Ohm's Law — Question 8

Question 8

An anisotropic ohmic medium can carry current away from the field direction

A planar medium obeys $J_x=2\sigma E_x$ and $J_y=\sigma E_y$. The imposed field is $\vec E=E_0(\hat x+\hat y)/\sqrt2$, with positive sigma and E0. Ignore boundaries.

1. Find current magnitude and its angle to x.

2. Find the component of J perpendicular to E, using $\hat t=(-\hat x+\hat y)/\sqrt2$, and assess whether response is linear.

Question 8 - Solution

An anisotropic ohmic medium can carry current away from the field direction

(a) Method and calculation

$$\vec J=\frac{\sigma E_0}{\sqrt2}(2\hat x+\hat y),\quad |\vec J|=\sigma E_0\sqrt{5/2}.$$$$\theta_J=\tan^{-1}(1/2)=26.565^\circ.$$

(b) Interpretation and further calculation

$$J_\perp=\vec J\cdot\hat t=-\sigma E_0/2.$$$$\vec J(a\vec E_1+b\vec E_2)=a\vec J(\vec E_1)+b\vec J(\vec E_2).$$A constant conductivity tensor is linear even when J and E are not parallel.

Physical check

Scalar resistivity presumes isotropy; linearity itself is a broader property.