Question

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 rho( x ) = a + bx^2, where a and b are constants. At the left end, the resistivity is 2.25 10^-8 Omega*m, while at the right end it is 8.50 x 10^-8 Omega*m.

What is the resistance of this rod?

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

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

Answer 1

2.25 10^-8 = a+b(0)^2
2.25 10^-8 = a

8.50 x 10^-8 = a +b(1.5)^2
8.50 x 10^-8 =2.25 10^-8 +b(1.5)^2
6.25 x 10^-8 =b(1.5)^2
6.25 x 10^-8/(1.5)^2 = b

now integrate:
\[R =\int\limits_{0}^{1.5} \frac{ (a+bx^2) dx }{ \pi(.011)^2}\]

Answer 2

I got 1.64 x 10^-4 and I got the answer wrong, did I mess up somewhere?

Answer 3

I got 1.23 E-4

Answer 4

Even with that, I'm still not getting the answer right

Answer 5

whoops used wrong 'a'

Answer 6

so, a isn't equal to 2.25*10^-8?

Answer 7

1.71E-4

Answer 8

it is, but I used 8.5E-8 accidentally

Answer 9

ok, so for E, I did IR/L where I = 1.75. is that correct?

Answer 10

no, to find R integrate from l=0 to .75m

Answer 11

that's the total resistance for the first half of the rod.

Answer 12

so did you use the same a & b?

Answer 13

yes.

Answer 14

then find the voltage drop between l=0 and l =.75
use that to find the field..

Answer 15

I got 4.11*10^-5 for R?

Answer 16

ok, nevermind, I got 5.47E-5. So, what do I do with this R to find E?

Answer 17

V=current*R

Answer 18

then do I divide that by 0.75? or is it 1.5?