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Protons delivered per second in a proton beam

  • Thread starter mjolnir80
  • Start date
1. Homework Statement
a 5.0 mm-diameter proton beam carries a total current of 1.5mA. the current density in the proton beam, which increases with distance from the center, is given by j=j(edge)(r/R), where R is the radius of the beam and j(edge) is the current density at the edge.
a) how many protons per second are delivered by this proton beam?
b)determine j(edge)


2. Homework Equations



3. The Attempt at a Solution
a) we know that J=I/A(area) and also that in this particular case J=[tex]\int[/tex](from 0 to R) J(edge)(r/R)dr which turns out to be J(edge)R/2
now we know that I=AJ(edge)R/2 and I=Q/t so we find Q and divide it by e to get the number of protons per second
im just not really sure if this is right or not cause the end result still has J(edge) in it which we dont know. can someone help me out here please.
 

Redbelly98

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3. The Attempt at a Solution
a) we know that J=I/A(area) and also that in this particular case J=[tex]\int[/tex](from 0 to R) J(edge)(r/R)dr which turns out to be J(edge)R/2
now we know that I=AJ(edge)R/2 and I=Q/t so we find Q and divide it by e to get the number of protons per second
im just not really sure if this is right or not cause the end result still has J(edge) in it which we dont know. can someone help me out here please.
(a)A good way to think about it is:
  • We know I, the charge per second
  • We know the charge per proton
From that we can figure out protons per second

(b)
Actually, the current is given by

I = Jedge (r/R) dA

I.e., it is an area-integral.
 

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