Ohm's second law: step by step to obtain this law

Question

Ohm's second law is valid only if the electric current is evenly distributed in the section of the conductor, i.e. in the case of direct electric current and its relation is the following as we all know:

$$R=\rho\,\frac{\ell}{S} \tag 1$$

where $R$ is the resistence knowing that the wire is characterized by a length $\ell$ and a cross-section of area $S$.

I ask you if exist a proof of the $(1)$ starting from Ohm's microscopic law:

$$\mathbf{J}=\sigma \mathbf{E}$$

Note: My question is not a duplicate or How can one derive Ohm's Law or affine questions.

It's useful in many cases to look at Ohm's [first, I assume] Law (V=IR) as a definition of resistance. Similarly, this law can be seen as a definition of resistivity ($\rho$). If you don't want to see this law as a definition, you'll first have to find a way to define $\rho$ that's independent of this law. — The Photon, Feb 03 '20 at 17:01
@hyportnex I'm asking only for the 2nd Ohm's law and not the first. — Sebastiano, Feb 03 '20 at 20:51
@ThePhoton Is there, hence, with the your words that "this law can be seen as a definition of resistivity with a proof? — Sebastiano, Feb 03 '20 at 20:52
@ThePhoton Ah, hence is it a definition? Any complete answer is appreciated and welcome. Otherwise I will ask for the question to be closed. — Sebastiano, Feb 03 '20 at 21:39
@Sebastiano, there's probably more to say about how Ohm developed his laws, but the modern interpretation of the law has evolved slightly since Ohm's time. — The Photon, Feb 03 '20 at 21:45
@JohnRennie Good morning and thank you. No, sorry. I'm interested to know if there's a proof to get the 1 and not the microscopic Ohm law $\mathbf{J}=\sigma\mathbf{E}$. If you want to close a question of mine for no reason, you're free to do so. — Sebastiano, Feb 04 '20 at 12:16
When you ask a proof, you should indicate the set of considerations that you set as the given facts. For example, you can derive that expression with the Drude model for metals. — Grego_gc, Feb 04 '20 at 13:55
@Sebastiano You obtain equation (1) by integrating an surface in a wire from the Ohm relation between electric field and current density — Grego_gc, Feb 04 '20 at 13:58
@Grego_gc I honestly don't even know what is "Drude model for metals". I have no such extensive knowledge. I will be happy to welcome a full response. Thank you for your comments. — Sebastiano, Feb 04 '20 at 20:29

score 1 · Accepted Answer · edited Jan 27 '22 at 21:44

The Ohm law give us a relation between current density and electric field.

$$\sigma E=J$$

or, if possible,

$$E=\rho J$$

with $\sigma=1/\rho$ as a definition.

If you apply a voltage across a wire, you will have $$V=\int_{r_{start}}^{r_{end}}E\cdot dr$$ The electric field is constant and you end with just the length of the wire times the electric field $V=E L$. Then, the current is defined as the current density crossing a surface. We just to that

$$I=\int J\cdot dA$$

and we expect that $J$ is constant across a section of a wire. So we integrate and we just end with $I=JA$ with $A$ the cross section of the wire.

We use the Ohm's Law to connect these two things

$$E=\rho J, \qquad \frac{V}{L}=\rho \frac{I}{A},$$

$$V=\frac{\rho\, L}{A}I,$$

and then you just want to give a name to this factor. We define the resistance $R$ as

$$R=\frac{\rho\, L}{A}.$$

So, what you call in the comments microscopic Ohm's law will, if you can do all the assumptions that I made, end in the relationship between resistivity (the property of a material) and resistance (a property of an object with certain geometry and made of a material).

$EA = I\rho$ eq.(i) . I have seen people wrongly answering that when area of cross section is increased the number of collison decreases ,which is not true. From eq. (i) we can say that number of electrons are increased and since $E$ is still same the current increases. Here the number of electrons in a cross scetion increased. — , Jan 08 '22 at 07:24

Ohm's second law: step by step to obtain this law

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