The Resistivity of a Wire

Q. A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 × 10⁻⁷ m², and its resistance is measured to be 5.0 Ω. What is the resistivity of the material at the temperature of the experiment, in Ω m? expects 2.0 × 10⁻⁷ Ω m

What the draft establishes

Fills in once a draft is checked.

The web of concepts

given read from the words equation asked

The equations it used

None yet.

Check an answer

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?

The playground

Inputs

length

l

15.00 m

area

A

6e-07 m²

resistance

R

5 Ω

Outputs

resistivity

ρ

The solution

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 Ruby

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

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

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

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)