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. I am assuming that integration will have to be used somewhere along the way. Thank you :)

Answer 1

radius of the ring \(\mathrm{R}=\frac{D}{2}=2.5 \mathrm{~mm}\)

Current density \(\mathrm{j}=\frac{I}{A}\)

Here \(\mathrm{A}=\) area of the circualr ring \(=\pi \mathrm{r}^{2}\) \(d A=2 \pi r d r\)

Current \(d I=J d A \Rightarrow J(2 \pi r d r)=J_{e d g}\left(\frac{r}{R}\right) 2 \pi r d r\)

\(I_{\text {tst }}=\frac{2 \pi J_{\text {edge }}}{R}\left(\frac{R^{3}}{3}\right)=\frac{2 \pi J_{\text {edge }} R^{2}}{3}\)

\(J_{\text {edge }}=\frac{3 I_{\text {tot }}}{2 \pi R^{2}}=\frac{3\left(1.5^{*} 10^{-3} \mathrm{~A}\right)}{2 \pi\left(2.5^{*} 10^{-3} \mathrm{~m}\right)^{2}}=115 \mathrm{~A} / \mathrm{m}^{2}\)