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Resistivity = (Resistance times area ) divided by length

$$\rho = (RA)/l$$

now my question is that in order to calculate the resistivity for some metal we have to take into account its length and area then why is it said that the resistivity of a metal is independent of its area or length . i.e. Why does every piece of copper have the same resistivity regardless of its area or length?

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marked as duplicate by sammy gerbil, John Rennie, Qmechanic Mar 1 '18 at 18:20

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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One way to think about it is as an empirical law. If you have a bunch of copper wires of varying lengths $l$, all with the same cross-sectional area, and you measure their resistances, you will find that the resistance is directly proportional to the length: $$ R \propto l \qquad \text{(constant $A$)}. $$ Similarly, if you have a bunch of wires with different cross-sectional areas $A$, all with the same length, you will find that their resistance is inversely proportional to their area: $$ R \propto \frac{1}{A} \qquad \text{(constant $l$)}. $$ Putting these two equations together, we have that $R \propto l/A$, or $$ R = \rho \frac{l}{A}. $$ for some proportionality constant $\rho$. This number is what we define to be the resistivity of copper.

If you turn it around, and calculate $\rho = RA/l$, you're right that you'll get different numbers for $R$, $A$, and $l$ depending on what wire you're talking about. But because of the above proportionalities, these differences will all cancel each other out. For example, if you had two copper wires with the same $A$, but one was twice as long as the other, then the resistance $R$ of the longer wire would be twice as high as well. The factors of two would cancel out, and you'd get the same value of $\rho$.

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Because $R$ also depends on the dimensions of wire. Recall that: $$ R=\dfrac{m_e l} {n e^2 A \tau} $$ So, the resistivity is $$\begin{align}\rho &=R\left(\dfrac {A}{l}\right)\\&=\left(\dfrac{m_e l}{n e^2 A \tau}\right )\left(\dfrac Al\right)\\&= \dfrac{m_e}{n e^2 \tau}\end{align}$$ This is independent of $A$ and $l$.

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Another way to think about resistivity, as sammy gerbil alluded, is that it is analogous to density. Density is a property of a material, but mass is a property of a particular specimen of that material. Similarly, resistivity is a property of a material, but resistance is a property of a particular specimen of that material. Think about that for a minute. To calculate density, we normally measure the mass of a specimen, then calculate the density by dividing mass by volume. So, we need the dimensions of that specimen (because we need the volume) to determine the density. Same with resistivity; we need the dimensions of the particular specimen to relate resistance to resistivity. This is one of those distinctions that you will understand suddenly, not gradually. When you think about it the right way, at some point you will say "Ah, of course!"

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