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JEE Main 2022 (28 Jul, Shift 2) · Biot-Savart Law & Circular Coils

Biot-Savart Law & Circular Coils question from JEE Main 2022 (28 Jul, Shift 2)

Q72JEE Main 2022 · 28 JUL · SHIFT 2Medium

Biot-Savart Law & Circular Coils

The magnetic field at the center of current carrying circular loop is B1B_1. The magnetic field at a distance of 3\sqrt{3} times radius of the given circular loop from the center on its axis is B2B_2. The value of B1/B2B_1/B_2 will be

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Answer: (c)

  1. B1B2=(R2+x2)3/2R3\dfrac{B_1}{B_2} = \dfrac{(R^2+x^2)^{3/2}}{R^3} with x=3Rx = \sqrt{3}R.
  2. (R2+3R2)3/2=(4R2)3/2=8R3(R^2+3R^2)^{3/2} = (4R^2)^{3/2} = 8R^3
  3. So the answer is (c) 8:18 : 1.
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