Question

A proton beam that carries a total current of 1.3 mA has 10.0 mm diameter. The current density in the proton beam increases linearly with distance from the center. This is expressed mathematically as J(r) = J0 (r/R), where R is the radius of the beam and J0 is the current density at the edge. Determine the value of J0.

Answer 1

The current density of the proton beam at the edge is 8.275 A/m².

The given parameters;

• total current carried by the beam, I = 1.3 mA
• diameter of the beam, d = 10 mm

The radius of the beam is calculated as follows;

$R = \frac{d}{2} \\\\R = \frac{10 \ mm}{2} \\\\R = 5 \ mm$ = 5 x 10⁻³ m

Current density is calculated as follows;

$J = \frac{I}{A}$

where;

A is the area of the beam

Area of the beam is calculated as follows;

A = πr²

A = π x (5 x 10⁻³)²

A = 7.855 x 10⁻⁵ m²

The current density is calculated as follows;

$J = \frac{I}{A} \\\\J = \frac{1.3 \times 10^{-3} }{7.855\times 10^{-5}} \\\\J = 16.55 \ A/m^2$

Since the current density increases with distance;

$J_0 = J \times \frac{R}{r} \\\\J_0 = 16.55 \times \frac{5 \times 10^{-3} }{10 \times 10^{-3}} \\\\J_0 = 8.275 \ A/m^2$

Thus, the current density of the proton beam at the edge is 8.275 A/m².

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