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A circular loop of radius R carries current I. The magnitude of the magnetic field at its centre is

Aμ₀I / (4πR)
Bμ₀I / R
Cμ₀I / (2R)
Dμ₀I / (4R)
Answer & Solution
Correct answer: C. μ₀I / (2R)
1. The Biot-Savart contribution from each element of the loop adds constructively at the centre. 2. Each element dl is perpendicular to the radius r̂, so |dl × r̂| = dl. 3. Integrating around the full loop: B = μ₀I/(4π) · ∫dl/R² = μ₀I/(4πR²) · 2πR. 4. Simplify to B = μ₀I/(2R). _Source: NCERT Class 12 Physics Ch 4 "Moving Charges and Magnetism", §4.5_
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