JEE Main 2024 Physics question
A coil having 100 turns, area of 5 × 10⁻³ m², carrying current of 1 mA is placed in uniform magnetic field of 0.20 T such a way that plane of coil is perpendicular to the magnetic field. The work done in turning the coil through 90° is μJ.
Answer: 100
Solution
M = NIA = 100 × 10⁻³ × 5 × 10⁻³ = 5 × 10⁻⁴ A m². With M along B, turning through 90° needs W = MB = 5 × 10⁻⁴ × 0.2 = 10⁻⁴ J = 100 μJ.