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: A Microscopic View of Resistivity

INTRO:
Recall that the density J of current flowing through a material can be written in terms of microscopic properties of the material: j=nqv_d, where n is the density of current carriers, q is the charge of one current carrier, and v_d is the drift velocity of a current carrier. In a metal, the current carriers are electrons.
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PART A:
The drift velocity is the component of the current-carrier's velocity due to acceleration from the electric field in the conductor. This corresponds to the average speed of all of the current carriers in the conductor. The current carriers also have random thermal motions, but the randomness causes the velocities due to thermal motion to cancel when averaged over a large number of current carriers. If the electric field inside of the conductor has magnitude E, and the charge q is accelerated from rest for a time τ, what is the final speed v of the charge?
Express the speed in terms of E, q, τ, and the mass m of the charge.

SOLUTION:
The equation given by the intro is 

J = nqv_d (also eq. (30.13) from the book)

The book uses e as the charge, but q is an arbitrary charge, and more applicable to situations not concerning electrons. so q^* will be subbed into other equations that have e. (The asterick demonstrates a substituted variable)

Section 30.2 Creating a Current has a subsection titled: A Model of Conduction. This section is vital to this problem.

eq. (30.7) states that 

v_d = q^*τE/m
and that's actually the answer too

v = Eqτ/m

NOTE: 
At every collision, the electron's motion is randomized, bringing the drift velocity back to zero. If τ is the mean time between collisions, then the speed v that you just calculated is equal to the net drift velocity vd, for all current carriers. This is true, because the mean time between collisions is equal to the mean time since the last collision. (To see how this is possible requires looking at the actual distribution of times between collisions and how the different averages are calculated, but this type of analysis lies beyond the scope of this problem.)
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PART B:
The magnitude of the drift velocity is very small compared to the speed of random electron motion in a metal. The mean time between collisions can be calculated from the mean free path d, which is the average distance that an electron can travel before colliding with one of the metal nuclei. Using these variables, what is the mean time between collisions τ? Let E_F be the energy of the electrons.
Express the mean time between collisions in terms of E_F, d, and m.

SOLUTION:
so... how long, τ, does it take to go a distance, d, at a speed, v_F (the speed of an electron with kinetic energy equal to E_F)?

What equation relates energy velocity and mass? The kinetic energy equation!

K = ½mv²

E_F = ½mv_F²
→ v_F = SQRT{2E_F/m}

velocity is distance traveled over a period of time : v = d/t

therefore, distance traveled is equal to the velocity times time: d = v⋅t
Finally, that means that time elapsed is equal to distance traveled divided by velocity: t = d/v

∴ τ = d/v_F =d⋅ 1/SQRT{2E_F/m}

τ = d⋅SQRT{½m/E_F} = d⋅√{m/(2E_F)}

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PART C:
Recall that the conductivity σ of a substance is defined by the relation J=σE and that the resistivity ρ is the inverse of the conductivity: ρ=1/σ. Give an expression for ρ in terms of microscopic properties of a metal. Use q_e for the charge on the electron.
Express your answer in terms of E_F, m, q_e, d, E, and n.

SOLUTION:
so, if J=σE & ρ=1/σ, then J = E/ρ
eq(30.13) states that J = nev_d

∴ J = E/ρ = nev_d
→ρ = E/[nev_d]

if we plug in the value we got in part A for v, and the instructions statement that e = q_e ... 

ρ = E_F/[nq_e⋅(E_Fq_eτ/m)]

= m/[nq_e²⋅d⋅SQRT{m2E_F}]

If we then plug in the value we got in part B for τ... we should have everything in the proper variables. 

ρ = m/[nq_e²⋅τ]

=m/[nq_e²⋅d⋅SQRT{m/(2E_F)}]
=m⋅SQRT{2E_F/m}/[nq_e²⋅d]
=SQRT{2m²E_F/m}/[nq_e²⋅d]
⇒ ρ = SQRT{2mE_F} / [nq_e²⋅d]

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PART D:
Find the resistivity of gold at room temperature. Use the following information:

• free electron density of gold = 5.90×10²⁸ m⁻³
• Fermi energy of gold = 8.86×10⁻¹⁹ J
• mass of electron = 9.11×10⁻³¹ kg
• charge of an electron = −1.60×10−¹⁹ C
• mean free path of electron in gold = 3.45×10⁻⁸ m

Express your answer in ohm-meters to three significant figures.

SOLUTION:

Just plug these values into the equation we determined in the previous part... 

so what we were given was:

n = 5.90×10²⁸ m⁻³
E_F = 8.86×10⁻¹⁹ J
m = 9.11×10⁻³¹ kg
q_e = −1.60×10−¹⁹ C
d = 3.45×10⁻⁸ m

So ρ = SQRT{2mE_F} / [nq_e²⋅d] 

= SQRT{2(9.11×10⁻³¹ kg)(8.86×10⁻¹⁹ J)} / [(5.90×10²⁸ m⁻³)(−1.60×10−¹⁹ C)²⋅(3.45×10⁻⁸ m)]

⇒ ρ = 2.44×10^-8 Ωm
Wolfram LINK

30: Introduction to Electric Current

INTRO:
To understand the nature of electric current and the conditions under which it exists.
Electric current is defined as the motion of electric charge through a conductor. Conductors are materials that contain movable charged particles. In metals, the most commonly used conductors, such charged particles are electrons. The more electrons that pass through a cross section of a conductor per second, the greater the current. The conventional definition of current is

I=Q_total/Δt

where I is the current in a conductor and Qtotalis the total charge passing through a cross section of the conductor during the time interval Δt.

The motion of free electrons in metals not subjected to an electric field is random: Even though the electrons move fairly rapidly, the net result of such motion is that Q_total=0 (i.e., equal numbers of electrons pass through the cross section in opposite directions). However, when an electric field is imposed, the electrons continue in their random motion, but in addition, they tend to move in the direction of the force applied by the electric field.

In summary, the two conditions for electric current in a material are the presence of movable charged particles in the material and the presence of an electric field.

Quantitatively, the motion of electrons under the influence of an electric field is described by the drift speed, which tends to be much smaller than the speed of the random motion of the electrons. The number of electrons passing through a cross section of a conductor depends on the drift speed (which, in turn, is determined by both the microscopic structure of the material and the electric field) and the cross-sectional area of the conductor.

In this problem, you will be offered several conceptual questions that will help you gain an understanding of electric current in metals.
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PART A:
You are presented with several long cylinders made of different materials. Which of them are likely to be good conductors of electric current?

• copper
• aluminum
• glass
• quartz
• cork
• plywood
• table salt
• gold

SOLUTION:
As stated in the intro, metals are most likely to be good conductors of electric current,
so:
copper, aluminum, and gold
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PART B:
Metals are good conductors of electric current for which of the following reasons?

• They possess high concentrations of protons
• They possess low concentrations of protons
• They possess high concentrations of free electrons
• They possess low concentrations of free electrons

SOLUTION:
The intro states that "In metals, the most commonly used conductors, such charged particles are electrons."
so, the third option: They possess high concentrations of free electrons
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PART C:
Which of the following is the most likely drift speed of the electrons in the filament of a light bulb?

• 10^-8 m/s
• 10^-4 m/s
• 10 m/s
• 10⁴ m/s
• 10⁸ m/s

SOLUTION:
The second option,  10^-4 m/s

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PART D:
You are presented with several wires made of the same conducting material. The radius and drift speed are given for each wire in terms of some unknown units r and v. Rank the wires in order of decreasing electron current.
Rank from most to least electron current. To rank items as equivalent, overlap them.

SOLUTION:
Since the wires are made of the same material, the charge carriers and their densities are the same for all the wires.

Other conditions being equal, the current is proportional to the product of the cross-sectional area of the wire and the drift velocity, that is,

I=n|q|v_dA,

where I is the current, v_d is the drift velocity, A is the cross-sectional area, n is the density of charge carriers, and q is the charge on the carriers.

Therefore:

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PART E:
The drift speed of the electrons in a wire depends strongly on which of the following factors?

• The cross-sectional area of the wire
• The mass of the wire
• The temperature of the wire
• The internal electric field in the wire

SOLUTION:
In the intro, it states "the motion of electrons under the influence of an electric field is described by the drift speed"
so, the final option is correct: The internal electric field in the wire
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PART F:
What quality must the charge density on the surface of a conducting wire possess if an electric field is to act on the negatively charged electrons inside the wire?

The charge density must be... 

• positive
• negative
• nonuniform
• uniform

SOLUTION:
nonuniform

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 30.10

INTRO:
The current in a 100 watt lightbulb is 0.870 A . The filament inside the bulb is 0.230 mm in diameter.
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PART A:
What is the current density in the filament?
Express your answer to three significant figures and include the appropriate units.

SOLUTION:
Current density is simply current divided by area...
The area = A = πr² = π/4⋅d²

⇒ J = (0.87 A)/[π/4⋅(2.3×10^-4 m)²]
⇒ J = 2.09×10⁷ A/m²
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PART B:
What is the electron current in the filament?
Express your answer using three significant figures.

SOLUTION:
current = charge/ time
1 Amp = 1 C/s

using equation (30.11) ... I = e⋅i_e

where e is the charge of an electron

∴ i_e = I/e
and I, in terms of C/s -> I = 0.87 C/s

→ i_e = (0.87 C/s)/(-1.602×10^-19 C/electron)
⇒ i_e = 5.43×10^-19 electron/s

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​