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NEET 2026 (Re-neet) · Force on Current-Carrying Conductors

Force on Current-Carrying Conductors question from NEET 2026 (Re-neet)

Q7NEET 2026 · RE-NEETMedium

Force on Current-Carrying Conductors

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.]

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Answer: (c)

  1. Same direction currents attract, so B (above) pulls A upward.
  2. Upward force per unit length on A: μ0I(2I)2πh=μ0I2πh\frac{\mu_0 I(2I)}{2\pi h} = \frac{\mu_0 I^2}{\pi h}
  3. A stays on the floor if μ0I2πh≤λg⇒h≥μ0I2πλg\frac{\mu_0 I^2}{\pi h} \le \lambda g \Rightarrow h \ge \frac{\mu_0 I^2}{\pi\lambda g}
  4. So the answer is (c) μ0I2πλg\frac{\mu_0 I^2}{\pi\lambda g}.
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