# 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.