An equilateral triangle of side length l is formed from a piece of wire of uniform resistance. The current I is fed as shown in the figure. The magnitude of the magnetic field at its centre O is

Explanation

The magnetic field induction at O due to current through PR is
`B_(1) = (mu_0)/(4pi) (2I//3)/(r) [ sin 60^@ + sin 60^@]`
`= (mu_0)/(4pi)(2I)/(3r)odot` (directed outward)
The megantic field induction at O due to current through PQR is
`B_(2) = 2xx(mu_0(I//3))/(4pir)[sin 60^@ + sin 60^2]`
`= (mu_(0)2I)/(2pi3r)ox` (directed inward)
`therefore` Resultant magnetic induction at O `rArr B_1 - B_2 = 0`