/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 A gold wire of 0.50 mm diameter ... [FREE SOLUTION] | 91影视

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Chapter 18: Problem 18

A gold wire of 0.50 mm diameter has \(5.90 \times 10^{28} \mathrm{con}\) duction electrons/m". If the drift speed is \(6.5 \mu \mathrm{m} / \mathrm{s}\), what is the current in the wire?

Short Answer

Expert verified
Answer: The current in the gold wire is approximately \(1.22 \times 10^{-3}\) A.

Step by step solution

01

Calculate the cross-sectional area

First, we need to find the cross-sectional area (A) of the gold wire. For a cylindrical wire, the area is computed using the formula: \(A = \pi r^2\), where r is the radius. The diameter of the wire is given as 0.50 mm, so the radius would be \(r = \frac{0.50}{2}\) mm = \(0.25\) mm =\(2.50 \times 10^{-4}\) m. Now calculating the area: \(A = \pi r^2 = \pi (2.50 \times 10^{-4})^2 = \pi (6.25 \times 10^{-8}) = 1.9635 \times 10^{-7} \, \mathrm{m}^2\)
02

Charge of one electron

The charge of one electron (q) is a constant value, which is approximately \(1.6 \times 10^{-19}\) Coulombs.
03

Calculate the number of conduction electrons

We are given that the concentration of conduction electrons is \(5.90 \times 10^{28}\, \mathrm{m}^{-3}\). We will use this value directly as the number of conduction electrons (n) per unit volume.
04

Calculate the current in the wire

Now that we have all the required values, we can use the formula for current in terms of drift speed, charge of an electron, cross-sectional area of the wire, and concentration of conduction electrons, which is: \(I = nqAv\) where: I = current n = number of conduction electrons per unit volume q = charge of one electron A = cross-sectional area of the wire v = drift speed Substituting the known values: \(I = (5.90 \times 10^{28} \, \mathrm{m}^{-3})(1.6 \times 10^{-19} \, \mathrm{C})(1.9635 \times 10^{-7} \, \mathrm{m}^2)(6.5 \times 10^{-6} \, \mathrm{m/s})\) \(I = 1.22 \times 10^{-3}\, \mathrm{A}\) So the current in the gold wire is approximately \(1.22 \times 10^{-3}\) A.

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