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magnetic field on centre of triangle
Asked by radharajendran5555 | 20 Dec, 2022, 06:16: PM
Expert Answer
Let us find magnetic field induction due to linear current segment of length L and carrying current i .
 
Let us find magnetic field induction at a point P which is at a distance d from mid point of segment
 
as shown in figure .
 
Let idl be a small current segment of length dl at a distance l from mid point .
 
Magnetic field induction dB at P due to this small segment idl is given by Biot-Savart's law
 
begin mathsize 14px style d B space equals space fraction numerator mu subscript o i over denominator 4 pi end fraction space cross times fraction numerator d l space sin theta over denominator d squared plus l squared end fraction end style
where θ  is angle between small current segment idl and line joining current segment to point P.
 
If we write sinθ in terms of d and l , then we get
 
begin mathsize 14px style d B space equals space fraction numerator mu subscript o i space over denominator 4 pi end fraction space cross times fraction numerator d l space over denominator d squared plus l squared end fraction space cross times fraction numerator d over denominator square root of d squared plus l squared end root end fraction end style

 
begin mathsize 14px style d B space equals space fraction numerator mu subscript o space i space d space over denominator 4 pi end fraction space cross times fraction numerator d l space over denominator open parentheses d squared plus l squared close parentheses to the power of 3 divided by 2 end exponent end fraction space end style
Magnetic field induction B due to full length is given as
 
begin mathsize 14px style B space equals space integral d B space equals space fraction numerator mu subscript o space i space d space over denominator 4 pi end fraction space cross times integral subscript fraction numerator negative L over denominator 2 end fraction end subscript superscript fraction numerator plus L over denominator 2 end fraction end superscript fraction numerator d l space over denominator open parentheses d squared plus l squared close parentheses to the power of 3 divided by 2 end exponent end fraction space end style
if we use the substitution begin mathsize 14px style l space equals space d space tan alpha end style, then above integration becomes
 
begin mathsize 14px style B space equals space space fraction numerator mu subscript o space i space space over denominator 4 pi space d end fraction space cross times integral subscript fraction numerator negative L over denominator 2 end fraction end subscript superscript fraction numerator plus L over denominator 2 end fraction end superscript cos alpha space d alpha space equals space space fraction numerator mu subscript o space i space space over denominator 4 pi space d end fraction space open square brackets sin alpha close square brackets subscript begin inline style fraction numerator negative L over denominator 2 end fraction end style end subscript superscript begin inline style fraction numerator plus L over denominator 2 end fraction end style end superscript end style

begin mathsize 14px style B space equals space fraction numerator mu subscript o space i space space over denominator 4 pi space d end fraction space open square brackets fraction numerator l over denominator square root of l squared plus d squared end root end fraction close square brackets subscript begin inline style fraction numerator negative L over denominator 2 end fraction end style end subscript superscript begin inline style fraction numerator plus L over denominator 2 end fraction end style end superscript space equals space fraction numerator mu subscript o space i space space over denominator 4 pi space d end fraction cross times fraction numerator 2 L over denominator square root of L squared plus 4 space d squared end root end fraction end style  .............................(1)
 
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Now Let us find the magnetic field induction at the centroid of equilateral triangle of side length L
 
Let i be the current passing through the sides in counterclockwise direction.
 
if we apply eqn.(1) , then we get magnetic field at centroid as
 
begin mathsize 14px style B space equals space 3 space cross times fraction numerator mu subscript o space i space space over denominator 4 pi space d end fraction cross times fraction numerator 2 L over denominator square root of L squared plus 4 space d squared end root end fraction end style   ................................. (2)
The factor 3 in above expression is due to superimposition of magnetic field induction
at centroid due to three sides .
 
Distance d from centroid to sides is given as
 
d = (1/3) ( √3/2) L  = L / ( 2 √3 )
 
If we substitute d as above in eqn.(2), then we get
 
begin mathsize 14px style B space equals space fraction numerator 9 over denominator 2 pi end fraction cross times fraction numerator mu subscript o space end subscript i over denominator L end fraction end style
 
 
Answered by Thiyagarajan K | 21 Dec, 2022, 12:31: AM
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