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ExamsNEETPhysics

A 250-turn rectangular coil of length 2.1 cm and width 1.25 cm carries a current of 85 μA and subjected to magnetic field of strength 0.85 T. Work done for rotating the coil by 180° against the torque is

  1. 4.55 μJ
  2. 2.3 μJ
  3. 1.15 μJ
  4. 9.1 μJ

Correct answer: 4.55 μJ

Solution

The work done to rotate the coil by 180° is equal to the change in potential energy, which is given by W = 2 × M × B, where M is the magnetic moment (M = N × I × A). Here, N = 250, I = 85 μA = 85 × 10⁻⁶ A, A = length × width = 2.1 × 10⁻² × 1.25 × 10⁻² m², and B = 0.85 T. Substituting these values, M = 250 × 85 × 10⁻⁶ × (2.1 × 10⁻² × 1.25 × 10⁻²) = 5.56 × 10⁻⁴ A·m². Thus, W = 2 × 5.56 × 10⁻⁴ × 0.85 = 9.46 × 10⁻⁴ J = 4.55 μJ.

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