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CBSE Class 12-science Answered

drive the expression for the magnetic field around a straight current carrying conductor of finite size?
please explain in detail. earlier because tomorrow is my exam.
Asked by modi72879 | 24 Nov, 2017, 09:29: AM
answered-by-expert Expert Answer
 
 
 
 
F r o m space B i o t minus space S a v a r t space l a w space f o r space a space s t r a i g h t space c o n d u c t o r dB space equals fraction numerator straight mu subscript 0 over denominator 4 straight pi end fraction fraction numerator straight I space dl with rightwards arrow on top straight x straight r with rightwards arrow on top over denominator straight r cubed end fraction angle space between space straight I space dl with rightwards arrow on top space and space straight r with rightwards arrow space on top is space left parenthesis 180 minus straight theta right parenthesis dB space equals fraction numerator straight mu subscript 0 over denominator 4 straight pi end fraction fraction numerator straight I space dl space sin space left parenthesis 180 minus straight theta right parenthesis over denominator straight r squared end fraction dB space equals fraction numerator straight mu subscript 0 over denominator 4 straight pi end fraction fraction numerator straight I space dl space sin space left parenthesis straight theta right parenthesis over denominator straight r squared end fraction E G space equals space space E F space space space sin space straight theta equals space dl space sin space straight theta space space space E G equals space E P space space sin space dϕ space equals straight r space sin space dϕ space dB space equals fraction numerator straight mu subscript 0 over denominator 4 straight pi end fraction fraction numerator straight I space dϕ space over denominator straight r end fraction from space increment space space E Q P space space space space space space space space space space space space space space space straight r equals fraction numerator straight R over denominator cos space straight ϕ end fraction space space semicolon space space dB space equals fraction numerator straight mu subscript 0 over denominator 4 straight pi end fraction fraction numerator straight I space space cos space left parenthesis straight theta right parenthesis space dϕ over denominator straight r end fraction magnetic space field space at space point space straight P space space becuase space of space entire space conductor space space can space be space found space by space integration straight B space equals space integral subscript negative straight ϕ subscript 1 end subscript superscript straight ϕ subscript 2 end superscript fraction numerator straight mu subscript 0 space end subscript straight x thin space straight I over denominator 4 πR end fraction space cosϕ space dϕ straight B equals fraction numerator straight mu subscript 0 space end subscript straight x thin space straight I over denominator 4 πR end fraction space left parenthesis sinϕ subscript 1 space plus space sinϕ subscript 2 right parenthesis for space 90 space degree straight B equals fraction numerator straight mu subscript 0 space end subscript straight x thin space straight I over denominator 4 πR end fraction space space NA over straight m direction space of space straight B space can space be space given space by space Right space thumb space rule
Answered by Gajendra | 24 Nov, 2017, 05:30: PM

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