An infinite wire has a circular bend of radius a, and carrying a current I as shown in figure. The magnitude of magnetic field at the origin O of the arc is given by :

$$\vec{B}_{\text{net}} = \vec{B}_{\text{top wire}} + \vec{B}_{\text{arc}} + \vec{B}_{\text{bottom wire}}$$

For top semi-infinite wire aligned with the center line: $$B_{\text{top wire}} = 0$$

For the circular arc subtending angle $$\theta = \frac{3\pi}{2}$$ into the page ($$\otimes$$):

$$B_{\text{arc}} = \frac{\mu_0 I}{4\pi a} \theta \implies B_{\text{arc}} = \frac{\mu_0 I}{4\pi a} \left(\frac{3\pi}{2}\right)$$

For bottom semi-infinite wire at perpendicular distance $$a$$ into the page ($$\otimes$$):

$$B_{\text{bottom wire}} = \frac{\mu_0 I}{4\pi a} (\sin 90^\circ + \sin 0^\circ) = \frac{\mu_0 I}{4\pi a} (1)$$

Total field magnitude: $$B_{\text{net}} = \frac{\mu_0 I}{4\pi a} \left(\frac{3\pi}{2}\right) + \frac{\mu_0 I}{4\pi a} (1) \implies B_{\text{net}} = \frac{\mu_0 I}{4\pi a} \left[\frac{3\pi}{2} + 1\right]$$