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MHT CET 2021 (22 Sep, Shift 2) · Biot-Savart Law & Circular Coils

Biot-Savart Law & Circular Coils question from MHT CET 2021 (22 Sep, Shift 2)

Q78MHT CET 2021 · 22 SEP · SHIFT 2Medium

Biot-Savart Law & Circular Coils

A particle having a charge 100e100e is revolving in a circular path of radius 0.8 m0.8\text{ m} with 1. r.p.s The magnetic field produced at the centre of the circle in SI unit is (μ0=(\mu_0 = permeability of vacuum, e=1.6×10−19 C)e = 1.6 \times 10^{-19}\text{ C})

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Answer: (a)

  1. I=qf=100×1.6×10−19×1=1.6×10−17 AI = qf = 100 \times 1.6 \times 10^{-19} \times 1 = 1.6 \times 10^{-17}\text{ A}.
  2. B=μ0I2r=μ0×1.6×10−171.6=10−17μ0B = \frac{\mu_0 I}{2r} = \frac{\mu_0 \times 1.6 \times 10^{-17}}{1.6} = 10^{-17}\mu_0
  3. So the answer is (a) 10−17μ010^{-17}\mu_0.
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