Homework Help: Finding the resistance of the halved rod

1. Oct 3, 2012

Northbysouth

1. The problem statement, all variables and given/known data
A 1.50 m cylinder of radius 1.10 cm is made of a complicated mixture of materials. Its resistivity depends on the distance x from the left end and obeys the formula ρ(x) = a + bx^2, where a and b are constants. At the left end, the resistivity is 2.25x10^-8 Ω m, while at the right end it is 8.50x10^-8 Ωm.

A) What is the resistance of this rod?

B) What is the electric field at its midpoint if it carries a 1.75 A current?

C) If we cut the rod into two 75.0 m halves, what is the resistance of each half?

2. Relevant equations

3. The attempt at a solution

I have solved parts A and B by doing the following:

A) R = ρL/A

First I found the constants a and b by solving for them, which I found to be:

a = 2.25x10^-8 Ωm
b = 2.78x10^-8 Ωm

A = πr^2
A = 3.8013x10-4 m2

Taking the integral of the Resistance equation gave me:

1/A*(ax + bx3/3)

I then took the integral from 0 to 1.50m, thereby giving me:

(1/3.801299x10-4)*3.38x10-8*1.50 + 2.78x10-8*1.53/3
= 1.71x10-4

This I know to be correct.

B)

∫E.dl = V
dV/dx = d/dx(IR)

1.75/A*(a + bx2)

Plugging in 0.75 for x gave me

E = 1.76x10^-4 V/m

This is also correct

C) This is where I'm stuck. I had thought that if I used the integral in part A to find the integral from 0 to 0.75 then that would be my resistance in the left hand side. Then I thought that I could subtract the left hand value from the value I found in part A to give me the right hand value, but this hasn't worked. My workings are as follow:

R = ρL/A

A = 3.8x10^-4 m2

Rleft hand1/A*(2.25x10^-8)(0.75) + (2.78x10^-8)(0.75^3/3)
= 4.43966x10^-5

Then Rright hand = 1.71x10^-4 - 4.43966x10^-5 = 1.266x10^-4

Unfortunately, this did not work and I' not sure what else to try.

2. Oct 3, 2012

Staff: Mentor

I'm not sure how you've ended up with the value you did for the left had side resistance; The numbers in the formula look okay, but the calculated result does not. Calculator finger problems?

3. Oct 3, 2012

Northbysouth

Yes, I've just gotten the right answer. The correct calculation is as follows:

I know that the Resistance of the entire (1.50m) rod is 1.71x10^-4

Thus I used the integral form part A to calculate the value of R from 0 to 0.75, which gave me 5.4677x10^-5. This is the Resistance of the left hand piece of rod. I then subtracted this value from the resistance of the entire rod, thereby giving me a resistance of 1.16323x10^-4 for the right hand rod.