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Chapter 18: Problem 14

A current of \(2.50 \mathrm{A}\) is carried by a copper wire of radius $1.00 \mathrm{mm} .\( If the density of the conduction electrons is \)8.47 \times 10^{28} \mathrm{m}^{-3},$ what is the drift speed of the conduction electrons?

Short Answer

Expert verified
Answer: The drift speed of the conduction electrons in the copper wire is approximately \(1.160 \times 10^{-5} \mathrm{m}\mathrm{s}^{-1}\).

Step by step solution

01

Write down the given values

We are given the following values: - Current, \(I = 2.50 \mathrm{A}\) - Radius of the wire, \(r = 1.00 \mathrm{mm} = 1.00 \times 10^{-3} \mathrm{m}\) - Electron density, \(n = 8.47 \times 10^{28} \mathrm{m}^{-3}\) - Elementary charge, \(q = 1.60 \times 10^{-19} \mathrm{C}\) (this is a known constant)
02

Calculate the cross-sectional area of the wire

The cross-sectional area of the wire, \(A\), can be found using the formula for the area of a circle: \(A=\pi r^2\). Substituting the given radius, we have: \(A = \pi (1.00 \times 10^{-3} \mathrm{m})^2 = 3.14 \times 10^{-6} \mathrm{m}^2\)
03

Rearrange the formula for current to find the drift speed

The formula for current is \(I=nqA\upsilon_d\). We want to find the drift speed, \(\upsilon_d\), so we need to rearrange the formula to isolate \(\upsilon_d\): \(\upsilon_d = \frac{I}{nqA}\)
04

Substitute given values and solve for the drift speed

Now we will substitute the given values for \(I\), \(n\), \(q\), and \(A\) into the formula and solve for \(\upsilon_d\): \(\upsilon_d = \frac{2.50 \mathrm{A}}{(8.47 \times 10^{28} \mathrm{m}^{-3})(1.60 \times 10^{-19} \mathrm{C})(3.14 \times 10^{-6} \mathrm{m}^2)}\) \(\upsilon_d \approx 1.160 \times 10^{-5} \mathrm{m}\mathrm{s}^{-1}\) So, the drift speed of the conduction electrons in the copper wire is approximately \(1.160 \times 10^{-5} \mathrm{m}\mathrm{s}^{-1}\).

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Most popular questions from this chapter

In the physics laboratory, Oscar measured the resistance between his hands to be \(2.0 \mathrm{k} \Omega .\) Being curious by nature, he then took hold of two conducting wires that were connected to the terminals of an emf with a terminal voltage of \(100.0 \mathrm{V} .\) (a) What current passes through Oscar? (b) If one of the conducting wires is grounded and the other has an alternate path to ground through a \(15-\Omega\) resistor (so that Oscar and the resistor are in parallel), how much current would pass through Oscar if the maximum current that can be drawn from the emf is \(1.00 \mathrm{A} ?\)
A \(12-\Omega\) resistor has a potential difference of \(16 \mathrm{V}\) across it. What current flows through the resistor?
The label on a \(12.0-\mathrm{V}\) truck battery states that it is rated at 180.0 A-h (ampere-hours). Treat the battery as ideal. (a) How much charge in coulombs can be pumped by the battery? [Hint: Convert A to A-s.] (b) How much electric energy can the battery supply? (c) Suppose the radio in the truck is left on when the engine is not running. The radio draws a current of 3.30 A. How long does it take to drain the battery if it starts out fully charged?
A silver wire of diameter 1.0 mm carries a current of \(150 \mathrm{mA} .\) The density of conduction electrons in silver is $5.8 \times 10^{28} \mathrm{m}^{-3} .$ How long (on average) does it take for a conduction electron to move \(1.0 \mathrm{cm}\) along the wire?
A \(500-\) W electric heater unit is designed to operate with an applied potential difference of \(120 \mathrm{V}\). (a) If the local power company imposes a voltage reduction to lighten its load, dropping the voltage to $110 \mathrm{V}$, by what percentage does the heat output of the heater drop? (Assume the resistance does not change.) (b) If you took the variation of resistance with temperature into account, would the actual drop in heat output be larger or smaller than calculated in part (a)?
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