Resistance of a holed cylinder — Phénomènes de transport 1 : Transport de charge

We study an ohmic conductor whose shape is described below in steady state¹.

𝛾

represents the conductivity of the material, and

𝑉_{1}

and

𝑉_{2} < 𝑉_{1}

the electrical potentials inside and outside the tube respectively.

𝐼

is the total current and

\vec{𝑗}

the current density vector.

Coup de pouce 1

Le courant va des potentiels les plus élevés vers les potentiels les moins élevés.

Corrigé

The current flows from the highest potential to the lowest potential, so

\vec{𝑗}

is oriented radially from the inside to the outside of the tube:

\vec{𝑗} = 𝑗 (𝑟) {\vec{𝑒}}_{𝑟}

with

𝑗 (𝑟) \geq 0

2/ Why is the flux of

\vec{𝑗}

conservative ? By applying this to cylinders of any radius

𝑟

, deduce that

\vec{𝑗} = \frac{𝐶}{𝑟} {\vec{𝑒}}_{𝑟}

where

𝐶

is a constant that you will express as a function of

𝐼

and

𝑙

Coup de pouce 1

Laquelle des hypothèses de l’énoncé implique-t-elle que le régime est stationnaire ?

Coup de pouce 2

Exprimer le courant à travers un cylindre de rayon

𝑟

et de hauteur

𝑙

en fonction de

𝑗

𝑙

𝑟

Corrigé

The regime is stationary. Thus, by conservation of charge, the flux of $\vec{𝑗}$ is conservative. Meaning that the flux of $\vec{𝑗}$ through the inner cylinder of radius $𝑅_{1}$ is equal to the flux through any cylinder of radius $𝑟$ between $𝑅_{1}$ and $𝑅_{2}$ :

\iint_{𝑆_{1}} \vec{𝑗} \cdot \vec{d 𝑆} = \iint_{𝑆_{𝑟}} \vec{𝑗} \cdot \vec{d 𝑆} = \iint_{𝑆_{𝑟}} 𝑗 (𝑟) {\vec{𝑒}}_{𝑟} \cdot d 𝑆 {\vec{𝑒}}_{𝑟} = 𝑗 (𝑟) \iint_{𝑆_{𝑟}} d 𝑆 = 𝑗 (𝑟) 2 𝜋 𝑟 𝑙

The first term is independent of $𝑟$ (and is in fact equal to $𝐼$ ). Thus, we have :

𝑗 (𝑟) = \frac{𝐼}{2 𝜋 𝑟 𝑙} = \frac{𝐶}{𝑟}

with $𝐶 = \frac{𝐼}{2 𝜋 𝑙}$ .

3/ By integrating the previous expression between

𝑅_{1}

and

𝑅_{2}

and using Ohm’s law, determine the resistance of the tube.

Corrigé

By Ohm’s law, we have :

\vec{𝑗} = 𝛾 \vec{𝐸} = - 𝛾 \vec{grad} (𝑉)

Thus :

𝑗 (𝑟) = - 𝛾 \frac{d 𝑉}{d 𝑟}

We have previously established that $𝑗 (𝑟) = \frac{𝐶}{𝑟}$ . Thus :

\frac{d 𝑉}{d 𝑟} = - (\frac{𝐶}{𝛾 𝑟})

By integrating this expression between $𝑅_{1}$ and $𝑅_{2}$ , we obtain :

𝑉_{2} - 𝑉_{1} = - \frac{𝐶}{𝛾} ln (\frac{𝑅_{2}}{𝑅_{1}})

Finally, the resistance of the tube is :

𝑅 = \frac{𝑉_{1} - 𝑉_{2}}{𝐼} = \frac{𝐶}{𝛾 𝐼} ln (\frac{𝑅_{2}}{𝑅_{1}}) = \frac{1}{2 𝜋 𝑙 𝛾} ln (\frac{𝑅_{2}}{𝑅_{1}})