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JEE Main 2024 (31 Jan, Shift 2) · Biot-Savart Law & Circular Coils

Biot-Savart Law & Circular Coils question from JEE Main 2024 (31 Jan, Shift 2)

Q45JEE Main 2024 · 31 JAN · SHIFT 2Medium

Biot-Savart Law & Circular Coils

Two circular coils PP and QQ of 100 turns each have same radius of π cm\pi\text{ cm}. The currents in PP and QQ are 1 A1\text{ A} and 2 A2\text{ A} respectively. PP and QQ are placed with their planes mutually perpendicular with their centers coincide. The resultant magnetic field induction at the center of the coils is x mT\sqrt{x}\text{ mT}, where x=x = __________.

[Use μ0=4π×10−7 T m A−1\mu_0 = 4\pi \times 10^{-7}\text{ T m A}^{-1} ]

Numerical answer type. Work it out, then open the answer.

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Answer: 20

  1. BP=μ0NI2R=4π×10−7×100×12×π×10−2=2×10−3B_P = \dfrac{\mu_0 NI}{2R} = \dfrac{4\pi\times10^{-7}\times100\times1}{2\times\pi\times10^{-2}} = 2\times10^{-3} T =2= 2 mT
  2. BQ=2×2=4B_Q = 2 \times 2 = 4 mT (double the current)
  3. Perpendicular fields: B=22+42=20B = \sqrt{2^2+4^2} = \sqrt{20} mT
  4. So x=20x = 20.
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