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​