Let B₁ be the magnetic field at the centre of a circular coil of radius R carrying current I, and B₂ the field at an axial distance x from the centre. For x:R = 3:4, the ratio B₂:B₁ is:
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Question

Let B₁ be the magnetic field at the centre of a circular coil of radius R carrying current I, and B₂ the field at an axial distance x from the centre. For x:R = 3:4, the ratio B₂:B₁ is:

G1 Physics · Accepted Answer

For a circular coil, B₁ = μ₀NI/(2R). At axial distance x, B₂ = μ₀NI R² / [2(R²+x²)^(3/2)]. Therefore, B₂/B₁ = R³/(R²+x²)^(3/2). Given x:R = 3:4, take x=3k and R=4k: B₂/B₁ = 4³/(4²+3²)^(3/2) = 64/(25)^(3/2) = 64/125. Hence B₂:B₁ = 64:125. Correct option: C.