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JEE Main 2021 (25 Feb, Shift 1) · Biot-Savart Law & Circular Coils

Biot-Savart Law & Circular Coils question from JEE Main 2021 (25 Feb, Shift 1)

Q83JEE Main 2021 · 25 FEB · SHIFT 1Difficult

Biot-Savart Law & Circular Coils

Magnetic fields at two points on the axis of a circular coil at a distance of 0.05 m and 0.2 m from the centre are in the ratio 8: 1. The radius of coil is ________.

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Answer: (d)

  1. B1B2=(R2+x22R2+x12)3/2=8⇒R2+0.04R2+0.0025=4\dfrac{B_1}{B_2} = \left(\dfrac{R^2+x_2^2}{R^2+x_1^2}\right)^{3/2} = 8 \Rightarrow \dfrac{R^2+0.04}{R^2+0.0025} = 4
  2. R2+0.04=4R2+0.01⇒3R2=0.03R^2 + 0.04 = 4R^2 + 0.01 \Rightarrow 3R^2 = 0.03
  3. R2=0.01⇒R=0.1R^2 = 0.01 \Rightarrow R = 0.1 m
  4. So the answer is (d) 0.1 m.
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