The magnetic field at the centre of a wire loop formed by two semicircular wires of radii $$R_1 = 2\pi$$ m and $$R_2 = 4\pi$$ m carrying current $$I = 4$$ A as per figure given below is $$\alpha \times 10^{-7}$$ T. The value of $$\alpha$$ is _____.
(Centre O is common for all segments)

$$\text{Magnetic field contribution from straight wire segments:}$$ $$B_{\text{straight}} = 0$$

$$\text{Total magnetic field at the centre O:}$$ $$B = B_1 + B_2 = \frac{\mu_0 I}{4R_1} + \frac{\mu_0 I}{4R_2}$$

$$B = \frac{\mu_0 I}{4} \left( \frac{1}{R_1} + \frac{1}{R_2} \right)$$

$$B = \frac{(4\pi \times 10^{-7})(4)}{4} \left( \frac{1}{2\pi} + \frac{1}{4\pi} \right)$$

$$B = (4\pi \times 10^{-7}) \left( \frac{3}{4\pi} \right) = 3 \times 10^{-7}\ \text{T}$$

$$\alpha \times 10^{-7} = 3 \times 10^{-7} \implies \alpha = 3$$