Question

A copper wire has a square cross section 2.3 mm on a side. The wire is 4.0 m long and carries a current of 3.6 A. The density of free electrons is 8.5 \(\times\) 10\(^{28}\)/m\({^3}\). Find the magnitudes of (a) the current density in the wire and (b) the electric field in the wire. (c) How much time is required for an electron to travel the length of the wire?

Short answer

(a) Current density is \( 6.81 \times 10^5 \text{ A/m}^2 \). (b) Electric field is \( 0.0114 \text{ V/m} \). (c) Time is approximately 7984 s.

Step-by-step solution

Calculate the cross-sectional area

The wire has a square cross-section with each side measuring 2.3 mm. Convert this measurement into meters: \[ 2.3 ext{ mm} = 2.3 imes 10^{-3} ext{ m} \]The area \( A \) is then given by:\[ A = ( ext{side})^2 = (2.3 imes 10^{-3} ext{ m})^2 = 5.29 imes 10^{-6} ext{ m}^2 \]

Calculate the current density

Current density \( J \) is defined as the current \( I \) divided by the cross-sectional area \( A \):\[ J = \frac{I}{A} = \frac{3.6 ext{ A}}{5.29 imes 10^{-6} ext{ m}^2} \approx 6.81 imes 10^5 ext{ A/m}^2 \]

Use Ohm's Law to find the electric field

Ohm's Law in terms of current density is \( J = \sigma E \), where \( \sigma \) is the conductivity and \( E \) is the electric field. Conductivity can be expressed as \( \sigma = n e \mu \), where \( n \) is the density of electrons, \( e \) is the elementary charge \( (1.6 imes 10^{-19} ext{ C}) \), and \( \mu \) is mobility:First, find the mobility using the relationship between resistivity \( \rho \), where \( \rho = 1/\sigma \) and standard values:For copper, \( \rho \approx 1.68 \times 10^{-8} \text{ Ohm} \cdot \text{m} \) Thus, \( \sigma = 1/\rho \approx 5.96 imes 10^7 \text{ S/m} \).Now using \( J = \sigma E \) we get:\[ E = \frac{J}{\sigma} = \frac{6.81 \times 10^5}{5.96 \times 10^7} \approx 0.0114 \text{ V/m} \]

Calculate the electron drift velocity

The drift velocity \( v_d \) is given by the relation \( J = n e v_d \), solving for \( v_d \):\[ v_d = \frac{J}{n e} = \frac{6.81 \times 10^5}{8.5 \times 10^{28} \times 1.6 \times 10^{-19}} \approx 5.01 \times 10^{-4} \text{ m/s} \]

Find time taken for electron to travel the wire length

Using the drift velocity \( v_d \) and wire length \( L = 4.0 \text{ m} \), time \( t \) is:\[ t = \frac{L}{v_d} = \frac{4.0}{5.01 \times 10^{-4}} \approx 7984.03 \text{ s} \]

Key concepts

Electric Field

The electric field in a conductor is a fundamental concept in electricity. It represents the force that sets the charge carriers, like electrons, in motion. This electric field arises due to the electric potential difference at the ends of the wire.
When a voltage is applied across a wire, it creates an electric field.

• The strength of this field in the wire is influenced by several factors, such as the material's conductivity and the current flowing through it.
• The electric field is calculated using the relationship with current density (Ohm's Law), where it is proportional to the current density and inversely proportional to the material's conductivity.

Understanding this concept is crucial because a stronger electric field means more force is exerted on the electrons, causing them to flow more rapidly through the conductor.

Ohm's Law

Ohm's Law is a cornerstone of electrical circuits, defining the relationship between voltage, current, and resistance in a conductor.
Expressed as \( V = IR \), where \( V \) is the voltage, \( I \) is the current, and \( R \) is the resistance. Ohm's Law can be adapted to include current density and electric field.

• In terms of current density, it uses the formula \( J = \sigma E \), where \( J \) is current density, \( \sigma \) is conductivity, and \( E \) is the electric field.
• Conductivity \( \sigma \) links the drift velocity of charge carriers to the efficiency of their movement within a material.

This law provides the framework for calculating the electric field in a given wire or component, considering its conductivity and the current running through it.

Drift Velocity

Drift velocity is a measure of how fast charge carriers, mainly electrons, are moving through a conductor.
Though it might seem electrons move fast, their drift velocity is quite slow because they constantly collide with atoms in the conductor.

• The drift velocity \( v_d \) is calculated by \( v_d = \frac{J}{n e} \), where \( J \) is the current density, \( n \) is the density of charge carriers, and \( e \) is the elementary charge.
• This velocity indicates how long it takes an electron to move from one end of the wire to the other.

Even though individual electrons move slowly, the overall effect produces a rapid propagation of electrical energy along the conductor.

Electron Mobility

Electron mobility represents how quickly an electron can move through a material when an electric field is applied. It is a significant property that affects a material's conductivity.

• Mathematically, electron mobility \( \mu \) is expressed in terms of conductivity and resistivity. Meeting the formula \( \sigma = n e \mu \), with \( n \) as the density of electrons and \( e \) as the charge of an electron.
• High electron mobility means that electrons can traverse the conductor with less resistance, making it a highly conductive material.

Copper, for instance, is known for its high electron mobility, contributing to its popularity as an efficient conductor in electrical systems.