30: Current in a Wire

INTRO:
A metallic wire has a diameter of 4.12 mm. When the current in the wire is 8.00 A, the drift velocity is 5.40×10⁻⁵ m/s.
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PART A:
What is the density of free electrons in the metal?
Express your answer numerically in m⁻³ to two significant figures.

SOLUTION:

Givens/ Conversions:​

wire diameter ≡ d = 4.12 mm = 4.12×10^-3 m
Current ≡ I = 8 A
drift velocity ≡ v_d = 5.40×10⁻⁵ m/s.

Using eq. (30.13) ... J = I/A = n⋅e⋅v_d

→ n = I/(A⋅e⋅v_d)​

e = 1.6×10^-19 C
& A = π/4⋅d²

∴ n = I/(π/4⋅d²⋅e⋅v_d)

= (8 A)/(π/4⋅(4.12×10^-3 m)²⋅(1.6×10^-19 C)⋅(5.40×10⁻⁵ m/s))
⇒ n = 6.9×10²⁸ m^-3
Wolfram LINK

30: Problem 30.45

INTRO:
The starter motor of a car engine draws a current of 180 A from the battery. The copper wire to the motor is 4.30 mm in diameter and 1.2 m long. The starter motor runs for 0.630 s until the car engine starts.
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PART A:
How much charge passes through the starter motor?
Express your answer with the appropriate units.

SOLUTION:

Givens/ Conversions:

I = 180 A
L = 1.2 m
d = 4.30 mm = 4.3×10^-3 m
t = 0.630 s

Using eq. (30.10): Q = I⋅Δt...

Q = (180 A)(0.630 s)
⇒ Q = 113.4 C (A⋅s = C)

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PART B:
How far does an electron travel along the wire while the starter motor is on?
Express your answer with the appropriate units.

SOLUTION:
Using eq. (30.13) ... J = I/A = n_e⋅e⋅v_d
rearranging... v_d = I/(A⋅n_e⋅e)

this equation has all knowns and solves for the speed. first step for determining distance traveled over time, right? right.
​

using table (30.1) ... n_ecopper = 8.5×10²⁸ m^-3

We know that the distance traveled = velocity⋅time
∴ distance = I⋅Δt/(A⋅n_e⋅e)

= (180 A)(0.63 s)/[(π/4⋅(4.3×10^-3 m)²)(8.5×10²⁸ m^-3)(1.602×10^-19C)]
= 5.735×10^-4 m
⇒ 0.574 mm

30: Problem 30.6

PART A:
How many conduction electrons are there in a 4.50 mm diameter gold wire that is 20.0 cm long?

SOLUTION:
signature first step...

Givens/ conversions:

Diameter ≡ d = 4.50 mm = 0.0045 m
Length ≡ l = 20.0 cm = 0.200 m

Using Table 30.1...

n_eGOLD = 5.9×10²⁸ m^-3

And we need to determine N_e... 

Equation (30.2) states that N_e = n_e⋅V
We can determine volume by V = A⋅l = (π/4)d²⋅l

∴ N_e = n_e⋅(π/4)d²⋅l

= (5.9×10²⁸ m^-3)(π/4)⋅(0.0045 m)²⋅(0.200 m)

⇒N_e = 1.88×10²³ electrons

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PART B:
How far must the sea of electrons in the wire move to deliver -38.0 nC of charge to an electrode?
Express your answer with the appropriate units.

SOLUTION:
So we know from part A that in 0.2 m, there are 1.877×10²³ electrons ...

each electron has a charge of -1 e, or -1.602×10^-19 C
and that 1 nC = 1×10^-9 C, or 1×10^-9 C/nC = 1×10⁹ C/nC....

this means that the desired charge = -38×10^-9 C

The total charge created by that many electrons is simply equal to the number of electrons multiplied by the charge or each individual electron, or Q_tot = Q_e⋅N_e

The charge per unit length can be determined by dividing the total charge by the length from part A.

Q/l = Q_e⋅N_e/l = (-1.602×10^-19 C)(1.877×10²³)/(0.2 m)
→ Q/l = -1.504×10⁵ C/m
lets call this Q_perLength
​

The total length (L) required to contain a total charge (Q_f) may then be solved for by ... L = Q_f/Q_perLength = l⋅Q_f/(Q_e⋅N_e) 

→ L = (0.2 m)(-38×10^-9 C)/[(-1.602×10^-19 C)(1.877×10²³)]
⇒ L = 2.528×10^-13 m

Which is a really small number... which brings us to the last aspect. Units... I think that the units (which are units of length, of course) are most likely going to be desired in the same units that the length in part A was expressed, so centimeters... 

10² cm/m... so

⇒ L = 2.53×10^-11 cm

30: Problem 30.2

INTRO:
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):

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 v_d.

The number of electrons (N_e) that pass through a cross section during a time interval (Δt) is defined in eq. (30.1) → N_e = i_e⋅Δt

where i_e 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' (n_e) 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) → N_e = n_e⋅V = n_e⋅A⋅Δx = n_e⋅A⋅v_d⋅Δt​

using equations (30.1) & (30.2)...

N_e = i_e⋅Δt = n_e A⋅v_d⋅Δt
→ N_e/Δt = i_e = n_e A⋅v_d

→ N_e/(Δt⋅n_e⋅A) = v_d​

lets call that eq. (A), for now​

Looking back at the initial givens, let's go ahead and do my signature first step...

Givens/ conversions:​

Number of electrons ≡ N_e = 1.00×10²⁰
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 ...

v_d = N_e/(Δt⋅n_e⋅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... 

n_e = 8.5×10²⁸ m^-3​

∴ v_d = N_e/(Δt⋅n_e⋅π/4⋅d²) = (1.00×10²⁰)/((6.00 s)⋅(8.5×10²⁸ m^-3)⋅(π/4)⋅(0.0045 m)²)

⇒v_d =1.233×10^-5 m/s​

They want units of μm/s, so multiply through by 10⁶ μm/m

⇒v_d =12.33 μm/s​