume in the beam and the volume of the cylinder containing those protons that will strike the target in an elapsed time t and solve for N : ( 29 vA n t N = t nvA N = Substitute for v and A to obtain: p 2 4 1 2 m K t n D N = Substitute numerical values and evaluate N : ( 29 ( 29 ( 29 ( 29 ( 29 17 27 19 3 13 2 4 1 10 7 . 3 kg 10 673 . 1 J/eV 10 602 . 1 MeV 20 2 min 1 m 10 21 . 3 mm 2 = = N ( c ) The magnitude of the current density in this beam is given by: ( 29 2 2 3 kA/m .32 m 10 . 1 mA .0 1 = = = A I J 31 [SSM] A 10gauge copper wire carries a current equal to 20 A. Assuming copper has one free electron per atom, calculate the drift speed of the free electrons in the wire. Picture the Problem We can relate the drift velocity of the electrons to the current density using A nev I d = . We can find the number density of charge carriers n using , M N n A = where is the mass density, N A Avogadros number, and M the molar mass. We can find the crosssectional area of 10gauge wire in Table 252....
View
Full
Document
This note was uploaded on 03/07/2011 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics, Current

Click to edit the document details