JEE MAIN - Physics (2024 - 27th January Evening Shift - No. 27)
The magnetic field at the centre of a wire loop formed by two semicircular wires of radii $$R_1=2 \pi \mathrm{m}$$ and $$R_2=4 \pi \mathrm{m}$$, carrying current $$\mathrm{I}=4 \mathrm{~A}$$ as per figure given below is $$\alpha \times 10^{-7} \mathrm{~T}$$. The value of $$\alpha$$ is ________. (Centre $$\mathrm{O}$$ is common for all segments)
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Explanation
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$$\begin{aligned} & \frac{\mu_0 \mathrm{i}}{2 \mathrm{R}_2}\left(\frac{\pi}{2 \pi}\right) \otimes+\frac{\mu_0 \mathrm{i}}{2 \mathrm{R}_1}\left(\frac{\pi}{2 \pi}\right) \otimes \\ & \left(\frac{\mu_0 \mathrm{i}}{4 \mathrm{R}_2}+\frac{\mu_0 \mathrm{i}}{4 \mathrm{R}_1}\right) \otimes \\ & \frac{4 \pi \times 10^{-7} \times 4}{4 \times 4 \pi}+\frac{4 \pi \times 10^{-7} \times 4}{4 \times 2 \pi} \\ & =3 \times 10^{-7}=\alpha \times 10^{-7} \\ & \alpha=3 \end{aligned}$$
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