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× 10- 8Ω m. (Density of free electron in gold wire is 5 . 90 × 1028 m- 3 .

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× 10- 8Ω m. (Density of free electron in gold wire is 5 . 90 × 1028 m- 3 . (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

t=(l/vd)=(l/I)× neA=(l/v)× ρ (l/A)× neA t=(ρ ℓ 2/v)⋅ ne =0.86=860milliseconds

t=vd​l​=Il​×neA=vl​×ρAl​×neA t=vρℓ2​⋅ne =0.86=860milliseconds

$t = \frac{l}{v _{d}} = \frac{l}{I} \times n e A = \frac{l}{v} \times ρ \frac{l}{A} \times n e A$
$t = \frac{ρ ℓ ^{2}}{v} \cdot n e$
$= 0.86 = 860 mi ll i seco n d s$