# 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.

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}