A circular loop of flexible iron wire has an initial circumference of 165cm , but its circumference is decreasing at a constant rate of 12.0 cm\/s due to a tangential pull on the wire. The loop is in a constant, uniform magnetic field oriented perpendicular to the plane of the loop of magnitude 0.500 T Part A Find the emf induced in the loop, at the instant when 9.0s have passed. And find the direction<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To find the electromotive force (emf) induced in the loop, we can apply Faraday’s Law of Induction<\/strong>, which states that the induced emf (( \\mathcal{E} )) in a loop is given by:<\/p>\n\n\n\n [ where ( \\Phi_B ) is the magnetic flux through the loop and ( \\frac{d\\Phi_B}{dt} ) is the rate of change of magnetic flux.<\/p>\n\n\n\n Magnetic flux (( \\Phi_B )) is defined as:<\/p>\n\n\n\n [ where:<\/p>\n\n\n\n The area of a circular loop is given by:<\/p>\n\n\n\n [ where ( r ) is the radius of the loop.<\/p>\n\n\n\n Since the circumference ( C ) of the loop is related to the radius by:<\/p>\n\n\n\n [ we can express the radius in terms of the circumference:<\/p>\n\n\n\n [ At ( t = 9.0 \\, \\text{s} ), the initial circumference of the loop is 165 cm, but it is decreasing at a rate of 12.0 cm\/s. Thus, after 9.0 seconds, the circumference will be:<\/p>\n\n\n\n [ Now, calculate the radius:<\/p>\n\n\n\n [ Now we can find the area ( A ) of the loop:<\/p>\n\n\n\n [ Next, we calculate the magnetic flux at this time:<\/p>\n\n\n\n [ The rate of change of magnetic flux is related to the rate of change of the area because the magnetic field ( B ) is constant. Since the circumference is decreasing at a constant rate, the area is also decreasing at a constant rate. The rate of change of the area ( \\frac{dA}{dt} ) is:<\/p>\n\n\n\n [ The rate of change of the radius is:<\/p>\n\n\n\n [ Thus, the rate of change of the area is:<\/p>\n\n\n\n [ Now, we can find the rate of change of the magnetic flux:<\/p>\n\n\n\n [ Finally, we can calculate the induced emf:<\/p>\n\n\n\n [ According to Lenz’s Law<\/strong>, the direction of the induced current will oppose the change in flux. Since the magnetic flux through the loop is decreasing as the loop’s area decreases, the induced current will flow in such a way as to try to increase the magnetic flux. Therefore, the induced current will generate a magnetic field that points in the same direction as the external magnetic field (into the page), meaning the current will flow in a counterclockwise direction when viewed from above.<\/p>\n\n\n\n A circular loop of flexible iron wire has an initial circumference of 165cm , but its circumference is decreasing at a constant rate of 12.0 cm\/s due to a tangential pull on the wire. The loop is in a constant, uniform magnetic field oriented perpendicular to the plane of the loop of magnitude 0.500 T […]<\/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-162378","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/162378","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=162378"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/162378\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=162378"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=162378"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=162378"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\\mathcal{E} = -\\frac{d\\Phi_B}{dt}
]<\/p>\n\n\n\nStep 1: Find the magnetic flux<\/h3>\n\n\n\n
\\Phi_B = B A
]<\/p>\n\n\n\n\n
A = \\pi r^2
]<\/p>\n\n\n\n
C = 2\\pi r
]<\/p>\n\n\n\n
r = \\frac{C}{2\\pi}
]<\/p>\n\n\n\n
C = 165 \\, \\text{cm} – 12.0 \\, \\text{cm\/s} \\times 9.0 \\, \\text{s} = 165 \\, \\text{cm} – 108 \\, \\text{cm} = 57.0 \\, \\text{cm}
]<\/p>\n\n\n\n
r = \\frac{57.0 \\, \\text{cm}}{2\\pi} = \\frac{57.0}{2\\pi} \\approx 9.07 \\, \\text{cm} = 0.0907 \\, \\text{m}
]<\/p>\n\n\n\nStep 2: Find the change in magnetic flux<\/h3>\n\n\n\n
A = \\pi r^2 = \\pi (0.0907 \\, \\text{m})^2 \\approx 0.0259 \\, \\text{m}^2
]<\/p>\n\n\n\n
\\Phi_B = B A = 0.500 \\, \\text{T} \\times 0.0259 \\, \\text{m}^2 = 0.01295 \\, \\text{Wb}
]<\/p>\n\n\n\nStep 3: Calculate the rate of change of magnetic flux<\/h3>\n\n\n\n
\\frac{dA}{dt} = \\frac{d}{dt}(\\pi r^2) = 2\\pi r \\frac{dr}{dt}
]<\/p>\n\n\n\n
\\frac{dr}{dt} = \\frac{1}{2\\pi} \\frac{dC}{dt} = \\frac{1}{2\\pi} \\times (-12.0 \\, \\text{cm\/s}) = -\\frac{12.0 \\, \\text{cm\/s}}{2\\pi} \\approx -0.191 \\, \\text{m\/s}
]<\/p>\n\n\n\n
\\frac{dA}{dt} = 2\\pi r \\times \\left(-0.191 \\, \\text{m\/s}\\right) \\approx 2\\pi \\times 0.0907 \\, \\text{m} \\times (-0.191 \\, \\text{m\/s}) \\approx -0.102 \\, \\text{m}^2\/\\text{s}
]<\/p>\n\n\n\n
\\frac{d\\Phi_B}{dt} = B \\frac{dA}{dt} = 0.500 \\, \\text{T} \\times (-0.102 \\, \\text{m}^2\/\\text{s}) = -0.051 \\, \\text{Wb\/s}
]<\/p>\n\n\n\nStep 4: Calculate the induced emf<\/h3>\n\n\n\n
\\mathcal{E} = -\\frac{d\\Phi_B}{dt} = -(-0.051 \\, \\text{V}) = 0.051 \\, \\text{V} = 51.0 \\, \\text{mV}
]<\/p>\n\n\n\nStep 5: Determine the direction of the induced current<\/h3>\n\n\n\n
Final Answer:<\/h3>\n\n\n\n
\n