Ohm's Law — Question 7

Section 26-4 · Physics pages 4224–4225

Question 7 A nonuniform ohmic bar remains globally linear when its profile stays fixed A bar has constant area A and fixed conductivity profile σ(x) = σ0ex/L on 0 ≤x ≤L. Its sidewalls are insulating and local transport is J = σE. x σ Axes and labels specify the model; the sketch is schematic. (a) Derive the end-to-end I-V relation. (b) Find the ratio of endpoint fields and the limiting result if the conductivity instead becomes uniform at sigma0.
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Question 7 - Solution A nonuniform ohmic bar remains globally linear when its profile stays fixed x σ (a) Method and calculation J = I/A, V = ∫L 0 I Aσ0 e−x/Ldx. I = Aσ0 L(1 −e−1)V. (b) Interpretation and further calculation E(L)/E(0) = e−1. σ(x) = σ0 ⇒I = (Aσ0/L)V. Physical check Spatial nonuniformity changes the conductance but does not alone destroy proportionality.

Question 7

A nonuniform ohmic bar remains globally linear when its profile stays fixed

A bar has constant area A and fixed conductivity profile σ(x)=σ0ex/L\sigma(x)=\sigma_0e^{x/L} on 0≤x≤L0\le x\le L. Its sidewalls are insulating and local transport is J=σEJ=\sigma E.

Diagram shown in the original SVG worksheet.

  1. Derive the end-to-end I-V relation.

  2. Find the ratio of endpoint fields and the limiting result if the conductivity instead becomes uniform at sigma0.

Question 7 - Solution

A nonuniform ohmic bar remains globally linear when its profile stays fixed

Diagram shown in the original SVG worksheet.

(a) Method and calculation

J=I/A,V=∫0LIAσ0e−x/Ldx.J=I/A,\quad V=\int_0^L\frac{I}{A\sigma_0}e^{-x/L}dx.I=Aσ0L(1−e−1)V.I=\frac{A\sigma_0}{L(1-e^{-1})}V.

(b) Interpretation and further calculation

E(L)/E(0)=e−1.E(L)/E(0)=e^{-1}.σ(x)=σ0⇒I=(Aσ0/L)V.\sigma(x)=\sigma_0\Rightarrow I=(A\sigma_0/L)V.

Physical check

Spatial nonuniformity changes the conductance but does not alone destroy proportionality.