The current of electrons in a certain cathode-ray tube forms a thin cylindrical beam of constant radius \(a\) in which the charges move parallel to the axis. The current density in a cross-section is found not to be constant but to vary with the distance \(r\) from the axis of the beam according to \(J=J_{0}(1-r / a)\). What is the total current in the beam, assuming that it can be treated as the flow of a continuous distribution of charge? \(J_{0}\) is clearly the current density on the axis at \(r=0\). What is the mean current density in any cross-section in terms of \(J_{0}\) only?

Understanding how to integrate current density within a cylindrical beam is crucial for calculating the total current in devices such as cathode-ray tubes. In our problem, the current density varies as a function of the distance from the axis, described by the equation \( J=J_{0}(1-r / a) \).

To find the total current, think of current density as the amount of charge passing through a unit area perpendicular to the direction of flow per unit time. By integrating this current density across the entire cross-sectional area of the cylinder, we capture the total flow of charge.

First, we express the infinitesimal area in polar coordinates, which for our cylindrical geometry is \( dA=r{dr}d\theta \). Next, we setup the double integral using these coordinates. What we're doing is summing up the contribution of each infinitesimal area to the total current. The integral takes into account that the current density is not uniform, but changes with distance \( r \) from the center. Finally, after evaluating the integral, we obtain the total current. This process of integration over a complex area exemplifies how calculus can be applied to physical problems to yield significant results.

The calculation of electric current in a conducting medium is a fundamental concept in physics and electrical engineering. We can calculate the total current flowing through the beam by integrating the current density over the entire cross-section of the beam.

As demonstrated in the solution, this involves setting up a double integral because we're dealing with a two-dimensional cross-section. After integrating with respect to the radius (\( r \)) and then the angle (\( \theta \)), we get the total current \( I \).

It's important to understand that the electric current we find, \( I = \frac{2}{3}J_0a^2\theta \), represents the entire flow of charge per unit time through the cross-sectional area of the beam. This current is what drives the cathode-ray tube's functionality, as it's the motion of these charges that interact with the electric and magnetic fields within the tube to produce the desired properties, like an image on a screen in old television sets.

Mean current density is an averaged value that gives us a simplified view of the current's distribution over a given area. After finding the total current, we divide this by the cross-sectional area of the beam to find the mean current density, which provides a measure of how much current flows through each unit area, on average. This is symbolized by \( \bar{J} = \frac{I}{A} \).

For our cylindrical beam, the step-by-step solution shows that dividing the total current by the cross-sectional area yields \( \bar{J} = \frac{2}{3}J_0 \). This mean value is less than the maximum current density at the center of the beam (\( J_0 \)) because the current density decreases as one moves away from the center.

Understanding the mean current density is crucial as it provides a way to describe the overall behavior of the electric current in materials or components where the current density is not uniform. It simplifies the analysis and design of electrical systems, such as determining the power dissipation across a cross-section of a conductor.