30: Problem 30.66

INTRO:
Household wiring often uses 2.0-mm-diameter copper wires. The wires can get rather long as they snake through the walls from the fuse box to the farthest corners of your house.
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PART A:
What is the potential difference across a 16-m-long, 2.0-mm-diameter copper wire carrying an 7.8 A current?
Express your answer to two significant figures and include the appropriate units.

SOLUTION:
Using eq. (30. 21)... I = A⋅ΔV/ρL

→ ΔV = IρL/A
​

using table (30.2) → ρ_copper = 1.7×10^-8 Ωm
& A = π/4⋅d²

→ ΔV = (7.8 A)(1.7×10^-8 Ωm)(16 m)/(π/4⋅d²)
⇒ ΔV = 0.675 V = 0.68 V​

30: Problem 30.55

INTRO:
The two wires in the figure (Figure 1) are made of the same material.

Figure 1

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PART A:
What is the current in the 2.0-mm-diameter segment of the wire?

SOLUTION:
Since both segments are connected in series, current is the same 
I₁ = I₂ = 2 A

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PART B:
What is the electron drift speed in the 2.0-mm-diameter segment of the wire?

SOLUTION:
Using eq. (30.13)... J = I/A = n_e⋅e⋅v_d
rearranging... I = A⋅n_e⋅e⋅v_d
I₁ = I₂ 

→ A₁⋅n_e1⋅e⋅v_d1 = A₂⋅n_e2⋅e⋅v_d2​

The electron charge cancels out, and also the intro says that the wires are the same material, so n_e1= n_e2 ... 

→ A₁⋅v_d1 = A₂⋅v_d2
→v_d2 = A₁⋅v_d1 / A₂

= (π/4⋅d₁²)(v_d1) / (π/4⋅d₂²)
= (d₁²)(v_d1) / (d₂²)
= (1×10^-3 m)²(2×10^-4 m/s) / (2×10^-3 m)²
⇒ v_d2 = 5×10^-5 m/s

= 50 μm/s​

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​

30: Problem 30.24

INTRO:
The two segments of the wire in the figure (Figure 1) have equal diameters but different conductivities σ₁ and σ₂. Current I passes through this wire.

Figure 1

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PART A:
If the conductivities have the ratio σ₂/σ₁=5, what is the ratio E₂/E₁ of the electric field strengths in the two segments of the wire?

- 1/25
- 5
- 25
- 1/5

SOLUTION:
Since the current I passes through the entire wire, and the diameters are equal... the current density for both sides is... 
J = I/A

This means that J₁ = J₂

using eq. (30.17) ... J = σE ...

J₁ = σ₁E₁
J₂ = σ₂E₂​

since J₁ = J₂... σ₁E₁ = σ₂E₂

by rearranging...
σ₂/σ₁ = E₁/E₂ = 5 (given)​

E₂/E₁ = [E₁/E₂]^-1

⇒ E₂/E₁ = 1/5​

30: Problem 30.19

INTRO:
The electric field in a 1.0mm×1.0mm square aluminum wire is 2.8×10⁻² V/m .
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PART A:
What is the current in the wire?
Express your answer to two significant figures and include the appropriate units.

SOLUTION:
E = 2.8×10⁻² V/m
b = h = 1.0 mm = 0.001 m

∴ A = 1×10^-6 m²​

using table 30.2... 

σ_Al = 3.5×10⁷ 1/(Ωm)​

Using eq. (30.17) & eq. (30.13)... 
J = I/A = σE

→ I = σEA

= (3.5×10⁷ 1/(Ωm))(2.8×10⁻² V/m)(1×10^-6 m²)
⇒ I = 0.98 A

* note 1 V/Ω = A ( since V=IR ) *​