Question

A gold wire has a $0.10mm$ diameter cross section. Opposite ends of this wire are connected to the terminals of a $1.5V$ battery. If the length of the wire is $7.5cm$ how much time (in milliseconds), on average, is required for electrons leaving the negative terminal of the battery to reach the positive terminal? Assume the resistivity of gold is $2.44\times 10^{- 8}\Omega m.$ (Density of free electron in gold wire is $5 . 90 \times 10^{28} m^{- 3} .$

Answer 1

$t=\frac{l}{v_{d}}=\frac{l}{I}\times neA=\frac{l}{v}\times \rho \frac{l}{A}\times neA$
$t=\frac{\rho \ell ^{2}}{v}\cdot ne$
$=0.86=860milliseconds$