{"id":165227,"date":"2024-11-11T14:15:26","date_gmt":"2024-11-11T14:15:26","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=165227"},"modified":"2024-11-11T14:15:30","modified_gmt":"2024-11-11T14:15:30","slug":"two-long-parallel-wires-carry-currents-of-10-a-in-opposite-directions-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2024\/11\/11\/two-long-parallel-wires-carry-currents-of-10-a-in-opposite-directions-2\/","title":{"rendered":"Two long parallel wires carry currents of 10 A in opposite directions."},"content":{"rendered":"\n<p>Two long parallel wires carry currents of 10 A in opposite directions. They are separated by 40 cm. What is the magnitude of the magnetic field in the plane of the wires at a point that is 20 cm from one wire and 60 cm from the other? (\u03bc 0 = 4\u03c0 \u00d7 10-7 T \u00b7 m\/A)<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To solve this problem, we will apply Amp\u00e8re&#8217;s Law for the magnetic field around a long straight current-carrying wire. The formula for the magnetic field at a distance ( r ) from a wire carrying a current ( I ) is:<\/p>\n\n\n\n<p>[<br>B = \\frac{\\mu_0 I}{2 \\pi r}<br>]<\/p>\n\n\n\n<p>where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>( B ) is the magnetic field strength,<\/li>\n\n\n\n<li>( \\mu_0 ) is the permeability of free space (( 4\\pi \\times 10^{-7} \\, \\text{T} \\cdot \\text{m\/A} )),<\/li>\n\n\n\n<li>( I ) is the current,<\/li>\n\n\n\n<li>( r ) is the distance from the wire.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Problem Setup<\/h3>\n\n\n\n<p>We have two wires with currents of 10 A flowing in opposite directions. The wires are separated by 40 cm (0.4 m), and the point of interest is located 20 cm (0.2 m) from one wire and 60 cm (0.6 m) from the other.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Calculate the Magnetic Field from Each Wire<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Magnetic field due to the first wire (20 cm away):<\/strong> Using the formula ( B_1 = \\frac{\\mu_0 I}{2 \\pi r_1} ), where ( r_1 = 0.2 \\, \\text{m} ): [<br>B_1 = \\frac{(4\\pi \\times 10^{-7} \\, \\text{T} \\cdot \\text{m\/A})(10 \\, \\text{A})}{2 \\pi (0.2 \\, \\text{m})} = \\frac{4 \\times 10^{-6}}{0.4} = 1 \\times 10^{-5} \\, \\text{T}<br>]<\/li>\n\n\n\n<li><strong>Magnetic field due to the second wire (60 cm away):<\/strong> Similarly, using the formula ( B_2 = \\frac{\\mu_0 I}{2 \\pi r_2} ), where ( r_2 = 0.6 \\, \\text{m} ): [<br>B_2 = \\frac{(4\\pi \\times 10^{-7} \\, \\text{T} \\cdot \\text{m\/A})(10 \\, \\text{A})}{2 \\pi (0.6 \\, \\text{m})} = \\frac{4 \\times 10^{-6}}{1.2} = 3.33 \\times 10^{-6} \\, \\text{T}<br>]<\/li>\n<\/ol>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Determine the Directions of the Magnetic Fields<\/h3>\n\n\n\n<p>Since the currents in the wires are opposite in direction, the magnetic fields generated by each wire at the point of interest will be in opposite directions. Using the right-hand rule:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The magnetic field due to the first wire (closer, 20 cm away) will be directed into the plane of the paper.<\/li>\n\n\n\n<li>The magnetic field due to the second wire (further, 60 cm away) will be directed out of the plane of the paper.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Calculate the Net Magnetic Field<\/h3>\n\n\n\n<p>The magnetic fields are in opposite directions, so we subtract the smaller field from the larger field:<\/p>\n\n\n\n<p>[<br>B_{\\text{net}} = B_1 &#8211; B_2 = (1 \\times 10^{-5} \\, \\text{T}) &#8211; (3.33 \\times 10^{-6} \\, \\text{T}) = 6.67 \\times 10^{-6} \\, \\text{T}<br>]<\/p>\n\n\n\n<p>Thus, the magnitude of the net magnetic field at the given point is:<\/p>\n\n\n\n<p>[<br>\\boxed{6.67 \\times 10^{-6} \\, \\text{T}}<br>]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Conclusion<\/h3>\n\n\n\n<p>The net magnetic field at the point is ( 6.67 \\times 10^{-6} \\, \\text{T} ), directed into the plane of the paper (towards the first wire). This result is derived by considering the contributions from both wires and applying the principle of superposition.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Two long parallel wires carry currents of 10 A in opposite directions. They are separated by 40 cm. What is the magnitude of the magnetic field in the plane of the wires at a point that is 20 cm from one wire and 60 cm from the other? (\u03bc 0 = 4\u03c0 \u00d7 10-7 T [&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-165227","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/165227","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=165227"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/165227\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=165227"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=165227"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=165227"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}