Answer
We can rank the three sections according to the magnitude of the current density:
$B \gt A \gt C$
Work Step by Step
In part (a), we ranked the three sections according to the magnitude of the electric field as follows:
$B \gt A \gt C$
We can write a general expression for the current density:
$J = \sigma E$
$\sigma$ is a characteristic of the copper wire which is the same for all sections.
We can see that there is a higher magnitude of current density when there is a higher magnitude of electric field.
We can rank the three sections according to the magnitude of the current density:
$B \gt A \gt C$
