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MHT CET 2023 (13 May, Shift 1) · Biot-Savart Law & Circular Coils

Biot-Savart Law & Circular Coils question from MHT CET 2023 (13 May, Shift 1)

Q60MHT CET 2023 · 13 MAY · SHIFT 1Medium

Biot-Savart Law & Circular Coils

A circular coil of radius 'rr' and number of turns ' nn ' carries a current 'II'. The magnetic fields at a small distance 'hh' along the axis of the coil (Ba)(B_a) and at the centre of the coil (Bc)(B_c) are measured. The relation between BcB_c and BaB_a is

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Answer: (c)

  1. Bc=μ0nI2rB_c = \dfrac{\mu_0 nI}{2r}, Ba=μ0nIr22(r2+h2)3/2B_a = \dfrac{\mu_0 nI r^2}{2(r^2+h^2)^{3/2}}
  2. BcBa=(r2+h2)3/2r3=(1+h2r2)3/2\dfrac{B_c}{B_a} = \dfrac{(r^2+h^2)^{3/2}}{r^3} = \left(1+\dfrac{h^2}{r^2}\right)^{3/2}
  3. So the answer is (c) Bc=Ba(1+h2r2)3/2B_c = B_a\left(1+\frac{h^2}{r^2}\right)^{3/2}.
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