A common flashlight connects the positive end of two 1.5[V] dry cells (in series

Question

A common flashlight connects the positive end of two 1.5[V] dry cells (in series) to its tungsten lamp filament and returns the current with 20-gauge copper wire (0.52[mm²] Area × 150[mm] long). Fresh batteries establish 2.47[V] across the filament, which conducts 0.3 Amperes of electrons.
Copper has (8920[kg/m³])*(1[mol]/0.06355[kg])*(6.022E23[e/mol])*(1.6E-19[C/e]) mobile charge density.

use ?q v · A to compute the (average) drift velocity for the conduction electrons in this copper wire.

{follow-up: how how does this compare with their average thermal speed v(kT/m)?}

Explanation / Answer

9860 = (0.52[mm²]/160[µm²]) ÷ (4.45[C/µm³]/13.5[C/µm³]) all of the electrons moving through the thicker wire must shoot through the thinner filament. The speed is going to be the inverse ratio of the cross sectional areas times the inverse ratio of the mobile electron density. There are more 3x mobile electrons per unit volume in the copper, so the electrons need to go 3x faster in the tungsten to prevent a bottleneck. The cross sectional area of the filament is 0.00016 mm^2/0.52mm^2 as large as that of the copper wire, so the electrons in tungsten need to go .52/.00016 faster to prevent a bottleneck. (.52/.00016)*3 = 9750 This is choice l. The correct answer is a little different since the electron density isn't exactly three. Data from the table 17.1 mentioned gives the correct ratio 13.5/4.45 The length of the wire and filament aren't part of this problem. We only care about the relative drift velocities.

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