A hollow cylindrical resistor with inner radius r1 and outer radius r2, and length l, is made of a material whose resistivity is rho.

A.Find the resistance for current that flows radially outward. Divide the resistor into concentric cylindrical shells and integrate. Express your answer in terms of the variables r1,r2, l,rho, and appropriate constants.

B.Evaluate the resistance R for such a resistor made of carbon whose inner and outer radii are 1.0mm and 3.0mm and whose length is 4.5cm. rho=15*10^-5 omega*m

C.What is the resistance in part B for current flowing parallel to the axis?

I think I got part A and B. I need help on part C. How do I do it?

A. To find the resistance for current that flows radially outward in the hollow cylindrical resistor, we can divide the resistor into concentric cylindrical shells and integrate.

The resistance of a cylindrical shell can be calculated using the formula:

dR = (ρ * dV) / A

Where dR is the resistance of the cylindrical shell, ρ is the resistivity of the material, dV is the volume of the cylindrical shell, and A is the cross-sectional area of the cylindrical shell.

The volume of a cylindrical shell is given by:

dV = π * (r2^2 - r1^2) * l

The cross-sectional area of a cylindrical shell can be calculated using the formula:

A = 2 * π * r * l

Where r is the average radius of the cylindrical shell, which can be calculated as:

r = (r1 + r2) / 2

Substituting these equations into the resistance formula, we get:

dR = (ρ * π * (r2^2 - r1^2) * l) / (2 * π * r * l)

Simplifying further, we get:

dR = ρ * (r2 + r1) / (2 * r)

To find the total resistance R, we need to integrate this expression over the range of r1 to r2. So the expression for R becomes:

R = ∫ [ρ * (r2 + r1) / (2 * r)] dr, integrated from r1 to r2

B. To evaluate the resistance R for a carbon resistor with given values, we substitute the values into the expression we derived in part A.

In this case, r1 = 1.0 mm = 0.001 m, r2 = 3.0 mm = 0.003 m, l = 4.5 cm = 0.045 m, and ρ = 15 * 10^(-5) Ω*m.

The expression for R becomes:

R = ∫ [15 * 10^(-5) * (0.003 + 0.001) / (2 * r)] dr, integrated from 0.001 to 0.003

Evaluating this integral will give us the resistance R for the given carbon resistor.

C. To find the resistance for current flowing parallel to the axis, we can use the same approach as part A, but with a different cross-sectional area.

The cross-sectional area for current flowing parallel to the axis is given by:

A = π * (r2^2 - r1^2)

Substituting this into the resistance formula, we get:

dR = (ρ * dV) / (π * (r2^2 - r1^2))

To find the total resistance for current flowing parallel to the axis, we integrate this expression over the range of r1 to r2. So the expression for the resistance R becomes:

R = ∫ [(ρ * π * (r2^2 - r1^2) * l) / (π * (r2^2 - r1^2))] dr, integrated from r1 to r2

Simplifying further, we get:

R = ρ * l * ∫ dr, integrated from r1 to r2

This integral simplifies to:

R = ρ * l * (r2 - r1)

Substituting the given values for the carbon resistor, we can evaluate this expression to find the resistance for current flowing parallel to the axis.

A. To find the resistance for current that flows radially outward, we can divide the resistor into concentric cylindrical shells and integrate. The resistance of each infinitesimally small shell can be represented by dR = (ρdl)/(2πr).

We can integrate this expression over the length l and radii r1 to r2 to find the total resistance R.

∫dR = ∫((ρdl)/(2πr))

R = ∫((ρdl)/(2πr))

B. To evaluate the resistance R for a carbon resistor with inner radius r1 = 1.0mm and outer radius r2 = 3.0mm, and length l = 4.5cm, we substitute the values into the integral.

R = ∫((ρdl)/(2πr))

However, we need to convert the dimensions to meters to match the unit of resistivity ρ.

r1 = 1.0mm = 0.001m
r2 = 3.0mm = 0.003m
l = 4.5cm = 0.045m

R = ∫((ρdl)/(2πr)) from l=0 to l=0.045, and r=r1 to r=r2

ρ = 15*10^-5 Ω*m

Substituting the values:

R = ∫((15*10^-5)*(0.045)/(2πr)) from r=0.001 to r=0.003

C. To find the resistance for current flowing parallel to the axis, we need to consider the effective length of the resistor. In this case, the length l will be the effective length, and the radii r1 and r2 will be irrelevant.

So the resistance R for current flowing parallel to the axis is the same as in part B:

R = ∫((15*10^-5)*(0.045)/(2πr)) from r=0.001 to r=0.003