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[sin60^(@)+sin60^(@)]`
`=(mu_(0))/(4pi)(2I)/(3r)odot` ( directed outside)
The magnetic fiel `B_(2)=2xx(mu_(0)(I//3))/(4pir)[sin60^(@)+sin60^(@)]`
`=(mu_(0)2I)/(4pi3r)` ( directed inside)
`therefore` Resultant magnetic induction at O
`rArrB_(1)-B_(2)=0`