play phryzby with physics

An Electron Drifting Three Metres

solved DriftVelocity

Q. The number density of free electrons in a copper conductor is 8.5 × 10²⁸ m⁻³n=8.5×1028m⁻³. A wire 3.0 ml=3.0m long has an area of cross-section of 2.0 × 10⁻⁶ m²A=2.0×10−6m² and is carrying a current of 3.0 AI=3.0A. How long does an electron take to drift from one end of the wire to its other end, in seconds?

RI = 3 Av_d = ?l = 3 m
Inputs
n
8.5e+28 m⁻³
l
3.00 m
A
2e-06 m²
I
3 A
Outputs
timet?

How it's solved

Solved with DriftVelocity

Given

  • n=8.5×1028m⁻³
  • l=3.0m
  • A=2.0×10−6m²
  • I=3.0A

Find t.

Working
  1. current from drift(wire)I=n⋅e⋅A⋅vd
  2. 3=8.5×1028⋅1.60217663×10−19⋅2×10−6⋅vd
  3. vd=1.10144278×10−4m/s
  4. drifting the length of the wire(wire)l=vd⋅t
  5. 3=1.10144278×10−4⋅t
  6. t=27237.00 s
  7. t≈2.7×104sto 2 significant figures
Conclusion

So many free electrons share the current that each drifts very slowly, taking hours to cross a few metres even though the current starts at once.

The web of concepts

given read from the words equation asked

The Ruby that solves it

The question and its solution are small Ruby files. The engine runs them.

question.rb
1require_relative "../physics"23question :an_electron_drifting_three_metres do4  kind :numerical5  describes "The number density of free electrons in a copper conductor is %{density}. A wire %{length} long has an " \6            "area of cross-section of %{area} and is carrying a current of %{current}. %{time} to drift from one " \7            "end of the wire to its other end, in seconds?"8  given density: "8.5e28 m-3"9  given length: "3.0 m"10  given area: "2.0e-6 m2"11  given current: "3.0 A"12  asks time: "How long does an electron take"13end
solution.rb
1require_relative "../questions/an_electron_drifting_three_metres"2require_relative "../science/current_electricity/drift_velocity"34solve :an_electron_drifting_three_metres, using: DriftVelocity do5  objects :wire67  take density: wire.free_electrons8  take length: wire.length_of_the_conductor9  take area: wire.cross_section10  take current: wire.current1112  time = wire.drift_time13  answer time:, in: "s", figures: 21415  conclusion "So many free electrons share the current that each drifts very slowly, taking hours to cross a " \16             "few metres even though the current starts at once."17end
playground.rb
1require_relative "../solutions/an_electron_drifting_three_metres"23play_with :an_electron_drifting_three_metres do4  title "An Electron Drifting Three Metres"5end
The laws it drew on
quantitywrittenunitswithinfrom
elementary chargeeC—DriftVelocity
avogadro constantNAmol⁻¹—DriftVelocity
currentIA—DriftVelocity
cross sectionAm²—DriftVelocity
densityρkg/m³0 … ∞DriftVelocity
molar massMkg/mol—DriftVelocity
free electronsnm⁻³—DriftVelocity
drift speedvdm/s—DriftVelocity
length of the conductorlm—DriftVelocity
drift timets—DriftVelocity
free electrons of groundngm⁻³—
density of groundρgkg/m³0 … ∞
molar mass of groundMgkg/mol—
free electrons of wirenm⁻³—
density of wireρkg/m³0 … ∞
molar mass of wireMkg/mol—
current of groundIgA—
cross section of groundAgm²—
drift speed of groundvd,gm/s—
current of wireIA—
cross section of wireAm²—
drift speed of wirevdm/s—
length of the conductor of groundlgm—
drift time of groundtgs—
length of the conductor of wirelm—
drift time of wirets—
nw=ρw⋅NAMw
one free electron an atom
Iw=nw⋅e⋅Aw⋅vd,w
current from drift
lw=vd,w⋅tw
drifting the length of the wire
ng=ρg⋅NAMg
one free electron an atom(ground)
n=ρ⋅NAM
one free electron an atom(wire)
Ig=ng⋅e⋅Ag⋅vd,g
current from drift(ground)
I=n⋅e⋅A⋅vd
current from drift(wire)
lg=vd,g⋅tg
drifting the length of the wire(ground)
l=vd⋅t
drifting the length of the wire(wire)