An Electron Drifting Three Metres

Q. The number density of free electrons in a copper conductor is 8.5 × 10²⁸ m⁻³. A wire 3.0 m long has an area of cross-section of 2.0 × 10⁻⁶ m² and is carrying a current of 3.0 A. How long does an electron take to drift from one end of the wire to its other end, in seconds? expects 2.7 × 10⁴ s

What the draft establishes

Fills in once a draft is checked.

The web of concepts

given read from the words equation asked

The equations it used

None yet.

Check an answer

Q. The number density of free electrons in a copper conductor is 8.5 × 10²⁸ m⁻³ $n = 8.5 \times 10^{28}$ m⁻³. A wire 3.0 m $l = 3.0$ m long has an area of cross-section of 2.0 × 10⁻⁶ m² $A = 2.0 \times 10^{−6}$ m² and is carrying a current of 3.0 A $I = 3.0$ A. How long does an electron take to drift from one end of the wire to its other end, in seconds?

The playground

Inputs

density

n

8.5e+28 m⁻³

length

l

3.00 m

area

A

2e-06 m²

current

I

3 A

Outputs

time

t

The solution

Solved with DriftVelocity

Given

$n = 8.5 \times 10^{28}$ m⁻³
$l = 3.0$ m
$A = 2.0 \times 10^{−6}$ m²
$I = 3.0$ A

Find $t$ .

Working

current from drift(wire) $I = n \cdot e \cdot A \cdot v_{d}$
$3 = 8.5 \times 10^{28} \cdot 1.60217663 \times 10^{−19} \cdot 2 \times 10^{−6} \cdot v_{d}$
$v_{d} = 1.10144278 \times 10^{−4}$ m/s
drifting the length of the wire(wire) $l = v_{d} \cdot t$
$3 = 1.10144278 \times 10^{−4} \cdot t$
$t = 27237.00 s$
$t \approx 2.7 \times 10^{4}$ sto 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 Ruby

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

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

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

quantity	written	units	within	from
elementary charge	$e$	C	—	DriftVelocity
avogadro constant	$N_{A}$	mol⁻¹	—	DriftVelocity
current	$I$	A	—	DriftVelocity
cross section	$A$	m²	—	DriftVelocity
density	$ρ$	kg/m³	0 … ∞	DriftVelocity
molar mass	$M$	kg/mol	—	DriftVelocity
free electrons	$n$	m⁻³	—	DriftVelocity
drift speed	$v_{d}$	m/s	—	DriftVelocity
length of the conductor	$l$	m	—	DriftVelocity
drift time	$t$	s	—	DriftVelocity
free electrons of ground	$n_{g}$	m⁻³	—
density of ground	$ρ_{g}$	kg/m³	0 … ∞
molar mass of ground	$M_{g}$	kg/mol	—
free electrons of wire	$n$	m⁻³	—
density of wire	$ρ$	kg/m³	0 … ∞
molar mass of wire	$M$	kg/mol	—
current of ground	$I_{g}$	A	—
cross section of ground	$A_{g}$	m²	—
drift speed of ground	$v_{d,g}$	m/s	—
current of wire	$I$	A	—
cross section of wire	$A$	m²	—
drift speed of wire	$v_{d}$	m/s	—
length of the conductor of ground	$l_{g}$	m	—
drift time of ground	$t_{g}$	s	—
length of the conductor of wire	$l$	m	—
drift time of wire	$t$	s	—

n_{w} = \frac{ρ_{w} \cdot N_{A}}{M_{w}}

one free electron an atom

I_{w} = n_{w} \cdot e \cdot A_{w} \cdot v_{d,w}

current from drift

l_{w} = v_{d,w} \cdot t_{w}

drifting the length of the wire

n_{g} = \frac{ρ_{g} \cdot N_{A}}{M_{g}}

one free electron an atom(ground)

n = \frac{ρ \cdot N_{A}}{M}

one free electron an atom(wire)

I_{g} = n_{g} \cdot e \cdot A_{g} \cdot v_{d,g}

current from drift(ground)

I = n \cdot e \cdot A \cdot v_{d}

current from drift(wire)

l_{g} = v_{d,g} \cdot t_{g}

drifting the length of the wire(ground)

l = v_{d} \cdot t

drifting the length of the wire(wire)