The Electrical Properties of Solids — Question 9

Section 41-1 · Physics pages 6808–6809

Question 9 A cubic anisotropic sample can resist differently along two axes A homogeneous solid obeys Jx = σxEx, Jy = σyEy, with σx = 4σ0 and σy = σ0. Apply Ex = Ey = E0. x y E J Schematic; labels, not drawing lengths, specify the data. (a) Find the angle of the current density relative to x. (b) For a cube contacted on opposite x or y faces, find Ry/Rx.
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Question 9 - Solution A cubic anisotropic sample can resist differently along two axes x y E J (a) Method and calculation J = σ0E0(4ˆx + ˆy). tan θJ = Jy/Jx = 1/4, θJ = 14.04◦. (b) Interpretation and further calculation Ri = L/(σiL2) = 1/(σiL). Ry/Rx = σx/σy = 4 . Physical check The applied field points at 45 degrees, while current does not. A scalar conductivity is insufficient for a general anisotropic crystal.

Question 9

A cubic anisotropic sample can resist differently along two axes

A homogeneous solid obeys Jx=σxExJ_x=\sigma_xE_x, Jy=σyEyJ_y=\sigma_yE_y, with σx=4σ0\sigma_x=4\sigma_0 and σy=σ0\sigma_y=\sigma_0. Apply Ex=Ey=E0E_x=E_y=E_0.

Diagram shown in the original SVG worksheet.

  1. Find the angle of the current density relative to x.

  2. For a cube contacted on opposite x or y faces, find Ry/RxR_y/R_x.

Question 9 - Solution

A cubic anisotropic sample can resist differently along two axes

Diagram shown in the original SVG worksheet.

(a) Method and calculation

𝑱=σ0E0(4x̂+ŷ).\boldsymbol J=\sigma_0E_0(4\hat x+\hat y).tan⁡θJ=Jy/Jx=1/4,θJ=14.04∘.\boxed{\tan\theta_J=J_y/J_x=1/4,\quad\theta_J=14.04^\circ}.

(b) Interpretation and further calculation

Ri=L/(σiL2)=1/(σiL).R_i=L/(\sigma_iL^2)=1/(\sigma_iL).Ry/Rx=σx/σy=4.\boxed{R_y/R_x=\sigma_x/\sigma_y=4}.

Physical check

The applied field points at 45 degrees, while current does not. A scalar conductivity is insufficient for a general anisotropic crystal.