# A copper wire of radius 0.1 mm and resistance 1 k\Omega is connected across a power supply of 20...

## Question:

A copper wire of radius 0.1 mm and resistance 1 k{eq}\Omega {/eq} is connected across a power supply of 20 V.

a) How many electrons are transferred per second between the supply and the wire at one end?

b) What is the current density in the wire?

## Current Density:

The current density in a wire of uniform cross-section is defined as the total charge conducted through a unit cross-section of the wire per unit time. If {eq}I {/eq} be current through a wire of uniform cross-section {eq}A {/eq}, then the current density can be given by:

{eq}\displaystyle{J = \dfrac{I}{A}} {/eq}.

The current density is a vector quality and its direction is the same as the drift velocity. The SI unit of the current density is {eq}\rm A/m^2 {/eq}.

We know the current through a wire is defined as:

{eq}\displaystyle{I = \dfrac{dq}{dt}} {/eq}.

If {eq}N {/eq} number of electrons is responsible for the charge {eq}q {/eq}, then we write:

{eq}\displaystyle{q = Ne} {/eq}.

({eq}e {/eq} is an electronic charge).

Therefore, {eq}\displaystyle{I = e\dfrac{dN}{dt}} {/eq}.

Therefore, the rate of the number of conducted electrons with respect to time can be given by:

{eq}\displaystyle{\dfrac{dN}{dt} = \dfrac{I}{e}} {/eq}.

If the current {eq}I {/eq} is caused, owing to the potential difference {eq}V {/eq} between the ends of a resistor of radius {eq}R {/eq}, we have the current through the wire given by:

{eq}\displaystyle{I = \dfrac{V}{R}} {/eq}.

Then, the rate of the number of electrons conducted across the resistor can be given by:

{eq}\displaystyle{\dfrac{dN}{dt} = \dfrac{V}{Re}} {/eq}.

The current density can be given as:

{eq}\displaystyle{J = \dfrac{V}{RA}} {/eq}.

{eq}A {/eq} is the cross-section of the resistor.

## Answer and Explanation:

**Given**

- The resistance of the copper wire: {eq}R = 1 \ \rm k\omega\times \dfrac{10^3 \omega}{1 \ \rm k\omega} = 10^3 \ \omega {/eq}.

- The potential difference across the copper wire: {eq}V = 20 \ \rm V {/eq}.

- Radius of the copper wire: {eq}r = 0.1 \ \rm mm\times \dfrac{1 \ \rm m}{10^{3} \ \rm mm} = 10^{-4} \ \rm m {/eq}.

Therefore, the area of the cross-section of the copper wire is given by: {eq}A = \pi r^2 {/eq}.

**Known**

- Electronic charge: {eq}e = 1.6\times 10^{-19} \ \rm C {/eq}.

**a)** The rate of the number of electrons conducted across the wire will be given by:

{eq}\displaystyle{\begin{align*} \dfrac{dN}{dt} &= \dfrac{V}{ R \ e}\\ &= \dfrac{20 \ \rm V}{(10^3 \ \Omega)\times (1.6\times 10^{-19} \ \omega)}\\ &= \boxed{{1.25\times 10^{17} /\rm s}}\\ \end{align*}} {/eq}

**b)** The current density in the wire can be given by:

{eq}\displaystyle{\begin{align*} J &= \dfrac{V}{R \ A}\\ &= \dfrac{V}{R\times \pi r^2}\\ &= \dfrac{(20 \ \rm V)}{(10^3 \ \Omega)\times \pi \times (10^{-4} \ \rm m)^2}\\ &\approx \boxed{{63661.98 \ \rm A/m^2}}\\ \end{align*}} {/eq}

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from CLEP Natural Sciences: Study Guide & Test Prep

Chapter 6 / Lesson 7