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Chapter 24: Problem 63

The current density in a particle beam with circular cross section of radius \(a\) points along the beam axis with a magnitude that decreases linearly from \(J_{0}\) at the center \((r=0)\) to half that value at the edge \((r=a) .\) Find an expression for the total current in the beam.

Short Answer

Expert verified
After performing the angular and radial integrals, the total current in the beam is found to be \(I = \frac{2}{3}\pi a^{2}J_{0}\), where \(a\) is the beam radius and \(J_{0}\) is the current density at the center of the beam.

Step by step solution

01

Understand the Current Density Distribution

The magnitude of the current density \(J\) is said to decrease linearly as we move from the center (where \(r = 0\) and \(J = J_{0}\)) to the edge (where \(r = a\) and \(J = 0.5*J_{0}\)). The rate of change for this linear decrease can be represented as: \(- \frac{J_{0}}{a}.\) Therefore the current density magnitude can be represented as a function of r: \(J(r) = J_{0} - \frac{J_{0}}{a}r .\) This formula indicates that \(J\) decreases as \(r\) increases, starting from \(J_{0}\) at the center (where \(r = 0\)) to \(0.5*J_{0}\) at the edge (where \(r = a\)).
02

Identify the Current Particle Element

The current flowing through a small area element \(dA\) of the beam cross section is \(dI = JdA\). In a polar coordinate system, \(dA = r dr d\theta\), where \(r\) is the radius, \(dr\) is a small increment of radius and \(d\theta\) is a small angle in radian measure.
03

Integrate to Find Total Current

To find the total current \(I\) in the beam, integrate the current element \(dI\) over the entire cross-sectional area. As this is a cylindrical system and the current distribution is symmetric about the beam axis, convert it to cylindrical coordinates and integrate over \(r\) and \(\theta\): \(I= \int_0^{2\pi}\int_0^{a} (J_{0} - \frac{J_{0}}{a}r) r dr d\theta = J_{0}\int_0^{2\pi}\int_0^{a}r(1 - \frac{r}{a}) dr d\theta\). Evaluating the radial and angular integrals yields the total current.

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