NEET Class neet Answered
magnetic field on centre of triangle
Asked by radharajendran5555 | 20 Dec, 2022, 06:16: PM

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



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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

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

Answered by Thiyagarajan K | 21 Dec, 2022, 12:31: AM
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