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Northbysouth
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Homework Statement
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?
Homework Equations
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.