{"id":228405,"date":"2025-06-07T09:06:11","date_gmt":"2025-06-07T09:06:11","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=228405"},"modified":"2025-06-07T09:06:15","modified_gmt":"2025-06-07T09:06:15","slug":"conducting-wire-formed-in-the-shape-of-a-right-triangle-with-base-b-28-cm-and-height-h-81-cm-and-having-resistance-r-2-7-rotates-uniformly-around-the-y-axis-in-the-direction-indicated-by-the-arr","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/07\/conducting-wire-formed-in-the-shape-of-a-right-triangle-with-base-b-28-cm-and-height-h-81-cm-and-having-resistance-r-2-7-rotates-uniformly-around-the-y-axis-in-the-direction-indicated-by-the-arr\/","title":{"rendered":"Conducting wire formed in the shape of a right triangle with base b = 28 cm and height h = 81 cm and having resistance R = 2.7 rotates uniformly around the Y-axis in the direction indicated by the arrow (clockwise as viewed from above, looking down in the negative Y-direction)."},"content":{"rendered":"\n<p>Conducting wire formed in the shape of a right triangle with base b = 28 cm and height h = 81 cm and having resistance R = 2.7 rotates uniformly around the Y-axis in the direction indicated by the arrow (clockwise as viewed from above, looking down in the negative Y-direction). The triangle makes one complete rotation in time seconds. A constant magnetic field B = 1.1 T pointing in the positive z-direction (out of the screen) exists in the region where the wire is rotating. 1) What is W, the angular frequency of rotation? radians \/ second Submit You currently have 0 submissions for this question. Only 10 submissions are allowed: You can make 10 more submissions for this question. 2) What is Imax, the magnitude of the maximum induced current in the loop? Submit You currently have 0 submissions for this question. Only 10 submissions are allowed: You can make 10 more submissions for this question. 3) At time t = 0, the wire is positioned as shown. What is the magnitude of the magnetic flux \u00ce\u00a6 at time 0.4125 T-m^2? Submit You currently have 0 submissions for this question. Only 10 submissions are allowed: You can make 10 more submissions for this question. 4) What is I1, the induced current in the loop at time 0.4125 s? I is defined to be positive if it flows in the negative y-direction in the segment of length h. Submit You currently have 0 submissions for this question. Only 10 submissions are allowed: You can make 10 more submissions for this question.<br>Conducting wire formed in the shape of a right triangle with base b = 28 cm and height h = 81 cm and having resistance R = 2.7 rotates uniformly around the Y-axis in the direction indicated by the arrow (clockwise as viewed from above, looking down in the negative Y-direction). The triangle makes one complete rotation in time seconds. A constant magnetic field B = 1.1 T pointing in the positive z-direction (out of the screen) exists in the region where the wire is rotating. 1) What is W, the angular frequency of rotation? radians \/ second Submit You currently have 0 submissions for this question. Only 10 submissions are allowed: You can make 10 more submissions for this question. 2) What is Imax, the magnitude of the maximum induced current in the loop? Submit You currently have 0 submissions for this question. Only 10 submissions are allowed: You can make 10 more submissions for this question. 3) At time t = 0, the wire is positioned as shown. What is the magnitude of the magnetic flux \u00ce\u00a6 at time 0.4125 T-m^2? Submit You currently have 0 submissions for this question. Only 10 submissions are allowed: You can make 10 more submissions for this question. 4) What is I1, the induced current in the loop at time 0.4125 s? I is defined to be positive if it flows in the negative y-direction in the segment of length h. Submit You currently have 0 submissions for this question. Only 10 submissions are allowed: You can make 10 more submissions for this question.<\/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\/image-233.png\" alt=\"\" class=\"wp-image-228406\"\/><\/figure>\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<h3 class=\"wp-block-heading\"><strong>Given:<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Triangle: Right triangle with base b=0.28\u2009mb = 0.28 \\, \\text{m}b=0.28m, height h=0.81\u2009mh = 0.81 \\, \\text{m}h=0.81m<\/li>\n\n\n\n<li>Resistance: R=2.7\u2009\u03a9R = 2.7 \\, \\OmegaR=2.7\u03a9<\/li>\n\n\n\n<li>Rotates about the <strong>y-axis<\/strong>, uniformly.<\/li>\n\n\n\n<li>Time for one rotation: T=1.1\u2009sT = 1.1 \\, \\text{s}T=1.1s<\/li>\n\n\n\n<li>Magnetic field: B=1.1\u2009TB = 1.1 \\, \\text{T}B=1.1T (constant, out of the screen, +z^+\\hat{z}+z^ direction)<\/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>1) What is \u03c9\\omega\u03c9, the angular frequency of rotation?<\/strong><\/h3>\n\n\n\n<p>Angular frequency: \u03c9=2\u03c0T=2\u03c01.1\u22485.712\u2009rad\/s\\omega = \\frac{2\\pi}{T} = \\frac{2\\pi}{1.1} \\approx 5.712 \\, \\text{rad\/s}\u03c9=T2\u03c0\u200b=1.12\u03c0\u200b\u22485.712rad\/s<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>2) What is ImaxI_{\\text{max}}Imax\u200b, the maximum induced current?<\/strong><\/h3>\n\n\n\n<p>Max EMF is: Emax=B\u22c5A\u22c5\u03c9\\mathcal{E}_{\\text{max}} = B \\cdot A \\cdot \\omegaEmax\u200b=B\u22c5A\u22c5\u03c9<\/p>\n\n\n\n<p>Area AAA of triangle: A=12bh=12(0.28)(0.81)=0.1134\u2009m2A = \\frac{1}{2} b h = \\frac{1}{2} (0.28)(0.81) = 0.1134 \\, \\text{m}^2A=21\u200bbh=21\u200b(0.28)(0.81)=0.1134m2 Emax=1.1\u22c50.1134\u22c55.712\u22480.712\u2009V\\mathcal{E}_{\\text{max}} = 1.1 \\cdot 0.1134 \\cdot 5.712 \\approx 0.712 \\, \\text{V}Emax\u200b=1.1\u22c50.1134\u22c55.712\u22480.712V Imax=EmaxR=0.7122.7\u22480.264\u2009AI_{\\text{max}} = \\frac{\\mathcal{E}_{\\text{max}}}{R} = \\frac{0.712}{2.7} \\approx 0.264 \\, \\text{A}Imax\u200b=REmax\u200b\u200b=2.70.712\u200b\u22480.264A<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>3) What is the magnetic flux \u03a61\\Phi_1\u03a61\u200b at t=0.4125\u2009st = 0.4125 \\, \\text{s}t=0.4125s?<\/strong><\/h3>\n\n\n\n<p>Flux: \u03a6(t)=BAcos\u2061(\u03c9t)\\Phi(t) = B A \\cos(\\omega t)\u03a6(t)=BAcos(\u03c9t)<\/p>\n\n\n\n<p>Using previously calculated values:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>\u03c9t=5.712\u22c50.4125\u22482.356\u2009rad\\omega t = 5.712 \\cdot 0.4125 \\approx 2.356 \\, \\text{rad}\u03c9t=5.712\u22c50.4125\u22482.356rad<\/li>\n<\/ul>\n\n\n\n<p>\u03a6=1.1\u22c50.1134\u22c5cos\u2061(2.356)\u22480.1247\u22c5(\u22120.707)\u2248\u22120.0882\u2009Wb&nbsp;(or&nbsp;T\\cdotpm2)\\Phi = 1.1 \\cdot 0.1134 \\cdot \\cos(2.356) \\approx 0.1247 \\cdot (-0.707) \\approx -0.0882 \\, \\text{Wb (or T\u00b7m}^2)\u03a6=1.1\u22c50.1134\u22c5cos(2.356)\u22480.1247\u22c5(\u22120.707)\u2248\u22120.0882Wb&nbsp;(or&nbsp;T\\cdotpm2)<\/p>\n\n\n\n<p>Magnitude: \u2223\u03a6\u2223=0.0882\u2009T\\cdotpm2|\\Phi| = 0.0882 \\, \\text{T\u00b7m}^2\u2223\u03a6\u2223=0.0882T\\cdotpm2<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>4) What is I1I_1I1\u200b at t=0.4125\u2009st = 0.4125 \\, \\text{s}t=0.4125s?<\/strong><\/h3>\n\n\n\n<p>Induced EMF: E=\u2212d\u03a6dt=BA\u03c9sin\u2061(\u03c9t)\\mathcal{E} = &#8211; \\frac{d\\Phi}{dt} = B A \\omega \\sin(\\omega t)E=\u2212dtd\u03a6\u200b=BA\u03c9sin(\u03c9t) E=1.1\u22c50.1134\u22c55.712\u22c5sin\u2061(2.356)\u22480.712\u22c50.707\u22480.503\u2009V\\mathcal{E} = 1.1 \\cdot 0.1134 \\cdot 5.712 \\cdot \\sin(2.356) \\approx 0.712 \\cdot 0.707 \\approx 0.503 \\, \\text{V}E=1.1\u22c50.1134\u22c55.712\u22c5sin(2.356)\u22480.712\u22c50.707\u22480.503V I1=ER=0.5032.7\u22480.186\u2009AI_1 = \\frac{\\mathcal{E}}{R} = \\frac{0.503}{2.7} \\approx 0.186 \\, \\text{A}I1\u200b=RE\u200b=2.70.503\u200b\u22480.186A<\/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 of Correct Answers:<\/strong><\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li>\u03c9=5.712\u2009rad\/s\\omega = 5.712 \\, \\text{rad\/s}\u03c9=5.712rad\/s<\/li>\n\n\n\n<li>Imax=0.264\u2009AI_{\\text{max}} = 0.264 \\, \\text{A}Imax\u200b=0.264A<\/li>\n\n\n\n<li>\u03a61=0.0882\u2009T\\cdotpm2\\Phi_1 = 0.0882 \\, \\text{T\u00b7m}^2\u03a61\u200b=0.0882T\\cdotpm2<\/li>\n\n\n\n<li>I1=0.186\u2009AI_1 = 0.186 \\, \\text{A}I1\u200b=0.186A<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation<\/strong><\/h3>\n\n\n\n<p>This problem illustrates electromagnetic induction due to a rotating wire loop in a magnetic field. The conducting wire is shaped into a right triangle and rotates uniformly about the y-axis. As it rotates, the area vector of the loop changes its orientation relative to the magnetic field, leading to a time-varying magnetic flux. According to Faraday\u2019s Law of Induction, a time-varying magnetic flux through a loop induces an electromotive force (EMF), which can drive a current if the loop has resistance.<\/p>\n\n\n\n<p>The angular frequency \u03c9\\omega\u03c9 is directly related to the period TTT of rotation. With T=1.1\u2009sT = 1.1 \\, \\text{s}T=1.1s, we calculate \u03c9=2\u03c0\/T\\omega = 2\\pi \/ T\u03c9=2\u03c0\/T. The maximum EMF is when the rate of change of flux is at its peak, corresponding to the sine function reaching its maximum value. We multiply the angular frequency \u03c9\\omega\u03c9, magnetic field strength BBB, and the area of the loop to get the peak EMF. Dividing by the resistance gives us the maximum current, ImaxI_{\\text{max}}Imax\u200b.<\/p>\n\n\n\n<p>At a specific time t=0.4125\u2009st = 0.4125 \\, \\text{s}t=0.4125s, the flux through the loop is found by evaluating \u03a6=BAcos\u2061(\u03c9t)\\Phi = BA \\cos(\\omega t)\u03a6=BAcos(\u03c9t). The instantaneous induced current is then computed using I=(BA\u03c9sin\u2061(\u03c9t))\/RI = (B A \\omega \\sin(\\omega t)) \/ RI=(BA\u03c9sin(\u03c9t))\/R. The sinusoidal nature of flux and current is a hallmark of rotating systems in uniform magnetic fields.<\/p>\n\n\n\n<p>This system models real-world applications like electric generators, where mechanical rotation converts into electric current using magnetic induction.<\/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-banner7-276.jpeg\" alt=\"\" class=\"wp-image-228407\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Conducting wire formed in the shape of a right triangle with base b = 28 cm and height h = 81 cm and having resistance R = 2.7 rotates uniformly around the Y-axis in the direction indicated by the arrow (clockwise as viewed from above, looking down in the negative Y-direction). The triangle makes one [&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-228405","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/228405","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=228405"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/228405\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=228405"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=228405"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=228405"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}