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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

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