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The free-electron density in a copper wire is \(8.5 \times 10^{28}\) electrons \(/ \mathrm{m}^{3} .\) The electric field in the wire is \(0.0600 \mathrm{~N} / \mathrm{C}\) and the temperature of the wire is \(20.0^{\circ} \mathrm{C}\). (a) What is the drift speed \(v_{\mathrm{d}}\) of the electrons in the wire? (b) What is the potential difference between two points in the wire that are separated by \(20.0 \mathrm{~cm} ?\)

Short Answer

Expert verified
The drift speed \(v_{\mathrm{d}}\) of the electrons in the copper wire is computed using the given variables and the appropriate formula. After that, the potential difference \(V\) between two points in the wire, separated by \(20.0 \, \mathrm{cm} = 0.2 \, \mathrm{m}\), is calculated using \(V = E * d\).

Step by step solution

01

Establish the Known Variables

The task provided these variables:\nElectric field \(E = 0.0600 \, \mathrm{N/C}\)\nFree-electron density \(n = 8.5 \times 10^{28} \, \mathrm{electrons/m}^{3}\)\nElectron charge \(q = 1.6 \times 10^{-19} \, \mathrm{C}\) (This is a constant value that we should memorize)\nDistance \(d = 20.0 \, \mathrm{cm} = 0.2 \, \mathrm{m}\) (Convert cm to m as standard SI unit of length is meter)
02

Solve for Drift Speed \(v_{d}\)

The formula for drift speed is \(v_{d} = J / (n*q)\), where \(J\) is the current density, \(n\) is the free-electron density and \(q\) is the electron charge. The current density (\(J\)) can be found by multiplying electric field (\(E\)) and electron mobility (\(\mu_{\mathrm{e}}\)). Since there's no temperature dependency, we use the constant value for electron mobility in copper, i.e., \(\mu_{\mathrm{e}} = 0.00320 \, \mathrm{m^2/Vs}\). Now, calculate \(J = E * \mu_{\mathrm{e}}\) and substitute into the drift speed formula to find \(v_{d}\).
03

Calculate Potential Difference

The relation between potential difference \(V\), electric field \(E\), and distance \(d\) is \(V = E * d\). Use the given values for \(E\) and \(d\) to calculate \(V\).

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Key Concepts

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

Electric Field
The electric field in a conductor, such as a copper wire, plays a pivotal role in how current flows through the material. An electric field is created when a voltage is applied across the conductor, which generates a force exerted on free charged particles, such as electrons. In simple terms, it's like an invisible push that moves the electrons along the wire.

The strength of this electric field, measured in newtons per coulomb (N/C), dictates how strongly the electrons will be 'pushed'. Higher electric field strengths mean a greater force on each electron, which can influence the drift speed, or the average velocity at which the electrons move through the conductor. This is crucial for understanding how quickly electrical signals can be transmitted through materials and for calculating the current flow in circuits.
Free-Electron Density
Free-electron density refers to the number of free electrons available per unit volume within a material, typically measured in electrons per cubic meter (\( \text{electrons/m}^3 \)). In conductors like copper, a large number of electrons are not bound to any particular atom and can move freely. This characteristic is essential for the material's ability to conduct electric current.

The higher the free-electron density, the more electrons are available to move or 'drift' under the influence of an electric field. This can result in a higher current for a given electric field strength. It's interesting to note that the free-electron density is inherent to the material and is often a fixed value, like the one given for copper in our exercise, at room temperature. Variations in electron density can occur due to changes in temperature or the presence of impurities and can affect electrical conductivity.
Potential Difference
Potential difference, also commonly known as voltage, is the measure of the work needed to move a charge from one point to another in an electric field. It's the 'electrical pressure' that drives the flow of electrons through a circuit and is measured in volts (V).

The potential difference between two points in a wire is the product of the electric field strength and the distance between those two points, as shown by the formula: \( V = E * d \). This relationship tells us that a higher electric field over the same distance will produce a greater voltage, and conversely, for a given electric field, increasing the distance will also increase the voltage. Understanding this concept is essential for circuit analysis and for designing electronics that function properly under certain voltage levels.

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Most popular questions from this chapter

A circular loop of wire with radius \(2.00 \mathrm{~cm}\) and resistance \(0.600 \Omega\) is in a region of a spatially uniform magnetic field \(\vec{B}\) that is perpendicular to the plane of the loop. At \(t=0\) the magnetic field has magnitude \(B_{0}=3.00 \mathrm{~T}\). The magnetic field then decreases according to the equation \(B(t)=B_{0} e^{-t / \tau},\) where \(\tau=0.500 \mathrm{~s}\). (a) What is the maximum magnitude of the current \(I\) induced in the loop? (b) What is the induced current \(I\) when \(t=1.50 \mathrm{~s} ?\)

BIO 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?

A magnetic field of \(0.080 \mathrm{~T}\) is in the \(y\) -direction. The velocity of wire segment \(S\) has a magnitude of \(78 \mathrm{~m} / \mathrm{s}\) and components of \(18 \mathrm{~m} / \mathrm{s}\) in the \(x\) -direction, \(24 \mathrm{~m} / \mathrm{s}\) in the \(y\) -direction, and \(72 \mathrm{~m} / \mathrm{s}\) in the \(z\) -direction. The segment has length \(0.50 \mathrm{~m}\) and is parallel to the \(z\) -axis as it moves. (a) Find the motional emf induced between the ends of the segment. (b) What would the motional emf be if the wire segment was parallel to the \(y\) -axis?

A single loop of wire with an area of \(0.0900 \mathrm{~m}^{2}\) is in a uniform magnetic field that has an initial value of \(3.80 \mathrm{~T},\) is perpendicular to the plane of the loop, and is decreasing at a constant rate of \(0.190 \mathrm{~T} / \mathrm{s}\)(a) What emf is induced in this loop? (b) If the loop has a resistance of \(0.600 \Omega,\) find the current induced in the loop.

The magnetic flux through a coil is given by \(\Phi_{B}=\alpha t-\beta t^{3}\) where \(\alpha\) and \(\beta\) are constants. (a) What are the units of \(\alpha\) and \(\beta ?\) (b) If the induced emf is zero at \(t=0.500 \mathrm{~s},\) how is \(\alpha\) related to \(\beta ?\) (c) If the emf at \(t=0\) is \(-1.60 \mathrm{~V},\) what is the \(\mathrm{emf}\) at \(t=0.250 \mathrm{~s} ?\)

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