A current of 1 A is flowing along the side of an equilateral triangle of side `4.5 xx 10^(-2)` m. the magnetic field at the centroid of the triangle is `(mu_(0) = 4 pi xx 10^(-7) H//m)`

Options

A. `4 xx 10^(-6) T`
B. `2 xx 10^(-8) T`
C. `4 xx 10^(-4) T`
D. `2 xx 10^(-4) T`

Answer

b

Explanation

Let the centroid of the traingle is O, magnetic fiedl due to side AB

`B_(1) = (mu_(0))/(4 pi) - (i)/(r ) (sin phi_(1) + sin phi_(2))`
Where, r is perpendicular distance of AB from centroid O. The magnetic field due to each side is same in magnitude and direction, so the resultant field
`B = 3B_(1)`
`= 3 (mu_(0))/(4 pi) , (i)/(r ) (sin phi_(1) + sin phi_(2))`
(Here, `mu_(0) = 4 pi xx 10^(-7) H//m`)
i = 1 A
In equilateral triangle, `phi_(1) = phi_(2) = 60^(@)`
Also, `r = 4.5 xx 10^(-2) m`
`implies B = (3 mu_(0))/(4 pi) , (1)/(4.5 xx 10^(-2)) ((sqrt(3))/(2) + (sqrt(3))/(2)) T`,
`= (3 sqrt(3). mu_(0))/(4 pi) . (1)/(4.5 xx 10^(-2)) T`
`= (3 sqrt(3) xx 10^(-7))/(4.5 xx 10^(-2)) T` `[:' (mu_(0))/(4 pi) = 10^(-7) N//A^(2)]`
`= (3 xx 1.73)/(4.5) xx 10^(-6) T`
`= 1.15 xx 10^(-8) T = 2 xx 10^(-5) T`