B_axis = μ₀IR²/(2(R²+x²)^(3/2)). At center B₀ = μ₀I/(2R). For B/8: (R²+x²)^(3/2) = 8R³. So R²+x² = 4R², x = √3R
JEE Main 2023 — Physics Electromagnetism
A current carrying circular loop of radius R produces a magnetic field B at its center. At what distance from center on its axis, the magnetic field becomes B/8?
Verified 30 May 2026.
R
√3R
2R
R/√3
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