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MHT CET 2024 (4 May, Shift 2) · Biot-Savart Law & Circular Coils

Biot-Savart Law & Circular Coils question from MHT CET 2024 (4 May, Shift 2)

Q33MHT CET 2024 · 4 MAY · SHIFT 2Easy

Biot-Savart Law & Circular Coils

A particle carrying a charge equal to 100 times the charge on an electron is rotating one rotation per second in a circular path of radius 0.8 m. The value of magnetic field produced at the centre will be ( μ0=\mu_0 = permeability of vacuum)

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Answer: (b)

  1. Equivalent current I=qf=100×1.6×10−19×1=1.6×10−17 AI = qf = 100\times1.6\times10^{-19}\times 1 = 1.6\times10^{-17}\text{ A}.
  2. B=μ0I2r=μ0×1.6×10−172×0.8B = \frac{\mu_0 I}{2r} = \frac{\mu_0 \times 1.6\times10^{-17}}{2\times 0.8}
  3. So the answer is (b) 10−17μ010^{-17}\mu_0.
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