Resistance of a Cylindrical Conductor with Variable Conductivity

Hi Rüzgar, let's solve this university-level electromagnetics problem together to find the resistance of a hollow cylinder with variable conductivity.

Let's list the geometric parameters given in the problem: the inner radius is a, the outer radius is b, and the length of the cylinder along the z-axis is capital L.

Two conducting plates are placed at z equals zero and z equals L, which means the current flows along the z direction, parallel to the axis of the cylinder.

Let's establish the relation between the electric field and the current density inside the conductor using Ohm's law in differential form.

Since the plates at z equals zero and z equals L are equipotential surfaces, a voltage difference V zero creates a uniform electric field in the negative z direction, or simply with a magnitude of V zero over L.

The conductivity is given as sigma of rho equals sigma zero times rho, which increases linearly with the radius.

Now, let's substitute the conductivity and electric field expressions into Ohm's law to get the current density vector.

To find the total current, we need to integrate this current density over the cross-sectional area of the hollow cylinder.

The differential area element perpendicular to the z-axis in cylindrical coordinates is rho times d rho times d phi, pointing in the z direction.

Substituting these terms, the dot product of the z unit vectors becomes one, and we get a double integral over phi from zero to two pi and rho from a to b.

Let's simplify the integrand by grouping the rho terms together, which gives us rho squared.