I need help in doing this problem An accelerator produces a beam of protons with

Question

I need help in doing this problem

An accelerator produces a beam of protons with a circular cross section that is 2.4 mm in diameter and has a current of 1.0 mA. The current density is uniformly distributed throughout the beam. The kinetic energy of each proton is 20 MeV. The beam strikes a metal target and is absorbed by the target. What is the number density of the protons in tire beam? How many protons strike the target in each minute? What Is the magnitude of the current density in this beam?

Explanation / Answer

a) kinetic energy is 0.5*m*v^2 = 20*10^6*1.6*10^-19 = 3.2*10^-12 J

m is the mass of the proton

0.5*1.67*10^-27*v^2 = 3.2*10^-12

v = 6.2*10^7 m/sec is the drift speed of proton

but actually drift speed is v = I/(n*e*A)

A is the area of cross section = pi*(d/2)^2 = 3.142*(2.4*10^-3/2)^2 = 4.52*10^-6 m^2

6.2*10^7 = (1*10^-3)/(n*1.6*10^-19*4.52*10^-6)

number density of protons is n = 2.23*10^13 protons/m^3

b) i = Q/t

Q = i*t = 1*10^-3*60 = 60*10^-3 C = 60*10^-3/(1.6*10^-19) = 3.75*10^17 protons per minute

C) current density is J = i/A = (1*10^-3)/(4.52*10^-6) = 221.3 A/m^2

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