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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?

Short Answer

Expert verified
The current in the wire is \(I = \frac{420C}{4800s}\) A. The magnitude of the drift velocity of the electrons in the wire is \(v = \frac{I}{5.8 x 10^{28} electrons/m^3 x A x 1.6 x 10^{-19} C}\) m/s.

Step by step solution

01

Find the Current

The formula for current is \(I = \frac{Q}{t}\). Given that \(Q = 420C\) and \(t = 80 min = 4800s\), we can plug these numbers into the formula: \(I = \frac{420C}{4800s}\)
02

Calculate the Area

The area of the wire can be calculated using the formula for the area of a cylinder \(A = πr^2\), where \(r\) is the radius of the wire. Given that the diameter of the wire is \(2.6mm\), the radius is \(\frac{2.6mm}{2} = 1.3mm = 1.3x10^-3m\). Substituting the radius into the formula gives us \(A = π(1.3x10^-3m)^2\).
03

Find the Drift Velocity

The formula for drift velocity is \(v = \frac{I}{nAe}\), where \(v\) is the drift velocity, \(I\) is the current, \(n\) is the number of free electrons per cubic meter, \(A\) is the area and \(e\) is the charge of an electron. Substituting in the given values of \(I\), \(n = 5.8 x 10^{28} electrons/m^3\), \(A\), and \(e = 1.6 x 10^{-19} C\), into the drift velocity formula gives \(v = \frac{I}{5.8 x 10^{28} electrons/m^3 x A x 1.6 x 10^{-19} C}\).

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Key Concepts

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

Electric Current
Imagine a river flowing, with its water particles moving in a certain direction - this is similar to the concept of electric current, which is the flow of electric charge. Electric current is measured in Amperes (A) and is calculated using the formula:
\( I = \frac{Q}{t} \)
where I is the current, Q is the charge in Coulombs (C), and t is the time in seconds (s) for which the charge flows. If we think of charge as water in our river analogy, then current tells us how much water flows past a point every second.

In our exercise, the current in a silver wire is determined by the charge passing through the wire over an 80-minute period. Thus, we can better understand electrical current not just as an abstract quantity but as a rate of charge transfer, much like water flowing through a channel.
Cross-Sectional Area of Wire
Now let's consider the size of the conduit or the path through which our electric current flows. This brings us to the cross-sectional area of the wire. Just like a larger pipe can allow more water to flow through it, a wire with a larger cross-sectional area can allow more charge to pass through. It's measured in square meters (m²). In analogy, a wire's cross-sectional area affects its ability to carry an electric current much like a wide highway allows more cars to travel side by side.

The cross-sectional area is critical when calculating the drift velocity of electrons, as it directly influences how tightly packed the electrons are when moving through the wire. The standard formula for the area of a cylinder, which is used for wires, is \( A = \pi r^2 \) where r is the radius of the wire.
Charge Transfer in Conductors
When we flip a switch, we expect an immediate response as the electric current flows, but have you ever wondered how the charge moves within the conductor? This is described by the concept of charge transfer in conductors. In conductors, such as the silver wire in the exercise, electrons are the primary charge carriers, and the speed at which they drift through a conductor is referred to as 'drift velocity'.

The transfer of charge is facilitated by free electrons moving through the network of atoms in the wire, much like cars navigating a network of roads. The numerous free electrons in a conductor, like silver, drift in response to an electric field. These electrons do not move rapidly; instead, they drift slowly but steadily. It's this coordinated movement that constitutes electric current, akin to traffic flowing smoothly in one direction. Understanding the mechanisms behind charge transfer helps students grasp how rapidly or slowly electrons navigate through conductors which is central to the study of electrical engineering and physics.

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Most popular questions from this chapter

In another experiment, a piece of the web is suspended so that it can move freely. When either a positively charged object or a negatively charged object is brought near the web, the thread is observed to move toward the charged object. What is the best interpretation of this observation? The web is (a) a negatively charged conductor; (b) a positively charged conductor; (c) either a positively or negatively charged conductor; (d) an electrically neutral conductor.

The current in a wire varies with time according to the relationship \(I=55 \mathrm{~A}-\left(0.65 \mathrm{~A} / \mathrm{s}^{2}\right) t^{2} .\) (a) How many coulombs of charge pass a cross section of the wire in the time interval between \(t=0\) and \(t=8.0 \mathrm{~s} ?(\mathrm{~b}) \mathrm{What}\) constant current would transport the same charge in the same time interval?

\(\mathrm{A}\) cell phone or computer battery has three ratings marked on it: a charge capacity listed in mAh (milliamp-hours), an energy capacity in Wh (watt-hours), and a potential rating in volts. (a) What are these three values for your cell phone? (b) Convert the charge capacity \(Q\) into coulombs. (c) Convert the energy capacity \(U\) into joules. (d) Multiply the charge rating \(Q\) by the potential rating \(V,\) and verify that this is equivalent to the energy capacity \(U\). (e) If the charge \(Q\) were stored on a parallel-plate capacitor with air as the dielectric, at the potential \(V,\) what would be the corresponding capacitance? (f) If the energy in the battery were used to heat \(1 \mathrm{~L}\) of water, estimate the corresponding change in the water temperature? (The heat capacity of water is \(4190 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K} .)\)

A dielectric of permittivity \(3.5 \times 10^{-11} \mathrm{~F} / \mathrm{m}\) completely fills the volume between two capacitor plates. For \(t>0\) the electric flux through the dielectric is \(\left(8.0 \times 10^{3} \mathrm{~V} \cdot \mathrm{m} / \mathrm{s}^{3}\right) t^{3} .\) The dielectric is ideal and nonmagnetic; the conduction current in the dielectric is zero. At what time does the displacement current in the dielectric equal \(21 \mu \mathrm{A} ?\)

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.

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