Solution

In 9.2.10 we obtained an expression for the magnetic induction at the center of the ring $$B_0 = \frac{\mu_0 I}{2R}\quad(1)$$ Due to superposition, the magnetic induction at the center of one semicircle will be half of this value: $$B_1 = B_2 = \frac{B_0}{2} = \frac{\mu_0 I}{4R}\quad(2)$$ From the drawing, it is clear that the angle between the vectors $\vec{B}_1$ and $\vec{B}_2$ is equal to $\alpha$ $$(\vec{B_1},\vec{B_2})=\alpha$$ The total vector of magnetic induction is equal to the vector sum of the components of magnetic induction of two rings $\vec{B}_1$ and $\vec{B}_2$ $$\vec{B} = \vec{B_1} + \vec{B_2}$$ We will find the modulus of this vector using the cosine law. $$B = \sqrt{{B_1}^2 + {B_2}^2 - 2B_1 B_2 \cos\alpha}$$ Taking into account the expression $(2)$, we find the total magnetic induction as $$B= \frac{\mu_0 I}{4R}\sqrt{2-2\cos\alpha} = \frac{\mu_0 I}{2R}\cdot\sin\frac{\alpha}{2}\quad(3)$$ Dividing the expression $(3)$ by $(1)$, we obtain the desired ratio $$\boxed{\frac{B}{B_0} = \sin\frac{\alpha}{2}}$$