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. `4xx10^(-5)T`
B. `2xx10^(-5)T`
C. `4xx10^(-4)T`
D. `2xx10^(-4)T`

Answer

B

Explanation

`B_(1) =(mu_0)/(4pi).(i)/ ( r ) sin theta_1 + sin theta_2)`
`B= 3 B_1`
`= 3(mu_0)/(4 pi) . (i)/( r) (sin theta_1 + sin theta_2)`
`(I =1A)`
`mu_0 = 4pi xx 10^(-7) H// m)`
`theta_1=theta_2=60^@`
`B=(3 mu_0)/(4 pi ) .(1)/( r) (sin 60^@+sin 60^@)`
`(therefore r = 4.5 xx 10^(-2))`
`rArr B=(3mu_0)/(4pi) .(1)/(4.5xx10^(-2))((sqrt(3))/(2).(sqrt(3))/(2))T`
`rArr B =(3 sqrt(3.mu_0))/(4pi)xx(4pixx10^(-7))/(4.5xx10^(-2))`
`(3sqrt(3)xx10^(-7))/(4.5xx10^(-2))`
`(3xx1.732)/(4.5)xx10^(-5)`
`1.15xx10^(-5)Tapprox 2 xx 10^(-5) T`.