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MHT CET 2025 (5 May, Shift 2) · Biot-Savart Law & Circular Coils

Biot-Savart Law & Circular Coils question from MHT CET 2025 (5 May, Shift 2)

Q26MHT CET 2025 · 5 MAY · SHIFT 2Medium

Biot-Savart Law & Circular Coils

A circular coil carrying current has radius ' RR '. The distance from the centre of the coil on the axis where the magnetic induction will be 127\frac{1}{27} th of its value at the centre of the coil is

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Answer: (c)

  1. BaxisBc=R3(R2+x2)3/2=127\dfrac{B_{axis}}{B_c} = \dfrac{R^3}{(R^2+x^2)^{3/2}} = \dfrac{1}{27}
  2. (R2+x2)3/2=27R3⇒R2+x2=9R2(R^2+x^2)^{3/2} = 27R^3 \Rightarrow R^2+x^2 = 9R^2
  3. x2=8R2⇒x=22Rx^2 = 8R^2 \Rightarrow x = 2\sqrt{2}R
  4. So the answer is (c) 22R2\sqrt{2}R.
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