A potential difference of 1.2 nV is set up across a 4.6 m length of copper wire that has a radius...

Question:

A potential difference of 1.2 nV is set up across a 4.6 m length of copper wire that has a radius of 2.00 mm. How much charge drifts through a cross-section in 1.1 ms? Express your answer in units of mega coulombs i.e. 106 C. (Specify your answer up to one decimal place)

Electric Current

Electric current is defined as the flow of electric charge with respect to time. Electric current {eq}I = \dfrac { q } { t } {/eq} . Here q and t are the amount of charge flowing in the given time t. According to Ohm's law current through a conductor of resistance R under a potential difference V is {eq}I = \dfrac { V } { R } {/eq}. The resistance of a conductor of length L and area of cross section A is {eq}R = \dfrac { \rho L } {A } {/eq}. Here {eq}\rho {/eq} is the resistivity of the material.

Answer and Explanation:

Given points

  • Potential difference {eq}V = 1.2 \times 10^{-9} \ V {/eq}
  • Length of the copper wire L = 4.6 \ m
  • Radius of the copper wire {eq}r = 2.00 \times 10^{-3 } \ m {/eq}
  • Given time duration {eq}t = 1.1 \times 10^{-3 } \ s {/eq}
  • Resistivity of copper {eq}\rho = 1.72 \times 10^{-8} \ \Omega. m {/eq}

Resistance of the copper wire {eq}R = \dfrac { \rho L } {A } \\ R = \dfrac { 1.72 \times 10^{-8} \times 4.6 } { \pi \times ( 2.0 \times 10^{-3 } )^2 } \\ R = 6.296 \times 10^{-3} \ \Omega {/eq}

So then current through the wire {eq}I = \dfrac { V } { R } \\ I = \dfrac { 1.2 \times 10^{-9} } { 6.296 \times 10^{-3} } \\ I = 1.906 \times 10^{-7} \ A {/eq}

So the charge flowing through the cross section in the given time {eq}q = I \times t \\ q = 1.906 \times 10^{-7} \times 1.1 \times 10^{-3 } \\ q = 2.1 \times 10^{-10} \ C {/eq}


Learn more about this topic:

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What is Electric Current? - Definition, Unit & Types

from CLEP Natural Sciences: Study Guide & Test Prep

Chapter 6 / Lesson 7
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