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A circular loop of radius R carries current I. The magnetic field at a point on its axis at distance x from the centre is
Aμ₀IR² / 2(R²+x²)^(3/2)
Bμ₀I(R²+x²)^(1/2) / 2R²
Cμ₀IR / 2(R²+x²)
Dμ₀I / 2R
Answer & Solution
Correct answer: A. μ₀IR² / 2(R²+x²)^(3/2)
1. Apply Biot-Savart to each current element on the loop; vertical components cancel by symmetry.
2. Each element contributes dB·sin α along the axis, where sin α = R/(R²+x²)^(1/2).
3. Integrating around the loop gives B = μ₀IR² / (2(R²+x²)^(3/2)).
4. At x = 0 the formula reduces to μ₀I/(2R), as expected for the centre.
_Source: NCERT Class 12 Physics Ch 4 "Moving Charges and Magnetism", §4.6_
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