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magnetic field on centre of triangle
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

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

Magnetic field induction B due to full length is given as

if we use the substitution , then above integration becomes

.............................(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

................................. (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

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