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MHT CET 2023 (10 May, Shift 2) · Biot-Savart Law & Circular Coils

Biot-Savart Law & Circular Coils question from MHT CET 2023 (10 May, Shift 2)

Q46MHT CET 2023 · 10 MAY · SHIFT 2Easy

Biot-Savart Law & Circular Coils

The magnetic field at the centre of a circular coil of radius 'RR', carrying current 2A2A is 'B1B_1 '. The magnetic field at the centre of another coil of radius '3R3R' carrying current 4A4A is 'B2B_2'. The ratio B1:B2B_1 : B_2 is

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Answer: (d)

  1. B=μ0I2RB = \frac{\mu_0 I}{2R}, so B∝IRB \propto \frac{I}{R}.
  2. B1B2=2/R4/(3R)=64=32\frac{B_1}{B_2} = \frac{2/R}{4/(3R)} = \frac{6}{4} = \frac{3}{2}
  3. So the answer is (d) 3:23:2.
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