1.00 × 1020 electrons flow through a cross section of a 4.50-mm-diameter iron wire in 6.00 s .
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PART A:
What is the electron drift speed?
SOLUTION:
Some basics that I learned (by reading! :D):
using equations (30.1) & (30.2)...
Looking back at the initial givens, let's go ahead and do my signature first step...
Diameter ≡ d = 4.50 mm = 0.0045 m
Time passed ≡ Δt = 6.00 s
We know the cross-sectional area ≡ A = πr2 = π/4⋅d2
so, we can simplify eq. (A) into almost all givens/ knowns ...
but what is this number density thing... well it's actually a constant determined by the type of metal that the electrons are flowing through! It's given in table 30.1 in the book but I also included it below, for easy reference.
We can cause a net motion of electrons through a metal by pushing on them with an electric field. This net motion is called the drift speed, symbolized by vd.
The number of electrons (Ne) that pass through a cross section during a time interval (Δt) is defined in eq. (30.1) → Ne = ie⋅Δt
The electrons that travel a distance Δx during the time interval, form a cylinder of charge with volume V = A Δx
if the 'number density' (ne) is an element describing the electrons per volume, or cubic meter, then the total number of electrons in the cylinder is given in eq. (30.2) → Ne = ne⋅V = ne⋅A⋅Δx = ne⋅A⋅vd⋅Δt
The number of electrons (Ne) that pass through a cross section during a time interval (Δt) is defined in eq. (30.1) → Ne = ie⋅Δt
where ie is the electron current
The electrons that travel a distance Δx during the time interval, form a cylinder of charge with volume V = A Δx
where A is the cross-sectional area of the cylinder
if the 'number density' (ne) is an element describing the electrons per volume, or cubic meter, then the total number of electrons in the cylinder is given in eq. (30.2) → Ne = ne⋅V = ne⋅A⋅Δx = ne⋅A⋅vd⋅Δt
using equations (30.1) & (30.2)...
Ne = ie⋅Δt = ne A⋅vd⋅Δt
→ Ne/Δt = ie = ne A⋅vd
→ Ne/Δt = ie = ne A⋅vd
→ Ne/(Δt⋅ne⋅A) = vd
lets call that eq. (A), for nowLooking back at the initial givens, let's go ahead and do my signature first step...
Givens/ conversions:
Number of electrons ≡ Ne = 1.00×1020Diameter ≡ d = 4.50 mm = 0.0045 m
Time passed ≡ Δt = 6.00 s
We know the cross-sectional area ≡ A = πr2 = π/4⋅d2
so, we can simplify eq. (A) into almost all givens/ knowns ...
vd = Ne/(Δt⋅ne⋅π/4⋅d2)
but what is this number density thing... well it's actually a constant determined by the type of metal that the electrons are flowing through! It's given in table 30.1 in the book but I also included it below, for easy reference.
The intro states that the wire's material is iron.
so, using table 30.1...
∴ vd = Ne/(Δt⋅ne⋅π/4⋅d2) = (1.00×1020)/((6.00 s)⋅(8.5×1028 m-3)⋅(π/4)⋅(0.0045 m)2)
They want units of μm/s, so multiply through by 106 μm/m
so, using table 30.1...
ne = 8.5×1028 m-3
∴ vd = Ne/(Δt⋅ne⋅π/4⋅d2) = (1.00×1020)/((6.00 s)⋅(8.5×1028 m-3)⋅(π/4)⋅(0.0045 m)2)
⇒vd =1.233×10-5 m/s
They want units of μm/s, so multiply through by 106 μm/m
⇒vd =12.33 μm/s
Awesome. Thank you so much
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