The Resistivity of a Wire

solved OhmsLaw

Q. A negligibly small current is passed through a wire of length 15 m $l = 15$ m and uniform cross-section 6.0 × 10⁻⁷ m² $A = 6.0 \times 10^{−7}$ m², and its resistance is measured to be 5.0 Ω $R = 5.0$ Ω. What is the resistivity of the material at the temperature of the experiment, in Ω m?

Inputs

length

l

15.00 m

area

A

6e-07 m²

resistance

R

5 Ω

Outputs

resistivity

ρ

How it's solved

Solved with OhmsLaw

Given

$l = 15$ m
$A = 6.0 \times 10^{−7}$ m²
$R = 5.0$ Ω

Find $ρ$ .

Working

resistance from resistivity(wire) $R = \frac{ρ \cdot l}{A}$
$5 = \frac{ρ \cdot 15}{6 \times 10^{−7}}$
$ρ = 2 \times 10^{−7}$ Ω m
$ρ \approx 2.0 \times 10^{−7}$ Ω mto 2 significant figures

Conclusion

Resistance grows with length and shrinks with cross-section; what is left over belongs to the material alone.

The web of concepts

given read from the words equation asked

The Ruby that solves it

The question and its solution are small Ruby files. The engine runs them.

question.rb

1require_relative "../physics"23question :the_resistivity_of_a_wire do4  kind :numerical5  describes "A negligibly small current is passed through a wire of length %{length} and uniform cross-section " \6            "%{area}, and its resistance is measured to be %{resistance}. %{resistivity}, in Ω m?"7  given length: "15 m"8  given area: "6.0e-7 m2"9  given resistance: "5.0 Ω"10  asks resistivity: "What is the resistivity of the material at the temperature of the experiment"11end

solution.rb

1require_relative "../questions/the_resistivity_of_a_wire"2require_relative "../science/current_electricity/ohms_law"34solve :the_resistivity_of_a_wire, using: OhmsLaw do5  objects :wire67  take length: wire.length_of_the_conductor8  take area: wire.cross_section9  take resistance: wire.resistance1011  resistivity = wire.resistivity12  answer resistivity:, in: "Ω m", figures: 21314  conclusion "Resistance grows with length and shrinks with cross-section; what is left over belongs to the " \15             "material alone."16end

playground.rb

1require_relative "../solutions/the_resistivity_of_a_wire"23play_with :the_resistivity_of_a_wire do4  title "The Resistivity of a Wire"5end

The laws it drew on

quantity	written	units	within	from
voltage	$V$	V	—	OhmsLaw
current	$I$	A	—	OhmsLaw
resistance	$R$	Ω	0 … ∞	OhmsLaw
length of the conductor	$l$	m	—	OhmsLaw
cross section	$A$	m²	—	OhmsLaw
resistivity	$ρ$	Ω m	—	OhmsLaw
voltage of ground	$V_{g}$	V	—
current of ground	$I_{g}$	A	—
resistance of ground	$R_{g}$	Ω	0 … ∞
voltage of wire	$V$	V	—
current of wire	$I$	A	—
resistance of wire	$R$	Ω	0 … ∞
resistivity of ground	$ρ_{g}$	Ω m	—
length of the conductor of ground	$l_{g}$	m	—
cross section of ground	$A_{g}$	m²	—
resistivity of wire	$ρ$	Ω m	—
length of the conductor of wire	$l$	m	—
cross section of wire	$A$	m²	—

V_{c} = I_{c} \cdot R_{c}

ohms law

R_{c} = \frac{ρ_{c} \cdot l_{c}}{A_{c}}

resistance from resistivity

V_{g} = I_{g} \cdot R_{g}

ohms law(ground)

V = I \cdot R

ohms law(wire)

R_{g} = \frac{ρ_{g} \cdot l_{g}}{A_{g}}

resistance from resistivity(ground)

R = \frac{ρ \cdot l}{A}

resistance from resistivity(wire)