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MHT CET 2025 (19 Apr, Shift 2) · Biot-Savart Law & Circular Coils

Biot-Savart Law & Circular Coils question from MHT CET 2025 (19 Apr, Shift 2)

Q25MHT CET 2025 · 19 APR · SHIFT 2Medium

Biot-Savart Law & Circular Coils

A particle carrying a charge equal to 1000 times the charge on an electron, is rotating one rotation per second in a circular path of radius ' rr ' mm. If the magnetic field produced at the centre of the path is xx times the permeability of vacuum, the radius ' r ' in m is [e=1.6×10−19 C] [x=2×10−16]\left[e = 1.6 \times 10^{-19}\text{ C}\right]\ \left[x = 2 \times 10^{-16}\right]

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Answer: (d)

  1. I=qf=1000×1.6×10−19×1=1.6×10−16 AI = qf = 1000 \times 1.6 \times 10^{-19} \times 1 = 1.6 \times 10^{-16}\text{ A}.
  2. B=μ0I2r=xμ0⇒r=I2x=1.6×10−164×10−16B = \frac{\mu_0 I}{2r} = x\mu_0 \Rightarrow r = \frac{I}{2x} = \frac{1.6 \times 10^{-16}}{4 \times 10^{-16}}
  3. So the answer is (d) 0.4 m0.4\text{ m}.
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