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
The current density of the proton beam that carries the given current is 114.58 A/m².
The given parameters;
The total current (I) flowing in the circuit is determined by integrating the current density (J) and the area of the beam (A), as follows;
[tex]I = \int\limits^R_0 J {(r)} \, dA\\\\A = 2\pi r\\\\I = \int\limits^R_0 J {(\frac{r}{R} )} .2\pi r\ dr\\\\I = \frac{J}{R} \int\limits^R_0 2\pi r^2 \ dr\\\\I = 2\pi \frac{J}{R} \int\limits^R_0 r^2 \ dr\\\\I = 2\pi \frac{J}{R}[\frac{r^3}{3} ]^R_0\\\\I = 2\pi \frac{J}{R} (\frac{R^3}{3} )\\\\I = \frac{2}{3} \pi R^2 J\\\\J = \frac{3I}{2\pi R^2} \\\\J = \frac{ 3\times (0.0015)}{2\pi \times (0.0025)^2} \\\\J = 114.58 \ A/m^2[/tex]
Thus, the current density of the proton beam that carries the given current is 114.58 A/m².
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