{"id":186357,"date":"2025-01-24T18:53:57","date_gmt":"2025-01-24T18:53:57","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=186357"},"modified":"2025-01-24T18:53:59","modified_gmt":"2025-01-24T18:53:59","slug":"a-solenoid-of-length-0-700m-having-a-circular-cross-section-of-radius-5-00cm-stores-6-00-%c2%b5j-of-energy","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/24\/a-solenoid-of-length-0-700m-having-a-circular-cross-section-of-radius-5-00cm-stores-6-00-%c2%b5j-of-energy\/","title":{"rendered":"A solenoid of length 0.700m having a circular cross-section of radius 5.00cm stores 6.00 \u00b5J of energy"},"content":{"rendered":"\n<p>A solenoid of length 0.700m having a circular cross-section of radius 5.00cm stores 6.00 \u00b5J of energy when a 0.400-A current runs through it. What is the winding density of the solenoid? (\u00b50 = 4p<em>10-7 T<\/em>m\/A)<\/p>\n\n\n\n<p>A) 865 turns\/m<br>B) 327 turns\/m<br>C) 1080 turns\/m<br>D) 104 turns\/m<br>E) 472 turns\/m&amp;nb\u2026<\/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 determine the winding density of the solenoid based on the given information.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Key Equations and Variables:<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Energy Stored in a Solenoid<\/strong>: U=12LI2U = \\frac{1}{2} L I^2 where UU is the energy stored, LL is the inductance, and II is the current.<\/li>\n\n\n\n<li><strong>Inductance of a Solenoid<\/strong>: L=\u03bc0n2A\u2113L = \\mu_0 n^2 A \\ell where nn is the winding density (turns per unit length), A=\u03c0r2A = \\pi r^2 is the cross-sectional area of the solenoid, \u2113\\ell is its length, and \u03bc0=4\u03c0\u00d710\u22127\u2009T\\cdotpm\/A\\mu_0 = 4\\pi \\times 10^{-7} \\, \\text{T\u00b7m\/A} is the permeability of free space.<\/li>\n<\/ol>\n\n\n\n<h3 class=\"wp-block-heading\">Known Values:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Length of the solenoid: \u2113=0.700\u2009m\\ell = 0.700 \\, \\text{m}<\/li>\n\n\n\n<li>Radius: r=5.00\u2009cm=0.050\u2009mr = 5.00 \\, \\text{cm} = 0.050 \\, \\text{m}<\/li>\n\n\n\n<li>Energy stored: U=6.00\u2009\u03bcJ=6.00\u00d710\u22126\u2009JU = 6.00 \\, \\mu\\text{J} = 6.00 \\times 10^{-6} \\, \\text{J}<\/li>\n\n\n\n<li>Current: I=0.400\u2009AI = 0.400 \\, \\text{A}<\/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: Calculate LL using the energy equation.<\/h3>\n\n\n\n<p>Rearranging the energy equation: L=2UI2L = \\frac{2U}{I^2}<\/p>\n\n\n\n<p>Substitute the values: L=2\u00d76.00\u00d710\u22126(0.400)2=1.20\u00d710\u221250.16=7.50\u00d710\u22125\u2009HL = \\frac{2 \\times 6.00 \\times 10^{-6}}{(0.400)^2} = \\frac{1.20 \\times 10^{-5}}{0.16} = 7.50 \\times 10^{-5} \\, \\text{H}<\/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: Solve for nn using the inductance equation.<\/h3>\n\n\n\n<p>Rearranging the inductance equation: n=L\u03bc0A\u2113n = \\sqrt{\\frac{L}{\\mu_0 A \\ell}}<\/p>\n\n\n\n<p>Substitute A=\u03c0r2=\u03c0(0.050)2=7.85\u00d710\u22123\u2009m2A = \\pi r^2 = \\pi (0.050)^2 = 7.85 \\times 10^{-3} \\, \\text{m}^2, \u03bc0=4\u03c0\u00d710\u22127\\mu_0 = 4\\pi \\times 10^{-7}, \u2113=0.700\u2009m\\ell = 0.700 \\, \\text{m}, and L=7.50\u00d710\u22125L = 7.50 \\times 10^{-5}: n=7.50\u00d710\u22125(4\u03c0\u00d710\u22127)(7.85\u00d710\u22123)(0.700)n = \\sqrt{\\frac{7.50 \\times 10^{-5}}{(4\\pi \\times 10^{-7}) (7.85 \\times 10^{-3}) (0.700)}}<\/p>\n\n\n\n<p>First, calculate the denominator: 4\u03c0\u00d710\u22127\u00d77.85\u00d710\u22123\u00d70.700=6.90\u00d710\u221294\\pi \\times 10^{-7} \\times 7.85 \\times 10^{-3} \\times 0.700 = 6.90 \\times 10^{-9}<\/p>\n\n\n\n<p>Now calculate nn: n=7.50\u00d710\u221256.90\u00d710\u22129=1.09\u00d7104\u22481044\u2009turns\/mn = \\sqrt{\\frac{7.50 \\times 10^{-5}}{6.90 \\times 10^{-9}}} = \\sqrt{1.09 \\times 10^{4}} \\approx 1044 \\, \\text{turns\/m}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Match to the closest answer.<\/h3>\n\n\n\n<p>The closest option is <strong>C) 1080 turns\/m<\/strong>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<p>The winding density of a solenoid represents the number of turns of wire per unit length. Using the energy stored in the solenoid, we calculated the inductance, which depends on the winding density, cross-sectional area, and length. By rearranging and solving, we found the winding density to be approximately 1044\u2009turns\/m1044 \\, \\text{turns\/m}, which aligns closely with option CC. Accurate understanding of energy storage and electromagnetic principles is key to solving this.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>A solenoid of length 0.700m having a circular cross-section of radius 5.00cm stores 6.00 \u00b5J of energy when a 0.400-A current runs through it. What is the winding density of the solenoid? (\u00b50 = 4p10-7 Tm\/A) A) 865 turns\/mB) 327 turns\/mC) 1080 turns\/mD) 104 turns\/mE) 472 turns\/m&amp;nb\u2026 The correct answer and explanation is: To solve [&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-186357","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186357","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=186357"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186357\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=186357"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=186357"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=186357"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}