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MHT CET 2019 (2 May, Shift 2) · Biot-Savart Law & Circular Coils

Biot-Savart Law & Circular Coils question from MHT CET 2019 (2 May, Shift 2)

Q86MHT CET 2019 · 2 MAY · SHIFT 2Easy

Biot-Savart Law & Circular Coils

A circular coil of wire consisting of 100 turns each of radius 9 cm carries a current of 0.4 A. The magnitude of the magnetic field at the centre of coil is [μ0=12.56×10−7 SI[\mu_0 = 12.56 \times 10^{-7}\text{ SI} Unit]

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Answer: (c)

  1. B=μ0nI2rB = \frac{\mu_0 n I}{2r}
  2. B=12.56×10−7×100×0.42×0.09=5.024×10−50.18B = \frac{12.56\times10^{-7}\times 100 \times 0.4}{2\times 0.09} = \frac{5.024\times10^{-5}}{0.18}
  3. So the answer is (c) 2.79×10−4 T2.79\times10^{-4}\text{ T}.
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