# Consider a wire of length 4 m and cross-sectional area 1 mm^2 carrying a current 2 A. If each...

## Question:

Consider a wire of length 4 m and cross-sectional area 1 mm{eq}^2 {/eq} carrying a current 2 A. If each cubic meter of the material contains {eq}10^{29} {/eq} free electrons, find the average time taken by an electron to cross the length of the wire.

## Electric current

Elecric current I is the amount of electrical charge Q passing at a given point per unit of time t. It is mathematically expressed as:

{eq}I = \dfrac{Q}{t} {/eq}

One electron has the charge

{eq}e = -1.602 \times 10^{-19} \ \rm C {/eq}

The total charge of n electrons.

{eq}Q = n e {/eq}

The number of free electrons is the product of the density times the volume.

{eq}n = V \rho {/eq}

The volume of a wire is the cross section times the length

{eq}V = A L {/eq}

Unit conversion

{eq}1 \rm \ mm^2 = 1 \times 10^{-6} \ \rm m^2 {/eq}

## Answer and Explanation:

The current in the wire is the charge of the free electrons go through the wire in opposite direction of the current. Thus we can ignore the negative sign of the electron charge. Suppose all the free electrons go through the wire within time t. We can have the equation and solve for t.

{eq}\begin{align} \dfrac{Q}{t} & = I \\ \dfrac{ne}{t} & = I\\ \dfrac{V\rho e}{t} & = I\\ \dfrac{A L \rho e}{t} & = I\\ \dfrac{A L \rho e}{I} & = t\\ t & = \dfrac{A L \rho e}{I}\\ t & = \dfrac{1\times 10^{-6} * 4 * 10^{29} * 1.602 \times 10^{-19}}{2}\\ t & = \boxed{\bf 3.204 \times 10^4 \ \rm s} \end{align} {/eq}

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from CLEP Natural Sciences: Study Guide & Test Prep

Chapter 6 / Lesson 7