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A copper wire has a square cross section 2.3 \(\mathrm{mm}\) on a side. The wire is 4.0 \(\mathrm{m}\) long and carries a current of 3.6 \(\mathrm{A}\) . The density of free electrons is \(8.5 \times 10^{28} / \mathrm{m}^{3} .\) 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

Expert verified
(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 \(8000\,\text{s}\).

Step by step solution

01

Calculate the cross-sectional area

Given that the cross-section of the wire is square with side length 2.3 mm, first convert the side length to meters: \(2.3\,\text{mm} = 2.3 \times 10^{-3}\,\text{m}\). The cross-sectional area \(A\) is calculated as: \(A = (2.3 \times 10^{-3})^2\,\text{m}^2 = 5.29 \times 10^{-6}\,\text{m}^2\).
02

Compute the current density

Current density \(J\) is calculated using the formula \(J = \frac{I}{A}\), where \(I\) is the current. Given \(I = 3.6\,\text{A}\) and \(A = 5.29 \times 10^{-6}\,\text{m}^2\), the current density is \(J = \frac{3.6}{5.29 \times 10^{-6}} = 6.81 \times 10^{5}\,\text{A/m}^2\).
03

Calculate the drift velocity

The drift velocity \(v_d\) of electrons is given by \(v_d = \frac{J}{nq}\), where \(n = 8.5 \times 10^{28}\,\text{electrons/m}^3\) is the electron density and \(q = 1.6 \times 10^{-19}\,\text{C}\) is the charge of an electron. Substitute to get \(v_d = \frac{6.81 \times 10^5}{8.5 \times 10^{28} \times 1.6 \times 10^{-19}} = 5.0 \times 10^{-4} \,\text{m/s}\).
04

Determine the electric field in the wire

Use Ohm’s law in the form \(E = \frac{J}{\sigma}\), where \(\sigma\) is the conductivity of copper \((\sigma \approx 5.96 \times 10^7\,\text{S/m})\). Compute \(E = \frac{6.81 \times 10^5}{5.96 \times 10^7} = 0.0114 \text{ V/m}\).
05

Calculate the time required for an electron to travel the length of the wire

The time \(t\) required is calculated by \(t = \frac{L}{v_d}\), where \(L = 4.0\,\text{m}\). With \(v_d = 5.0 \times 10^{-4}\,\text{m/s}\), we find \(t = \frac{4.0}{5.0 \times 10^{-4}} = 8000\,\text{s}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Drift Velocity
Drift velocity is the average velocity that a particle, such as an electron, attains due to an electric field in a conductor. In simple terms, it's the speed at which electrons move through the wire when a current is flowing. The concept is crucial for understanding how quickly or slowly electric current travels through materials.
To calculate the drift velocity, we use the formula:
  • \( v_d = \frac{J}{nq} \)
Here, \( J \) represents the current density, \( n \) is the density of free electrons, and \( q \) is the charge of an electron. In the given example, with \( J = 6.81 \times 10^5 \, \text{A/m}^2 \), \( n = 8.5 \times 10^{28} \, \text{electrons/m}^3 \), and \( q = 1.6 \times 10^{-19} \, \text{C} \), the drift velocity is calculated as:\[ v_d = \frac{6.81 \times 10^5}{8.5 \times 10^{28} \times 1.6 \times 10^{-19}} = 5.0 \times 10^{-4} \, \text{m/s} \]Despite how fast we perceive electricity to be, the actual drift velocity of electrons is quite slow due to their chaotic, random motion in the conductor.
Electric Field
The electric field within a conductor is a vector quantity describing the force per unit charge. It acts on charged particles and causes them to move, resulting in an electric current. In everyday language, it's the push that moves electrons in a wire, making electrical devices work.
In conductors like copper, the electric field is generally weak because metals are excellent conductors. To find the electric field in a wire, we apply Ohm's law in a unique form:
  • \( E = \frac{J}{\sigma} \)
where \( J \) is the current density and \( \sigma \) is the conductivity of the material. For copper, with \( \sigma \approx 5.96 \times 10^7 \, \text{S/m} \), and the already calculated \( J = 6.81 \times 10^5 \, \text{A/m}^2 \), the electric field \( E \) is derived as:\[ E = \frac{6.81 \times 10^5}{5.96 \times 10^7} = 0.0114 \, \text{V/m} \]
Ohm's Law
Ohm's Law is a foundational principle in electrical engineering and physics that relates voltage, current, and resistance in a conductor. Ohm's Law states that the current passing through a conductor between two points is directly proportional to the voltage across the two points, provided the temperature remains constant.
This fundamental law is expressed mathematically as:
  • \( V = IR \)
where \( V \) is the voltage, \( I \) is the current, and \( R \) is the resistance. In the context of an electric field, we can rearrange Ohm's Law to use current density (\( J \)) and conductivity (\( \sigma \)), as shown in the calculation of the electric field:\[ E = \frac{J}{\sigma} \]Ohm's Law is practical because it helps us understand how electrical circuits behave. It allows us to calculate one of the three variables—voltage, current, or resistance—as long as we know the other two. This simplification is especially useful for designing and troubleshooting electrical circuits.

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Most popular questions from this chapter

A 5.00 -A current runs through a 12 -gauge copper wire (diameter 2.05 \(\mathrm{mm}\) ) and through a light bulb. Copper has \(8.5 \times\) \(10^{28}\) free electrons per cubic meter, (a) How many electrons pass through the light bulb each second? (b) What is the current density in the wire? (c) At what speed does a typical electron pass by any given point in the wire? (d) If you were to use wire of twice the diameter, which of the above answers would change? Would they increase or decrease?

A person with body resistance between his hands of 10 \(\mathrm{k} \Omega\) accidentally grasps the terminals of a \(14-\mathrm{kV}\) power supply. (a) If the internal resistance of the power supply is \(2000 \Omega,\) what is the current through the person's body? (b) What is the power dissipated in his body? (c) If the power supply is to be made safe by increasing its internal resistance, what should the internal resistance be for the maximum current in the above situation to be 1.00 \(\mathrm{mA}\) or less?

Electric Eels. Electric eels generate electric pulses along their skin that can be used to stun an enemy when they come into contact with it. Tests have shown that these pulses can be up to 500 \(\mathrm{V}\) and produce currents of 80 \(\mathrm{mA}\) (or even larger). A typical pulse lasts for 10 \(\mathrm{ms}\) . What power and how much energy are delivered to the unfortunate enemy with a single pulse, assuming a steady current?

Copper has \(8.5 \times 10^{28}\) free electrons per cubic meter. A 71.0 -cm length of 12 -gauge copper wire that is 2.05 \(\mathrm{mm}\) in diameter carries 4.85 A of current. (a) How much time does it take for an electron to travel the length of the wire? (b) Repeat part (a) for 6-gauge copper wire (diameter 4.12 \(\mathrm{mm}\) ) of the same length that carries the same current.(c) Generally speaking, how does changing the diameter of a wire that carries a given amount of current affect the drift velocity of the electrons in the wire?

A wire 6.50 \(\mathrm{m}\) long with diameter of 2.05 \(\mathrm{mm}\) has a resistance of 0.0290\(\Omega .\) What material is the wire most likely made of?

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