A current of `1*0A` flowing in the sides of an equilateral triangle of side `4*5xx10^-2m`. Find the magnetic fied at the centroid of the triangle.

Answer

`(4)`

Explanation

The magnetic field induction at the centroid O due to current I through one side BC of the triangle will be
`B_1=(mu_0)/(4pi)I/r(sin theta_1+sin theta_2)`
It will be acting perpendicular to the plane of triangle upwards.
Total magnetic field induction at O due to current through all the three sides of the triangle will be
`B=3B_1=(3mu_0)/(4pi)I/r[sin theta_1+sin theta_2]`

Here, `I=1A`, `theta_1=60^@=theta_2`
and `r=OD=(BD)/(tan 60^@)=(a//2)/(sqrt3)`
`=(a)/(2sqrt3)=(4*5xx10^-2)/(2sqrt3)m`
`:. B=3xx10^-7xx(1)/((4*5xx10^-2//2sqrt3))xx[sin 60^@+sin 60^@]`
On solving, `B=4xx10^-5T`