A current I flows along a triangle loop having sides of equal length a. The strength of magnetic field at the centre of the loop is :

Options

A. `(3mu_(0)I)/(2pi a)`
B. `(9mu_(0)I)/(2 pi a)`
C. `(3sqrt(3)mu_(0)I)//(2pia)`
D. `(3mu_(0)I)/(4pi a)`

Answer

B

Explanation

The magnetic field at the cenetoid (or circumcentre) because of side AB of triangular loop is
`(mu_(0))/(4pi)=(i(sin 60^(@)+sin 60^(@)))/((a)/(2sqrt(3)))=(3mu_(0)I)/(2pia)`

hence the net magnetic field due to all 3 sides is
`3xx(9mu_(0)I)/(2pia)`