Change in Resistance as a Wire is Stretched

[cep]

Homework Statement

A copper wire that is 1 m long and has a radius of 0.5 mm is stretched to a length of 2m. What is the fractional change in resistance, [delta]R/R, as the wire is stretched?

Homework Equations

R=[rho]*L/A, V=[pi]*r^2*L

The Attempt at a Solution

I have a sense that this is a stupid and obvious question, but I'm having a hard time relating the radius and the length as the wire is being stretched. I found the total volume to be 7.85x10^-7 m^3, and the final radius to be 3.54x10^-4 m. I am just not really sure how to get the "fractional" change in resistance from this information. Thanks for the help :)

 

[gneill]

Volume is a constant, so perhaps you should start by insinuating it into the expression for resistance. In this case V = A*L, so you might replace A in the resistance formula with V/L.

Next you'll have to look at how you might find a ∆R.

 

[cep]

Ahhh okay, it's definitely easy :). Thank you!

 

[cep]

So I expressed R as [rho]*L^2/V (eq 1).

Differentiating with respect to L, as V is a constant, gives dR = 2*[rho]*L/V.

Dividing this by the initial expression for R (eq 1). gives dR/R = 2/L.

Does this look right to you? I'm not sure if it's correct, as then the answer is 1 when L is equal to 2.

 

[gneill]

So I expressed R as [rho]*L^2/V (eq 1).

Differentiating with respect to L, as V is a constant, gives dR = 2*[rho]*L/V.

Dividing this by the initial expression for R (eq 1). gives dR/R = 2/L.

Does this look right to you? I'm not sure if it's correct, as then the answer is 1 when L is equal to 2.

Wouldn't that be

\frac{dR}{dL} = 2*\frac{\rho}{V} L

so that

\frac{dR}{R} = \frac{2}{L} dl

or

\frac{\Delta R}{R} = 2 \frac{\Delta L}{L}

 

[tedbradly]

Let us say that 1 stands for values prior to stretching, and that 2 stands for values after stretching. Then, the end goal becomes:
\frac{\Delta R}{R_1} = \frac{R_2 - R_1}{R_1}
And we can then describe the journey toward this goal:
R_1 = \rho \frac{L_1}{A_1} \to \rho = \frac{R_1 A_1}{L_1}
V_1 = V_2 = L_1 A_1
\therefore A_2 = \frac{L_1 A_1}{L_2}
R_2 = \rho \frac{L_2}{A_2} = \frac{R_1 A_1}{L_1} L_2 \frac{L_2}{L_1 A_1}
And finally
\frac{\Delta R}{R_1} = \frac{R_2 - R_1}{R_1} = \frac{\frac{R_1 A_1}{L_1} L_2 \frac{L_2}{L_1 A_1} - R_1}{R_1}=\frac{R_1\frac{L_2^2}{L_1^2}-R_1}{R_1}= \left ( \frac{L_2}{L_1}\right )^2-1

Looking back on it, a calculus solution is probably required since it says, "as the wire is stretched."

 

[gneill]

Heh. Nice.

Expanding L₂ as L₁ + ∆L,

\left( \frac{L_2}{L_1} \right) - 1 = \frac{2 L_1 \Delta L + \Delta L^2}{L_1^2}

And letting L₁ = L,

= 2\frac{\Delta L}{L} + \frac{\Delta L^2}{L^2}

So the differential route provides a first order approximation.

 

[Dickfore]

A wire under stress does not have a constant volume. See http://en.wikipedia.org/wiki/Poisson%27s_ratio" .

Also, I don't know if the resistivity of a metal, a quantity that is essentially dependent on the scattering rate and the effective mass of the conduction electrons, remains the same under stress/strain. This is because these two quantities depend on the phonon spectrum and the band dispersion of the metal, and both of these depend on the inter atomic distance, a quantity which inevitably changes when the material is under strain.

 

[tedbradly]

A wire under stress does not have a constant volume. See http://en.wikipedia.org/wiki/Poisson%27s_ratio" .

Also, I don't know if the resistivity of a metal, a quantity that is essentially dependent on the scattering rate and the effective mass of the conduction electrons, remains the same under stress/strain. This is because these two quantities depend on the phonon spectrum and the band dispersion of the metal, and both of these depend on the inter atomic distance, a quantity which inevitably changes when the material is under strain.

Since none of the parameters needed for that equation are stated in the problem, why are you even talking about it?