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Q

Asked by Arushi Juyal 24th December 2017, 11:57 PM
Answered by Expert
Answer:
 
Let us calculate the field at O due to the length AB
Let a small current wire element dl which is at the distance r from the point O as shown in figure. As per Biot-Savart's law the magnetic field at O due to this dl is given by

begin mathsize 14px style b space equals space fraction numerator mu subscript 0 over denominator 4 pi r cubed end fraction space stack i d l with rightwards arrow on top space cross times space r with rightwards arrow on top space equals space fraction numerator mu subscript 0 space end subscript i over denominator 4 pi r squared end fraction space d l space sin space alpha space equals space fraction numerator mu subscript 0 space i over denominator 4 pi r squared end fraction space d l space sin space left parenthesis 90 plus beta right parenthesis space equals fraction numerator mu subscript 0 space i over denominator 4 pi r squared end fraction space d l space cos space beta end style
Magnetic field due to wire length AB is given by
begin mathsize 12px style B subscript 1 space equals space integral subscript 0 superscript theta subscript 1 end superscript fraction numerator mu subscript 0 space i over denominator 4 pi space r squared end fraction d l space cos space beta space l space equals d cross times tan beta comma space h e n c e space d l space equals d cross times s e c squared beta space cross times space d beta  r space equals space d space cross times space s e c space beta space b y space s u b s t i t u t i n g space f o r space d l space a n d space r space i n space t h e space a b o v e space i n t e g r a t i o n space w e space g e t  B subscript 1 equals fraction numerator mu subscript 0 space i over denominator 4 pi d end fraction integral subscript 0 superscript theta subscript 1 end superscript cos beta space d beta space equals space fraction numerator mu subscript 0 space i over denominator 4 pi d end fraction sin space theta subscript 1 space equals space fraction numerator mu subscript 0 space i over denominator 4 pi d end fraction sin space left parenthesis 90 minus ϕ subscript 1 right parenthesis space equals space fraction numerator mu subscript 0 space i over denominator 4 pi d end fraction cos ϕ subscript 1 space end style
in same way we can calculate the magnetic field B2 due to length BC as
 
begin mathsize 12px style B subscript 2 space equals space fraction numerator mu subscript 0 space i over denominator 4 pi d end fraction cos ϕ subscript 2 end style
total field B = B1+B2 = begin mathsize 12px style fraction numerator mu subscript 0 space i over denominator 4 pi d end fraction open parentheses cos ϕ subscript 1 space plus space cos ϕ subscript 2 close parentheses end style
Answered by Expert 26th December 2017, 8:59 AM
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