# Ohm's second law: step by step to obtain this law [duplicate]

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. Every times that I ask a question there is always a problem (or downvote, or duplicate, or unclear, or bad English language).

• What is "Ohm's first law"? – hyportnex Feb 3 at 16:55
• 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 3 at 17:01
• @Sebastiano, you can't prove a definition. – The Photon Feb 3 at 21:38
• 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 4 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 4 at 13:58

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

$$E=\rho J,$$

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

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

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).