A current of 2A is flowing in an equilateral triangle of side a = 1m. Find the magnetic flux density at the centroid of the triangle.

Explanation

In the given triangle ABC, O is centroid,

Length, `OD=(1)/(3) xx (AD) =(1)/(3) sin 60^(@) =(1)/(2sqrt(3))`
Magnetic flux density due to branch CB at O,
`B_(CD)= (mu_(0))/(4pi) (i)/((OD)) (sin 60^(@) +sin 60^(@))`
`=10^(-7) xx (2)/(((1)/(2sqrt(3)))) xx ((2sqrt(3))/(2))`
`=12 xx 10^(-7)` in upward direction
Similarly, due to AC and AB, the magnetic flux density `B_(AC)=12 xx 10^(-7), B_(AB)=12 xx 10^(-7)` both in upward direction at O.
So, net magnetic field at O
`=B_(AB) +B_(AC)+B_(CB)=36 xx 10^(-7)T`
Its direction is perpendicular to plane of paper.