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JEE Main 2026 (6 Apr, Shift 2) · Magnetic Field of Straight Wires

Magnetic Field of Straight Wires question from JEE Main 2026 (6 Apr, Shift 2)

Q7JEE Main 2026 · 6 APR · SHIFT 2Easy

Magnetic Field of Straight Wires

A current of 30 A each flows in opposite directions in two conducting wires, placed parallel to each other at a distance of 8 cm. The magnetic field at the mid point between the two wires is ‾ μT⋅(μ04π=10−7 N/A2)\underline{\hspace{3em}}\ \mu\text{T} \cdot \left(\frac{\mu_0}{4\pi} = 10^{-7}\text{ N/A}^2\right)

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Answer: (b)

  1. Opposite currents: fields at the midpoint add; r=4 cm=0.04 mr = 4\text{ cm} = 0.04\text{ m}.
  2. Each: B=μ04π2Ir=10−7×2×300.04=1.5×10−4 TB = \frac{\mu_0}{4\pi}\frac{2I}{r} = 10^{-7} \times \frac{2 \times 30}{0.04} = 1.5 \times 10^{-4}\text{ T}
  3. Total =3×10−4 T=300 μT= 3 \times 10^{-4}\text{ T} = 300\ \mu\text{T}
  4. So the answer is (b) 300.
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