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MHT CET 2026 (14 May, Shift 1) · Biot-Savart Law & Circular Coils

Biot-Savart Law & Circular Coils question from MHT CET 2026 (14 May, Shift 1)

Q2MHT CET 2026 · 14 MAY · SHIFT 1Easy

Biot-Savart Law & Circular Coils

The magnitude of magnetic induction at a point on the axis at a large distance rr from a centre of a circular coil of nn turns and area AA carrying current ii is (μ0\mu_0 = permeability of free space)

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Answer: (b)

  1. Axial field: B=μ0niR22(R2+r2)3/2B = \frac{\mu_0 n i R^2}{2(R^2+r^2)^{3/2}}; for r≫Rr \gg R, B≈μ0niR22r3B \approx \frac{\mu_0 n i R^2}{2r^3}.
  2. With A=πR2A = \pi R^2: B=μ0niA2πr3=2μ0niA4πr3B = \frac{\mu_0 n i A}{2\pi r^3} = \frac{2\mu_0 n i A}{4\pi r^3}
  3. So the answer is (b).
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