{"id":39423,"date":"2025-06-27T07:25:48","date_gmt":"2025-06-27T07:25:48","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=39423"},"modified":"2025-06-27T07:25:56","modified_gmt":"2025-06-27T07:25:56","slug":"a-rectangular-shaped-core-is-made-up-of-steel-plate-15mm-and-20mm-cross-section","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/a-rectangular-shaped-core-is-made-up-of-steel-plate-15mm-and-20mm-cross-section\/","title":{"rendered":"A rectangular shaped core is made up of Steel Plate 15mm and 20mm cross section"},"content":{"rendered":"\n<p>A rectangular shaped core is made up of Steel Plate 15mm and 20mm cross section. The mean length Path of magnetic path is 20cm. The existing coil has 300 turns and current 0.7A. Calculate: A) Magnetizing force B) Flux density C) Reluctance D) Flux of magnetic circuits<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The correct answer and explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To solve this magnetic circuit problem, we will go through each part step-by-step.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Given Data:<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Cross-sectional dimensions: 15 mm \u00d7 20 mm<\/li>\n\n\n\n<li>Mean magnetic path length: 20 cm = 0.2 m<\/li>\n\n\n\n<li>Number of turns (N): 300<\/li>\n\n\n\n<li>Current (I): 0.7 A<\/li>\n\n\n\n<li>Material: Steel (assume relative permeability \u03bcr\\mu_r \u2248 1000)<\/li>\n\n\n\n<li>Permeability of free space \u03bc0\\mu_0 = 4\u03c0\u00d710\u22127\u2009H\/m4\\pi \\times 10^{-7} \\, \\text{H\/m}<\/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\"><strong>A) Magnetizing Force (H):<\/strong><\/h3>\n\n\n\n<p>The magnetizing force is given by: H=N\u22c5IlH = \\frac{N \\cdot I}{l} H=300\u00d70.70.2=2100.2=1050\u2009A\/mH = \\frac{300 \\times 0.7}{0.2} = \\frac{210}{0.2} = 1050 \\, \\text{A\/m}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>B) Flux Density (B):<\/strong><\/h3>\n\n\n\n<p>First, calculate the magnetic field strength BB: B=\u03bc\u22c5H=\u03bc0\u22c5\u03bcr\u22c5HB = \\mu \\cdot H = \\mu_0 \\cdot \\mu_r \\cdot H \u03bc=4\u03c0\u00d710\u22127\u00d71000=1.2566\u00d710\u22123\u2009H\/m\\mu = 4\\pi \\times 10^{-7} \\times 1000 = 1.2566 \\times 10^{-3} \\, \\text{H\/m} B=1.2566\u00d710\u22123\u00d71050=1.3194\u2009TB = 1.2566 \\times 10^{-3} \\times 1050 = 1.3194 \\, \\text{T}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>C) Reluctance (\u211c):<\/strong><\/h3>\n\n\n\n<p>Reluctance is given by: \u211c=l\u03bc\u22c5A\\mathcal{\u211c} = \\frac{l}{\\mu \\cdot A}<\/p>\n\n\n\n<p>Cross-sectional area: A=15\u00d710\u22123\u00d720\u00d710\u22123=3\u00d710\u22124\u2009m2A = 15 \\times 10^{-3} \\times 20 \\times 10^{-3} = 3 \\times 10^{-4} \\, \\text{m}^2 \u211c=0.21.2566\u00d710\u22123\u22c53\u00d710\u22124=0.23.7698\u00d710\u22127\u2248530,576\u2009A\/Wb\\mathcal{\u211c} = \\frac{0.2}{1.2566 \\times 10^{-3} \\cdot 3 \\times 10^{-4}} = \\frac{0.2}{3.7698 \\times 10^{-7}} \\approx 530,576 \\, \\text{A\/Wb}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>D) Magnetic Flux (\u03a6):<\/strong><\/h3>\n\n\n\n<p>Flux is given by: \u03a6=MMF\u211c=N\u22c5I\u211c\\Phi = \\frac{MMF}{\\mathcal{\u211c}} = \\frac{N \\cdot I}{\\mathcal{\u211c}} \u03a6=300\u00d70.7530576\u2248210530576=3.96\u00d710\u22124\u2009Wb\\Phi = \\frac{300 \\times 0.7}{530576} \\approx \\frac{210}{530576} = 3.96 \\times 10^{-4} \\, \\text{Wb}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Summary:<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>A) Magnetizing Force (H)<\/strong> = <strong>1050 A\/m<\/strong><\/li>\n\n\n\n<li><strong>B) Flux Density (B)<\/strong> = <strong>1.32 T<\/strong><\/li>\n\n\n\n<li><strong>C) Reluctance (\u211c)<\/strong> = <strong>530,576 A\/Wb<\/strong><\/li>\n\n\n\n<li><strong>D) Magnetic Flux (\u03a6)<\/strong> = <strong>3.96 \u00d7 10\u207b\u2074 Wb<\/strong><\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation:<\/strong><\/h3>\n\n\n\n<p>A magnetic circuit is similar to an electric circuit, but it deals with magnetic flux instead of electric current. The coil on the core creates magnetomotive force (MMF), which drives magnetic flux through the steel. The magnetizing force (H) is the intensity of the magnetic field created along the core\u2019s length. The flux density (B) shows how concentrated the magnetic field is in the core, depending on the material\u2019s permeability. Reluctance acts like resistance, opposing the magnetic flux. Magnetic flux (\u03a6) is the total magnetic field passing through the core\u2019s area. By knowing the number of turns, current, core dimensions, and material properties, all these quantities can be calculated using magnetic circuit laws.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>A rectangular shaped core is made up of Steel Plate 15mm and 20mm cross section. The mean length Path of magnetic path is 20cm. The existing coil has 300 turns and current 0.7A. Calculate: A) Magnetizing force B) Flux density C) Reluctance D) Flux of magnetic circuits The correct answer and explanation is: To solve [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-39423","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/39423","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=39423"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/39423\/revisions"}],"predecessor-version":[{"id":39426,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/39423\/revisions\/39426"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=39423"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=39423"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=39423"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}