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what is the potential difference across a 2.00 mm length of the wire (for copper rhorho = 1.72×10−81.72×10−8 ω⋅mω⋅m )? express your answer with the appropriate units.

Answer :

The potential difference across a 2.00 m length of the silver wire would also be [tex]\( 1.31 \, \text{V} \)[/tex], assuming all other parameters remain the same.

To find the potential difference across a length of wire, we can use Ohm's Law, [tex]\( V = IR \)[/tex], where [tex]\( V \)[/tex] is the potential difference, [tex]\( I \)[/tex] is the current, and [tex]\( R \)[/tex] is the resistance.

The resistance [tex](\( R \))[/tex] of a wire can be calculated using the formula [tex]\( R = \frac{\rho L}{A} \)[/tex], where:

[tex]\( \rho \)[/tex] is the resistivity of the material

[tex]\( L \)[/tex] is the length of the wire

[tex]\( L \)[/tex] is the cross-sectional area of the wire

The cross-sectional area [tex](\( A \))[/tex] of the wire can be calculated using the formula [tex]\( A = \pi r^2 \)[/tex], where [tex]\( r \)[/tex] is the radius of the wire.

First, let's find the resistance [tex](\( R \))[/tex] of the copper wire:

Given:

Gauge: 14.0

Diameter: [tex]\( 1.628 \, \text{mm} \)[/tex] (which we'll convert to meters)

Current: [tex]\( 12.5 \, \text{mA} \)[/tex] (which we'll convert to Amperes)

Resistivity of copper [tex](\( \rho_{\text{Cu}} \)): \( 1.72 \times 10^{-8} \, \Omega \cdot \text{m} \)[/tex]

We can use the gauge to find the radius of the wire and then use that to find the cross-sectional area [tex](\( A \))[/tex].

The formula to find the radius of a wire given the gauge is:

[tex]\[ r = \frac{0.127 \times 92^{\frac{36 - \text{gauge}}{39}}}{1000} \][/tex]

Substituting the given values:

[tex]\[ r = \frac{0.127 \times 92^{\frac{36 - 14}{39}}}{1000} \][/tex]

[tex]\[ r = \frac{0.127 \times 92^{\frac{22}{39}}}{1000} \][/tex]

Now, we can calculate the radius and then the cross-sectional area [tex](\( A \))[/tex]:

[tex]\[ r = \frac{0.127 \times 92^{0.564}}{1000} \][/tex]

[tex]\[ r = \frac{0.127 \times 2.549}{1000} \][/tex]

[tex]\[ r = \frac{0.323}{1000} \][/tex]

[tex]\[ r = 3.23 \times 10^{-4} \, \text{m} \][/tex]

Now, the cross-sectional area [tex](\( A \))[/tex]:

[tex]\[ A = \pi r^2 \][/tex]

[tex]\[ A = \pi (3.23 \times 10^{-4} \, \text{m})^2 \][/tex]

[tex]\[ A = \pi \times 1.044 \times 10^{-7} \, \text{m}^2 \][/tex]

[tex]\[ A = 3.28 \times 10^{-7} \, \text{m}^2 \][/tex]

Now, we can calculate the resistance [tex](\( R \))[/tex]:

[tex]\[ R = \frac{\rho_{\text{Cu}} L}{A} \][/tex]

[tex]\[ R = \frac{1.72 \times 10^{-8} \, \Omega \cdot \text{m} \times 2.00 \, \text{m}}{3.28 \times 10^{-7} \, \text{m}^2} \][/tex]

[tex]\[ R = \frac{3.44 \times 10^{-8} \, \Omega \cdot \text{m}}{3.28 \times 10^{-7} \, \text{m}} \][/tex]

[tex]\[ R = 0.105 \, \Omega \][/tex]

Now, we can find the potential difference [tex](\( V \))[/tex]:

[tex]\[ V = IR \][/tex]

[tex]\[ V = (12.5 \times 10^{-3} \, \text{A}) \times 0.105 \, \Omega \][/tex]

[tex]\[ V = 1.31 \, \text{V} \][/tex]

So, the potential difference across a 2.00 m length of the copper wire is [tex]\( 1.31 \, \text{V} \)[/tex].

Now, let's find the potential difference if the wire were silver instead of copper. The only thing that changes is the resistivity [tex](\( \rho_{\text{Ag}} \))[/tex].

Given:

Resistivity of silver [tex](\( \rho_{\text{Ag}} \))[/tex]: [tex]\( 1.59 \times 10^{-8} \, \Omega \cdot \text{m} \)[/tex]

We'll use the same resistance value [tex](\( R \))[/tex] calculated for copper and Ohm's Law to find the potential difference [tex](\( V \))[/tex] for silver:

[tex]\[ V = IR \][/tex]

[tex]\[ V = (12.5 \times 10^{-3} \, \text{A}) \times 0.105 \, \Omega \][/tex]

[tex]\[ V = 1.31 \, \text{V} \][/tex]

So, the potential difference across a 2.00 m length of the silver wire would also be [tex]\( 1.31 \, \text{V} \)[/tex], assuming all other parameters remain the same.

Complete Correct Question:

A 14.0 gauge copper wire of diameter 1.628 mm carries a current of 12.5 mA. What is the potential difference across a 2.00 m length of the wire (for copper rho = 1.72 times 10^-8 Ohm middot m)? What would the potential difference in part (a) be if the wire were silver instead of copper, but all else was the same (for silver rho = 1.59 times 10^-8 Ohm middot m)?

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Rewritten by : Barada

To find the potential difference across a length of copper wire, you need to calculate the resistance of the wire using its dimensions and resistivity, then apply Ohm's Law to determine the voltage.

To determine the potential difference across a 2.00 mm length of copper wire (ρ = 1.72×10⁻⁸ ω⋅m), we would need to know the current flowing through the wire.

Without this information, it is not possible to calculate the potential difference using Ohm's law (V = IR).

However, if we assume that the wire is part of a circuit with a known current, we can calculate the potential difference using Ohm's law.

For example, if the circuit has a current of 1.0 A flowing through the wire, the potential difference across the 2.00 mm length of wire would be:

V = IR

V = (1.0 A)(2.00×10⁻³ m)(1.72×10⁻⁸ Ω⋅m)

V = 3.44×10⁻¹¹ V

This is the required potential difference.