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: Resistance from Microscopic Ohm's Law

INTRO:
Your task is to calculate the resistance of a simple cylindrical resistor with wires connected to the ends, such as the carbon composition resistors that are used on electronic circuit boards. Imagine that the resistor is made by squirting material whose conductivity is σ into a cylindrical mold with length L and cross-sectional area A as shown in (Figure 1) . Assume that this material satisfies Ohm's law. (It should if the resistor is operated within its power dissipation limits.)

Figure 1

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PART A:
What is the resistance R of this resistor?
Express the resistance in terms of variables L, A and σ given in the introduction. Do not use V, I, E or J in your answer.

SOLUTION:
There are two equations that are necessary to complete this problem. 

Eq. (30.18) states that resistivity ρ = 1/σ

and

Eq. (30.22) states that resistance R = ρL/A

plugging (30.18) into (30.22) gives... 

R = L/σA

NOTE

Real resistors vary tremendously in overall size. The larger the size, the more power the resistor can dissipate without heating to the point that it is dangerous to nearby components or that the material of which it is constructed begins to change its conductivity (i.e., so that the resistance would no longer be constant). The amount of resistance is determined by the conductivity of the material of the resistor, which can vary over more than 20 orders of magnitude. Commercially available resistors vary from 0.1 ohm or less to more than 10⁷ ohm.

30: Problem 30.51

INTRO:
A hollow metal sphere has inner radius a, outer radius b, and conductivity σ. The current I is radially outward from the inner surface to the outer surface.
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PART A:
Find an expression for the electric field strength inside the metal as a function of the radius r from the center.
Express your answer in terms of the variables I, σ, r, and appropriate constants.

SOLUTION:
If you were to consider a thin radial section of the cylinder (so that the cross section would be almost uniform throughout), what would be its resistance? 

The radial area, not cross-sectional this time, is A = 2πrL
the L = 2r ∴ A = 4πr²

J = I/A = σE 

→ E = I/(σA) = I/(σ4πr²)

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PART B:
Evaluate the electric field strength at the inner surface of a copper sphere if a = 1.0 cm , b = 3.0 cm , and I = 26 A .
Express your answer to two significant figures and include the appropriate units.

SOLUTION:

Givens/ Conversions:

a = 1.0 cm = 1×10^-2 m
b = 3.0 cm = 3×10^-2 m
I = 26 A
material = copper,

table (30.2) → σ = 6.0×10⁷ 1/(Ωm)

they want the strength at the inner surface, so use a as r...

→ E = (26 A)/[4π(6.0×10⁷ 1/(Ωm))(1×10^-2 m)²]
⇒ E = 3.45×10^-4 V/m

They want E = 3.4×10^-4 V/m though... for some reason.  It'll automatically correct it though.

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PART C:
Evaluate the electric field strength at the outer surface of a copper sphere if a = 1.0 cm , b = 3.0 cm , and I = 26 A .
Express your answer to two significant figures and include the appropriate units.

SOLUTION:

Givens/ Conversions:

a = 1.0 cm = 1×10^-2 m
b = 3.0 cm = 3×10^-2 m
I = 26 A
material = copper,

table (30.2) → σ = 6.0×10⁷ 1/(Ωm)

they want the strength at the inner surface, so use a as r...

→ E = (26 A)/[4π(6.0×10⁷ 1/(Ωm))(3×10^-2 m)²]
⇒ E = 3.83×10^-4 V/m

They want E = 3.8×10^-4 V/m ... At least that rounding makes sense :) 

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 30.32

INTRO:
The terminals of a 0.70 V watch battery are connected by a 70.0-m-long gold wire with a diameter of 0.200 mm .

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PART A:
What is the current in the wire?
Express your answer using three significant figures.

SOLUTION:

Givens/ Conversions: ​

V = 0.70 V
L = 70.0 m
d = 0.200 mm = 2×10^-4 m
also, Ω = V/A
​

So.. we have a voltage, a length, a diameter, and the material. 
We know that the cross-sectional area : A = π/4⋅d²
From table (30.2), Gold has a resistivity of: 

ρ = 2.4×10^-8 Ωm​

By using eq. (30.22), we can determine the resistance.

R = ρL/A = ρL/(π/4⋅d²)​

Finally, we know V=IR, by know, that means that I = V/R

or... I = V/[ρL/(π/4⋅d²)] = πVd²/(4ρL)
→ I = π(0.70 V)(2×10^-4 m)²/[4(2.4×10^-8 Ωm)(70.0 m)]
⇒ I = 0.0131 A​

They want it in mA, though. 1 mA = 10^-3 A

⇒ I = 13.1 mA​