An accelerator produces a beam of protons with a circular cross section that is

Question

An accelerator produces a beam of protons with a circular cross section that is 2.3 mm in diameter and has a current of 0.8 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. 1) What is the number density of the protons in the beam? 31 Submit You currently have 0 submissions for this question. Only 5 submission are allowed. You can make 5 more submissions for this question. 2) How many protons strike the target in each minute? Submit You currently have 0 submissions for this question. Only 5 submission are allowed. You can make 5 more submissions for this question. 3) What is the magnitude of the current density J in this beam? /m2 Submit

Explanation / Answer

Area of beam produced

A = pi*d^2 / 4

A = 3.14*(2.3*10^(-3))^2 / 4 = 4.15*10^(-6) m^2

Kinetic energy of each proton,

E = 20 MeV = 3.20*10^(-12)

E = (1/2)mv^2

3.20*10^(-12) = (1/2)*1.67*10^(-27)*v^2

so, v = 6.19*10^7 m/s

Currrent l = neAv

number density of proton,

n = l / eAv

n = 0.8*10^(-3) / (1.6*10^(-19)* 4.15*10^(-6)*6.19*10^7)

n = 1.94*10^13

(b)

current l = Q / t = Ne / t

so, N = 0.8*10^(-3)*60 / 1.6*10^(-19)

protons strike each minute, N = 3*10^17

(c)

current density, J = l / A

J = 0.8 / 4.15*10^(-6)

J = 192.65 kA/m^2

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