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MHT CET 2024 (16 May, Shift 1) · Biot-Savart Law & Circular Coils

Biot-Savart Law & Circular Coils question from MHT CET 2024 (16 May, Shift 1)

Q37MHT CET 2024 · 16 MAY · SHIFT 1Easy

Biot-Savart Law & Circular Coils

Two coils P and Q each of radius R carry currents I and 8 I\sqrt{8}\, I respectively in same direction. Those coils are lying in perpendicular planes such that they have a common centre. The magnitude of the magnetic field at the common centre of the two coils is ( μ0=\mu_0 = permeability of free space)

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Answer: (b)

  1. BP=μ0I2RB_P = \frac{\mu_0 I}{2R}, BQ=8 μ0I2RB_Q = \frac{\sqrt{8}\,\mu_0 I}{2R}; the coils are in perpendicular planes, so the fields are perpendicular.
  2. B=BP2+BQ2=μ0I2R1+8=3μ0I2RB = \sqrt{B_P^2+B_Q^2} = \frac{\mu_0 I}{2R}\sqrt{1+8} = \frac{3\mu_0 I}{2R}
  3. So the answer is (b).
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