{"id":3104,"date":"2018-06-09T17:23:43","date_gmt":"2018-06-10T00:23:43","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=3104"},"modified":"2018-06-09T17:23:43","modified_gmt":"2018-06-10T00:23:43","slug":"magnetic-field-at-the-center-of-an-arc-of-current-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/magnetic-field-at-the-center-of-an-arc-of-current-problems-and-solutions.htm","title":{"rendered":"Magnetic field at the center of an arc of current – problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

1. Based on the figure below, if the radius of the curvature of the wire is 50 cm, determine the magnitude of the magnetic field<\/a> at the center of curvature (at point 0, see figure below). (\u00b5<\/span><\/span>o<\/span><\/span><\/sub> = 4\u03c0.10<\/span><\/span>-7<\/span><\/span><\/sup> Wb.A<\/span><\/span>-1<\/span><\/span><\/sup> m<\/span><\/span>-1<\/span><\/span><\/sup>)<\/span><\/span><\/p>\n

Known :\"Magnetic<\/u><\/span><\/span><\/p>\n

Radius (r) = 50 cm = 0.5 m<\/span><\/span><\/p>\n

Electric current<\/a> (I) = 1.5 Ampere<\/span><\/span><\/p>\n

The\u00a0vacuum permeability <\/span><\/span><\/span>(\u00b5<\/span><\/span>o<\/span><\/span><\/sub>) = 4\u03c0.10<\/span><\/span>-7 <\/span><\/span><\/sup>Wb.A<\/span><\/span>-1<\/span><\/span><\/sup> m<\/span><\/span>-1<\/span><\/span><\/sup><\/p>\n

Wanted:<\/u><\/span><\/span> T<\/span><\/span>he magnitude of the magnetic field<\/span><\/span><\/p>\n

Solution :<\/u><\/span><\/span><\/span><\/p>\n

360<\/span><\/span><\/span>o<\/span><\/span><\/sup><\/span> = 1 <\/span><\/span><\/span>circumference<\/span><\/span><\/span><\/em>\u00a0<\/span><\/span><\/span>of a circle<\/span><\/span><\/span><\/a>. 120<\/span><\/span><\/span>o<\/span><\/span><\/sup><\/span> \/ 360<\/span><\/span><\/span>o<\/span><\/span><\/sup><\/span> = 1\/3 then 120<\/span><\/span><\/span>o<\/span><\/span><\/sup><\/span> = 1\/3 x <\/span><\/span><\/span>circumference<\/span><\/span><\/span><\/em>\u00a0<\/span><\/span><\/span>of a circle.<\/span><\/span><\/span><\/p>\n

The equation of the magnetic field at the center of the coil with several loops :<\/u><\/span><\/span><\/p>\n

\"Magnetic<\/p>\n

B = the magnitude of the magnetic field, N = number of loops, I = electric current, r = radius of curvature<\/i><\/span><\/span><\/p>\n

In the above problem, there is only one loop so that N is eliminated from the equation. The wire coil on the above problem is not 1 circle but 1\/3 circle :<\/span><\/span><\/p>\n

\"Magnetic<\/p>\n

The magnitude of the magnetic field at the center of curvature :<\/u><\/span><\/span><\/p>\n

\"Magnetic<\/p>\n

2. Based on the figure below, the electric current flows in the wire is 6-A and radius of curvature is R = 3\u03c0 cm, to determine the magnitude of the magnetic field at point P.<\/span><\/span><\/p>\n

Known :\"Magnetic<\/u><\/span><\/span><\/p>\n

Radius of curvature (r) = 3\u03c0 cm = (3\u03c0\/100) m<\/span><\/span><\/p>\n

= 3\u03c0\/10<\/span><\/span>2<\/span><\/span><\/sup> m = 3\u03c0.10<\/span><\/span>-2 <\/span><\/span><\/sup>m<\/span><\/span><\/p>\n

Electric current (I) = 6 A<\/span><\/span><\/p>\n

The\u00a0vacuum permeability <\/span><\/span><\/span>(\u00b5<\/span><\/span>o<\/span><\/span><\/sub>) = 4\u03c0.10<\/span><\/span>-7 <\/span><\/span><\/sup>Wb.A<\/span><\/span>-1<\/span><\/span><\/sup> m<\/span><\/span>-1<\/span><\/span><\/sup><\/p>\n

Wanted :<\/u><\/span><\/span> The magnitude of the magnetic field <\/span><\/span><\/p>\n

Solution :<\/u><\/span><\/span><\/p>\n

360<\/span><\/span>o<\/span><\/span><\/sup> = 1 <\/span><\/span>circumference<\/span><\/span><\/span><\/em>\u00a0<\/span><\/span><\/span>of a circle. <\/span><\/span><\/span>45<\/span><\/span>o<\/span><\/span><\/sup> \/ 360<\/span><\/span>o<\/span><\/span><\/sup> = 1\/8 then 45<\/span><\/span>o<\/span><\/span><\/sup> = 1\/8 <\/span><\/span>x <\/span><\/span><\/span>circumference<\/span><\/span><\/span><\/em>\u00a0<\/span><\/span><\/span>of a circle.<\/span><\/span><\/span><\/p>\n

The magnitude of the magnetic field at the center of curvature :<\/u><\/span><\/span><\/p>\n

\"Magnetic<\/p>\n

3. Electric current flows in wire = 9-A, the radius of curvature (R) = 2\u03c0 cm and \u00b5<\/span><\/span>o<\/span><\/span><\/sub> = 4\u03c0.10<\/span><\/span>-7<\/span><\/span><\/sup> Wb.A<\/span><\/span>-1<\/span><\/span><\/sup>.m<\/span><\/span>-1<\/span><\/span><\/sup>, determine the magnitude of the magnetic field at point P.<\/span><\/span><\/p>\n

Known :<\/u><\/span><\/span><\/p>\n

Radius of curvature (r) = 2\u03c0 cm = (2\u03c0\/100) m\"Magnetic<\/span><\/span><\/p>\n

= 2\u03c0\/10<\/span><\/span>2<\/span><\/span><\/sup> m = 2\u03c0.10<\/span><\/span>-2 <\/span><\/span><\/sup>m <\/span><\/span><\/p>\n

Electric current (I) = 9 A<\/span><\/span><\/p>\n

The\u00a0vacuum permeability <\/span><\/span><\/span>(\u00b5<\/span><\/span>o<\/span><\/span><\/sub>) = 4\u03c0.10<\/span><\/span>-7 <\/span><\/span><\/sup>Wb.A<\/span><\/span>-1<\/span><\/span><\/sup> m<\/span><\/span>-1<\/span><\/span><\/sup><\/p>\n

Wanted :<\/u><\/span><\/span> The magnitude of the magnetic field at point P<\/span><\/span><\/p>\n

Solution :<\/u><\/span><\/span><\/p>\n

360<\/span><\/span>o<\/span><\/span><\/sup> \u2013 120<\/span><\/span>o<\/span><\/span><\/sup> = 240<\/span><\/span>o<\/span><\/span><\/sup>. 240<\/span><\/span>o<\/span><\/span><\/sup> \/ 360<\/span><\/span>o<\/span><\/span><\/sup> = 2\/3 then 240<\/span><\/span>o<\/span><\/span><\/sup> = 2\/3 x <\/span><\/span>circumference<\/span><\/span><\/span><\/em>\u00a0<\/span><\/span><\/span>of a circle.<\/span><\/span><\/span><\/p>\n

The magnitude of the magnetic field at the center of curvature :<\/u><\/span><\/span><\/p>\n

\"Magnetic<\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

1. Based on the figure below, if the radius of the curvature of the wire is 50 cm, determine the magnitude of the magnetic field at the center of curvature (at point 0, see figure below). (\u00b5o = 4\u03c0.10-7 Wb.A-1 m-1) Known : Radius (r) = 50 cm = 0.5 m Electric current (I) = … Read more<\/a><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_seopress_titles_title":"","_seopress_titles_desc":"","_seopress_robots_index":"","_seopress_robots_follow":"","_seopress_robots_imageindex":"","_seopress_robots_snippet":"","_seopress_robots_primary_cat":"","_seopress_robots_breadcrumbs":"","_seopress_robots_freeze_modified_date":"","_seopress_robots_custom_modified_date":"","_seopress_robots_canonical":"","_seopress_social_fb_title":"","_seopress_social_fb_desc":"","_seopress_social_fb_img":"","_seopress_social_fb_img_attachment_id":0,"_seopress_social_fb_img_width":0,"_seopress_social_fb_img_height":0,"_seopress_social_twitter_title":"","_seopress_social_twitter_desc":"","_seopress_social_twitter_img":"","_seopress_social_twitter_img_attachment_id":0,"_seopress_social_twitter_img_width":0,"_seopress_social_twitter_img_height":0,"_seopress_redirections_value":"","_seopress_redirections_enabled":"","_seopress_redirections_enabled_regex":"","_seopress_redirections_logged_status":"","_seopress_redirections_param":"","_seopress_redirections_type":0,"_seopress_analysis_target_kw":"Magnetic field at the center of an arc of current - problems and solutions","_seopress_news_disabled":"","_seopress_video_disabled":"","_seopress_video":[],"_seopress_pro_schemas_manual":[],"_seopress_pro_rich_snippets_disable_all":"","_seopress_pro_rich_snippets_disable":[],"_seopress_pro_schemas":[],"footnotes":""},"categories":[3],"tags":[],"class_list":["post-3104","post","type-post","status-publish","format-standard","hentry","category-solved-problems-in-basic-physics"],"gt_translate_keys":[{"key":"link","format":"url"}],"_links":{"self":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/3104","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/comments?post=3104"}],"version-history":[{"count":0,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/3104\/revisions"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=3104"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=3104"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=3104"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}