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JEE Main 2005 (AIEEE) · Biot-Savart Law & Circular Coils

Biot-Savart Law & Circular Coils question from JEE Main 2005 (AIEEE)

Q105JEE Main 2005 · AIEEEEasy

Biot-Savart Law & Circular Coils

Two concentric coils each of radius equal to 2π cm2\pi\text{ cm} are placed at right angles to each other. 33 ampere and 44 ampere are the currents flowing in each coil respectively. The magnetic induction in Weber / m2\text{m}^2 at the center of the coils will be

(μ0=4π×10−7 Wb/A m)\left(\mu_0 = 4\pi \times 10^{-7}\text{ Wb/A m}\right)

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Answer: (d)

  1. B=μ0I2RB = \frac{\mu_0 I}{2R} with R=2π cm=0.02π mR = 2\pi\text{ cm} = 0.02\pi\text{ m}.
  2. B1=4π×10−7×32×0.02π=3×10−5 TB_1 = \frac{4\pi \times 10^{-7} \times 3}{2 \times 0.02\pi} = 3 \times 10^{-5}\text{ T}, B2=4π×10−7×40.04π=4×10−5 TB_2 = \frac{4\pi \times 10^{-7} \times 4}{0.04\pi} = 4 \times 10^{-5}\text{ T}.
  3. Fields are perpendicular: B=32+42×10−5=5×10−5 TB = \sqrt{3^2 + 4^2} \times 10^{-5} = 5 \times 10^{-5}\text{ T}.
  4. So the answer is (d) 5×10−55 \times 10^{-5}.
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