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JEE Main 2026 · 6 Apr · Shift 2, Physics question 38

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.

(μo4π=10−7 N/A2)\left(\dfrac{\mu_o}{4\pi}=10^{-7}\text{ N/A}^2\right)

Choose one option
Answer and worked solution

Official answer: B

Opposite currents. At the midpoint the two fields point the same way and add:
B=2⋅μ0I2πd=2⋅2×10−7×300.04=3×10−4 T=300 μT.B=2\cdot\frac{\mu_0I}{2\pi d}=2\cdot\frac{2\times10^{-7}\times30}{0.04}=3\times10^{-4}\text{ T}=300\ \mu\text{T}.

(B)

Topics:Biot-Savart law
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