{"id":42597,"date":"2025-06-29T06:56:52","date_gmt":"2025-06-29T06:56:52","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=42597"},"modified":"2025-06-29T06:56:53","modified_gmt":"2025-06-29T06:56:53","slug":"a-student-takes-a-strand-of-copper-wire-and-forms-it-into-a-circular-loop-with-a-circumference-of-0-335-m","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/a-student-takes-a-strand-of-copper-wire-and-forms-it-into-a-circular-loop-with-a-circumference-of-0-335-m\/","title":{"rendered":"A student takes a strand of copper wire and forms it into a circular loop with a circumference of 0.335 m."},"content":{"rendered":"\n<p>A student takes a strand of copper wire and forms it into a circular loop with a circumference of 0.335 m. The student then places the loop in a uniform, constant magnetic field of magnitude 0.00313 T that is oriented perpendicular to the face of the loop. Pulling on the ends of the wire, they reduce the circumference of the loop to 0.167 m in a time interval of 0.682 s. Assuming that the loop remains circular as it shrinks, what is the magnitude of the average emf induced around the loop during this time interval?<\/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 calculate the magnitude of the average induced electromotive force (emf) around the loop, we can apply <strong>Faraday\u2019s Law of Induction<\/strong>. This law states that the induced emf is equal to the rate of change of the magnetic flux through the loop:E=\u2212\u0394\u03a6\u0394t\\mathcal{E} = &#8211; \\frac{\\Delta \\Phi}{\\Delta t}E=\u2212\u0394t\u0394\u03a6\u200b<\/p>\n\n\n\n<p>where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>E\\mathcal{E}E is the induced emf,<\/li>\n\n\n\n<li>\u0394\u03a6\\Delta \\Phi\u0394\u03a6 is the change in magnetic flux,<\/li>\n\n\n\n<li>\u0394t\\Delta t\u0394t is the time interval over which the change occurs.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Magnetic Flux<\/h3>\n\n\n\n<p>The magnetic flux (\u03a6\\Phi\u03a6) through the loop is given by the formula:\u03a6=B\u22c5A\\Phi = B \\cdot A\u03a6=B\u22c5A<\/p>\n\n\n\n<p>where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>BBB is the magnetic field strength (0.00313 T),<\/li>\n\n\n\n<li>AAA is the area of the loop.<\/li>\n<\/ul>\n\n\n\n<p>Since the loop is circular, the area AAA can be calculated using the formula:A=\u03c0r2A = \\pi r^2A=\u03c0r2<\/p>\n\n\n\n<p>where rrr is the radius of the loop. The radius can be derived from the circumference CCC, using the formula C=2\u03c0rC = 2\\pi rC=2\u03c0r. Therefore, the radius is:r=C2\u03c0r = \\frac{C}{2\\pi}r=2\u03c0C\u200b<\/p>\n\n\n\n<p>For the initial circumference C1=0.335\u2009mC_1 = 0.335 \\, \\text{m}C1\u200b=0.335m, the initial radius is:r1=0.3352\u03c0\u22480.0533\u2009mr_1 = \\frac{0.335}{2\\pi} \\approx 0.0533 \\, \\text{m}r1\u200b=2\u03c00.335\u200b\u22480.0533m<\/p>\n\n\n\n<p>Now, the initial area A1A_1A1\u200b is:A1=\u03c0(0.0533)2\u22480.00896\u2009m2A_1 = \\pi (0.0533)^2 \\approx 0.00896 \\, \\text{m}^2A1\u200b=\u03c0(0.0533)2\u22480.00896m2<\/p>\n\n\n\n<p>For the final circumference C2=0.167\u2009mC_2 = 0.167 \\, \\text{m}C2\u200b=0.167m, the final radius is:r2=0.1672\u03c0\u22480.0266\u2009mr_2 = \\frac{0.167}{2\\pi} \\approx 0.0266 \\, \\text{m}r2\u200b=2\u03c00.167\u200b\u22480.0266m<\/p>\n\n\n\n<p>Now, the final area A2A_2A2\u200b is:A2=\u03c0(0.0266)2\u22480.00222\u2009m2A_2 = \\pi (0.0266)^2 \\approx 0.00222 \\, \\text{m}^2A2\u200b=\u03c0(0.0266)2\u22480.00222m2<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Change in Magnetic Flux<\/h3>\n\n\n\n<p>The change in magnetic flux \u0394\u03a6\\Delta \\Phi\u0394\u03a6 is:\u0394\u03a6=B\u22c5(A2\u2212A1)\\Delta \\Phi = B \\cdot (A_2 &#8211; A_1)\u0394\u03a6=B\u22c5(A2\u200b\u2212A1\u200b)<\/p>\n\n\n\n<p>Substituting the values:\u0394\u03a6=0.00313\u22c5(0.00222\u22120.00896)\u22480.00313\u22c5(\u22120.00674)\u2248\u22122.11\u00d710\u22125\u2009Wb\\Delta \\Phi = 0.00313 \\cdot (0.00222 &#8211; 0.00896) \\approx 0.00313 \\cdot (-0.00674) \\approx -2.11 \\times 10^{-5} \\, \\text{Wb}\u0394\u03a6=0.00313\u22c5(0.00222\u22120.00896)\u22480.00313\u22c5(\u22120.00674)\u2248\u22122.11\u00d710\u22125Wb<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Calculate the Induced emf<\/h3>\n\n\n\n<p>The time interval over which the change occurs is \u0394t=0.682\u2009s\\Delta t = 0.682 \\, \\text{s}\u0394t=0.682s. Now we can calculate the induced emf:E=\u2212\u0394\u03a6\u0394t=2.11\u00d710\u221250.682\u22483.09\u00d710\u22125\u2009V\\mathcal{E} = &#8211; \\frac{\\Delta \\Phi}{\\Delta t} = \\frac{2.11 \\times 10^{-5}}{0.682} \\approx 3.09 \\times 10^{-5} \\, \\text{V}E=\u2212\u0394t\u0394\u03a6\u200b=0.6822.11\u00d710\u22125\u200b\u22483.09\u00d710\u22125V<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer:<\/h3>\n\n\n\n<p>The magnitude of the average induced emf is approximately <strong>3.09 \u00d7 10\u207b\u2075 V<\/strong>, or <strong>30.9 \u00b5V<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Faraday\u2019s Law<\/strong> tells us that a changing magnetic flux induces an emf in the loop.<\/li>\n\n\n\n<li>The loop\u2019s area changes as it shrinks, causing the magnetic flux to change, which leads to the induction of emf.<\/li>\n\n\n\n<li>The rate at which the magnetic flux changes over time gives us the average induced emf.<\/li>\n<\/ul>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-337.jpeg\" alt=\"\" class=\"wp-image-42598\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-337.jpeg 1024w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-337-300x300.jpeg 300w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-337-150x150.jpeg 150w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-337-768x768.jpeg 768w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>A student takes a strand of copper wire and forms it into a circular loop with a circumference of 0.335 m. The student then places the loop in a uniform, constant magnetic field of magnitude 0.00313 T that is oriented perpendicular to the face of the loop. Pulling on the ends of the wire, they [&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-42597","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/42597","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=42597"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/42597\/revisions"}],"predecessor-version":[{"id":42599,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/42597\/revisions\/42599"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=42597"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=42597"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=42597"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}