Question

Consider a wire of length 4m and cross-sectional area 1mm2 carrying a current of 2A. If each cubic meter of the material contains 1029 free electrons, calculate the average time taken by an electron to cross the length of the wire.

Answer 1

first of all, find drift velocity.

$i=neAv_d$

where, i is current, n is number of electrons per m³ , A is cross sectional area of wire and vd is drift velocity.

so, vd = i/neA

here, i = 2A, n = 10^29 electrons per m³, e = 1.6 × 10^-19C and A = 1mm² = 10^-6 m²

so, vd = 2/(10^29 × 1.6 × 10^-19 × 10^-6)

= 2/(1.6 × 10⁴)

= 1/8000

time taken to cross the length of wire, t = l/vd

where, l is length of wire i.e., l = 4m

so, t = 4/(1/8000) = 32000 sec = 3.2 × 10⁴ sec

hence, 3.2 × 10⁴ sec time taken by an electron to cross the length of wire.

Answer 2

Answer:

l=4ml=4m

A=1mm2=10−6m2A=1mm2=10−6m2

i=2Ai=2A

n=1029n=1029

τ=?τ=?

vd=ineA=21029×1.6×10−19×10−6vd=ineA=21029×1.6×10−19×10−6

=1.25×10−4 m/s=1.25×10−4 m/s

vd=lτvd=lτ

τ=lvd=41.25×10−7τ=lvd=41.25×10−7

τ=3.2×104sec