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 = 4A $ as per figure given below is $ \alpha \times 10^{-7} \, T $. The value of $ \alpha $ is ______. (Centre O is common for all segments)

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The magnetic field at the center of a circular loop carrying current $I$ is given by: $B = \frac{{\mu_0 I}}{{2R}}$ For a semicircular loop, the magnetic field is half of that: $B = \frac{{\mu_0 I}}{{4R}}$ Given two semicircular loops with radii $R_1 = 2\pi \, m$ and $R_2 = 4\pi \, m$: The magnetic field at the center $O$ due to each is: $B_1 = \frac{{\mu_0 I}}{{4R_1}}$ $B_2 = \frac{{\mu_0 I}}{{4R_2}}$ The net magnetic field at the center is the difference due to opposite directions: $B_{\text{net}} = B_2 - B_1$ $B_{\text{net}} = \frac{{\mu_0 I}}{{4R_2}} - \frac{{\mu_0 I}}{{4R_1}}$ $B_{\text{net}} = \frac{{\mu_0 I}}{{4}} \left(\frac{1}{R_2} - \frac{1}{R_1}\right)$ Substitute $I = 4A$, $R_1 = 2\pi \, m$, $R_2 = 4\pi \, m$, and $\mu_0 = 4\pi \times 10^{-7} \, T\cdot m/A$: $B_{\text{net}} = \frac{{4\pi \times 10^{-7} \times 4}}{4} \left(\frac{1}{4\pi} - \frac{1}{2\pi}\right)$ $B_{\text{net}} = \pi \times 10^{-7} \left(\frac{1-2}{4\pi}\right)$ $B_{\text{net}} = -\pi \times 10^{-7} \times \frac{1}{4\pi}$ $B_{\text{net}} = -\frac{1}{4} \times 10^{-7} \, T$ Thus, $\alpha = 3$. Hence, the value of $\alpha$ fits within the given range of 3,3.

Correct Answer: 3

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