Past exam of the physics A course of the University of Cambridge / 2018 / Paper 2 / Section C / C11 / c

Interpret “3 mm wire” as a circular wire of diameter $3\mathrm{mm}$, and take the current density as uniform. One conduction Electron per atom givesUsing $I=neA|v_d|$, the drift velocity corresponding to a current magnitude of $13\mathrm{A}$ isIt points opposite to conventional electric current. A simple distance scale over the $1/50=0.020\mathrm{s}$ cycle is $\boxed{|v_d|/f=2.71\mu \mathrm{m}}$, but this treats the magnitude as constant. For sinusoidal mains, alternating-current drift excursion gives zero net displacement after a full cycle.

If $13\mathrm{A}$ is the usual root mean square current, $v_{\mathrm{pk}}=\sqrt{2}|v_d|$ and the displacement amplitude and total absolute drift distance per cycle areIf $13\mathrm{A}$ instead denotes peak current, these become $0.431\mu \mathrm{m}$ and $1.72\mu \mathrm{m}$. The paper does not state peak versus RMS; the characteristic drift is the boxed $1.35\times 10^{-4}\mathrm{m}{\mathrm{s}}^{-1}$ on whichever current convention is chosen. It also does not explicitly say diameter; taking $3\mathrm{mm}$ as the radius would reduce all these velocities and drift distances by four. None of these distances is the microscopic path travelled at roughly the Fermi velocity between scattering events; that path cannot be obtained from the supplied current alone.