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An electrical conductor designed to carry large currents has a circular cross section \(2.50 \mathrm{~mm}\) in diameter and is \(14.0 \mathrm{~m}\) long. The resistance between its ends is \(0.104 \Omega\). (a) What is the resistivity of the material? (b) If the electric-field magnitude in the conductor is \(1.28 \mathrm{~V} / \mathrm{m},\) what is the total current? (c) If the material has \(8.5 \times 10^{28}\) free electrons per cubic meter, find the average drift speed under the conditions of part (b).

Short Answer

Expert verified
The resistivity of the material is \(1.76 \times 10^{-8} \Omega.m\), the total current is \(172.65 A\), and the average drift speed is \(1.01 \times 10^{-4} m/s\).

Step by step solution

01

Calculate the resistivity of the material

The resistance of a cylindrical conductor is given by the formula: \(R = \frac{\rho L}{A}\), where \(R\) is resistance, \(\rho\) is resistivity, \(L\) is the length of the conductor and \(A\) is cross-sectional area. The cross-sectional area can be calculated as \(A = \pi r^2\), where \(r\) is the radius of the conductor (half of the diameter, which is given). Plug the given values into these formulas and isolate \(\rho\) to calculate the resistivity: \(\rho = R\frac{A}{L}\).
02

Calculate the total current

The electric field \(E\) in a conductor is defined as the voltage \(V\) divided by the length of the conductor \(L\), therefore \(V = E \times L\). Ohm's law states that \(I = \frac{V}{R}\), where \(I\) is the current (which we need to find), \(V\) is voltage and \(R\) is resistance. Substituting the first equation into the second, one gets \(I = \frac{E \times L}{R}\). Plug in the given values to find the current.
03

Calculate the average drift speed

The drift speed \(v\) can be calculated using the formula: \(I = nqAv\), where \(I\) is the current, \(n\) is the number of free charge carriers per unit volume, \(q\) is the charge of the carriers and \(A\) is the cross-sectional area. In this case, \(q\) is the charge of an electron, which is \(1.6 \times 10^{-19}\) C. Solve this formula for \(v\) to get \(v = \frac{I}{nqA}\). Plug the given values, the cross-sectional area from step 1, and the acquired current from step 2 into this formula to get the drift speed.

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Key Concepts

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

Circular Cross-Section Conductor
The concept of a circular cross-section conductor is important in understanding how electricity flows through a wire. When we say a conductor has a circular cross-section, we are referring to its shape as a cylinder. This is commonly the shape for wires, rods, and cables that conduct electricity. The diameter is key in determining the cross-sectional area, which affects the conductor’s resistance and overall performance. To calculate the cross-sectional area, use the formula
  • Let the radius be half of the diameter: \(r = \frac{d}{2}\).
  • Calculate the area as:\(A = \pi r^2\).
Thus, an accurate calculation of the radius and area helps in determining properties like resistance and ampacity of the circular conductor. Since these factors are essential for electrical applications, understanding this concept is crucial.
Electric Field
The electric field in a conductor is a measure of force that acts on a charge. It’s this field that drives the free electrons through the conductor, allowing current to flow. The strength of the electric field is directly related to the potential difference (voltage) and inversely related to the length of the conductor. According to the formula
  • Electric field, \(E = \frac{V}{L}\),
  • where \(V\) is the voltage and \(L\) is the length.
A higher electric field means that charges are pushed more forcefully, resulting in greater current flow. It's important to consider the material of the conductor to determine how it will react to a given electric field.
Drift Speed
Drift speed describes how fast the electrons in a conductor are moving due to an electric field. Although electrons are always moving randomly, the presence of an electric field creates a small net velocity in one direction. We calculate drift speed using the formula
  • Drift speed \(v = \frac{I}{nqA}\),
where \(I\) is current, \(n\) is the density of free charge carriers, \(q\) is the charge of an electron, and \(A\) is the cross-sectional area. Although drift speed is generally quite low, it is essential for understand how quickly electricity can be transmitted through a conductor. Factors like the density of free electrons and the current affect the drift speed.
Ohm's Law
Ohm's Law is a fundamental principle in electronics and electrical engineering. It describes the relationship between voltage, current, and resistance in a simple equation:
  • \(V = IR\),
where \(V\) is voltage, \(I\) is the current, and \(R\) is the resistance. This law allows us to understand how changing one of these three quantities affects the other two. It also helps in designing circuits to ensure they function correctly. For instance, by knowing two variables, you can solve for the third, allowing you to predict how a component will behave in a circuit. Ohm's Law serves as a critical tool in both theoretical and applied physics.
Cylindrical Conductor Resistance Formula
Resistivity and resistance are two related but distinct concepts. Resistivity refers to the inherent property of a material that hinders electric current, whereas resistance is a measure of how much a particular conductor opposes the electric current.
For cylindrical conductors, you can calculate the resistance using
  • \(R = \frac{\rho L}{A}\),
where \(\rho\) is the resistivity, \(L\) is the length, and \(A\) is the cross-sectional area. A larger cross-sectional area or a shorter length results in less resistance, making it easier for current to flow. Choosing the right materials and dimensions for your conductor is key to achieving desired electrical properties. Understanding this formula is essential when dealing with electrical design and problem-solving.

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Most popular questions from this chapter

The voltage drop \(V_{a b}\) across each of resistors \(A\) and \(B\) was measured as a function of the current \(I\) in the resistor. The results are shown in the table: $$ \begin{array}{l|llll} \text { Resistor } A & & & & \\ I(\mathrm{~A}) & 0.50 & 1.00 & 2.00 & 4.00 \\ V_{a b}(\mathrm{~V}) & 2.55 & 3.11 & 3.77 & 4.58 \\ & & & & \\ \begin{array}{l} \text { Resistor } B \\ I(\mathrm{~A}) \end{array} & 0.50 & 1.00 & 2.00 & 4.00 \\ V_{a b}(\mathrm{~V}) & 1.94 & 3.88 & 7.76 & 15.52 \end{array} $$ (a) For each resistor, graph \(V_{a b}\) as a function of \(I\) and graph the resistance \(R=V_{a b} / I\) as a function of \(I\). (b) Does resistor \(A\) obey Ohm's law? Explain. (c) Does resistor \(B\) obey Ohm's law? Explain. (d) What is the power dissipated in \(A\) if it is connected to a \(4.00 \mathrm{~V}\) battery that has negligible internal resistance? (e) What is the power dissipated in \(B\) if it is connected to the battery?

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 \(t=1.50 \mathrm{~s} ?\)

A silver wire \(2.6 \mathrm{~mm}\) in diameter transfers a charge of \(420 \mathrm{C}\) in 80 min. Silver contains \(5.8 \times 10^{28}\) free electrons per cubic meter. (a) What is the current in the wire? (b) What is the magnitude of the drift velocity of the electrons in the wire?

Current passes through a solution of sodium chloride. In \(1.00 \mathrm{~s}, 2.68 \times 10^{16} \mathrm{Na}^{+}\) ions arrive at the negative electrode and \(3.92 \times 10^{16} \mathrm{Cl}^{-}\) ions arrive at the positive electrode. (a) What is the current passing between the electrodes? (b) What is the direction of the current?

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?

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