{"id":218348,"date":"2025-05-23T16:44:56","date_gmt":"2025-05-23T16:44:56","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=218348"},"modified":"2025-05-23T16:44:58","modified_gmt":"2025-05-23T16:44:58","slug":"energy-to-spin-up-a-charged-cylinder-the-angular-velocity-%cf%89tz%cb%86-of-the-cylindrical-shell-shown-below-increases-from-zero-and-smoothly-approaches-the-steady-value-%cf%890","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/23\/energy-to-spin-up-a-charged-cylinder-the-angular-velocity-%cf%89tz%cb%86-of-the-cylindrical-shell-shown-below-increases-from-zero-and-smoothly-approaches-the-steady-value-%cf%890\/","title":{"rendered":"Energy to Spin Up a Charged Cylinder The angular velocity \u03c9(t)z\u02c6 of the cylindrical shell shown below increases from zero and smoothly approaches the steady value \u03c90"},"content":{"rendered":"\n<p>Energy to Spin Up a Charged Cylinder The angular velocity \u03c9(t)z\u02c6 of the cylindrical shell shown below increases from zero and smoothly approaches the steady value \u03c90. The shell has infinitesimal thickness and carries a uniform charge per unit length \u03bb = 2\u03c0R\u03c3, where \u03c3 is a uniform charge per unit area. Assume that the shell radius R L and that the spin-up is very slow so the displacement current may be neglected.<\/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>energy required to spin up<\/strong> a charged cylindrical shell from rest to an angular velocity $\\omega_0$, we need to calculate the <strong>energy stored in the magnetic field<\/strong> generated during the process. Since the shell carries charge and rotates, it produces a current, which in turn creates a magnetic field. The energy used to spin it up is stored in this magnetic field.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Answer:<\/strong><\/h3>\n\n\n\n<p>$$<br>\\boxed{U = \\frac{\\mu_0 \\lambda^2 \\omega_0^2 R^2 L}{4\\pi}}<br>$$<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>$\\mu_0$: permeability of free space<\/li>\n\n\n\n<li>$\\lambda = 2\\pi R \\sigma$: charge per unit length<\/li>\n\n\n\n<li>$\\omega_0$: final angular velocity<\/li>\n\n\n\n<li>$R$: radius of the shell<\/li>\n\n\n\n<li>$L$: length of the cylinder<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\"><\/h3>\n\n\n\n<p>We are dealing with a <strong>cylindrical shell<\/strong> of radius $R$, length $L$, and negligible thickness, carrying a <strong>uniform charge per unit area $\\sigma$<\/strong>. As it spins up slowly (quasistatically) to a final angular velocity $\\omega_0$, each bit of charge moves in a circle, thus constituting a <strong>circular current<\/strong>. The result is a magnetic field inside the shell.<\/p>\n\n\n\n<p>Because the spin-up is slow, <strong>displacement currents are neglected<\/strong>, and the system can be analyzed using <strong>magnetostatics<\/strong>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Step 1: Current due to rotation<\/strong><\/p>\n\n\n\n<p>The line charge density is:<\/p>\n\n\n\n<p>$$<br>\\lambda = \\sigma \\cdot 2\\pi R<br>$$<\/p>\n\n\n\n<p>Since this line charge is rotating with angular velocity $\\omega$, the current per unit length (along the cylinder&#8217;s axis) is:<\/p>\n\n\n\n<p>$$<br>I = \\lambda \\cdot v = \\lambda \\cdot R \\omega = 2\\pi R \\sigma \\cdot R \\omega = 2\\pi R^2 \\sigma \\omega<br>$$<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Step 2: Magnetic field inside a spinning charged cylinder<\/strong><\/p>\n\n\n\n<p>Using Amp\u00e8re\u2019s law for a long cylinder, the magnetic field inside (along the axis) due to a surface current is:<\/p>\n\n\n\n<p>$$<br>B = \\frac{\\mu_0 I}{2\\pi R} = \\frac{\\mu_0 \\lambda \\omega R}{2\\pi}<br>$$<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Step 3: Magnetic energy stored<\/strong><\/p>\n\n\n\n<p>The energy density in a magnetic field is:<\/p>\n\n\n\n<p>$$<br>u = \\frac{B^2}{2\\mu_0}<br>$$<\/p>\n\n\n\n<p>Total magnetic energy stored:<\/p>\n\n\n\n<p>$$<br>U = u \\cdot \\text{Volume} = \\frac{B^2}{2\\mu_0} \\cdot \\pi R^2 L<br>$$<\/p>\n\n\n\n<p>Plugging in $B$:<\/p>\n\n\n\n<p>$$<br>U = \\frac{1}{2\\mu_0} \\left( \\frac{\\mu_0 \\lambda \\omega_0 R}{2\\pi} \\right)^2 \\cdot \\pi R^2 L = \\frac{\\mu_0 \\lambda^2 \\omega_0^2 R^2 L}{4\\pi}<br>$$<\/p>\n\n\n\n<p>This is the energy needed to spin up the cylinder.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/image-218.png\" alt=\"\" class=\"wp-image-218349\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Energy to Spin Up a Charged Cylinder The angular velocity \u03c9(t)z\u02c6 of the cylindrical shell shown below increases from zero and smoothly approaches the steady value \u03c90. The shell has infinitesimal thickness and carries a uniform charge per unit length \u03bb = 2\u03c0R\u03c3, where \u03c3 is a uniform charge per unit area. Assume that the [&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-218348","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/218348","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=218348"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/218348\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=218348"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=218348"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=218348"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}