Find resultant magnetic field at 'C' in the figure shown.

Explanation

It is clear that ‘B’ at ‘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
`therefore B_(res)=2(B_(PQ)+B_(SP))`
`B_(PQ)=(mu_(0)i)/(4pi(a)/(2))(sin60^(@)+sin60^(@))`
`B_(SP)=(mu_(0)i)/(4pi(sqrt(3)a)/(2))(sin30^(@)+sin30^(@))`
`B_(res)=2((sqrt(3)mu_(0)i)/(2pia)+(mu_(0)i)/(2piasqrt(3)))=(4mu_(0)i)/(sqrt(3)pia)`