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MHT CET 2025 (19 Apr, Shift 1) · Magnetic Field of Straight Wires

Magnetic Field of Straight Wires question from MHT CET 2025 (19 Apr, Shift 1)

Q18MHT CET 2025 · 19 APR · SHIFT 1Medium

Magnetic Field of Straight Wires

Two long parallel wires carrying currents 4 A and 3 A in opposite directions are placed at a distance of 5 cm from each other. A point P is at equidistance from both the wires such that the line joining the point P to the wires are perpendicular to each other. The magnitude of magnetic field at point PP is ( μ0=\mu_0 = permeability of free space =4π×10−7= 4\pi \times 10^{-7} SI unit)

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Answer: (d)

  1. Equidistant with a right angle at P: r=52 cm=0.052 mr = \frac{5}{\sqrt{2}}\text{ cm} = \frac{0.05}{\sqrt{2}}\text{ m}.
  2. B1=2×10−7×420.05=1.62×10−5 TB_1 = \frac{2 \times 10^{-7} \times 4\sqrt{2}}{0.05} = 1.6\sqrt{2} \times 10^{-5}\text{ T}, B2=1.22×10−5 TB_2 = 1.2\sqrt{2} \times 10^{-5}\text{ T}
  3. The two fields are perpendicular: B=2×10−51.62+1.22=22×10−5 TB = \sqrt{2} \times 10^{-5}\sqrt{1.6^2 + 1.2^2} = 2\sqrt{2} \times 10^{-5}\text{ T}
  4. So the answer is (d) 22×10−5 T2\sqrt{2} \times 10^{-5}\text{ T}.
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