Find resultant magnetic field at C in Fig

Explanation

It is clear that `B` and `C` due all the wires is directed `ox`.Also `B` at `C` due `PQ` and `SR` is same.Also due to `QR` and `PS` is same
`B_(res)=2(B_(PO)+B_(SP))`
`B_(PQ)=(mu_(0)l)/(4pia/2)(sin 60^(@)+sin 60^(@))`.
`B_(sp)=(mu_(0)l)/(4pi(sqrt3a)/2)(sin 30^(@)+sin 30^(@)) rArr B_(res)=2((sqrt3mu_(0)i)/(2pia)+(sqrtmu_(0)i)/(2piasqrt3))=(4mu_(0)i)/(sqrt3pia)`