30: Stretching the A-Rod

INTRO:
To learn to apply the concept of current density and microscopic Ohm's law.
A slab of metal of volume V is made into a rod of length L. The rod carries current I when the electric field inside is E.
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PART A:
Find the resistivity of the metal ρ.
Expess your answer in terms of the given quantities.

SOLUTION:
So, first off, V is volume, not voltage. So be careful to read the instructions. V = l⋅w⋅h = A⋅L & A = V/L

eq. (30.13) & eq. (30.17):
J = I/A = σE = E/ρ
→ ρ = E⋅A/I
sub A = V/L since those are given quantities... 
⇒ ρ = EV/IL

NOTE: I think it is hilarious that this solution spells EVIL. 
It's the little things... 
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PART B:
The rod is now stretched so that its length is doubled. If the electric field remains the same, what is the new current I′ in the rod?
Express your answer in terms of some or all of the quantities given in the problem introduction.

SOLUTION:

So, L₂ = 2L₁ 
if the electric field remains the same, then that means that the volume remains constant, or V₁ = V₂ (V = A⋅L)
∴ A₂⋅L₂ = A₁⋅L₁

→ A₂⋅2L₁ = A₁⋅L₁
→ A₂ = ½A₁​

but J₂ = J₁ 

so... I₂/A₂ = I₁/A₁

I₂ = I₁½A₁/A₁

⇒ I₂ = I₁/2, or

I' = I/2       (note this is capital i, as in current, and not one, 1)

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PART C:

A piece of copper is made into a rod with a square cross-section. The side of the square is 2.00 centimeters. The resistivity of copper is 1.7⋅10⁻⁸ Ω⋅m. An unknown electric field E, directed along the rod, creates a current of 12.0 amperes through the rod. Find the magnitude of E.
Use two significant figures in your answer. Express your answer in newtons per coulomb.

SOLUTION:

Givens/ Conversions:​

b = h = 2 cm = 0.02 m ∴ A = 4×10^-4 m²
ρ = 1.7×10^-8 Ωm
I = 12 A

using eq. (30.20) ... I = AE/ρ ... 

E = Iρ/A

=(12 A)(1.7×10^-8 Ωm)/(4×10^-4 m²)
⇒ E = 5.10×10^-4 N/C
Wolfram LINK​

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.