{"id":197839,"date":"2025-03-07T18:19:40","date_gmt":"2025-03-07T18:19:40","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=197839"},"modified":"2025-03-07T18:19:42","modified_gmt":"2025-03-07T18:19:42","slug":"if-the-magnitude-of-the-drift-velocity-of-free-electrons-in-a-copper-wire-is-7-84-x-10-4-m-s","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/07\/if-the-magnitude-of-the-drift-velocity-of-free-electrons-in-a-copper-wire-is-7-84-x-10-4-m-s\/","title":{"rendered":"If the magnitude of the drift velocity of free electrons in a copper wire is 7.84 x 10-4 m\/s"},"content":{"rendered":"\n<p>If the magnitude of the drift velocity of free electrons in a copper wire is 7.84 x 10-4 m\/s, what is the electric field in the conductor?<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n<p>To calculate the electric field in a conductor given the drift velocity, we can use the relationship between the drift velocity and the electric field in a conducting wire. The formula to calculate the electric field (E) is:<\/p>\n\n\n\n<p>[<br>E = \\frac{v_d}{\\mu}<br>]<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>( E ) is the electric field,<\/li>\n\n\n\n<li>( v_d ) is the drift velocity,<\/li>\n\n\n\n<li>( \\mu ) is the mobility of the electrons.<\/li>\n<\/ul>\n\n\n\n<p>However, we need to use another approach since the mobility of electrons is typically not given directly in the problem statement. For a better understanding, we can also use a relationship between the drift velocity and the current, which involves the number of free electrons, the charge, and the cross-sectional area of the wire.<\/p>\n\n\n\n<p>The drift velocity ( v_d ) is given by:<\/p>\n\n\n\n<p>[<br>v_d = \\frac{J}{nq}<br>]<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>( J ) is the current density (current per unit area),<\/li>\n\n\n\n<li>( n ) is the number of free electrons per unit volume (for copper, ( n \\approx 8.5 \\times 10^{28} \\, \\text{electrons\/m}^3 )),<\/li>\n\n\n\n<li>( q ) is the charge of an electron (( q = 1.6 \\times 10^{-19} \\, \\text{C} )).<\/li>\n<\/ul>\n\n\n\n<p>Now, the current density ( J ) is related to the electric field ( E ) by:<\/p>\n\n\n\n<p>[<br>J = \\sigma E<br>]<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>( \\sigma ) is the electrical conductivity of the material (for copper, ( \\sigma = 5.8 \\times 10^7 \\, \\text{S\/m} )).<\/li>\n<\/ul>\n\n\n\n<p>So, we can combine these relationships to find the electric field ( E ). By rearranging the equations:<\/p>\n\n\n\n<p>[<br>E = \\frac{v_d}{\\mu}<br>]<\/p>\n\n\n\n<p>Now, plugging the known values:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Drift velocity ( v_d = 7.84 \\times 10^{-4} \\, \\text{m\/s} ),<\/li>\n\n\n\n<li>Conductivity of copper ( \\sigma = 5.8 \\times 10^7 \\, \\text{S\/m} ),<\/li>\n<\/ul>\n\n\n\n<p>and calculating the electric field, we find:<\/p>\n\n\n\n<p>[<br>E \\approx 0.0000135 \\, \\text{V\/m}<br>]<\/p>\n\n\n\n<p>Thus, the electric field in the conductor is approximately <strong>1.35 x 10\u207b\u2075 V\/m<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<p>The drift velocity represents the average velocity of free electrons in a conductor under the influence of an electric field. The electric field causes these free electrons to move in a direction opposite to the field, and their velocity is called drift velocity. The relationship between drift velocity and electric field is influenced by the material&#8217;s properties (such as conductivity), the density of free electrons, and the applied electric field.<\/p>\n\n\n\n<p>In copper, the electric field required to produce such a small drift velocity of ( 7.84 \\times 10^{-4} \\, \\text{m\/s} ) is tiny because of the high conductivity of the material, which allows the electrons to move easily even with a small applied field. This illustrates how efficient conductors like copper have very low resistive effects when a small voltage is applied.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>If the magnitude of the drift velocity of free electrons in a copper wire is 7.84 x 10-4 m\/s, what is the electric field in the conductor? The correct answer and explanation is : To calculate the electric field in a conductor given the drift velocity, we can use the relationship between the drift velocity [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-197839","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/197839","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=197839"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/197839\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=197839"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=197839"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=197839"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}