{"id":168673,"date":"2024-11-17T20:30:27","date_gmt":"2024-11-17T20:30:27","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=168673"},"modified":"2024-11-17T20:30:31","modified_gmt":"2024-11-17T20:30:31","slug":"three-infinite-straight-wires-are-fixed-in-place-and-aligned-parallel-to-the-z-axis-as-shown","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2024\/11\/17\/three-infinite-straight-wires-are-fixed-in-place-and-aligned-parallel-to-the-z-axis-as-shown\/","title":{"rendered":"Three infinite straight wires are fixed in place and aligned parallel to the z-axis as shown."},"content":{"rendered":"\n

Three infinite straight wires are fixed in place and aligned parallel to the z-axis as shown. The wire at (x,y) – (-15.5 cm, 0) carries current l1-3.4 A in the negative z-direction. The wire at (x,y) (15.5 cm, 0) carries current 12-0.5 A in the positive z-direction. The wire at (x.y) -(0, 26.8 cm) carries current 13 -5.2 A in the positive z-direction. 1What is Bx0,0), the x-component of the magnetic field produced by these three wires at the origin? Submit 2)What is 8yl0,0), the y-component of the magnetic field produced by these three wires at the origin? Submit 3)What is Fx(1), the x-component of the force exerted on a one meter length of the wire carrying current 1? 4) What is Fyl), the y-component of the force exerted on a one meter length of the wire carrying current 1? 5)What is Fx(2), the x-component of the force exerted on a one meter length of the wire carrying current l2? Submit 6)Another wire is now added, also aligned with the z-axis at (x,y) (0, -26.8 cm) as shown. This wire carries current 14 A. Which of the following statements is true If 14 is directed along the positive z-axis, then it is possible to make the y-component of the magnetic field equal to zero at the origin If 14 is directed along the negative z-axis, then it is possible to make the y-component of the magnetic field equal to zero at the origin If 14 is directed along the positive z-axis, then it is possible to make the x-component of the magnetic field equal to zero at the origin If 14 is directed along the negative z-axis, then it is possible to make the x- component of the magnetic field equal to zero at the origin<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Let’s solve the problem step by step, starting with the individual components for each question.<\/p>\n\n\n\n

1) Bx(0,0), the x-component of the magnetic field produced by these three wires at the origin<\/strong><\/h3>\n\n\n\n

The magnetic field due to a current-carrying wire at a distance rrr from the wire is given by Amp\u00e8re\u2019s law:B=\u03bc0I2\u03c0rB = \\frac{\\mu_0 I}{2\\pi r}B=2\u03c0r\u03bc0\u200bI\u200b<\/p>\n\n\n\n

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

    \n
  • BBB is the magnetic field,<\/li>\n\n\n\n
  • \u03bc0=4\u03c0\u00d710\u22127\u2009T\\cdotpm\/A\\mu_0 = 4\\pi \\times 10^{-7} \\, \\text{T\u00b7m\/A}\u03bc0\u200b=4\u03c0\u00d710\u22127T\\cdotpm\/A (permeability of free space),<\/li>\n\n\n\n
  • III is the current,<\/li>\n\n\n\n
  • rrr is the perpendicular distance from the wire to the point where the magnetic field is measured.<\/li>\n<\/ul>\n\n\n\n

    For the three wires, we will calculate the magnetic field at the origin (0,0) due to each wire.<\/p>\n\n\n\n

      \n
    • Wire 1 at (-15.5 cm, 0), I1=3.4\u2009AI_1 = 3.4 \\, \\text{A}I1\u200b=3.4A, direction: negative z-axis<\/strong>: The magnetic field produced by a current-carrying wire has a direction given by the right-hand rule. Since this wire is to the left of the origin, the field at the origin will point in the negative yyy-direction.The distance r1=15.5\u2009cm=0.155\u2009mr_1 = 15.5 \\, \\text{cm} = 0.155 \\, \\text{m}r1\u200b=15.5cm=0.155m, and the magnetic field due to this wire at the origin is:B1=\u03bc0I12\u03c0r1=(4\u03c0\u00d710\u22127)\u00d73.42\u03c0\u00d70.155\u22484.4\u00d710\u22126\u2009TB_1 = \\frac{\\mu_0 I_1}{2\\pi r_1} = \\frac{(4\\pi \\times 10^{-7}) \\times 3.4}{2\\pi \\times 0.155} \\approx 4.4 \\times 10^{-6} \\, \\text{T}B1\u200b=2\u03c0r1\u200b\u03bc0\u200bI1\u200b\u200b=2\u03c0\u00d70.155(4\u03c0\u00d710\u22127)\u00d73.4\u200b\u22484.4\u00d710\u22126TSince this field points in the negative yyy-direction, the contribution to Bx(0,0)Bx(0, 0)Bx(0,0) is 0.<\/li>\n\n\n\n
    • Wire 2 at (15.5 cm, 0), I2=0.5\u2009AI_2 = 0.5 \\, \\text{A}I2\u200b=0.5A, direction: positive z-axis<\/strong>: This wire is to the right of the origin. Using the right-hand rule, the magnetic field at the origin will point in the positive yyy-direction.The distance r2=0.155\u2009mr_2 = 0.155 \\, \\text{m}r2\u200b=0.155m, and the magnetic field due to this wire at the origin is:B2=\u03bc0I22\u03c0r2=(4\u03c0\u00d710\u22127)\u00d70.52\u03c0\u00d70.155\u22486.4\u00d710\u22127\u2009TB_2 = \\frac{\\mu_0 I_2}{2\\pi r_2} = \\frac{(4\\pi \\times 10^{-7}) \\times 0.5}{2\\pi \\times 0.155} \\approx 6.4 \\times 10^{-7} \\, \\text{T}B2\u200b=2\u03c0r2\u200b\u03bc0\u200bI2\u200b\u200b=2\u03c0\u00d70.155(4\u03c0\u00d710\u22127)\u00d70.5\u200b\u22486.4\u00d710\u22127TSince the field is in the positive yyy-direction, the contribution to Bx(0,0)Bx(0,0)Bx(0,0) is 0.<\/li>\n\n\n\n
    • Wire 3 at (0, 26.8 cm), I3=5.2\u2009AI_3 = 5.2 \\, \\text{A}I3\u200b=5.2A, direction: positive z-axis<\/strong>: This wire is directly above the origin. The magnetic field produced by this wire at the origin will point in the negative xxx-direction (by the right-hand rule).The distance r3=26.8\u2009cm=0.268\u2009mr_3 = 26.8 \\, \\text{cm} = 0.268 \\, \\text{m}r3\u200b=26.8cm=0.268m, and the magnetic field due to this wire at the origin is:B3=\u03bc0I32\u03c0r3=(4\u03c0\u00d710\u22127)\u00d75.22\u03c0\u00d70.268\u22482.4\u00d710\u22126\u2009TB_3 = \\frac{\\mu_0 I_3}{2\\pi r_3} = \\frac{(4\\pi \\times 10^{-7}) \\times 5.2}{2\\pi \\times 0.268} \\approx 2.4 \\times 10^{-6} \\, \\text{T}B3\u200b=2\u03c0r3\u200b\u03bc0\u200bI3\u200b\u200b=2\u03c0\u00d70.268(4\u03c0\u00d710\u22127)\u00d75.2\u200b\u22482.4\u00d710\u22126TThis field is in the negative xxx-direction, so it contributes to Bx(0,0)Bx(0,0)Bx(0,0) with a value of \u22122.4\u00d710\u22126\u2009T-2.4 \\times 10^{-6} \\, \\text{T}\u22122.4\u00d710\u22126T.<\/li>\n<\/ul>\n\n\n\n

      Thus, the x-component of the magnetic field at the origin is:Bx(0,0)=0+0\u22122.4\u00d710\u22126=\u22122.4\u00d710\u22126\u2009TBx(0,0) = 0 + 0 – 2.4 \\times 10^{-6} = -2.4 \\times 10^{-6} \\, \\text{T}Bx(0,0)=0+0\u22122.4\u00d710\u22126=\u22122.4\u00d710\u22126T<\/p>\n\n\n\n

      2) By(0,0), the y-component of the magnetic field produced by these three wires at the origin<\/strong><\/h3>\n\n\n\n
        \n
      • Wire 1<\/strong> contributes a magnetic field in the negative yyy-direction with magnitude 4.4\u00d710\u22126\u2009T4.4 \\times 10^{-6} \\, \\text{T}4.4\u00d710\u22126T.<\/li>\n\n\n\n
      • Wire 2<\/strong> contributes a magnetic field in the positive yyy-direction with magnitude 6.4\u00d710\u22127\u2009T6.4 \\times 10^{-7} \\, \\text{T}6.4\u00d710\u22127T.<\/li>\n\n\n\n
      • Wire 3<\/strong> contributes no magnetic field in the yyy-direction (it contributes only in the xxx-direction).<\/li>\n<\/ul>\n\n\n\n

        Thus, the y-component of the magnetic field at the origin is:By(0,0)=\u22124.4\u00d710\u22126+6.4\u00d710\u22127=\u22123.76\u00d710\u22126\u2009TBy(0,0) = -4.4 \\times 10^{-6} + 6.4 \\times 10^{-7} = -3.76 \\times 10^{-6} \\, \\text{T}By(0,0)=\u22124.4\u00d710\u22126+6.4\u00d710\u22127=\u22123.76\u00d710\u22126T<\/p>\n\n\n\n

        3) Fx(1), the x-component of the force exerted on a one meter length of wire carrying current I1<\/strong><\/h3>\n\n\n\n

        The force per unit length on a current-carrying wire in a magnetic field is given by:F\/L=I\u22c5B\u22c5sin\u2061(\u03b8)F\/L = I \\cdot B \\cdot \\sin(\\theta)F\/L=I\u22c5B\u22c5sin(\u03b8)<\/p>\n\n\n\n

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

          \n
        • I=3.4\u2009AI = 3.4 \\, \\text{A}I=3.4A,<\/li>\n\n\n\n
        • B=2.4\u00d710\u22126\u2009TB = 2.4 \\times 10^{-6} \\, \\text{T}B=2.4\u00d710\u22126T (from the magnetic field calculation),<\/li>\n\n\n\n
        • \u03b8=90\u2218\\theta = 90^\\circ\u03b8=90\u2218 (since the magnetic field is perpendicular to the wire).<\/li>\n<\/ul>\n\n\n\n

          Thus, the force per unit length on wire 1 is:F1=3.4\u22c52.4\u00d710\u22126=8.16\u00d710\u22126\u2009N\/mF_1 = 3.4 \\cdot 2.4 \\times 10^{-6} = 8.16 \\times 10^{-6} \\, \\text{N\/m}F1\u200b=3.4\u22c52.4\u00d710\u22126=8.16\u00d710\u22126N\/m<\/p>\n\n\n\n

          This force is directed in the positive yyy-direction (because wire 1\u2019s current is in the negative zzz-direction and the magnetic field produced by wire 3 is in the negative xxx-direction).<\/p>\n\n\n\n

          So, the x-component of the force on wire 1 is 0, and the y-component of the force on wire 1 is 8.16\u00d710\u22126\u2009N\/m8.16 \\times 10^{-6} \\, \\text{N\/m}8.16\u00d710\u22126N\/m.<\/p>\n\n\n\n

          4) Fy(1), the y-component of the force exerted on a one meter length of wire carrying current I1<\/strong><\/h3>\n\n\n\n

          The magnetic field at wire 1 is primarily in the yyy-direction, causing a force in the xxx-direction. Since the force on wire 1 is perpendicular to both the wire and the magnetic field, the yyy-component of the force is 0.<\/p>\n\n\n\n

          5) Fx(2), the x-component of the force exerted on a one meter length of the wire carrying current I2<\/strong><\/h3>\n\n\n\n

          For wire 2, the magnetic field at its location is in the yyy-direction. Since the current in wire 2 is in the positive zzz-direction, the force is directed in the positive xxx-direction.F2=0.5\u00d72.4\u00d710\u22126=1.2\u00d710\u22126\u2009N\/mF_2 = 0.5 \\times 2.4 \\times 10^{-6} = 1.2 \\times 10^{-6} \\, \\text{N\/m}F2\u200b=0.5\u00d72.4\u00d710\u22126=1.2\u00d710\u22126N\/m<\/p>\n\n\n\n

          Thus, Fx(2)=1.2\u00d710\u22126\u2009N\/mFx(2) = 1.2 \\times 10^{-6} \\, \\text{N\/m}Fx(2)=1.2\u00d710\u22126N\/m.<\/p>\n\n\n\n

          6) Effect of adding a new wire at (0, -26.8 cm)<\/strong><\/h3>\n\n\n\n

          Adding a new wire at (0,\u221226.8\u2009cm)(0, -26.8 \\, \\text{cm})(0,\u221226.8cm) with a current I4=4\u2009AI_4 = 4 \\, \\text{A}I4\u200b=4A will influence the magnetic field at the origin. The magnetic field produced by this wire will interact with the fields from the other wires.<\/p>\n\n\n\n

            \n
          • If I4I_4I4\u200b is directed along the positive<\/strong> zzz-axis, it will create a magnetic field that increases the yyy-component, and it will be possible to make By(0,0)=0B_y(0,0) = 0By\u200b(0,0)=0.<\/li>\n\n\n\n
          • If I4I_4I4\u200b is directed along the negative<\/strong> zzz-axis, it will create a magnetic field that decreases the yyy-component, potentially making it zero.<\/li>\n<\/ul>\n\n\n\n

            The correct answer is:
            “If I4I_4I4\u200b is directed along the positive z-axis, then it is possible to make the y-component of the magnetic field equal to zero at the origin.”<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"

            Three infinite straight wires are fixed in place and aligned parallel to the z-axis as shown. The wire at (x,y) – (-15.5 cm, 0) carries current l1-3.4 A in the negative z-direction. The wire at (x,y) (15.5 cm, 0) carries current 12-0.5 A in the positive z-direction. The wire at (x.y) -(0, 26.8 cm) carries […]<\/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-168673","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/168673","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=168673"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/168673\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=168673"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=168673"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=168673"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}