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MHT CET 2026 (11 Apr, Shift 1) · Magnetic Field of Straight Wires

Magnetic Field of Straight Wires question from MHT CET 2026 (11 Apr, Shift 1)

Q9MHT CET 2026 · 11 APR · SHIFT 1Medium

Magnetic Field of Straight Wires

Two long parallel wires carrying currents 88A and 1515A in opposite directions are placed at a distance of 77 cm from each other. A point P is at equidistant from both the wires such that the lines joining the point P to the wires are perpendicular to each other. The magnetic field at P is ___ ×10−6\times 10^{-6} T. (Given: 2=1.4, μ0=4π×10−7\sqrt{2} = 1.4,\ \mu_0 = 4\pi \times 10^{-7} SI unit)

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Answer: (c)

8 A15 APB₁B₂7 cm5 cm5 cm
  1. P is equidistant with ∠APB=90∘\angle APB = 90^\circ, so r=72=5 cmr = \frac{7}{\sqrt{2}} = 5\text{ cm} (using 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. Each field is perpendicular to its own line to P, so B1⊥B2B_1 \perp B_2: B=322+602=68 μTB = \sqrt{32^2 + 60^2} = 68\ \mu\text{T}
  4. So the answer is (c) 68.
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