Question

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 = Jedge(r/R), where R is the radius of the beam and Jedge is the current density at the edge. I calculated that 9.375 x 10^15 protons are delivered by the proton beam each second, but I do not know where to start in calculating the value of Jedge.

Answer 1

Answer: 114.6 A/m²

Explanation:

The total current, I is ∫I(r).dA, where dA = 2πr.dr

Therefore, if we substitute the relation for dr, we have

I = ∫J(r).2πr.dr [r = 0 to R]

I = ∫J(edge).(r/R).2πr.dr, if we rearrange this, we would have something like this

I = J(edge)2π ∫(r²/R).dr, and on integration, we have

I = J(edge).2πr³/(3R) [r = 0 to R]

I = J(edge).2πR²/3, of we make J(edge) subject of formula by rearranging, we have

J(edge) = (3/2).I/(π.R²), now, we solve for J(edge), where

I = 0.0015 A, R = 2.5*10^-3 m

J(edge) = [(3/2) * 0.0015] / [3.142 * 2.5*10^-3)²]

J(edge) = 0.00225 / (3.142 * 6.25*10^-6)

J(edge) = 0.00225 / 1.96*10^-5

J(edge) = 114.6 A/m²

Answer 2

Answer:

Jedge = 0.076A/cm^2

Explanation:

If you want to know what is Jedge is convenient to use an integral. The total density current is given by:

$J_T=\int_{0}^{R}J_{edge}(\frac{r}{R})dr=J_{edge}(\frac{1}{R})(\frac{R^2}{2})=\frac{J_{edge}R}{2}$  (1)

But also, we have that the total current is

$I_T=J_TA\\\\J_T=\frac{I_T}{A}=\frac{1.5*10^{-3}A}{\pi(5*10^{-3})^2}=19.09\frac{A}{m^2}$

where we have used that 5mm=5*10^{-3}m.

By replacing (1) we obtain:

$J_{edge}=\frac{2J_T}{R}=\frac{2(19.09A/m^2)}{5*10^{-3}m}=7639.4\frac{A}{m^2}=0.076\frac{A}{cm^2}$

hope this helps!!