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.
-----------------------------------------------------------------------------------------------------
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)

-----------------------------------------------------------------------------------------------------
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 .

-----------------------------------------------------------------------------------------------------
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​