A circular coil of 100 turns and radius 10 cm carries a current of 5 A. The magnetic field at its centre is (\(\mu_0=4\pi\times10^{-7}\) T m/A)

A. \(2\pi\times10^{-3}\) T

B. \(\frac{\pi}{2}\times10^{-3}\) T

C. \(\pi\times10^{-4}\) T

D. \(\pi\times10^{-3}\) T

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Answer: D. \(\pi\times10^{-3}\) T

\(B=\frac{\mu_0NI}{2R}=\frac{4\pi\times10^{-7}\times100\times5}{0.2}=\pi\times10^{-3}\) T.

Physics Tutor AI · 23 Oct 2026 AI-verified AI-generated Report