Find resultant magnetic field at C in Fig

Explanation

It is clear that 'B' at 'C' due to all the wires is directed `ox`. Also
B at 'C' due PQ and SR is same.
Also the magnetic field due to QR and PS is same
`:. B_(res)=2(B_(PQ)+B_(SP))`
`B_(PQ)=(mu_0i)/(4pia/2)(sin60^@+sin60^@)=(sqrt3mu_0i)/(2pia)`
`B_(SP)=(mu_oi)/(4pi(sqrt3a)/2)(sin30^@+sin30^@)=(mu_0i)/(2pisqrt3a)`
`B_(res)=2((sqrt3mu_0i)/(2pia)+(mu_0i)/(2pia))=(4mu_0i)/(sqrt3pia)`