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MHT CET 2025 (22 Apr, Shift 1) · Biot-Savart Law & Circular Coils

Biot-Savart Law & Circular Coils question from MHT CET 2025 (22 Apr, Shift 1)

Q19MHT CET 2025 · 22 APR · SHIFT 1Easy

Biot-Savart Law & Circular Coils

A circular coil of wire consisting of ' nn ' tums each of radius 8 cm carries a current of 0.4 A. The magnitude of the magnetic field at the centre of coil is 3.14×10−4 T3.14 \times 10^{-4}\text{ T}. The value of ' nn ' is [Take μ0=12.56×10−7\mu_0 = 12.56 \times 10^{-7} SI unit]

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Answer: (c)

  1. Field at centre: B=μ0nI2rB = \frac{\mu_0 n I}{2r}, so n=2rBμ0In = \frac{2rB}{\mu_0 I}.
  2. n=2×0.08×3.14×10−412.56×10−7×0.4=5.024×10−55.024×10−7n = \frac{2 \times 0.08 \times 3.14\times10^{-4}}{12.56\times10^{-7}\times 0.4} = \frac{5.024\times10^{-5}}{5.024\times10^{-7}}
  3. So the answer is (c) n=100n = 100.
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