{"id":223012,"date":"2025-06-01T07:14:15","date_gmt":"2025-06-01T07:14:15","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=223012"},"modified":"2025-06-01T07:14:18","modified_gmt":"2025-06-01T07:14:18","slug":"a-solenoid-of-length-0-700-m-having-a-circular-cross-section-of-radius-5-00-cm-stores-6-00-i%c2%bcj-of-energy-when-a-0-400-a-current-runs-through-it","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/01\/a-solenoid-of-length-0-700-m-having-a-circular-cross-section-of-radius-5-00-cm-stores-6-00-i%c2%bcj-of-energy-when-a-0-400-a-current-runs-through-it\/","title":{"rendered":"A solenoid of length 0.700 m having a circular cross-section of radius 5.00 cm stores 6.00 \u00ce\u00bcJ of energy when a 0.400 A current runs through it"},"content":{"rendered":"\n<p>A solenoid of length 0.700 m having a circular cross-section of radius 5.00 cm stores 6.00 \u00ce\u00bcJ of energy when a 0.400 A current runs through it. What is the winding density of the solenoid? (\u00ce\u00bc\u00e2\u201a\u20ac = 4\u00cf\u20ac<em>10\u00e2\u0081\u00bb\u00e2\u0081\u00b7 T<\/em>m\/A) A) 865 turns\/m B) 327 turns\/m C) 1080 turns\/m D) 104 turns\/m E) 472 turns\/m<br>A solenoid of length 0.700 m having a circular cross-section of radius 5.00 cm stores 6.00 \u00ce\u00bcJ of energy when a 0.400 A current runs through it. What is the winding density of the solenoid? (\u00ce\u00bc\u00e2\u201a\u20ac = 4\u00cf\u20ac<em>10\u00e2\u0081\u00bb\u00e2\u0081\u00b7 T<\/em>m\/A) A) 865 turns\/m B) 327 turns\/m C) 1080 turns\/m D) 104 turns\/m E) 472 turns\/m<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To find the <strong>winding density<\/strong> (number of turns per unit length, <strong>n<\/strong>) of the solenoid, we use the formula for the <strong>energy stored in the magnetic field<\/strong> of an ideal solenoid: U=12\u03bc0n2AlI2U = \\frac{1}{2} \\mu_0 n^2 A l I^2<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>U=6.00\u00a0\u03bcJ=6.00\u00d710\u22126\u00a0JU = 6.00 \\ \\mu\\text{J} = 6.00 \\times 10^{-6} \\ \\text{J} (energy stored)<\/li>\n\n\n\n<li>\u03bc0=4\u03c0\u00d710\u22127\u00a0T\\cdotpm\/A\\mu_0 = 4\\pi \\times 10^{-7} \\ \\text{T\u00b7m\/A} (permeability of free space)<\/li>\n\n\n\n<li>nn = winding density (turns per meter) <strong>[what we want to find]<\/strong><\/li>\n\n\n\n<li>A=\u03c0r2A = \\pi r^2 = cross-sectional area<\/li>\n\n\n\n<li>l=0.700\u00a0ml = 0.700 \\ \\text{m} = length of solenoid<\/li>\n\n\n\n<li>r=5.00\u00a0cm=0.0500\u00a0mr = 5.00 \\ \\text{cm} = 0.0500 \\ \\text{m}<\/li>\n\n\n\n<li>I=0.400\u00a0AI = 0.400 \\ \\text{A} = current<\/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-by-step Calculation:<\/h3>\n\n\n\n<p><strong>1. Calculate cross-sectional area AA:<\/strong> A=\u03c0r2=\u03c0(0.0500)2=\u03c0\u00d72.5\u00d710\u22123=7.854\u00d710\u22123&nbsp;m2A = \\pi r^2 = \\pi (0.0500)^2 = \\pi \\times 2.5 \\times 10^{-3} = 7.854 \\times 10^{-3} \\ \\text{m}^2<\/p>\n\n\n\n<p><strong>2. Rearranging the energy formula to solve for nn:<\/strong> U=12\u03bc0n2AlI2\u21d2n2=2U\u03bc0AlI2U = \\frac{1}{2} \\mu_0 n^2 A l I^2 \\Rightarrow n^2 = \\frac{2U}{\\mu_0 A l I^2}<\/p>\n\n\n\n<p><strong>3. Plug in values:<\/strong> n2=2\u00d76.00\u00d710\u22126(4\u03c0\u00d710\u22127)\u00d7(7.854\u00d710\u22123)\u00d70.700\u00d7(0.400)2n^2 = \\frac{2 \\times 6.00 \\times 10^{-6}}{(4\\pi \\times 10^{-7}) \\times (7.854 \\times 10^{-3}) \\times 0.700 \\times (0.400)^2}<\/p>\n\n\n\n<p>First, calculate the denominator: 4\u03c0\u00d710\u22127\u00d77.854\u00d710\u22123\u00d70.700\u00d70.16\u22481.105\u00d710\u221294\\pi \\times 10^{-7} \\times 7.854 \\times 10^{-3} \\times 0.700 \\times 0.16 \\approx 1.105 \\times 10^{-9} n2=1.2\u00d710\u221251.105\u00d710\u22129\u224810859.73n^2 = \\frac{1.2 \\times 10^{-5}}{1.105 \\times 10^{-9}} \\approx 10859.73 n\u224810859.73\u2248104.2n \\approx \\sqrt{10859.73} \\approx 104.2<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer:<\/h3>\n\n\n\n<p>104&nbsp;turns\/m(Choice&nbsp;D)\\boxed{104 \\ \\text{turns\/m}} \\quad \\text{(Choice D)}<\/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>A solenoid stores magnetic energy when current flows through it, and this stored energy depends on the winding density \u2014 the number of turns per meter of the coil. The magnetic field inside an ideal solenoid is uniform and given by B=\u03bc0nIB = \\mu_0 n I, and the energy stored is proportional to the square of this field.<\/p>\n\n\n\n<p>The energy formula for a solenoid is: U=12\u03bc0n2AlI2U = \\frac{1}{2} \\mu_0 n^2 A l I^2<\/p>\n\n\n\n<p>This equation reflects how energy increases with greater winding density nn, current II, cross-sectional area AA, and solenoid length ll. Given the energy stored (6.00 \u00b5J), current (0.400 A), and the solenoid\u2019s dimensions, we isolated nn and substituted all known values. Calculating the cross-sectional area from the radius (5.00 cm = 0.0500 m), and converting microjoules to joules, we carefully handled the powers of 10 to maintain precision.<\/p>\n\n\n\n<p>After solving for nn, we found the winding density to be approximately 104.2 turns\/m. This matches <strong>option D (104 turns\/m)<\/strong>. Small variations due to rounding are common in such physics problems, but the answer remains accurate within significant figures.<\/p>\n\n\n\n<p>This result tells us that the solenoid has about 104 loops of wire per meter, allowing it to generate the required magnetic field to store the given energy with the specified current.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner4-22.jpeg\" alt=\"\" class=\"wp-image-223013\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>A solenoid of length 0.700 m having a circular cross-section of radius 5.00 cm stores 6.00 \u00ce\u00bcJ of energy when a 0.400 A current runs through it. What is the winding density of the solenoid? (\u00ce\u00bc\u00e2\u201a\u20ac = 4\u00cf\u20ac10\u00e2\u0081\u00bb\u00e2\u0081\u00b7 Tm\/A) A) 865 turns\/m B) 327 turns\/m C) 1080 turns\/m D) 104 turns\/m E) 472 turns\/mA solenoid [&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-223012","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/223012","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=223012"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/223012\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=223012"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=223012"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=223012"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}