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NEETPhysicsMagnetic Effects of Current

A horizontal wire of total mass m and length L carrying a current I is suspended by two identical vertical non-conducting strings attached to its ends. A second infinitely long straight wire is fixed at a distance a directly below the suspended wire. The fixed wire carries a current 3I in a direction opposite to the current in the suspended wire. In the state of equilibrium, what is the tension in each string? [ g is

Options

  1. Amg 2 - 3 ₀ I^2 L 4 a
  2. Bmg - 3 ₀ I^2 L 2 a
  3. Cmg 2 + 3 ₀ I^2 L 4 a
  4. Dmg 2 - 3 ₀ I^2 4 a

Correct answer

A. mg 2 - 3 ₀ I^2 L 4 a

Step-by-step solution

The two wires carry currents in opposite directions, so the magnetic force between them is repulsive. Since the fixed wire is below the suspended wire, the magnetic force on the suspended wire acts vertically upwards. The magnitude of the total upward magnetic force on the suspended wire of length L is: F_m = ₀ I₁ I₂ 2 a L = ₀ (I)(3I) 2 a L = 3 ₀ I^2 L 2 a The downward gravitational force on the suspended wire is its weight: W = mg The wire is supported by two identical vertical strings, so let the tension in each

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