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ExamsJEE AdvancedPhysics

Twelve current-carrying wires are wound around a sphere of radius R. Each wire carries the same current I and is separated from the adjacent wire by an angle of 30 degrees. All wires carry current moving upward (along the same direction). A uniform horizontal magnetic field B exists throughout the region. The net magnetic force on the entire set of wires is F = alpha * beta * B * I * R. Find the product alpha * beta.

  1. 12
  2. 24
  3. 36
  4. 48

Correct answer: 24

Solution

Each of the 12 wires carries current I upward. In a horizontal magnetic field B, the force on a curved wire equals I * (vector displacement from start to end) cross B. For each wire going from the bottom to the top of the sphere, the displacement vector is 2R (vertical, upward). The force on each wire is F_one = I * 2R * B (horizontal, perpendicular to B). For 12 wires symmetrically arranged, the horizontal force components that are perpendicular to B add up (those parallel to B cancel by symmetry). Total effective force magnitude: net F = 12 * I * 2R * B = 24 * B * I * R. So alpha * beta = 24.

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