{"id":16875,"date":"2025-06-11T20:49:54","date_gmt":"2025-06-11T20:49:54","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=16875"},"modified":"2025-06-11T20:49:55","modified_gmt":"2025-06-11T20:49:55","slug":"the-12-gauge-copper-wire-in-typical-residential-building-has-cross-sectional-area-of-3-31-x-10-6-m","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/the-12-gauge-copper-wire-in-typical-residential-building-has-cross-sectional-area-of-3-31-x-10-6-m\/","title":{"rendered":"&#8216;The 12-gauge copper wire in typical residential building has cross-sectional area of 3.31 X 10-6 m"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">&#8216;The 12-gauge copper wire in typical residential building has cross-sectional area of 3.31 X 10-6 m? If it carries current of 10.0 A what is the drift speed of the electrons? Assume that each copper atom contributes one free electron to the current The density of copper is 8.95 g\/cm?_&#8217;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\"><strong>The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">To find the <strong>drift speed<\/strong> of electrons in a copper wire, we use the formula: vd=InAev_d = \\frac{I}{nAe}<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>vdv_d = drift speed (m\/s)<\/li>\n\n\n\n<li>II = current (A) = 10.0 A<\/li>\n\n\n\n<li>AA = cross-sectional area (m\u00b2) = 3.31\u00d710\u22126\u2009m23.31 \\times 10^{-6} \\, \\text{m}^2<\/li>\n\n\n\n<li>ee = elementary charge = 1.6\u00d710\u221219\u2009C1.6 \\times 10^{-19} \\, \\text{C}<\/li>\n\n\n\n<li>nn = number density of free electrons (electrons\/m\u00b3)<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Find the number density nn<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">To find nn, we need the number of free electrons per unit volume. First, use the <strong>density of copper<\/strong> and its <strong>molar mass<\/strong>:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Density of copper: \u03c1=8.95\u2009g\/cm3=8950\u2009kg\/m3\\rho = 8.95 \\, \\text{g\/cm}^3 = 8950 \\, \\text{kg\/m}^3<\/li>\n\n\n\n<li>Molar mass of copper: M=63.5\u2009g\/mol=0.0635\u2009kg\/molM = 63.5 \\, \\text{g\/mol} = 0.0635 \\, \\text{kg\/mol}<\/li>\n\n\n\n<li>Avogadro\u2019s number: NA=6.022\u00d71023\u2009atoms\/molN_A = 6.022 \\times 10^{23} \\, \\text{atoms\/mol}<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">Now compute number density: n=\u03c1NAM=8950\u00d76.022\u00d710230.0635\u22488.49\u00d71028\u2009electrons\/m3n = \\frac{\\rho N_A}{M} = \\frac{8950 \\times 6.022 \\times 10^{23}}{0.0635} \\approx 8.49 \\times 10^{28} \\, \\text{electrons\/m}^3<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Plug into the drift velocity formula<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">vd=10.0(8.49\u00d71028)(3.31\u00d710\u22126)(1.6\u00d710\u221219)v_d = \\frac{10.0}{(8.49 \\times 10^{28})(3.31 \\times 10^{-6})(1.6 \\times 10^{-19})} vd\u224810.04.496\u00d7104=2.22\u00d710\u22124\u2009m\/sv_d \\approx \\frac{10.0}{4.496 \\times 10^{4}} = 2.22 \\times 10^{-4} \\, \\text{m\/s}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\u2705 Final Answer:<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">vd\u22482.22\u00d710\u22124\u2009m\/s\\boxed{v_d \\approx 2.22 \\times 10^{-4} \\, \\text{m\/s}}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\ud83e\udde0 Explanation (300 Words):<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Drift speed refers to the <strong>average velocity<\/strong> that free electrons attain in a conductor when subjected to an electric field. Although electrical energy moves at nearly the speed of light in a wire, individual electrons drift very slowly\u2014on the order of fractions of a millimeter per second.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">In a typical copper wire (like 12-gauge used in homes), the current of 10 A might seem large, but copper has a <strong>very high electron density<\/strong>, meaning there are trillions of free electrons available in every cubic centimeter to carry the current.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Copper has one free electron per atom. Using its density and molar mass, we calculated that there are about 8.49\u00d710288.49 \\times 10^{28} free electrons in each cubic meter of copper. When current flows, each of these electrons contributes just a tiny bit, moving slowly but steadily in the direction opposite to the electric field.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">The drift speed calculation uses the idea that the total current is the product of the number of charge carriers, the charge per carrier, the cross-sectional area, and their average velocity. Even though electrons are moving slowly (here, \u22480.22\\approx 0.22 mm\/s), their vast numbers ensure that a substantial current is maintained.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">This low drift speed emphasizes that the <strong>signal propagation<\/strong> in a circuit is not due to electrons zooming from point A to B, but rather due to a chain reaction\u2014like pushing a line of marbles\u2014where energy is transferred almost instantaneously, even if each electron moves slowly.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>&#8216;The 12-gauge copper wire in typical residential building has cross-sectional area of 3.31 X 10-6 m? If it carries current of 10.0 A what is the drift speed of the electrons? Assume that each copper atom contributes one free electron to the current The density of copper is 8.95 g\/cm?_&#8217; The correct answer and explanation [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-16875","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/16875","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=16875"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/16875\/revisions"}],"predecessor-version":[{"id":16876,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/16875\/revisions\/16876"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=16875"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=16875"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=16875"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}