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MHT CET 2023 (14 May, Shift 2) · Biot-Savart Law & Circular Coils

Biot-Savart Law & Circular Coils question from MHT CET 2023 (14 May, Shift 2)

Q62MHT CET 2023 · 14 MAY · SHIFT 2Medium

Biot-Savart Law & Circular Coils

Figure shows two semicircular loops of radii R1R_1 and R2R_2 carrying current II. The magnetic field at the common centre 'OO' is

Figure

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Answer: (a)

  1. Straight parts lie along a line through O, so they give no field at O.
  2. Going round the single loop, the upper arc (R1R_1) and lower arc (R2R_2) are both traversed clockwise, so both fields at O are into the page.
  3. B=μ0I4R1+μ0I4R2=μ0I4(1R1+1R2)B = \dfrac{\mu_0 I}{4R_1} + \dfrac{\mu_0 I}{4R_2} = \dfrac{\mu_0 I}{4}\left(\dfrac{1}{R_1}+\dfrac{1}{R_2}\right)
  4. So the answer is (a).
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