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MHT CET 2023 (10 May, Shift 1) · Magnetic Field of Straight Wires

Magnetic Field of Straight Wires question from MHT CET 2023 (10 May, Shift 1)

Q39MHT CET 2023 · 10 MAY · SHIFT 1Medium

Magnetic Field of Straight Wires

Two long parallel wires carrying currents 8 A8\text{ A} and 15 A15\text{ A} in opposite directions are placed at a distance of 7 cm7\text{ cm} from each other. A point 'P\text{P}' is at equidistant from both the wires such that the lines joining the point to the wires are perpendicular to each other. The magnitude of magnetic field at point 'P\text{P}' is (2=1.4)(μ0=4π×10−7(\sqrt{2} = 1.4)(\mu_0 = 4\pi \times 10^{-7} SI units)

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Answer: (a)

  1. r=72=5 cmr = \frac{7}{\sqrt{2}} = 5\text{ cm} (with 2=1.4\sqrt{2} = 1.4).
  2. B1=2×10−7×80.05=32 μTB_1 = \frac{2 \times 10^{-7} \times 8}{0.05} = 32\ \mu\text{T}, B2=2×10−7×150.05=60 μTB_2 = \frac{2 \times 10^{-7} \times 15}{0.05} = 60\ \mu\text{T}
  3. Fields are perpendicular: B=322+602=68 μTB = \sqrt{32^2 + 60^2} = 68\ \mu\text{T}
  4. So the answer is (a) 68×10−6 T68 \times 10^{-6}\text{ T}.
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