B due to square part :- B due to side OA and OC will be zero at point O B due to side AB and BC will be equal so B1​​=2[4πbμ0​i​(sin45∘+0)]=22​πbμ0​i​⊗ B due to circular part B2​​=2aμ0​i​[2π(23π​)​]=8a3μ0​i​⊗ Bnet ​​=B1​​+B2​​=μ0​[8a3​+22​πb1​] =4πμ0​i​[2a3π​+b2​​]