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JEE Main 2023 (24 Jan, Shift 1) · Biot-Savart Law & Circular Coils

Biot-Savart Law & Circular Coils question from JEE Main 2023 (24 Jan, Shift 1)

Q47JEE Main 2023 · 24 JAN · SHIFT 1Easy

Biot-Savart Law & Circular Coils

A circular loop of radius rr is carrying current I A. The ratio of magnetic field at the center of circular loop and at a distance r from the center of the loop on its axis is:

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Answer: (c)

  1. BcBx=(r2+x2)3/2r3\frac{B_c}{B_x} = \frac{(r^2+x^2)^{3/2}}{r^3}
  2. With x=rx = r: (2r2)3/2r3=22\frac{(2r^2)^{3/2}}{r^3} = 2\sqrt{2}
  3. So the answer is (c) 22:12\sqrt{2}:1.
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