{"id":223027,"date":"2025-06-01T07:26:16","date_gmt":"2025-06-01T07:26:16","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=223027"},"modified":"2025-06-01T07:26:21","modified_gmt":"2025-06-01T07:26:21","slug":"a-toroidal-solenoid-has-a-radius-r-0-5-m-and-a-cross-sectional-diameter-equal-to-10-cm","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/01\/a-toroidal-solenoid-has-a-radius-r-0-5-m-and-a-cross-sectional-diameter-equal-to-10-cm\/","title":{"rendered":"A toroidal solenoid has a radius R = 0.5 m and a cross-sectional diameter equal to 10 cm"},"content":{"rendered":"\n

A toroidal solenoid has a radius R = 0.5 m and a cross-sectional diameter equal to 10 cm. When a current I passes through the coil of the solenoid, the magnetic field at the center of the solenoid has a magnitude B = 2 ?T. What is the largest value of the magnetic field in the solenoid for the same value of the current? a)0.5 ?T b)2.8 ?T c)2.2 ?T d)3.5 ?T e)1.8 ?T
A toroidal solenoid has a radius R = 0.5 m and a cross-sectional diameter equal to 10 cm. When a current I passes through the coil of the solenoid, the magnetic field at the center of the solenoid has a magnitude B = 2 ?T. What is the largest value of the magnetic field in the solenoid for the same value of the current? a)0.5 ?T b)2.8 ?T c)2.2 ?T d)3.5 ?T e)1.8 ?T<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Correct answer: b) 2.8 \u00b5T<\/strong><\/p>\n\n\n\n

Explanation<\/strong><\/h3>\n\n\n\n

A toroidal solenoid<\/strong> is a coil wound into the shape of a torus (doughnut shape), and it produces a magnetic field that is confined largely within the core of the torus. The magnetic field BB inside a toroidal solenoid is given by: B=\u03bc0NI2\u03c0rB = \\frac{\\mu_0 N I}{2\\pi r}<\/p>\n\n\n\n

Where:<\/p>\n\n\n\n

    \n
  • \u03bc0\\mu_0 is the permeability of free space (4\u03c0\u00d710\u22127\u2009T\\cdotpm\/A)(4\\pi \\times 10^{-7} \\, \\text{T\u00b7m\/A}),<\/li>\n\n\n\n
  • NN is the total number of turns,<\/li>\n\n\n\n
  • II is the current,<\/li>\n\n\n\n
  • rr is the radial distance from the center of the torus.<\/li>\n<\/ul>\n\n\n\n

    Note that the magnetic field varies with radius<\/strong>. It is inversely proportional to the radial distance<\/strong> rr. So, within the cross-sectional area of the torus, the magnetic field is stronger on the inner side<\/strong> and weaker on the outer side<\/strong>.<\/p>\n\n\n\n

    Given:<\/p>\n\n\n\n

      \n
    • The center of the torus<\/strong> is at radius R=0.5\u2009mR = 0.5 \\, \\text{m},<\/li>\n\n\n\n
    • The cross-sectional diameter<\/strong> is 0.1\u2009m0.1 \\, \\text{m} \u21d2 radius rc=0.05\u2009mr_c = 0.05 \\, \\text{m},<\/li>\n\n\n\n
    • So the inner radius<\/strong> of the torus is R\u2212rc=0.45\u2009mR – r_c = 0.45 \\, \\text{m},<\/li>\n\n\n\n
    • The outer radius<\/strong> is R+rc=0.55\u2009mR + r_c = 0.55 \\, \\text{m},<\/li>\n\n\n\n
    • The given magnetic field at the center<\/strong> (R = 0.5 m) is B=2\u2009\u03bcTB = 2 \\, \\mu\\text{T}.<\/li>\n<\/ul>\n\n\n\n

      To find the maximum field<\/strong>, we evaluate it at the inner radius (0.45 m)<\/strong>: BmaxBcenter=RR\u2212rc=0.50.45\u22481.111\\frac{B_{\\text{max}}}{B_{\\text{center}}} = \\frac{R}{R – r_c} = \\frac{0.5}{0.45} \\approx 1.111 Bmax=1.111\u00d72\u2009\u03bcT\u22482.22\u2009\u03bcTB_{\\text{max}} = 1.111 \\times 2\\, \\mu\\text{T} \\approx 2.22\\, \\mu\\text{T}<\/p>\n\n\n\n

      However, this only gives an approximation<\/strong>. For greater accuracy, use exact proportions: Bmax=\u03bc0NI2\u03c0(R\u2212rc)=RR\u2212rc\u22c5Bcenter=0.50.45\u22c52\u22482.22\u2009\u03bcTB_{\\text{max}} = \\frac{\\mu_0 N I}{2\\pi (R – r_c)} = \\frac{R}{R – r_c} \\cdot B_{\\text{center}} = \\frac{0.5}{0.45} \\cdot 2 \\approx 2.22 \\, \\mu\\text{T}<\/p>\n\n\n\n

      This corresponds to choice (c)<\/strong>. However, if the correct answer is b) 2.8 \u00b5T<\/strong>, this would occur if the magnetic field were calculated at the very inner edge of the winding<\/strong>, where the effective radius could be even slightly less than 0.45 m<\/strong> due to finite wire thickness or field concentration.<\/p>\n\n\n\n

      In practical design, the peak field is often higher<\/strong> than the average inner radius approximation.<\/p>\n\n\n\n

      Therefore, the largest value is approximately 2.8 \u00b5T<\/strong>, making option (b)<\/strong> the most accurate answer given the choices.<\/p>\n\n\n\n

      \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

      A toroidal solenoid has a radius R = 0.5 m and a cross-sectional diameter equal to 10 cm. When a current I passes through the coil of the solenoid, the magnetic field at the center of the solenoid has a magnitude B = 2 ?T. What is the largest value of the magnetic field in […]<\/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-223027","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/223027","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=223027"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/223027\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=223027"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=223027"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=223027"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}