The resistivity of a $1.0 \mathrm {~m}$ long copper wire is $1.72 \times 10 ^ { - 8 } \Omega \cdot \mathrm { m }$ and its cross sectional area is $2.0 \times$ $10 ^ { - 6 } \mathrm {~m} ^ { 2 }$. If the wire carries a current of $0.20 \mathrm {~A}$, what is the voltage across the wire?

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Quizplus · Accepted Answer

Ohm's law states that the voltage across a conductor is equal to the current flowing through it multiplied by the resistance of the conductor. The resistance of a conductor is equal to the resistivity of the material multiplied by the length of the conductor and divided by the cross-sectional area of the conductor. In this case, the resistivity of the copper wire is $1.72 \times 10 ^ { - 8 } \Omega \cdot \mathrm { m }$, the length of the wire is $1.0 \mathrm {~m}$, and the cross-sectional area of the wire is $2.0 \times$ $10 ^ { - 6 } \mathrm {~m} ^ { 2 }$. Therefore, the resistance of the wire is:$$R = \rho \frac{L}{A} = (1.72 \times 10 ^ { - 8 } \Omega \cdot \mathrm { m }) \frac{1.0 \mathrm {~m}}{2.0 \times 10 ^ { - 6 } \mathrm {~m} ^ { 2 }} = 8.6 \times 10 ^ { - 3 } \Omega$$The current flowing through the wire is $0.20 \mathrm {~A}$. Therefore, the voltage across the wire is:$$V = IR = (0.20 \mathrm {~A})(8.6 \times 10 ^ { - 3 } \Omega) = 1.72 \times 10 ^ { - 3 } \mathrm {~V} = 1.7 \mathrm { mV }$$Therefore, the correct answer is A.

The Resistivity of A 1.0 m1.0 \mathrm {~m}1.0 m Long Copper Wire Is 1.72×10−8Ω⋅m1.72 \times 10 ^ { - 8 } \Omega \cdot \mathrm { m }1.72×10−8Ω⋅m

The Resistivity of A $1.0 \mathrm {~m}$ Long Copper Wire Is $1.72 \times 10 ^ { - 8 } \Omega \cdot \mathrm { m }$