a b c d

b

Magnetic field due to circular arc AD $B_1 = \frac{\mu_0 I}{2\pi r} \times \frac{\pi/6}{2\pi} = \frac{\mu_0 I}{24\pi a} outward$ Magnetic field due to circular arc BC $B_2 = \frac{\mu_0 I}{2\pi b} \times \frac{\pi/6}{2\pi} = \frac{\mu_0 I}{2\pi r} inward$ Net magnetic field $\Rightarrow B_1 - B_2 = \frac{\mu_0I(a -b)}{24ab}$