A solenoid of length 0.700 m having a circular cross-section of radius 5.00 cm stores 6.00 \u00b5J of energy when a 0.400-A current runs through it. What is the winding density of the solenoid? (\u00b50 = 4p \u00d7 10-7 T \u00b7 m\/A)<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n The winding density ((n)) of a solenoid refers to the number of turns of wire per unit length. The energy stored in a solenoid can be expressed as:<\/p>\n\n\n\n [ where:<\/p>\n\n\n\n The inductance of a solenoid is given by:<\/p>\n\n\n\n [ where:<\/p>\n\n\n\n [ [ [ The winding density is approximately 104 turns per meter<\/strong>.<\/p>\n\n\n\n The winding density of a solenoid directly influences its inductance and magnetic field properties. In this problem, the solenoid stores a known amount of energy when a current passes through it. The relationship between the stored energy, current, and inductance allows us to determine the inductance first. Using the formula (U = \\frac{1}{2} L I^2), the inductance is calculated as (7.50 \\times 10^{-5} \\, \\text{H}).<\/p>\n\n\n\n Next, the inductance formula for a solenoid, (L = \\mu_0 n^2 A l), connects the physical properties of the solenoid (cross-sectional area, length, and winding density) with its inductance. By rearranging this formula, the winding density ((n)) can be isolated. The solenoid’s cross-sectional area is derived from its radius, and the given permeability of free space ((\\mu_0)) and length ((l)) are used.<\/p>\n\n\n\n After substituting the values into the formula, the result is approximately (104 \\, \\text{turns\/m}). This indicates that for every meter of the solenoid’s length, there are 104 wire loops. This density ensures the solenoid’s ability to store the specified energy and produce the required inductance for the given current.<\/p>\n\n\n\n Understanding winding density is essential in designing electromagnetic devices, as it determines the strength and efficiency of the magnetic field generated.<\/p>\n","protected":false},"excerpt":{"rendered":" A solenoid of length 0.700 m having a circular cross-section of radius 5.00 cm stores 6.00 \u00b5J of energy when a 0.400-A current runs through it. What is the winding density of the solenoid? (\u00b50 = 4p \u00d7 10-7 T \u00b7 m\/A) The Correct Answer and Explanation is : Solution: The winding density ((n)) of […]<\/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-180635","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/180635","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=180635"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/180635\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=180635"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=180635"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=180635"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Solution:<\/h3>\n\n\n\n
U = \\frac{1}{2} L I^2
]<\/p>\n\n\n\n\n
L = \\mu_0 n^2 A l
]<\/p>\n\n\n\n\n
Step 1: Solve for (L) using the energy equation.<\/h3>\n\n\n\n
L = \\frac{2U}{I^2}
]
[
L = \\frac{2(6.00 \\times 10^{-6})}{(0.400)^2} = 7.50 \\times 10^{-5} \\, \\text{H}
]<\/p>\n\n\n\nStep 2: Rearrange the inductance formula to solve for (n).<\/h3>\n\n\n\n
L = \\mu_0 n^2 A l \\implies n^2 = \\frac{L}{\\mu_0 A l}
]
[
n = \\sqrt{\\frac{L}{\\mu_0 A l}}
]<\/p>\n\n\n\nStep 3: Substitute values.<\/h3>\n\n\n\n
A = \\pi (0.0500)^2 = 7.854 \\times 10^{-3} \\, \\text{m}^2
]
[
n = \\sqrt{\\frac{7.50 \\times 10^{-5}}{(4\\pi \\times 10^{-7})(7.854 \\times 10^{-3})(0.700)}}
]
[
n = \\sqrt{\\frac{7.50 \\times 10^{-5}}{6.902 \\times 10^{-9}}} = \\sqrt{1.087 \\times 10^4} = 104.3 \\, \\text{turns\/m}
]<\/p>\n\n\n\nFinal Answer:<\/h3>\n\n\n\n
\n\n\n\nExplanation (300 words):<\/h3>\n\n\n\n