Question bank/NEET UG/Physics

NEET UGPhysicsModerateRe-NEET 2026

Two infinitely long parallel conducting wires AA and BB carry currents II and 2I2I, respectively, in the same direction. The wire AA has uniform mass per unit length λ\lambda and lies on an insulated floor. The wire BB is kept fixed at a height hh above the floor. The minimum magnitude of hh so that the wire AA does not rise from the floor is:
[
gg is the acceleration due to gravity and μ0\mu_0 is the permeability of free space.]

Show answer and solution
Sol.
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Force per unit length on wire
AA by wire BB
F=μ04π2I⋅2Ih=μ04π4I2hF = \frac{\mu_0}{4\pi} \frac{2I \cdot 2I}{h} = \frac{\mu_0}{4\pi} \frac{4I^2}{h}

Source: geekyrakshit/indian-exam-questions / reneet2026-code50

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