{"id":38409,"date":"2025-06-26T09:41:17","date_gmt":"2025-06-26T09:41:17","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=38409"},"modified":"2025-06-26T09:41:18","modified_gmt":"2025-06-26T09:41:18","slug":"a-cylindrical-wire-of-radius-2-0-mm-carries-a-current-of-2-5-a","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/a-cylindrical-wire-of-radius-2-0-mm-carries-a-current-of-2-5-a\/","title":{"rendered":"A cylindrical wire of radius 2.0 mm carries a current of 2.5 A"},"content":{"rendered":"\n

A cylindrical wire of radius 2.0 mm carries a current of 2.5 A. The potential 4) difference between points on the wire that are 46 m apart is 3.7 V.<\/p>\n\n\n\n

The correct answer and explanation is:<\/mark><\/strong><\/p>\n\n\n\n

To determine the resistivity<\/strong> of the cylindrical wire, we will use Ohm’s Law<\/strong> and the formula for the resistance of a cylindrical conductor<\/strong>.<\/p>\n\n\n\n

\n\n\n\n

Correct Answer<\/strong>:<\/h3>\n\n\n\n

The resistivity of the wire is<\/strong> approximately 4.03 \u00d7 10\u207b\u2077 \u03a9\u00b7m<\/strong><\/p>\n\n\n\n

\n\n\n\n

Explanation (300 words)<\/strong>:<\/h3>\n\n\n\n

Electric current flowing through a conductor encounters resistance. This resistance causes a voltage drop along the length of the conductor. In this problem, a cylindrical wire has a known current, length, radius, and voltage drop. Using these values, the resistivity of the material can be calculated.<\/p>\n\n\n\n

First, recall Ohm\u2019s Law<\/strong>: V=IRV = IR<\/p>\n\n\n\n

Where:<\/p>\n\n\n\n

    \n
  • VV is the potential difference (3.7 V),<\/li>\n\n\n\n
  • II is the current (2.5 A),<\/li>\n\n\n\n
  • RR is the resistance over the 46 m section.<\/li>\n<\/ul>\n\n\n\n

    Rearranging the equation: R=VI=3.72.5=1.48 \u03a9R = \\frac{V}{I} = \\frac{3.7}{2.5} = 1.48\\ \\Omega<\/p>\n\n\n\n

    Next, use the formula for resistance<\/strong> of a cylindrical conductor: R=\u03c1LAR = \\frac{\\rho L}{A}<\/p>\n\n\n\n

    Where:<\/p>\n\n\n\n

      \n
    • \u03c1\\rho is the resistivity,<\/li>\n\n\n\n
    • LL is the length (46 m),<\/li>\n\n\n\n
    • AA is the cross-sectional area of the wire,<\/li>\n\n\n\n
    • A=\u03c0r2=\u03c0(0.002)2=1.2566\u00d710\u22125\u00a0m2A = \\pi r^2 = \\pi (0.002)^2 = 1.2566 \\times 10^{-5}\\ m^2<\/li>\n<\/ul>\n\n\n\n

      Now, rearranging to solve for resistivity: \u03c1=RAL\\rho = \\frac{RA}{L} \u03c1=1.48\u00d71.2566\u00d710\u2212546\\rho = \\frac{1.48 \\times 1.2566 \\times 10^{-5}}{46} \u03c1\u22481.860\u00d710\u2212546\u22484.04\u00d710\u22127 \u03a9\u22c5m\\rho \\approx \\frac{1.860 \\times 10^{-5}}{46} \\approx 4.04 \\times 10^{-7}\\ \\Omega\\cdot m<\/p>\n\n\n\n

      So, the resistivity of the wire material is approximately 4.04 \u00d7 10\u207b\u2077 ohm-meters<\/strong>.<\/p>\n\n\n\n

      This value is consistent with the resistivity of materials like aluminum<\/strong>, indicating the wire might be made from that. Knowing resistivity helps identify materials and evaluate their effectiveness in conducting electricity. Lower resistivity means the material is a better conductor.<\/p>\n","protected":false},"excerpt":{"rendered":"

      A cylindrical wire of radius 2.0 mm carries a current of 2.5 A. The potential 4) difference between points on the wire that are 46 m apart is 3.7 V. The correct answer and explanation is: To determine the resistivity of the cylindrical wire, we will use Ohm’s Law and the formula for the resistance […]<\/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-38409","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/38409","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=38409"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/38409\/revisions"}],"predecessor-version":[{"id":38410,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/38409\/revisions\/38410"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=38409"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=38409"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=38409"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}