Question 26.70: A caterpillar of length 4.0 cm crawls in the direction of el......
A caterpillar of length 4.0 cm crawls in the direction of electron drift along a 5.2-mm-diameter bare copper wire that carries a uniform current of 12 A. (a) What is the potential difference between the two ends of the caterpillar? (b) Is its tail positive or negative relative to its head? (c) How much time does the caterpillar take to crawl 1.0 cm if it crawls at the drift speed of the electrons in the wire? (The number of charge carriers per unit volume is 8.49 × 10^{28}\,\mathrm{m}^{-3}.)
(a) The potential difference between the two ends of the caterpillar is
V=i R=i\rho\frac{L}{A}=\frac{\left(12\mathrm{~A}\right)\left(1.69\times10^{-8}\,\Omega\!\cdot\!\mathrm{m}\right)\left(4.0\times10^{-2}\,\mathrm{m}\right)}{\pi\left(5.2\times10^{-3}\,\mathrm{m}/2\right)^{2}}=3.8\times10^{-4}\,\mathrm{V}.
(b) Since it moves in the direction of the electron drift, which is against the direction of the current, its tail is negative compared to its head.
(c) The time of travel relates to the drift speed:
t={\frac{L}{\nu_{d}}}={\frac{L A n e}{i}}={\frac{\pi L d^{2}n e}{4i}}={\frac{\pi\left(1.0\times10^{-2}\,{\mathrm{m}}\right)\left(5.2\times10^{-3}\,{\mathrm{m}}\right)^{2}\left(8.47\times10^{28}\,{/{\mathrm{m}}^{3}}\right)\left(1.60\times10^{-19}\,{\mathrm{C}}\right)}{4(12\,{\mathrm{A}})}}
=238 \ s=3~\mathrm{min}~58\ s.