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MHT CET 2023 (13 May, Shift 2) · Biot-Savart Law & Circular Coils

Biot-Savart Law & Circular Coils question from MHT CET 2023 (13 May, Shift 2)

Q59MHT CET 2023 · 13 MAY · SHIFT 2Medium

Biot-Savart Law & Circular Coils

A circular current carrying coil has radius RR. The magnetic induction at the centre of the coil is BCB_C. The magnetic induction of the coil at a distance 3R\sqrt{3}R from the centre along the axis is BAB_A. The ratio BA:BCB_A : B_C is

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Answer: (b)

  1. BABC=R3(R2+x2)3/2\dfrac{B_A}{B_C} = \dfrac{R^3}{(R^2+x^2)^{3/2}} with x=3Rx = \sqrt{3}R.
  2. R2+x2=4R2⇒(4R2)3/2=8R3R^2+x^2 = 4R^2 \Rightarrow (4R^2)^{3/2} = 8R^3
  3. BABC=18\dfrac{B_A}{B_C} = \dfrac{1}{8}
  4. So the answer is (b) 1:81 : 8.
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