A Microscopic View of Ohm's Law — Question 6

Question 6

A sinusoidal field exposes the finite response time of drift

The current satisfies $\tau\,dJ/dt+J=\sigma_0E_0\cos\omega t$. Consider only its long-time periodic response, with $\omega>0$.

1. Use $J=A\cos\omega t+B\sin\omega t$ to solve for A and B.

2. Find the amplitude and phase lag, and the frequency at which the amplitude is reduced to $1/\sqrt2$ of its low-frequency value.

Question 6 - Solution

A sinusoidal field exposes the finite response time of drift

(a) Method and calculation

$$A+\omega\tau B=\sigma_0E_0,\quad B-\omega\tau A=0.$$$$A=\frac{\sigma_0E_0}{1+(\omega\tau)^2},\quad B=\frac{\sigma_0E_0\omega\tau}{1+(\omega\tau)^2}.$$

(b) Interpretation and further calculation

$$J=\frac{\sigma_0E_0}{\sqrt{1+(\omega\tau)^2}}\cos(\omega t-\phi),\quad\phi=\tan^{-1}(\omega\tau).$$$$\omega\tau=1\Rightarrow\omega=1/\tau,\quad\phi=\pi/4.$$

Physical check

The algebra is a supplied relaxation equation, so no AC circuit theory is required.